perfect codes and generalized ordinal sum of posets · luciano panek state university of west...
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Perfect Codes and Generalized Ordinal Sum of Posets
Luciano PanekState University of West Parana - Brazil
Joint work with
Jerry Anderson Pinheiros - UNICAMP
Marcelo Firer - UNICAMP
Marcelo Muniz Alves - UFPR
Latin American Week on Coding and Information - LAWCI
UNICAMP - Campinas - Brazil
July 2018
Luciano Panek (UNIOESTE) LAWCI 2018 1 19
Basic Definitions Posets Poset Metrics and Poset Codes
bull [n] is the finite set 1 2 n called n-set
bull P = ([n]leP) is a partial order (or poset) on [n]
bull I sube [n] is an ideal of P if whenever b isin I and a leP b then a isin I
bull 〈X 〉 is the ideal generated by X sube [n] the smallest ideal containing X
Luciano Panek (UNIOESTE) LAWCI 2018 2 19
Basic Definitions Posets Poset Metrics and Poset Codes
bull [n] is the finite set 1 2 n called n-set
bull P = ([n]leP) is a partial order (or poset) on [n]
bull I sube [n] is an ideal of P if whenever b isin I and a leP b then a isin I
bull 〈X 〉 is the ideal generated by X sube [n] the smallest ideal containing X
Luciano Panek (UNIOESTE) LAWCI 2018 2 19
Basic Definitions Posets Poset Metrics and Poset Codes
bull [n] is the finite set 1 2 n called n-set
bull P = ([n]leP) is a partial order (or poset) on [n]
bull I sube [n] is an ideal of P if whenever b isin I and a leP b then a isin I
bull 〈X 〉 is the ideal generated by X sube [n] the smallest ideal containing X
Luciano Panek (UNIOESTE) LAWCI 2018 2 19
Basic Definitions Posets Poset Metrics and Poset Codes
bull [n] is the finite set 1 2 n called n-set
bull P = ([n]leP) is a partial order (or poset) on [n]
bull I sube [n] is an ideal of P if whenever b isin I and a leP b then a isin I
bull 〈X 〉 is the ideal generated by X sube [n] the smallest ideal containing X
Luciano Panek (UNIOESTE) LAWCI 2018 2 19
Basic Definitions Posets Poset Metrics and Poset Codes
bull [n] is the finite set 1 2 n called n-set
bull P = ([n]leP) is a partial order (or poset) on [n]
bull I sube [n] is an ideal of P if whenever b isin I and a leP b then a isin I
bull 〈X 〉 is the ideal generated by X sube [n] the smallest ideal containing X
Luciano Panek (UNIOESTE) LAWCI 2018 2 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
The Y order P = ([5]leP)
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 3 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
X = 2 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 4 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
〈X 〉 = 1 2 3 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 5 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
Let P = ([5]leP) be the Y order If u = 11100 and v = 10110 we havethat u minus v = 01010 and so supp(u minus v) = 2 4
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3
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Luciano Panek (UNIOESTE) LAWCI 2018 7 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
Let P = ([5]leP) be the Y order If u = 11100 and v = 10110 we havethat u minus v = 01010 and so supp(u minus v) = 2 4 Therefore dP(u v) = 4
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Luciano Panek (UNIOESTE) LAWCI 2018 8 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows
i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order
1
2
3
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Luciano Panek (UNIOESTE) LAWCI 2018 12 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order If R = [3]times [2]
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 13 19
Generalized Ordinal Sum of Posets
Example
Let P = ([3]leP) be the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) be the antichain order If R = [3]times [2] then theR-sum is the Y order
1
2
3
5 (5-3)(4-3) 4
Luciano Panek (UNIOESTE) LAWCI 2018 14 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Basic Definitions Posets Poset Metrics and Poset Codes
bull [n] is the finite set 1 2 n called n-set
bull P = ([n]leP) is a partial order (or poset) on [n]
bull I sube [n] is an ideal of P if whenever b isin I and a leP b then a isin I
bull 〈X 〉 is the ideal generated by X sube [n] the smallest ideal containing X
Luciano Panek (UNIOESTE) LAWCI 2018 2 19
Basic Definitions Posets Poset Metrics and Poset Codes
bull [n] is the finite set 1 2 n called n-set
bull P = ([n]leP) is a partial order (or poset) on [n]
bull I sube [n] is an ideal of P if whenever b isin I and a leP b then a isin I
bull 〈X 〉 is the ideal generated by X sube [n] the smallest ideal containing X
Luciano Panek (UNIOESTE) LAWCI 2018 2 19
Basic Definitions Posets Poset Metrics and Poset Codes
bull [n] is the finite set 1 2 n called n-set
bull P = ([n]leP) is a partial order (or poset) on [n]
bull I sube [n] is an ideal of P if whenever b isin I and a leP b then a isin I
bull 〈X 〉 is the ideal generated by X sube [n] the smallest ideal containing X
Luciano Panek (UNIOESTE) LAWCI 2018 2 19
Basic Definitions Posets Poset Metrics and Poset Codes
bull [n] is the finite set 1 2 n called n-set
bull P = ([n]leP) is a partial order (or poset) on [n]
bull I sube [n] is an ideal of P if whenever b isin I and a leP b then a isin I
bull 〈X 〉 is the ideal generated by X sube [n] the smallest ideal containing X
Luciano Panek (UNIOESTE) LAWCI 2018 2 19
Basic Definitions Posets Poset Metrics and Poset Codes
bull [n] is the finite set 1 2 n called n-set
bull P = ([n]leP) is a partial order (or poset) on [n]
bull I sube [n] is an ideal of P if whenever b isin I and a leP b then a isin I
bull 〈X 〉 is the ideal generated by X sube [n] the smallest ideal containing X
Luciano Panek (UNIOESTE) LAWCI 2018 2 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
The Y order P = ([5]leP)
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 3 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
X = 2 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 4 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
〈X 〉 = 1 2 3 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 5 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
Let P = ([5]leP) be the Y order If u = 11100 and v = 10110 we havethat u minus v = 01010 and so supp(u minus v) = 2 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 7 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
Let P = ([5]leP) be the Y order If u = 11100 and v = 10110 we havethat u minus v = 01010 and so supp(u minus v) = 2 4 Therefore dP(u v) = 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 8 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows
i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 12 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order If R = [3]times [2]
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 13 19
Generalized Ordinal Sum of Posets
Example
Let P = ([3]leP) be the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) be the antichain order If R = [3]times [2] then theR-sum is the Y order
1
2
3
5 (5-3)(4-3) 4
Luciano Panek (UNIOESTE) LAWCI 2018 14 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Basic Definitions Posets Poset Metrics and Poset Codes
bull [n] is the finite set 1 2 n called n-set
bull P = ([n]leP) is a partial order (or poset) on [n]
bull I sube [n] is an ideal of P if whenever b isin I and a leP b then a isin I
bull 〈X 〉 is the ideal generated by X sube [n] the smallest ideal containing X
Luciano Panek (UNIOESTE) LAWCI 2018 2 19
Basic Definitions Posets Poset Metrics and Poset Codes
bull [n] is the finite set 1 2 n called n-set
bull P = ([n]leP) is a partial order (or poset) on [n]
bull I sube [n] is an ideal of P if whenever b isin I and a leP b then a isin I
bull 〈X 〉 is the ideal generated by X sube [n] the smallest ideal containing X
Luciano Panek (UNIOESTE) LAWCI 2018 2 19
Basic Definitions Posets Poset Metrics and Poset Codes
bull [n] is the finite set 1 2 n called n-set
bull P = ([n]leP) is a partial order (or poset) on [n]
bull I sube [n] is an ideal of P if whenever b isin I and a leP b then a isin I
bull 〈X 〉 is the ideal generated by X sube [n] the smallest ideal containing X
Luciano Panek (UNIOESTE) LAWCI 2018 2 19
Basic Definitions Posets Poset Metrics and Poset Codes
bull [n] is the finite set 1 2 n called n-set
bull P = ([n]leP) is a partial order (or poset) on [n]
bull I sube [n] is an ideal of P if whenever b isin I and a leP b then a isin I
bull 〈X 〉 is the ideal generated by X sube [n] the smallest ideal containing X
Luciano Panek (UNIOESTE) LAWCI 2018 2 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
The Y order P = ([5]leP)
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 3 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
X = 2 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 4 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
〈X 〉 = 1 2 3 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 5 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
Let P = ([5]leP) be the Y order If u = 11100 and v = 10110 we havethat u minus v = 01010 and so supp(u minus v) = 2 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 7 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
Let P = ([5]leP) be the Y order If u = 11100 and v = 10110 we havethat u minus v = 01010 and so supp(u minus v) = 2 4 Therefore dP(u v) = 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 8 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows
i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 12 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order If R = [3]times [2]
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 13 19
Generalized Ordinal Sum of Posets
Example
Let P = ([3]leP) be the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) be the antichain order If R = [3]times [2] then theR-sum is the Y order
1
2
3
5 (5-3)(4-3) 4
Luciano Panek (UNIOESTE) LAWCI 2018 14 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Basic Definitions Posets Poset Metrics and Poset Codes
bull [n] is the finite set 1 2 n called n-set
bull P = ([n]leP) is a partial order (or poset) on [n]
bull I sube [n] is an ideal of P if whenever b isin I and a leP b then a isin I
bull 〈X 〉 is the ideal generated by X sube [n] the smallest ideal containing X
Luciano Panek (UNIOESTE) LAWCI 2018 2 19
Basic Definitions Posets Poset Metrics and Poset Codes
bull [n] is the finite set 1 2 n called n-set
bull P = ([n]leP) is a partial order (or poset) on [n]
bull I sube [n] is an ideal of P if whenever b isin I and a leP b then a isin I
bull 〈X 〉 is the ideal generated by X sube [n] the smallest ideal containing X
Luciano Panek (UNIOESTE) LAWCI 2018 2 19
Basic Definitions Posets Poset Metrics and Poset Codes
bull [n] is the finite set 1 2 n called n-set
bull P = ([n]leP) is a partial order (or poset) on [n]
bull I sube [n] is an ideal of P if whenever b isin I and a leP b then a isin I
bull 〈X 〉 is the ideal generated by X sube [n] the smallest ideal containing X
Luciano Panek (UNIOESTE) LAWCI 2018 2 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
The Y order P = ([5]leP)
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 3 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
X = 2 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 4 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
〈X 〉 = 1 2 3 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 5 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
Let P = ([5]leP) be the Y order If u = 11100 and v = 10110 we havethat u minus v = 01010 and so supp(u minus v) = 2 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 7 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
Let P = ([5]leP) be the Y order If u = 11100 and v = 10110 we havethat u minus v = 01010 and so supp(u minus v) = 2 4 Therefore dP(u v) = 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 8 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows
i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 12 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order If R = [3]times [2]
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 13 19
Generalized Ordinal Sum of Posets
Example
Let P = ([3]leP) be the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) be the antichain order If R = [3]times [2] then theR-sum is the Y order
1
2
3
5 (5-3)(4-3) 4
Luciano Panek (UNIOESTE) LAWCI 2018 14 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Basic Definitions Posets Poset Metrics and Poset Codes
bull [n] is the finite set 1 2 n called n-set
bull P = ([n]leP) is a partial order (or poset) on [n]
bull I sube [n] is an ideal of P if whenever b isin I and a leP b then a isin I
bull 〈X 〉 is the ideal generated by X sube [n] the smallest ideal containing X
Luciano Panek (UNIOESTE) LAWCI 2018 2 19
Basic Definitions Posets Poset Metrics and Poset Codes
bull [n] is the finite set 1 2 n called n-set
bull P = ([n]leP) is a partial order (or poset) on [n]
bull I sube [n] is an ideal of P if whenever b isin I and a leP b then a isin I
bull 〈X 〉 is the ideal generated by X sube [n] the smallest ideal containing X
Luciano Panek (UNIOESTE) LAWCI 2018 2 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
The Y order P = ([5]leP)
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 3 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
X = 2 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 4 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
〈X 〉 = 1 2 3 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 5 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
Let P = ([5]leP) be the Y order If u = 11100 and v = 10110 we havethat u minus v = 01010 and so supp(u minus v) = 2 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 7 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
Let P = ([5]leP) be the Y order If u = 11100 and v = 10110 we havethat u minus v = 01010 and so supp(u minus v) = 2 4 Therefore dP(u v) = 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 8 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows
i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 12 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order If R = [3]times [2]
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 13 19
Generalized Ordinal Sum of Posets
Example
Let P = ([3]leP) be the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) be the antichain order If R = [3]times [2] then theR-sum is the Y order
1
2
3
5 (5-3)(4-3) 4
Luciano Panek (UNIOESTE) LAWCI 2018 14 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Basic Definitions Posets Poset Metrics and Poset Codes
bull [n] is the finite set 1 2 n called n-set
bull P = ([n]leP) is a partial order (or poset) on [n]
bull I sube [n] is an ideal of P if whenever b isin I and a leP b then a isin I
bull 〈X 〉 is the ideal generated by X sube [n] the smallest ideal containing X
Luciano Panek (UNIOESTE) LAWCI 2018 2 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
The Y order P = ([5]leP)
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 3 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
X = 2 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 4 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
〈X 〉 = 1 2 3 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 5 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
Let P = ([5]leP) be the Y order If u = 11100 and v = 10110 we havethat u minus v = 01010 and so supp(u minus v) = 2 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 7 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
Let P = ([5]leP) be the Y order If u = 11100 and v = 10110 we havethat u minus v = 01010 and so supp(u minus v) = 2 4 Therefore dP(u v) = 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 8 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows
i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 12 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order If R = [3]times [2]
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 13 19
Generalized Ordinal Sum of Posets
Example
Let P = ([3]leP) be the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) be the antichain order If R = [3]times [2] then theR-sum is the Y order
1
2
3
5 (5-3)(4-3) 4
Luciano Panek (UNIOESTE) LAWCI 2018 14 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
The Y order P = ([5]leP)
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 3 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
X = 2 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 4 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
〈X 〉 = 1 2 3 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 5 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
Let P = ([5]leP) be the Y order If u = 11100 and v = 10110 we havethat u minus v = 01010 and so supp(u minus v) = 2 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 7 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
Let P = ([5]leP) be the Y order If u = 11100 and v = 10110 we havethat u minus v = 01010 and so supp(u minus v) = 2 4 Therefore dP(u v) = 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 8 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows
i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 12 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order If R = [3]times [2]
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 13 19
Generalized Ordinal Sum of Posets
Example
Let P = ([3]leP) be the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) be the antichain order If R = [3]times [2] then theR-sum is the Y order
1
2
3
5 (5-3)(4-3) 4
Luciano Panek (UNIOESTE) LAWCI 2018 14 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
X = 2 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 4 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
〈X 〉 = 1 2 3 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 5 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
Let P = ([5]leP) be the Y order If u = 11100 and v = 10110 we havethat u minus v = 01010 and so supp(u minus v) = 2 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 7 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
Let P = ([5]leP) be the Y order If u = 11100 and v = 10110 we havethat u minus v = 01010 and so supp(u minus v) = 2 4 Therefore dP(u v) = 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 8 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows
i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 12 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order If R = [3]times [2]
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 13 19
Generalized Ordinal Sum of Posets
Example
Let P = ([3]leP) be the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) be the antichain order If R = [3]times [2] then theR-sum is the Y order
1
2
3
5 (5-3)(4-3) 4
Luciano Panek (UNIOESTE) LAWCI 2018 14 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
〈X 〉 = 1 2 3 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 5 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
Let P = ([5]leP) be the Y order If u = 11100 and v = 10110 we havethat u minus v = 01010 and so supp(u minus v) = 2 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 7 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
Let P = ([5]leP) be the Y order If u = 11100 and v = 10110 we havethat u minus v = 01010 and so supp(u minus v) = 2 4 Therefore dP(u v) = 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 8 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows
i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 12 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order If R = [3]times [2]
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 13 19
Generalized Ordinal Sum of Posets
Example
Let P = ([3]leP) be the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) be the antichain order If R = [3]times [2] then theR-sum is the Y order
1
2
3
5 (5-3)(4-3) 4
Luciano Panek (UNIOESTE) LAWCI 2018 14 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
Let P = ([5]leP) be the Y order If u = 11100 and v = 10110 we havethat u minus v = 01010 and so supp(u minus v) = 2 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 7 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
Let P = ([5]leP) be the Y order If u = 11100 and v = 10110 we havethat u minus v = 01010 and so supp(u minus v) = 2 4 Therefore dP(u v) = 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 8 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows
i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 12 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order If R = [3]times [2]
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 13 19
Generalized Ordinal Sum of Posets
Example
Let P = ([3]leP) be the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) be the antichain order If R = [3]times [2] then theR-sum is the Y order
1
2
3
5 (5-3)(4-3) 4
Luciano Panek (UNIOESTE) LAWCI 2018 14 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
Let P = ([5]leP) be the Y order If u = 11100 and v = 10110 we havethat u minus v = 01010 and so supp(u minus v) = 2 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 7 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
Let P = ([5]leP) be the Y order If u = 11100 and v = 10110 we havethat u minus v = 01010 and so supp(u minus v) = 2 4 Therefore dP(u v) = 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 8 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows
i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 12 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order If R = [3]times [2]
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 13 19
Generalized Ordinal Sum of Posets
Example
Let P = ([3]leP) be the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) be the antichain order If R = [3]times [2] then theR-sum is the Y order
1
2
3
5 (5-3)(4-3) 4
Luciano Panek (UNIOESTE) LAWCI 2018 14 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
Let P = ([5]leP) be the Y order If u = 11100 and v = 10110 we havethat u minus v = 01010 and so supp(u minus v) = 2 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 7 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
Let P = ([5]leP) be the Y order If u = 11100 and v = 10110 we havethat u minus v = 01010 and so supp(u minus v) = 2 4 Therefore dP(u v) = 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 8 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows
i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 12 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order If R = [3]times [2]
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 13 19
Generalized Ordinal Sum of Posets
Example
Let P = ([3]leP) be the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) be the antichain order If R = [3]times [2] then theR-sum is the Y order
1
2
3
5 (5-3)(4-3) 4
Luciano Panek (UNIOESTE) LAWCI 2018 14 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
bull Given u = (u1 un) isin Fnq the support of u is the set
supp(u) = i isin [n] ui ∕= 0
bull The P-distance of u and v is the cardinality
dP(u v) = |〈supp(u minus v)〉|
bull BP(u r) is the P-sphere with center u and radius r ge 0
BP(u r) = v isin Fnq dP(u v) le r
Luciano Panek (UNIOESTE) LAWCI 2018 6 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
Let P = ([5]leP) be the Y order If u = 11100 and v = 10110 we havethat u minus v = 01010 and so supp(u minus v) = 2 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 7 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
Let P = ([5]leP) be the Y order If u = 11100 and v = 10110 we havethat u minus v = 01010 and so supp(u minus v) = 2 4 Therefore dP(u v) = 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 8 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows
i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 12 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order If R = [3]times [2]
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 13 19
Generalized Ordinal Sum of Posets
Example
Let P = ([3]leP) be the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) be the antichain order If R = [3]times [2] then theR-sum is the Y order
1
2
3
5 (5-3)(4-3) 4
Luciano Panek (UNIOESTE) LAWCI 2018 14 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
Let P = ([5]leP) be the Y order If u = 11100 and v = 10110 we havethat u minus v = 01010 and so supp(u minus v) = 2 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 7 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
Let P = ([5]leP) be the Y order If u = 11100 and v = 10110 we havethat u minus v = 01010 and so supp(u minus v) = 2 4 Therefore dP(u v) = 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 8 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows
i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 12 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order If R = [3]times [2]
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 13 19
Generalized Ordinal Sum of Posets
Example
Let P = ([3]leP) be the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) be the antichain order If R = [3]times [2] then theR-sum is the Y order
1
2
3
5 (5-3)(4-3) 4
Luciano Panek (UNIOESTE) LAWCI 2018 14 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Basic Definitions Codes Posets Poset Metrics andPoset Codes
Let P = ([5]leP) be the Y order If u = 11100 and v = 10110 we havethat u minus v = 01010 and so supp(u minus v) = 2 4 Therefore dP(u v) = 4
1
2
3
54
Luciano Panek (UNIOESTE) LAWCI 2018 8 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows
i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 12 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order If R = [3]times [2]
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 13 19
Generalized Ordinal Sum of Posets
Example
Let P = ([3]leP) be the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) be the antichain order If R = [3]times [2] then theR-sum is the Y order
1
2
3
5 (5-3)(4-3) 4
Luciano Panek (UNIOESTE) LAWCI 2018 14 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows
i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 12 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order If R = [3]times [2]
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 13 19
Generalized Ordinal Sum of Posets
Example
Let P = ([3]leP) be the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) be the antichain order If R = [3]times [2] then theR-sum is the Y order
1
2
3
5 (5-3)(4-3) 4
Luciano Panek (UNIOESTE) LAWCI 2018 14 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows
i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 12 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order If R = [3]times [2]
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 13 19
Generalized Ordinal Sum of Posets
Example
Let P = ([3]leP) be the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) be the antichain order If R = [3]times [2] then theR-sum is the Y order
1
2
3
5 (5-3)(4-3) 4
Luciano Panek (UNIOESTE) LAWCI 2018 14 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows
i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 12 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order If R = [3]times [2]
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 13 19
Generalized Ordinal Sum of Posets
Example
Let P = ([3]leP) be the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) be the antichain order If R = [3]times [2] then theR-sum is the Y order
1
2
3
5 (5-3)(4-3) 4
Luciano Panek (UNIOESTE) LAWCI 2018 14 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows
i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 12 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order If R = [3]times [2]
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 13 19
Generalized Ordinal Sum of Posets
Example
Let P = ([3]leP) be the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) be the antichain order If R = [3]times [2] then theR-sum is the Y order
1
2
3
5 (5-3)(4-3) 4
Luciano Panek (UNIOESTE) LAWCI 2018 14 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Basic Definitions Codes Posets Poset Metrics and PosetCodes
bull A poset code is a linear subspace C of the metric space (Fnq dP)
bull dP(C ) is the minimum distance of a poset code C
dP(C ) = mindP(c c prime) c c prime isin C c ∕= c prime
bull RP(C ) is the packing radius of a poset code C
RP(C ) = maxr BP(c r) cap BP(cprime r) = empty c ∕= c prime isin C
bull C sube Fnq is a P-perfect code if
983134
cisinCBP(c RP(C )) = Fn
q
Luciano Panek (UNIOESTE) LAWCI 2018 9 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows
i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 12 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order If R = [3]times [2]
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 13 19
Generalized Ordinal Sum of Posets
Example
Let P = ([3]leP) be the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) be the antichain order If R = [3]times [2] then theR-sum is the Y order
1
2
3
5 (5-3)(4-3) 4
Luciano Panek (UNIOESTE) LAWCI 2018 14 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows
i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 12 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order If R = [3]times [2]
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 13 19
Generalized Ordinal Sum of Posets
Example
Let P = ([3]leP) be the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) be the antichain order If R = [3]times [2] then theR-sum is the Y order
1
2
3
5 (5-3)(4-3) 4
Luciano Panek (UNIOESTE) LAWCI 2018 14 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows
i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 12 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order If R = [3]times [2]
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 13 19
Generalized Ordinal Sum of Posets
Example
Let P = ([3]leP) be the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) be the antichain order If R = [3]times [2] then theR-sum is the Y order
1
2
3
5 (5-3)(4-3) 4
Luciano Panek (UNIOESTE) LAWCI 2018 14 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows
i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 12 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order If R = [3]times [2]
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 13 19
Generalized Ordinal Sum of Posets
Example
Let P = ([3]leP) be the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) be the antichain order If R = [3]times [2] then theR-sum is the Y order
1
2
3
5 (5-3)(4-3) 4
Luciano Panek (UNIOESTE) LAWCI 2018 14 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Examples of P-Perfect Codes
C = 000 111 sube F32
H anti-chain orderdH(u v) = |i ui ∕= vi|
Hamming distance
RdH (C ) = lfloordH(C)minus12 rfloor = 1
P usual total orderdP(u v) = maxi ui ∕= vi
NRT distanceRdP (C ) = dP(C )minus 1 = 2
Luciano Panek (UNIOESTE) LAWCI 2018 10 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows
i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 12 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order If R = [3]times [2]
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 13 19
Generalized Ordinal Sum of Posets
Example
Let P = ([3]leP) be the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) be the antichain order If R = [3]times [2] then theR-sum is the Y order
1
2
3
5 (5-3)(4-3) 4
Luciano Panek (UNIOESTE) LAWCI 2018 14 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows
i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 12 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order If R = [3]times [2]
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 13 19
Generalized Ordinal Sum of Posets
Example
Let P = ([3]leP) be the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) be the antichain order If R = [3]times [2] then theR-sum is the Y order
1
2
3
5 (5-3)(4-3) 4
Luciano Panek (UNIOESTE) LAWCI 2018 14 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows
i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 12 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order If R = [3]times [2]
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 13 19
Generalized Ordinal Sum of Posets
Example
Let P = ([3]leP) be the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) be the antichain order If R = [3]times [2] then theR-sum is the Y order
1
2
3
5 (5-3)(4-3) 4
Luciano Panek (UNIOESTE) LAWCI 2018 14 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows
i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 12 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order If R = [3]times [2]
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 13 19
Generalized Ordinal Sum of Posets
Example
Let P = ([3]leP) be the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) be the antichain order If R = [3]times [2] then theR-sum is the Y order
1
2
3
5 (5-3)(4-3) 4
Luciano Panek (UNIOESTE) LAWCI 2018 14 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Generalized Ordinal Sum of Posets
bull Given A sube [n] and B sube [m] let
R sube Atimes B
be a relation such that Im(R) = B
bull The R-sum of posets P = ([n] leP) and Q = ([m] leQ) is the posetP oplusR Q over [n +m] determined as follows i lePoplusRQ j if and only if
i j le n and i leP j
orn lt i j and (i minus n) leQ (j minus n)
ori le n lt j and iR (j minus n)
Luciano Panek (UNIOESTE) LAWCI 2018 11 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 12 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order If R = [3]times [2]
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 13 19
Generalized Ordinal Sum of Posets
Example
Let P = ([3]leP) be the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) be the antichain order If R = [3]times [2] then theR-sum is the Y order
1
2
3
5 (5-3)(4-3) 4
Luciano Panek (UNIOESTE) LAWCI 2018 14 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 12 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order If R = [3]times [2]
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 13 19
Generalized Ordinal Sum of Posets
Example
Let P = ([3]leP) be the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) be the antichain order If R = [3]times [2] then theR-sum is the Y order
1
2
3
5 (5-3)(4-3) 4
Luciano Panek (UNIOESTE) LAWCI 2018 14 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Generalized Ordinal Sum of Posets
Example
If P = ([3]leP) is the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) is the antichain order If R = [3]times [2]
1
2
3
21
Luciano Panek (UNIOESTE) LAWCI 2018 13 19
Generalized Ordinal Sum of Posets
Example
Let P = ([3]leP) be the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) be the antichain order If R = [3]times [2] then theR-sum is the Y order
1
2
3
5 (5-3)(4-3) 4
Luciano Panek (UNIOESTE) LAWCI 2018 14 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Generalized Ordinal Sum of Posets
Example
Let P = ([3]leP) be the usual chain order
1 ltP 2 ltP 3
and Q = ([2]leQ) be the antichain order If R = [3]times [2] then theR-sum is the Y order
1
2
3
5 (5-3)(4-3) 4
Luciano Panek (UNIOESTE) LAWCI 2018 14 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111
RdPoplusRQ(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Generalized Ordinal Sum of Posets
Example - to be continued
Let C1 = 000 111 C2 = 00 10 and C3 = 00 01 10 11 = F22 be the
binary codes Then
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus C3 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 15 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 16 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Generalized Ordinal Sum of Posets
Example - to be continued
Considering the binary codes C1 = 000 111 C2 = 00 10 andC3 = 00 01 10 11 = F2
2 we have that
RdP (C1) = 2 and C1 is a perfect code
|BP(000RP(C1))| = |BPoplusRQ(00000RP(C1))| = 4
BPoplusRQ(00000RPoplusRQ(C1)) = 00000 10000 01000 11000
C1 oplus C2 = 00000 00010 11100 11110
00110 isin BPoplusRQ(00010 3) cap BPoplusRQ(11110 3)
C1 oplus C3 = 00000 00010 11100 11110 00001 00011 11101 11111RdPoplusRQ
(C1 oplus C3) = RdP (C1) = 2 and C1 oplus F22 is a perfect code
Luciano Panek (UNIOESTE) LAWCI 2018 17 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
Generalized Ordinal Sum of Posets
Theorem
Let P = ([n] leP) and Q = ([m] leQ) be posets C1 oplus C2 sube Fnq oplus Fm
q be acode where C1 is a P-perfect code RPoplusRQ(C1 oplus C2) = RP (C1) and|BP(0RP(C1))| = |BPoplusRQ(0RP(C1))| The following items areequivalent
(i) C2 ≃ Fmq
(ii) C1 oplus C2 is P oplusR Q-perfect and for each minimal element j isin Q thevector x (j) = ej +
983123iisin[n]iRj ei has packing radius
RPoplusRQ (x (j)) ge RP (C1)
Corollary
If C1 is P-perfect then C1 oplus C2 is P oplus Q-perfect withRPoplusQ(C1 oplus C2) = RP (C1) for every poset Q = ([m] leQ) if and only ifC2 = Fm
q
Luciano Panek (UNIOESTE) LAWCI 2018 18 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19
References
J Ahn H K Kim J S Kim and M Kim Classification of perfectlinear codes with crown poset structure Discrete Mathematics 268(2003) 21-30
R Brualdi J S Graves and M Lawrence Codes with a poset metric Discrete Mathematics 147 (1995) 57-72
J Y Hyun and H K Kim The poset structures admitting theextended binary Hamming code to be a perfect code DiscreteMathematics 288 (2004) 37-47
J G Lee Perfect codes on some ordered sets Bulletin of the KoreanMathematical Society 43(2) (2006) 293-297
R G L DrsquoOliveira and M Firer The packing radius of a code andpartitioning problems the case for poset metrics on finite vectorspaces Discrete Mathematics 338 (2015) 2143-2167
Luciano Panek (UNIOESTE) LAWCI 2018 19 19