chuong 6. ly thuyet do thi

7/31/2019 Chuong 6. Ly Thuyet Do Thi

1/178

Bi 1

NH NGHA V PHN LOI

B mn: Khoa hc my tnhKhoa: Cng ngh thng tin - SPHN


2/178

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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1.1.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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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


21/178

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



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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.



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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



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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


1.2. 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

V d:


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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


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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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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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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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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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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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



80/178

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?



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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



92/178


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



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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



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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.



95/178


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


NI DUNG


97/178

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


102/178

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


105/178

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


1. BI TON THC T


116/178

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


119/178

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)


125/178

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)


126/178

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)


127/178

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


133/178

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


135/178

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


136/178

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.


137/178

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


141/178

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


142/178

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


144/178

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



148/178


php trung t

1712 16141087 11

11

8

7 10

12

16

14 17



149/178


php hu t

17 12161410 87 11

11

8

7 10

12

16

14 17

Bi 9


150/178

CY KHUNG CA TH


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


152/178

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


155/178

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


160/178

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


NI DUNG

1 Bc ca nh?


165/178

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


166/178

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:


167/178

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


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



171/178

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.



173/178

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



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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:


178/178

V d:

DA

C E

FB

h h l

1

2

3

4

5

6

chuong 6. ly thuyet do thi

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