lecture 12-13 complete problems in ph
DESCRIPTION
LECTURE 12-13 Complete Problems in PH. THEOREM. PROOF. BOUNDED HALTING PROBLEM (BHP). THEOREM. PROOF. A. BHP. LEMMA. PROOF. LEMMA. PROOF. DEFINITION. THEOREM. PROOF by induction on k. Natural Problem. (Machine unrelated). Machine Related Problem. SAT k. THEOREM. - PowerPoint PPT PresentationTRANSCRIPT
.for -complete is Clique-Exact 2pp
T
Clique.-ExactClique Therefore,
][ ))((
.Clique-Exact
Clique is Clique-Exact
2
2
pT
pm
pT
pT
p
pNP
pT
BA
BANPBA
P
THEOREM
PROOF
moves? within accept can ,0 string a
and input an , NTMan of code Given the
tyM
yMxt
BOUNDED HALTING PROBLEM (BHP)
PROOF
THEOREM
complete.-NP is BHP
NP.)(BHPThen moves. within input on
simulate to NTM universalan Construct
u
u
MLty
MM
.0,,construct , of input For
. bound timepolynomial
with NTMan for )(NP
|)(|ypyMMy
p
MMLAA
BHPA
moves. within accept can
,0 string a and input an , oracle with NTMan Given t
tyM
yAMA
LEMMA
.NPfor -complete is BHP Apm
A
.NP)(BHPThen moves. in input on
simulate to oracle with NTM universal aConstruct A
uA
u
MLtyM
AM
.0,,construct , of input For
. bound timepolynomial
and oracle with NTMan for )(NP
|)(|yp
A
yMMy
p
AMMLBB
PROOF
LEMMA
.NP then ,for -complete is If 1pk
Apk
pmA
PROOF
.NPP
.NP,NP
]NP)[)((
.NP have we, Since
1
1
AApm
Apm
C
Cpk
pk
pk
Apk
CCAC
BACB
BCB
A
THEOREM
DEFINITION
.BHP, 10kA
kAA
.for -complete is pk
pmkA
PROOF by induction on k
.NPfor -complete is
lemma,by Then .for -complete is Suppose
.NPfor -complete is BHPBHP
11
11
pk
Apmk
pk
pmk
ppm
kA
A
A
Natural Problem
Machine Related Problem
(Machine unrelated)
problems. natural containsit ift significan is class complexityA
SAT k
.even for and , oddfor denotes where
1|})1,0{,(
})1,0{,})(1,0{,(
: trueis following he whether terminedet
,over formulaBoolean a
and ,,...,, variablesof sets Given
,...,,
2211
21
21
21
kkQ
FXQ
XX
XXXXF
XXXk
k
kkk
k
k
k
THEOREM
.for -complete is SAT pk
pmk
PROOF (odd k)
Byyxnqyy
nqyynqyyAx
nxx
qBA
kkk
pk
,...,,))(||,(
))(||,))((||,(
,|| with *any for such that
polynomial and Pexist there
1
2211
1|})1,0{,(
})1,0{,})(1,0{,(
such that ,..., and construct we*,each for
,SAT show To
,...,,
2211
1
21
kFX
XXAx
XXFx
A
kk
k
kpm
; (c)
state; final thecontains (b)
ion;configurat initial theis (a)
1
0
ii
m
such that
|),,,(|||
|),,,,(| ),,...,,(
, ofpath n computatio a is there))[(||,(
))(||,))((||,( iff
. bound time-polynomial has where)(Let
1
110
2211
ki
km
kk
yyxp
yyxpm
Mnqyy
nqyynqyyAx
pMMLB
such that ,,, formulasBoolean Construct
otherwise 0
symbol holds ),( cell if 1
).()1)((),,,(Then
.###,,,Let
4321
,,
1
11
ffff
ajiz
nrnqknyyxp
yyxyyx
aji
k
kk
; i.e., ,holds (c)1
state; final thecontains i.e., holds, (b)1
ion;configurat initial theis i.e., holds, (a)1
symbol. oneexactly holds cellEach 1
14
3
02
1
ii
m
f
f
f
f
BByyxxsx kn ###]#[ 1210
Bnr
nqnynqnynnxnxsx
z
zzzzzzzfnqn
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#),(2,1),(1,1,2,1#,1,1,1,12#,112
)(,11121
).)((
witheach replacingby 'Obtain
for variablegintroducinby
aconstruct want to we,each From
0,,1,1,,1,
,,12
,,
)(,11,1
,
jhijhi
yj
hihi
inqi
zxzx
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Xyyy
hi
1|})1,0{,(
})1,0{,})(1,0{,(
Then .'Set
,...,,
2211
4321
21
kFX
XXAx
ffffF
kk
3-CNF SAT k
.even for and , oddfor denotes where
1|})1,0{,(
})1,0{,})(1,0{,(
: trueis following he whether terminedet
,over CNF-3 a
and ,,...,, variablesof sets Given
,...,,11
111
21
21
21
kkQ
FX
XQXQ
XXXXF
XXXk
k
kkkkkk
k
k
k
THEOREM
.even is when for -complete is SAT 3CNF
.odd is when for -complete is SAT 3CNF
k
kpk
pmk
pk
pmk
Generalized Ramsey Number
. size of
clique blue aor size of clique red aeither exists there
, from extended coloring-2any for that it true is ,0
integeran given and blue,*})red,{:( colored-2
partially edges with ),(graph complete aGiven
K
K
cK
Ec
EVG
THEOREM
.for -complete is GRN 2pp
m PROOF
GRN SAT CNF-3 2pm
KcGYXF ,,),( formulaBoolean CNF-3
Such that
Construction
]color same in the size of clique a has )[','(
1]|})[1,0{,})(1,0{,( ,
KGccc
FYX
12x11x
12x11x 21x
22x21x
22x
Variables
Clauses
111 : xC121 : xC
213 : xC112 : yC
R
))(( 211211 yyxxyx
223 : yC122 : yC
12x11x
12x11x 21x
22x21x
22x
Variables
Clauses
111 : xC121 : xC
213 : xC112 : yC
))(( 211211 yyxxyx
223 : yC122 : yC
12x11x
12x11x 21x
22x21x
22x
Variables
Clauses
111 : xC121 : xC
213 : xC112 : yC
R
))(( 211211 yyxxyx
223 : yC122 : yC
12x11x
12x11x 21x
22x21x
22x
111 : xC121 : xC
213 : xC112 : yC
R
223 : yC122 : yC
color. samein , size of clique a , ofextention as ' coloring
]1|))(})[(1,0{:})(1,0{:( ,211211
Kcc
yyxxyxYX
12x11x
12x11x 21x
22x21x
22x
111 : xC121 : xC
213 : xC112 : yC
R
. somefor red),('),(' 2121 ixxcxxc iiii
223 : yC122 : yC
Case 1
clique. red required a is ThereAnswer:
4|| KR
12x11x
12x11x 21x
22x21x
22x
111 : xC121 : xC
213 : xC112 : yC
R
occur.not does 1 Case
223 : yC122 : yC
Case 2
clique. blue required a is ThereAnswer:
4|| KR
12x11x
12x11x 21x
22x21x
22x
111 : xC121 : xC
213 : xC112 : yC
occur.not does 1 Case
223 : yC122 : yC
Case 2
clique. blue required a is ThereAnswer:
1
1
21
21
yy
xx
clauses of #
of #
2
k
xn
knK
i
12x11x
12x11x 21x
22x21x
22x
111 : xC121 : xC
213 : xC112 : yC
R
223 : yC122 : yC
color. samein , size of clique a , ofextention as ' coloring
]1|))(})[(1,0{:})(1,0{:( ,211211
Kcc
yyxxyxYX
color. samein , size of clique a , ofextention as ' coloring
]1|))(})[(1,0{:})(1,0{:( ,211211
Kcc
yyxxyxYX
]1|))(})[(1,0{:( ,211211 yyxxyxY
.),(',),(' then ,1)( If
.),(',),(' then ,1)( If
2121
2121
bluexxcredxxcx
redxxcbluexxcx
iiiii
iiiii
Consider
Define a coloring c’:
No red clique of size K exists.
}1 ,0{: XTo prove
12x11x
12x11x 21x
22x21x
22x
111 : xC121 : xC
213 : xC112 : yC
R
223 : yC122 : yC
121 xx
color. samein , size of clique a , ofextention as ' coloring
]1|))(})[(1,0{:})(1,0{:( ,211211
Kcc
yyxxyxYX
)])()[(( 211211 yyxxyxY
.),(',),(' then ,1)( If
.),(',),(' then ,1)( If
2121
2121
bluexxcredxxcx
redxxcbluexxcx
iiiii
iiiii
Consider
Define a coloring c’:
No red clique of size K exists.
}1,0{: XTo prove
There must exist a blue clique Q of size K.
color. samein , size of clique a , ofextention as ' coloring
]1|))(})[(1,0{:})(1,0{:( ,211211
Kcc
yyxxyxYX
12x11x
12x11x 21x
22x21x
22x
111 : xC121 : xC
213 : xC112 : yC
R
223 : yC122 : yC
121 xx
color. samein , size of clique a , ofextention as ' coloring
]1|))(})[(1,0{:})(1,0{:( ,211211
Kcc
yyxxyxYX
otherwise 0
, size of clique blue in the appearsit if 1 Kyi
PPROPERTY OF BLUE CLIQUE Q
.in appears and of onemost At .3
. clauseeach for oneexactly contains .2
.0)( if },{
1)( if },{ contains .1
21
21
Qyy
CCQ
xxx
xxxQ
ii
iij
iii
iii
Assignment for y
Then F(X,Y) is satisfied
1. equals at literal then , If ijij CQC
12x11x
12x11x 21x
22x21x
22x
111 : xC121 : xC
213 : xC112 : yC
R
223 : yC122 : yC
121 xx
color. samein , size of clique a , ofextention as ' coloring
]1|))(})[(1,0{:})(1,0{:( ,211211
Kcc
yyxxyxYX