chuong 6. ly thuyet do thi
TRANSCRIPT
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Bi 1
NH NGHA V PHN LOI
B mn: Khoa hc my tnhKhoa: Cng ngh thng tin - SPHN
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NI DUNG
1. nh ngha th?1.1. th v hng
1.2. th c hng
2. Phn loi th?
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1. NH NGHA TH
Khi nim th l mt m hnh ton hcdng gii quyt rt nhiu bi ton v cc vn ton hc.
Mt th c th hiu mt cch n gin lmt h thng cc nh v cc cnhni cc nhny vi nhau.
V d: Mt bn giao thng l mt thvi h thng nh l cc ng ba, ng t. Ccng i l cc cnh ca th.
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1. NH NGHA TH
1.1. th v hngVd: Cho tp V = {2, 3, 4, 5,6}. Hybiudinquan h nguyn t cng nhau catp trn.
2 3
4 5
6
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1. NH NGHA TH
1.1. th v hngthv hngG = (V, E). Trong :
+ V l tphp, ccphntca n cgil nh.
+ E l tp hp, miphn t l mt cpkhng tht(v, w)ca 2 nhthuc V.
(v, w) cgi l cnhni v v w. (v, w ) (w, v)
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1. NH NGHA TH1.1. th v hngVd: thcnh tranh trong sinh thi hcMi loi l cbiudinbngmtnh.Nu hai loi cnh tranh thcnvi nhau th hai
nhtngng c cnhnichut
chut tr
sc
chim qu
c
ct
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1. NH NGHA TH
1.2. th c hngVd: Cho tp V = {2, 3, 4, 5,6}. Hybiudinquan h: aRb a l cca b v a b
2 3
4 5
6
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1. NH NGHA TH
1.2. th c hngthc hngG = [V, E]. Trong :
+ V l tphp, ccphntca n c
gi l nh.+ E l tphp,miphn t l mtcp
c tht[v, w]ca hai nhcatp V.
[v, w] gi l cungt v n w.
[v, w ] [w, v]
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1. NH NGHA TH1.2. th c hngVd:th thi u vng trn.[a, b] c ngha l i a thngi b
i 1
i 2i 3
i 4 i 5
3 2
4 5
1
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1. NH NGHA TH
Mt s thut ng: Cnh e=(v,w)E, vV, wV,khi :+ e l cnh lin thuc v, w.
+ v, w c gi l k nhau+ v, w gi l nh u mtca cnh e.+ Nu e=[v,w] th v gi l nh u(nh xutpht), w l nh cui(nh ch) ca cung e.+ Nu v w th e c gi l khuyn.+ Nu c e = (v,w) th e v e c gi l haicnh song song(cng lin thuc mt cp nh).
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1. NH NGHA TH
V d:+ Cnh a lin thuc 2 nh 1
v 2.
+ nh 1 v 2 gi l hai nhk nhau.+ Cnh c l khuyn+ Cnh d v e song song
21
3
a
bd
c
e
21
3
a
bd
c
e
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2. PHN LOI TH
Phn loi theo tnh cht cnh ca th:+ th v hng l th m tt c cc cnh lcnh v hng.
+ th c hng l th m tt c cc cnh lc hng.+ th hn hp l th c c cnh v hngv cnh c hng.
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2. PHN LOI TH
V d: th hn hpMt bn giao thng ca H Ni l mt thhn hp. Trong :
+ Cc nh biu din cc nt giao thng (ng ba,ng t ng)+ Cc cnh biu din cc con ng ni cc ntgiao thng .Cnh c hng nu l ng mt chiuCnh v hng nu l ng hai chiu.
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2. PHN LOI TH
Ngoi ra, ta cn c:+ th n l th m khng c khuyn vcnh song song.
+ th im l th ch c mt nh v khngc cnh no.+ th rng l th khng c nh v khngc cnh.
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2. PHN LOI TH
Cc bi sau y chng ta ch n v th vhng. bi cui cng chng ta s ni v th chng.
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Bi 2
CC YU T C BN CA
TH
B mn: Khoa hc my tnh
Khoa: Cng ngh thng tin - SPHN
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NI DUNG
1. th con v th thnh phn?
1.1. th con1.2. th thnh phn
2. Bc ca nh?3. ng i v chu trnh?4. Lin thng
4.1. Khi nim?
4.2. Thnh phn lin thng?4.3. nh ct, cnh cu?4.4. Ch s lin thng?
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1. TH CON V TH THNH PHN
1.1.thconnhngha:
Cho G = (V, E).
G = (V,E)
NuV V
E E
Th Ggi l th con ca G.
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1. TH CON V TH THNH PHN
1.1.thcon
Vd:th con
G1l th con ca G
A D
B C
F H
E G
G
A D
B C E
1
G
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1. TH CON V TH THNH PHN
G2
l th con ca G
A D
B C
F H
E G
G
D
C
H
E G2G
1.1. th con
V d: th con
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G3l khng l th con ca G
A D
B C
F H
E G
G
A D
B C
F
E
3G
1.1. th con
V d: th con
1. TH CON V TH THNH PHN
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1.2.ththnhphnnhngha:
Cho G = (V, E).
G = (V,E)Nu:+ V V
+ v, w V v (v, w) E th (v, w) E
Th Ggi l th thnhphnca G.
1. TH CON V TH THNH PHN
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1.2. thnhphnVd:
A D
B C
F H
E GG
D
B C
F
E G4G
G4l th thnh phn ca G
1. TH CON V TH THNH PHN
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A D
B C
F H
E G
G
G5
l th con nhng khng l th phn ca G
A D
B C
F
E G5G
1. TH CON V TH THNH PHN
1.2. thnh phn
V d:
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2. BC CA NH
Cho G = (V, E), v
V.+ Bc ca v bng s cnh lin thuc vi n.
+ Ti v c khuyn th thm 2 n v.
K hiu l deg(v).Trng hp c bit:
Deg(v) = 0v l nh c lp
Deg(v) = 1v l nh treo
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2. BC CA NH
V d:deg(A) = 0; deg(B) = 1
deg(C) = 1; deg(D) = 4
deg(E) = 2; deg(F) = 3
deg(G) = 4; deg(H) = 1
A l nh c lp.
B, C, H l nh treo.
C D E
F G H
BA
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2. BC CA NH
nh l: Cho G = (V,E) c m cnh. Khi :
Vv vm )deg(2
Hqu: G = (V,E) c snh cbcl l schn.
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3. NG I V CHU TRNHng il dy cc cnh e
i
= (vi
, vi+1
) (i=1, 2,, m)dy cc nh trn ng i v1, v2, , vm, vm +1.
K hiu: H = (v1, e1, v2, e2, , em, vm+1)
ng i H = (v1, e
1, v
2, , e
m, v
m+1)
m v1vm+1c gi l chu trnh.
di ng i (chu trnh) bng s cnh ca n.
Khi th l n th ng i (chu trnh) c khiu n gin bng dy cc nh.
H = (v1, v2, , vm, vm+1)
C = (v1, v2, , vm, v1)
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3. NG I V CHU TRNHV d
A, D, C, G, E - ng i di 4
D, E, C, Akhng l ng i
B, C, G, E, Bchu trnh di 4
A B C
D E G
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4. LIN THNG
4.1. Khi nimHai nh lin thng: v v w c gi llin thng vi nhau nu c mt dy cnh ktip ni v vi w.V d: A D
B C
H
E
A v D, E v H l cc cp nh lin thng
A v E, B v H l cc cp nh khng lin thng
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4. LIN THNG
4.1. Khi nimTnh cht c bn ca quan h lin thng
hai nh:
+ v V th v lin thng vi chnh n.+ v lin thng vi w th w lin thng vi v+ Nu v lin thng w v w lin thng u th
v v u lin thng.
Quan h lin thng hai nh l quan htng ng.
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4. LIN THNG4.1. Khi nim
th lin thngl mt th m hai nhbt k lin thng vi nhau.
V d:
A D
B C
H
E
2 3
4 5
6
Khng l th lin thngL th lin thng
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4. LIN THNG
4.2. Thnh phn lin thngQuan h lin thng gia cc nh phn hoch tp
nh V thnh cc tp con tho mn hai nh bt k:
+ Nu thuc cng mt tp con th lin thng vinhau.
+ Nu thuc hai tp con khc nhau th khng linthng vi nhau.
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4. LIN THNG
4.2. Thnh phn lin thngMi tp con trong phn hoch cng vi cc
cnh ni cc nh ca chng to thnh mt
th thnh phn. th thnh phn ny c gi l thnh phn
lin thngca th cho.
Mt th khng lin thng c chiathnh cc th thnh phn lin thng.
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4. LIN THNG
4.2. Thnh phn lin thngV d:
A D
B C
H
E
C hai thnh phn lin thng
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4. LIN THNG
4.3. nhct,cnhcunh ct: v c gi l nh ct nub ncng cc cnh lin thuc s lm tng s thnh
phn lin thngcath con.Cnhcu: e cgi l cnhcunu xo n
th s lm tngs thnhphn lin thngca
th con.
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4. LIN THNG4.3. nhct,cnhcu
V d:A D
B C
F H
E G
A khng l nh ct
D
B C
F H
E G
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4. LIN THNG
A D
B C
F H
E G
A D
C
F H
E G
B l nh ct
4.3. nhct,cnhcu
V d:
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4. LIN THNG
A D
B C
F H
E G
C l nh ct
A D
B
F H
E G
4.3. nh ct, cnh cu
V d:
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4. LIN THNG
A D
B C
F H
E G
B
C
E
l nh ct
4.3. nh ct, cnh cu
V d:
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4. LIN THNG
A D
B C
F H
E G
(A, B) l cnh cu
A D
B C
F H
E G
4.3. nh ct, cnh cu
V d:
4 LIN THNG
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A D
B C
F H
E G
(B, C) khng l cnh cu
A D
B C
F H
E G
4. LIN THNG4.3. nh ct, cnh cu
V d:
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A D
B C
F H
E G
(A, B)
(C, E)l cnh cu
4. LIN THNG
4.3. nh ct, cnh cu
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4.4. Ch s lin thngCho G = (V,E) lin thng, k N, k 2.
Nu xo i t (t< k) nh bt k th thu c
vn l lin thng th ni G l th k-lin thngS t nhin ln nht k tho mn iu kin:
- G l k - lin thng.
-Nhng khng c (k + 1) lin thngKhi k c gi l ch s lin thngca G.
4. LIN THNG
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d
a
b e
g
c fG
4.4. Ch s lin thngV d:
a
b e
g
c
fH
4. LIN THNG
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da
b e
g
c fG
da
e
g
fK
G l 2-lin thng
4. LIN THNG4.4. Ch s lin thng
V d:
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Bi 3
MT S TH N V
HNG C BIT
B mn: Khoa hc my tnhKhoa: Cng ngh thng tin SPHN
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NI DUNG
1. th y ?2. th vng?3. th hnh bnh xe?4. th hnh khi?
5. th u?6. th lng phn?
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1. TH Y
Trong i sng chng ta gp nhiu m hnhca th y :
- Biu din quan h quen bit ca cc hc sinhtrong mt lp hc.
- Biu din cc cp u trong mt gii thi um cc i thi u vng trn mt lt
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2. TH VNG
th vng Cn(n 3) l mt th c n nhv1, v2, , vnv c n cnh (v1, v2), (v2, v3), ,(vn-1, vn), (vn, v1).
V d:
2 3
1
3C
1 4
2 3
4C
1 4
2 3
5C
5
Cn c:
-n nh, deg(v) = 2, n cnh.
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3. TH HNH BNH XE th hnh bnh xe: Cho chu trnh Cn(n3) vthc hin:
-Thm mt nh vnew.-Thm cc cnh ni v
newvi nh ca chu trnh
Ta s c th hnh bnh xe. K hiu l Wn(n3).
Wn c:- n + 1 nh.
- deg(vnew) = n, deg(v) = 3, v vnew
- 2n cnh.
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3. TH HNH BNH XE
V d:
1 4
2 3
4W
5
2 3
3W
1
4
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4. TH HNH KHI
th khi n chiu (n1), k hiu Qnl thc 2nnh, mi nh biu din bng mt xu nhphn di n. Hai nh k nhau khi v ch khicc xu nh phn biu din chng khc nhau
ng 1 bt.Bc ca mi nh bng nS cnh l n.2n-1
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4. TH HNH KHI
10 11
0100
2Q
110 111
011
001000
100 101
3Q
1
1Q
0
V d:
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5. TH U
th u bc k:l n th m mi nhu c bc k. th u bc k c n nh, khi :
-
Deg(vi) = k, i = 1, ,n- C (n*k)/2 cnh.
V d:
- Chu trnh Cnl th u bc 2.- th y Knl u bc n 1.
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6. TH LNG PHN
G = (V, E) c gi l th lng phnnutp nh c phn hoch thnh V1 v V2 saocho mi cnh ca th ch ni mt nh ca V1
vi mt nh ca V2. K hiu G = (V1, V2, E)
v1
v3v5
v2
v4
v6
V1 V2
V d:
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6. TH LNG PHN
Hai tnh cht ca th lng phn:
(1) Mi th con ca th lng phn cng l mt th lng phn
(2) th lng phn khng c khuynV d: Xt th biu din quan h hn nhn ca mt
lng.
- Mi nh biu din mt ngi- Mi cnh biu din quan h v chng gia 2 ngi
th hai pha vi V1l tp gm n ng, V2 ltp gm n b.
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6. TH LNG PHN
nh l: Mt th G l th lng phn khiv ch khi mi chu trnh ca n c di chn.
Chng minh:
iu kin cn: Gi s G = (V1, V2, E) l thlng phn.
Xt mt chu trnh bt k ca G (v1, v2, , vn,v1). Nu v1V1 th v2V2, c nh vy vnV2 vn l s chn. Tc l chu trnh c di chn.
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6. TH LNG PHN
iu kin : Gi s G = (V, E) l th mtt c cc chu trnh c di chn.
Ta chng minh rng mi thnh phn lin thng
ca G l mt th lng phn.Xt thnh phn G1c nh v0.
Vi mi nh v ca G1ta chn ng i D niv0vi v.
Nu di ng i D l chn th vV1, nu di l th v V2.
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6. TH LNG PHN
Vi hai tp nh V1 v V2c thit lp nhtrn phn hoch tp nh ca G1thnh 2 phntho mn mi cnh ch ni mt nh ca V1vi
mt nh ca V2.Nh vy, G1l th lng phn
G l th lng phn.
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6. TH LNG PHN
th G = (V, E) l th lng phn y ,k hiu l Km,n, nu G l th hai pha, tp nhV phn hoch thnh 2 tp V1, V2 m |V1| = m v
|V2| = n. Gia hai nh bt k khng cng trongmt lp nh th lun c ng mt cnh ni.Vi Km,nnh trn, c:
- m + n nh.- deg(v) = n v V1, deg(w) = m w V2.
- (m.n)/2 cnh.
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6. TH LNG PHN
V d:
v1
v3
v2
v4
v6
V1 V2K2,3
v3
v2
v4
v6
V1 V2K3,3
v5
v1
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Bi 4
BIU DIN TH TRN MY TNH
S NG CU HAI TH
B mn: Khoa hc my tnhKhoa: Cng ngh thng tin SPHN
NI DUNG
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NI DUNG
1. Ma trn k?2. Ma trn trng s?3. Ma trn lin thuc?4. S ng cu gia hai th?
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1. MA TRN K
Cho G = (V, E), V = {v1, v2, , vn }.Ma trn k biu din G l A = [aij]n x n.
aijl s cnh lin thuc hai nh vi v vj.
V d: 1 2 3 4 5 6123
456 001000
001011
110100
001010
010101
010011
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1. MA TRN K
Nhn xt:- Ma trn k ca th v hng l ma trn
i xng
- Tng cc phn t dng (ct) ca ma trn kchnh bng bc ca nh tng ng (khng ckhuyn).
- Xt ma trn tch Ap= A.AA (p tha s).
Khi aijpl s ng i khc nhau di p tnh i n nh j.
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2. MA TRNG S
th c trng s:Cho G = (V, E) c:- V = {v1, v2, , vn }-e E, e c gn trng s c(e).
Ma trn trng s biu din G l C = [cij]n x n
E)v,(vnu
E)v,(vnu
j
j
i
iji
ij
)v,v(c
c
Trong : {0, +, -}
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2. MA TRNG S
V d:Cij = 0 khi (vi, vj) EA B
D
C
E
F
1
2
3
4
5
6
6
2
4
A B C D E F
A
BC
D
E
F
026400
200603
600150
461002
005004
030240
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3. MA TRN LIN THUC
Cho G = (V, E), vi:- V = {v1, v2, , vn }
- E = {e1, e2, , em }
Ma trn lin thuc biu din G l M = [mij ]n x m
ij
i
vthucnlikhngenu
vthucnlienu
0
1m
j
ij
3 MA LIN THUC
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3. MA LIN THUCV d:
21
3
a
bd
c
e
Nhn xt:
- Hai ct ging nhau tng ng vi hai cnh song song
- Ct m ch c mt phn t bng 1 tng ng vikhuyn.
- Nu th khng c khuyn th tng cc phn t theohng chnh bng bc ca nh tng ng.
a b c d e
12
3 1101000111
11001
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4. S NG CU GIA HAI TH
Trong ho hc, cc th c dng to mhnh cc hp cht. C nhiu cht c cng cng thcphn t nhng cu trc khc nhau. Chng c biudin bng cc th khc nhau.
Cc th c cng cu trc c gi l cc thng cu biu din m hnh ca cng mt cht.
V d: Xt cng thc phn t C2H4O2.
C
OH
O
H
H C
H H C
O
O
C
H
H
H
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4. S NG CU GIA HAI TH
Cho hai n th G1 = (V1, E1) v G2 = (V2, E2)Nu tn ti mt song nh
f: V1 V2
Sao chofbo ton quan h lin k gia cc cpnh, tc l:
(v, w) E1khi v ch khi (f(v),f(w)) E2.
Khi : G1 v G2c gi l ng cuvi nhau.fc gi l mt php ng cu.
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4. S NG CU GIA HAI TH
V d: 1 4
2 3
G
a b
c d
H
f: {1, 2, 3, 4} {a, b, c, d}
f(1) = a f(2) = c
f(3) = b f(4) = d
(1, 2)(a, c) (1, 4)(a, d)
(2, 3) (c, b) (3, 4) (b, d)
G v Hng cuvi nhau
4 S NG CU GIA HAI TH
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4. S NG CU GIA HAI TH
Nhn xt:
- xc nh s ng cu gia hai th lmt bi ton khng ngin, v gia hai thc n nh tn ti ti n! song nh gia hai tp
nh.- ch ra hai th khng ngcuvi nhau
ta ch ra chng khng c mt tnh cht m hai
thngcuphi c:+ S lng nh
+ S lng cnh
+ Bc ca cc nh
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4. S NG CU GIA HAI TH
V d: 1 4
2 3
G
a b
c d
H
+ S nh: cng 4 nh.+ S cnh: cng 4 cnh.
+ Bc ca nh:G: mi nh c bc 2.H: 2 nh bc 2, 1 nh bc 3,
1 nh bc 1
G v H khng
ng cu vi nhau
4 S NG CU GIA HAI TH
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4. S NG CU GIA HAI TH
V d: 1 4
2 3
G H
1 4
2 3
+ S nh: cng 4 nh.+ S cnh: cng 4 cnh.+ Bc ca nh:
G: 2 nh bc 2, 2 nh bc 3 (mi nh bc 2k vi 2 nh bc 3).
H: 2 nh bc 2, 2 nh bc 3 (mi nh bc 2
k vi mt nh bc 3 v mt nh bc 2
G v H khng
ng cu vi nhau
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Bi 5
NG MT NT EULER V
CHU TRNH HAMILTON
B mn: Khoa hc my tnhKhoa: Cng ngh thng tin SPHN
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1. NG MT NT EULER
1 23
4
765
IV
I
III
II
By cy cu thnh ph
Konigsberg
I
IV
II
III
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1. NG MT NT EULER
Cho mt th v hng c n nh, m cnh.Mt dy cha ttc m cnhca th v c
dng:
v1, e1, v2, e2, , vm, em, vm+1Sao cho cnh eini hai nh vi v vi+1.
Dy cnh gi l ng mt nt Euler.
Nu v1 vm+1 th gi ng mt nt Eulerkhp kn.
Nu v1 vm+1th gi ng mt nt Euler m.
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1. NG MT NT EULER
nh l 1: G = (V, E) c ngmt nt Eulerkhp kn khi v ch khi minhca V c bcchn.
nh l 2: G = (V, E) c ngmt nt Eulermkhi v ch khi snhbclca G l 2.
th Euler cngdng trong cc bi tonthc tnh tm hnh trnh ngnnht cho ngiath, xe thu rc, cnh st tun tra.
1 G
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1. NG MT NT EULER
V d:a b
d c
e
1G
G1c ng mt nt Euler kn
1
2
4
35
6
1 NG MT NT EULER
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1. NG MT NT EULER
V d:
2G
a
d
b
ce
G2c ng mt nt Euler m
1
24
356
7
1 NG MT NT EULER
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1. NG MT NT EULER
V d:
2
G
a
d
b
c
e
G2khng c ng mt nt Euler
1 NG MT NT EULER
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1. NG MT NT EULER
Thut ton tm ng mt nt Euler khp knBc 1: Chn nh a lm nhbt u. Xydngngmt nt kp kn con C.
Bc 2: Loib cc cnh trong Ckhith.Loib cc nh c lp(nu c).
Bc 3: Lymtnh chung caC v th
cn li xy dngngmt nt con tip theoC. Ri khp vo C v quay libc 2. Lpcho n khi cch cnhcaht vo C.
1 NG MT NT EULER
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1. NG MT NT EULER
V d: A B
E F
DC
ng mt nt khp kn:
ACEA
1 NG MT NT EULER
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1. NG MT NT EULER
V d:
A B
E F
D
ng mt nt khp kn:ACEA
ng mt nt khp kn:
ACEADB - A
1 NG MT NT EULER
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1. NG MT NT EULER
V d:
E F
D
ng mt nt khp kn:ACEADB - A
ng mt nt khp kn Euler:
ACEAD EFDB - A
2 CHU TRNH HAMILTON
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2. CHU TRNH HAMILTON
Cho G = (V, E), nutntimt chu trnh C i qua ttc cc nh v minhngmtln. Khi C gi lchu trnh Hamilton.
Nu tn ti ng i H c tnh cht nh trn th H
cgi l ngi Hamilton.Vn tm chu trnh Hamilton trong thc nh
ton hc Anh l Hamilton nu ra nm 1858.
n nay, vic tm tiu chun nhnbit thHamilton vn l mv cngcha c thut ton hiuqukim tra mtth c l Hamilton hay khng?
2. CHU TRNH HAMILTON
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2. CHU TRNH HAMILTON
Whiney (1931): Trong thphng c mimin l mt tam gic viiukin 3 cnhbtkca n lp thnh mt tam gic ch khi tam gic l mtmincath th th l Hamilton
Tutte: Mi thphng 4-lin thng nhuc mt chu trnh Hamilton
Dirac (1952): nth v hng G c n > 2nh,minh c bc khng nhhn n/2 l th Hamilton
2. CHU TRNH HAMILTON
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2. CHU TRNH HAMILTON
V d:
a b
d c
1G
a b
d c
a b
d c
2G 3G
G1 khng cha ng i v chu trnh Hamilton
G2cha ng i HamiltonG3cha chu trnh Hamilton
2. CHU TRNH HAMILTON
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2. CHU TRNH HAMILTON
Thutton litk ttc cc chu trnh Hamilton- Thut ton c xy dng trn csthut ton quaylui cho php lit k ttc cc chu trnh Hamilton.
- Pht trinmt dy nh x[1], x[2], , x[k], ca
th G = (V, E) chobi danh sch k Ke(v), vV
2. CHU TRNH HAMILTON
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2. CHU TRNH HAMILTONThutton litk ttc cc chu trnh HamiltonProcedure Hamilton(k);
BeginFor yKe(x[k-1]) do
if (k=n+1)and(y=v0) then GhiNhan(x[1], , x[n], v0)else if ChuaXet[y] then
Begin
x[k] := y;
ChuaXet[y] := False;
Hamilton(k+1);
ChuaXet[y] = True;
End;
End;
BeginFor vV do ChuaXet[v] = True;
x[1] := v0;
ChuaXet[v0] := False;
Hamilton(2);
End.
2. CHU TRNH HAMILTON
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2. CHU TRNH HAMILTON
Vd:
1 35
2
4
1
1 1 1 1
2 4
4
3 5
55 5
5
3
3
33144 25
23
21
4 2
1
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Bi 6
TH PHNG
V BI TON T MU TH
B mn: Khoa hc my tnhKhoa: Cng ngh thng tin SPHN
NI DUNG
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1. Khi nim th phng?2. Cng thc Euler?3. nh l Kuratowski?4. T mu th?
1 KHI NIM TH PHNG
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1. KHI NIM TH PHNG
Bi ton:Xy ng i t ba ngi nh vi baging ring r nh hnh sau.
Ging 1 Ging 2 Ging 3
C cch nom khng c cc
ng ct nhau.
1. KHI NIM TH PHNG
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M hnh ho bi ton bng th phn i y
K3.3.
K3,3
Khi cu hi trn c din t nh sau:
Tn ti hay khng cch v th phn iy K3,3trn mt mt phng sao cho khng chai cnh no ct nhau?
1. KHI NIM TH PHNG
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Mtthcgi l thphngnu ta cthv n trn mtmtphng sao cho khng ccc cnh no ct nhau mtim khngphi lnhcath.
Ch : mt th c thcvbngnhiucch khc nhau. Nu tn ti mt cch v thomn nhngha trn th n l thphng.
1. KHI NIM TH PHNG
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V d 1:
K4 K4
K4l th phng.
1. KHI NIM TH PHNG
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V d 2:
Q3l th phng.
Q3 Q3
2 CNG THC EULER
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2. CNG THC EULER
Min phngBiu dinphng ca th chia mtphngthnh cc min khc nhau (kcmin v hn).
AB
CD
Vd: thphngsau chia mtphngthnh cc min A, B,C, D nh hnh v.
2 CNG THC EULER
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2. CNG THC EULERCng thc EulerCho G l mtnthphng lin thng vi
m l s cnh v n l s nh. Gi f l s mintrnbiudinphngca G.
Khi :f + n = m + 2
Tng qut ha: G l mt th phng vi nnh, m cnh, f min phng, (G) l s thnhphn lin thng. Khi ta c:
f + n = m + (G) + 1
2 CNG THC EULER
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2. CNG THC EULER
V d 1: th bn c:e = 7
v = 5Ta c:
f = ev + 2
= 75 + 2= 4
AB
C D
2. CNG THC EULER
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V d 2: Chng minh rng K3,3khng phi l th phng.
Gii: Gi s K3,3l th phng.Mi chu trnh ca K3,3gm t nht l 4 cnh.
Nn mi min b gii hn bi t nht 4 cnh.Tng s cnh gii hn cc mt l 4f.M mt cnh thuc vo hai min.Do : 2e 4f 2e 4(e v +2)Ta c: 2e = 2.9 = 18
4(ev + 2) = 4(9 - 6 + 2) = 20V l.Nh vy K3,3khng l th phng.
K3,3
3 NH L KURATOWSKI
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3. NH L KURATOWSKI
Ta c: K3,3 v K5khng l th phng.Nu K3,3hoc K5 l th con ca G th G l th khngphng.
Php nhn chia scpcnh (u, v) trn th l:+ Loi b cnh ny khi th.
+ Thm vo nh w v hai cnh (u, w), (w,v).
Hai th G v H c gi l ng phi nuchng c thnhnc t cng mt thbngmt dy cc php nhn chia scp.
3 NH L KURATOWSKI
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3. NH L KURATOWSKI
nh l Kuratowski: Mtth l thphng khi v chkhi n khng chath con ng phi vi K3,3hoc K5
a b
d e
1G
a b
c d e
f
h
g
2G
a b
c d e
ki
g j
3G
Vd: G2 v G3 l hai thng phi v chng c th
nhnctth G1bng cc php phn chia scp.
4 T MU TH
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4. T MU TH
Bi ton: phn bit cc min trn bn taphi t mu chngbng cc mu khc nhau.
Hicn t nht bao nhiu mu t mtbn
btk sao cho cc mink nhau khng cng mtmu.
B
B
C
DEF
C
A
D
E
FG
4 T MU TH
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4. T MU THM hnh ho bi ton:
+ Mi min tng ng mt nh ca th.+ Hai nh c cnh ni nu chng l hai min cchung bin gii.
th nhn c gi l th i ngu ca bn . th i ngu ca bn l th phng.Bi ton tngng: t mu cc nhca thsao cho hai nh k nhau th c t bi hai mukhc nhau.
4 T MU TH
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4. T MU TH
V d:
AB
C
DE
A
B
D
EC
4 T MU TH
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4. T MU THnh ngha: T mu th l vic gn mu chocc nh ca th sao cho khng c hai nh knhau c gn cng mt mu.nh ngha: s mu ca mt th l s mu ti
thiu cn t mu th ny.nh l 4 mu: s mu ca mt th phng btk l mt s khng ln hn 4.
Nhn xt:-S mu ca th lng phn l 2 mu.-S mu ca th y Kn l n mu
4. T MU TH
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4. T MU TH
V d:
D
A E
B
C
AB
C
DE
4. T MU TH
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4. T MU THV d: Bi ton lp lch thiHy lp lch thi trong trng i hc sao chokhng c sinh vin no phi thi ng thi haimn cng mt lc
Gii: M hnh ha bi ton nh sau:-Minh l mt mn thi- Hai nh c cnhninu l hai mn m mt
sinh vin no phi thi.Thi gian mi mn thi ngvimt mu.Bi ton trthnh bi ton t mu cho th trn
sao cho hai nh k nhau c mu khc nhau
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Bi 7
BI TON TM NG I
NGN NHT
B mn: Khoa hc my tnhKhoa: Cng ngh thng tin SPHN
1. BI TON THC T
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1. BI TON THC T
C 6 im du lch trong mt khu sinh thi l a,b, c, d, e, z. Gia hai im c th c hoc khngc ngitrctip.
Hy tm ngi c khong cch ngnnhttim a n z.
1. BI TON THC T
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1. BI TON THC T
Bi ton c m hnh hobngth c trngsnh sau:
+ Mi nh biu din mt im du lch.
+ Hai nh c cnh ni nu c ng i trc tip.+ Trngscacnhc gn l khong cch t
im ny sang im kia.
1. BI TON THC T
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1. BI TON THC T
th m hnh ho bi tonc
e
b
d
a z
2
3
5
2
1
4
2
1
ngingnnht l ngi c tngtrngs cch cnhca n l nhnht.
2. THUT TON DIJKSTRA
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Bi ton: Tm ngingnnhttnh a n z
cath c trngs lin thng G = (V, E).Thut ton Dijkstra ( xut nm 1959 do nhton hc H Lan E.Dijkstra)
Gi L(v) l di ngingnnht tnh annh v.
S l tp cc nh tm cngingnnht
t a n n.Pr(v) l nh ngay trcnh v trn ngingn
nht
2. THUT TON DIJKSTRA
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Thut ton:
+Bc 1: L(a) = 0, S = , vS, v a: L(v) =
+Bc 2: Nu z S th kt thc.
+Bc 3: Chn v S sao cho L(v) l nh nht.
a v vo S.
+Bc 4: vi mi nh x lin k v v x S:
Nu L(x) > L(v) + c(v, x) thL(x) = L(v) + c(v,x) v Pr(x) = v
Quay li bc 2.
3. V D cb25
2
a e
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3. V
B1: L(a) = 0, S =
L(b) = L(c) = L(d) = L(e) = L(z) =
ed
a z
2
3
2
1
4
2
1
B2: v = a, S = {a}.L(b) = min{, 2 + 0} = 2
L(d) = min{, 3 + 0} = 3
B3: v = b, S = {a,b}, L(c) = 7, L(e) = 4, L(d) = 3, L(z)=
B4: v = d, S = {a,b,d}, L(c) = 7, L(e) = 4, L(z) =
B5: v = e, S = {a,b,d,e}, L(c) = 5, L(z) = 8
B6: v = c, S = {a,b,d,e}, L(z) = 7
c
b
a
3. V D a e
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c
e
b
d
a z2
3
5
2
1
4
21 c
b
a
a e
ng i ngn nht t a n z l:
abec - z
3. V D
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Tm ng i ngn nht t A n cch nhcn li trong th sau:
A B
E F
DC
1
2
3
54
5
2 1
4
6
3. V D(0 ) (3,A)
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A B
E F
DC
1
2
3
54
5
2 1
4
6
(0,) (3,A)
(2,A)
(4,A)
(6,A)
3. V D(0 ) (3,A)
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A B
E F
DC
1
2
3
54
5
2 1
4
6
(0,) (3,A)
(2,A)
(6,A)
(4,A)(3,C)
3. V D(0 ) (3,A)
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A B
E F
DC
1
2
3
54
5
2 1
4
6
(0,) (3,A)
(2,A)
(6,A)
(3,C) (8,B)
3. V D(0,) (3,A)
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A B
E F
DC
1
2
3
54
5
2 1
4
6
(0,) (3, )
(2,A)
(6,A)
(3,C)
(5,E)
(8,B)(7,E)
3. V D A B3(0,) (3,A)
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E F
DC
1
2
54
5
2 1
4
6
(2,A)
(3,C)
(5,E)
(7,E)
AB (3)
AC (2)
ACE (3)
ACED (5)ACEF (7)
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Bi 8
CY V NG DNG
B mn: Khoa hc my tnh
Khoa: Cng ngh thng tin SPHN
NI DUNG
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1. Cy?
2. Cy nh phn tm kim?3. Cy biu thc s hc?4. Duyt cy nh phn?
1. CY
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nhngha: Cy l mtn th v hng,lin thng v khng c chu trnh.
Cy cgc: nu ta chnmtnhcbitgi lgcca cy v nhhng cc cnh trn cy tgci ra th ta cmtth c hnggi lcy c gc.
Chnnh lm gc khc nhau sto ra cc cykhc nhau.
1. CY
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V d:
a
b
d fe g
c h h
a
b
e c
d f g g h
e
fd
b
a c
1. CY
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Mtskhi nim: Cho T l mt cy c gc, v
l mtnh khc gcca T.+ Cha ca v l nh u nu c mt cnh c
hng duy nhtt u v. Khi , u cgi l
chaca
v; v l conca
u.+Anh eml cc nh c cng cha.+ Ttincamtnh khc gc l cc nh trn
ngitgcnnh.+ Con chu ca v l cc nh c v l t tin.+Ll nh khng c con.+nh trongl cc nh c con.
1. CY
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+ Cy con vigc a l th con ca cy ang xt,
bao gm a v cc con chu ca n cng ttc cccnh lin thucvi cc con chu ca a.
+ Mc ca mt nh: l khong cch t gc n
n.+ Chiu cao ca cy: mc lnnhtcamtnh
btk trong cy gi l chiu cao ca cy.
1. CYa
d
d
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V d:
- Conca a l b, c, d
- Chaca b, c, d l a
- b, c, d l anh em.- Con chuca d l g,h, i, l, m
- T tinca g l a, d.- a, b, d, f, h l nhtrong
- e, c, j, k, l, m, i l l
- Cy bn phi l cy congc d ca cy bn tri
-Mcca b, c, d l 1.Mc ca e, f, g, h, i l 2.
- Chiu caoca cy l 3
b cd
ef g h i
j k lm l m
ig h
1. CYnh l 1: Mt cy bt k vi t nht 2 nh th c t
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ynht hai nh treo.
nh l 2: Mt cy c n nh c ng n1 cnh.
1. CYnh l Daisy Chain: Cho T l mtth c n nh.
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yKhi cc khngnh sau l tngng:
(1): T l mt cy.(2): T khng c chu trnh v c n - 1 cnh.
(3): T l mtth lin thng v nuhybtkmtcnh no ca n cng lm mt tnh lin thng.
(4): Gia 2 nhbtkca T, lun tntimtngin duy nhtni 2 nh ny.
(5): T khng c chu trnh v nu thm mtcnhmini 2 nhbtkca T th sto ra mt chu trnh.
(6): T lin thng v c n - 1 cnh.
2. CY NH PHN TM KIMnh ngha cy nh phn: l mt cy c gc
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nh ngha cy nhphn: l mt cy c gc.
Mi nh trong ca cy khng c qu hai con,trong c mt con bn tri v con bnphi.
Cy nh phn cgi l ynuminhtrong ca n c ng 2 con.
Cy nh phn tm kim: l mt cy nh phntho mn:
(1) Mi nh c gn mt kho khc nhau.(2) Kho camtnh trong lnhn kho contri v nhhn khoa conphica n.
2. CY NH PHN TM KIMThut ton xy dng cy nh phn tm kim cho mt dy
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Thut ton xy dng cy nh phn tm kim cho mt dykho.
(1) Chn kho v0lm gc.
(2) [Lp] vi mi kho v cn li thc hin nh sau:
+ v0
gc.
+ v > v0: Nu v0 c con phi th v0 con phi
nu khng th v l con phi ca v0.
+ v < v0: Nu v c con tri th v
0 con tri
nu khng th v l con tri ca v0.
Thut ton kt thc khi mi kho ca vo cy.
2. CY NH PHN TM KIM
d 11 8 10 12 16 14 1
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V d: 11, 8, 7, 10, 12, 16, 14, 17
11
8
7 10
12
16
14 17
2. CY NH PHN TM KIM
V d 11 8 7 10 12 16 14 17
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V d: 11, 8, 7, 10, 12, 16, 14, 17
11
8
7 10
12
16
14 17
Tm kho 16:
+ So snh 11.
+ so snh 12.
+ So snh 16.
16 c trongdy cho
2. CY NH PHN TM KIM
V d 11 8 7 10 12 16 14 17
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V d: 11, 8, 7, 10, 12, 16, 14, 17
11
8
7 10
12
16
14 17
Tm kho 15:
+ So snh 11.
+ so snh 12.
+ So snh 16.
+ So snh 14.
15 khng ctrong dy cho
3. CY BIU THC S HCCy biu din biu thc s hc: l mt cy nh
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Cy biu din biu thc s hc: l mt cy nhphn tho mn:
(1) Nt trong biu din ton t 2 ngi (+ - * / ^).
(2) L biu din cho mt ton hng ca biu thc.
Biu thc E1 E
2c biu din bi cy c:
+ Gc biu din .
+ Con tri biu din cho biu thc E1.
+ Con phi biu din cho biu thc E2.
3. CY BIU THC S HCV d: 5/15*142^32E
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V d: 5/15*142^32 E
/
1 52 3 4 1
*4-1 15/5
(2+3)^2 (4-1)*(15/5)
^
2
2+3
4. DUYT CY NH PHN3 phng php duyt cy nh phn
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3 phng php duyt cy nh phn
+ Duyt tin t (PreOrder):(1): Duyt nt gc.
(2): Duyt con tri theo phng php tin t.
(3): Duyt con phi theo phng php tin t.+ Duyt trung t (InOrder):
(1): Duyt con tri theo phng php trung t.
(2): Duyt nt gc.
(3): Duyt con phi theo phng php trung t.
4. DUYT CY NH PHN
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+Duyt hu t (PostOrder):
(1): Duyt con tri theo phng php hu t.
(2): Duyt con phi theo phng php hu t.
(3): Duyt nt gc.
4. DUYT CY NH PHNV d: Duyt cy nh phn sau theo phng
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V d: Duyt cy nh phn sau theo phng
php tin t
1712 16 14108 711
11
8
7 10
12
16
14 17
4. DUYT CY NH PHNV d: Duyt cy nh phn sau theo phng
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V d: Duyt cy nh phn sau theo phng
php trung t
1712 16141087 11
11
8
7 10
12
16
14 17
4. DUYT CY NH PHNV d: Duyt cy nh phn sau theo phng
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V d: Duyt cy nh phn sau theo phng
php hu t
17 12161410 87 11
11
8
7 10
12
16
14 17
Bi 9
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CY KHUNG CA TH
B mn: Khoa hc my tnhKhoa: Cng ngh thng tin - SPHN
NI DUNG
1 h h
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1. nh ngha?
2. Xc nh cy khung?2.1. Theo chiu rng2.2. Theo chiu su
3. Cy khung nh nht?
1. NH NGHA
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nhngha cy khung: Cho th G = (V, E) l
mtth v hng lin thng.Cy T = (V, F) vi F l tp con ca E cgi
l cy khung cath G.
V d:
a
b
c
e dG1
a
b
c
edG2
a
b
c
e dG3
a
b
c
e Gd
2. XC NH CY KHUNG
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Xc nh cy khung l vic xy dng mt cychattc cc nhcath.
Hai thut ton xc nh cy khung l:
+ Xc nh u tin theo chiu rng.
+ Xc nh u tin theo chiu su.
2. XC NH CY KHUNG2.1. Theo chiu su.
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+Bc 1: Ly mt nh a lm gc ca cy khung.+ Bc 2: Xy dng ng i t nh ny bng
cch ghp lnlt cc cnh vo. Micnhc ghpvo ni nh cui cng ca ng i v mt nhcha c trong cy. Thc hin n khi khng ghpc thm cnh no na.
+Bc 3:Nungichattc cc nhcath th chnh l cy khung. Nu khng th chuyn
sangbc 4.+ Bc 4: Quay lui li nh ngay trc nh cuicng cangi v xy dngngimibtutnh ny.Nu khng c th li tipnhna.
2. XC NH CY KHUNG2.1. Theo chiu su
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V d:
C
A
B
D
E
F GTT Cnh
ghp vo
1 (A,B)
2 (B,F)
3 (F,G)
4 (G,E)
5 (E,D)
6 (D,C)
C
A
B
D
E
F G
2. XC NH CY KHUNG2.2. Theo chiu rng
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2.2. Theo chiu rng
Cc bc thc hin ca BFS(v)+Bc 1: Chn nh a lm gc ca cy.
+ Bc 2: ghp cc cnh lin thuc vi gc.
Cc nhkvigc trongbc ny c mc l 1.+ Bc 3: tip tc ghp cc cnh lin thuc
nh mc 1 sao cho khng to chu trnh. Cc
nhca vo bc ny c mc l 2.+Bc 4: Tip tc qu trnh khi tt c cc nh
c ghp vo cy.
2. XC NH CY KHUNG2.1. Theo chiu su
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V d:
C
A
B
D
E
FG
TT Cnhghp vo
1 (A,B)
2 (A,C)
3 (A,D)
4 (B,F)
5 (D,G)
6 (D,E)
C
A
B
D
E
F G
2. XC NH CY KHUNG
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ng dng hai thut ton trn:
- Tm ng i gia hai nh. ng i xydng khi u tin theo chiu rng l ngingnnht.
- Tm cc thnhphn lin thng cath.
- Thut ton theo chiu su tm chu trnh.
3. CY KHUNG NH NHT
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nh ngha cy khung nh nht:
Cy khung nh nht trong mt th linthng, c trng s l mt cy khung c tngtrngs trn cc cnhca n l nhnht.
Thut ton tm cy khung nh nht:
+ Prim (Robert Prim - 1957)
+ Kruskal (Joseph Kruskal1965)
3. CY KHUNG NH NHT
Th t t P i
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Thut ton Prim:
th G = (V, E) lin thng, c n nh.(1): Chn mt cnh bt k c trng s nh nht, t
n vo cy khung.
(2): Ln lt ghp vo cy cc cnh c trng snhnhtlin thucvimtnhca cy v khngto ra chu trnh trong cy.
Thut ton dngli khi n 1 cnhc ghp vocy.
3. CY KHUNG NH NHT
V d: Dng thut ton Prim tm cy khung nh
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V d: Dng thut ton Prim tm cy khung nh
nhtcath sau:A B
E
F
D
C 1
2
3
4
5
6
6
2
3D
C 1
A
2
F
3E
2
B4
Cy khung ca G
Trng s: 12
GCnh c chn
Cnh c th chn
3. CY KHUNG NH NHT
Th t t K k l
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Thut ton Kruskal:
th G = (V, E) lin thng, c n nh.(1): Chnmtcnhbtk c trngsnhnht,t
n vo cy khung.
(2): Ln lt ghp vo cy cc cnh c trng snhnht m khng to ra chu trnh trong cy.
Thut ton dngli khi n 1 cnhc ghp vocy.
3. CY KHUNG NH NHT
V d: Dng thut ton Kruskal tm cy khung nh
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Vd: Dng thut ton Kruskal tm cy khung nh
nhtcath sau:A B
E
F
D
C 1
2
3
4
5
6
6
2
3D
C 1
A
2
3E
F
2
B4
Cy khung ca G
Trng s: 12
GCnh c chn
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Bi 10
TH C HNG
B mn: Khoa hc my tnhKhoa: Cng ngh thng tin - SPHN
NI DUNG
1 Bc ca nh?
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1. Bc ca nh?
2. Biu din th trn my tnh?3. Dy chuyn, bng chuyn v chu trnh?4. th c hng lin thng?
5. th phn chu trnh?
1. BC CA NH h Ch G
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Bc ca nh: Cho G = [V, E], v V. Ta c:+ deg-(v) (bc vo) l scnhnhn v l nh
cui(nh vo)+ deg+(v) (bc ra) l scnhnhn v l nhu
(nh ra).Mt khuyn trn th c hngsng gp 2
nv vobc vo vbc ra tngngcanh.
1. BC CA NHV d:
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V d:
21
3 4
5
deg+(1) = 2, deg-(1) = 0
deg+(2) = 1, deg-(2) = 2
deg+
(3) = 1, deg-
(3) = 1deg+(4) = 0, deg-(4) = 2
deg+(5) = 2, deg-(5) = 1
Ch :
deg+(v) = 0v l nh h lu.
deg-(v) = 0v l nh ngun.
1. BC CA NH
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Vv Vv
vvm )(deg)(deg
nh l: Gi s G = [V, E]l mt th c hngc m cung khi :
2. BIU DIN TH TRN MY TNH
Ma trn k: Cho n th G = [V E]
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Ma trn k: Cho n th G = [V, E]
vi V = {v1, , vn}Ma trn k A = [aij]n x nbiu din G.
aijl s cung i t vin vjV d:
21
3 4
5
01000
00000
00010
01000
00110
5
4
3
2
1
54321
2. BIU DIN TH TRN MY TNH
Ma trn k lin thuc:
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Ma trn k lin thuc:
Cho n th G = [V,E]vi V = {v1, , vn}
E = {e1, e2, , em}
Ma trnk M = [mij]n x mbiudin G c xcnhnh sau:
+ mij= 1 nu vil nh ra ej
+ mij = -1 nu vil nh vo ej
+ mij= 0 nu ejl khng lin thuc vi
2. BIU DIN TH TRN MY TNH
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V d:
21
3 4
5
4,52,34,23,12,1
10000
10100
01010
01101
00011
5
4
3
2
1
3. DY CHUYN, BNG CHUYNV CHU TRNH
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Dy chuyn: l mt dy cc cung e1, e
2, , e
k
sao cho khibhngi th n trthnh mt dycnhktip trong th v hng thu c
Bngchuyn: l mt dy chuyn e1, e2, , ekm nh ch ca cung ei l nhxut pht cacung ei+1vi i = 1,, k-1.
Chu trnh: l mtbngchuyn khp kn.
3. DY CHUYN, BNG CHUYNV CHU TRNH
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V d:
[1,2]-[2,4]-[5,4] l dy chuyn.
[1,3]-[3,2]-[2,4] l bng chuyn.
[3,2]-[2,4]-[4,3] l chu trnh
21
3 4
5
3. DY CHUYN, BNG CHUYNV CHU TRNH
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Dy chuyn (bng chuyn) c bn l nhngdy chuyn(bngchuyn) m minhchxuthinnhiunhtmtln.
Bng chuyn cbn cha tt c cc nh cathcgi lbngchuyn Hamilton.
Chu trnh cbnchattc cc cnhcath
cgi l chu trnh Euler.Khuyn c coi l chu trnh c di 1.
4. TH C HNG LIN THNG
Cho G [V E] khi :
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Cho G = [V, E], khi :
+ 2 nh v v w cgi l lin thng yunutnti dy truynni chng.
+ G l lin thng yunu hai nhbtkulin thng yu.
+ G l lin thng mnh nu hai nhbt kcnivi nhaubimtbngchuyn.
4. TH C HNG LIN THNG
11V d:
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3 2
4 5
1
H
3 2
4 5
1
G
V d:
G l lin thng yu
H l lin thng mnh
5. TH PHN CHU TRNH
th phn chu trnh: l mt th c hng
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thphn chu trnh: l mtth c hng
G khng cha chu trnh.nh l 1: trong mithphn chu trnh lun
tn ti t nht mt nh ngun v mt nh h
lu.nh l 2: th c hng G = [V, E] vi nnh l thphn chu trnh khi v ch khi chngta c thnhs cc nhcathbi 1, 2, ,n sao cho micnh [i, j] tho mn quan h i < j.
5. TH PHN CHU TRNHV d:
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V d:
DA
C E
FB
h h l
1
2
3
4
5
6