7/23/2019 LTDT- C1.pdf

1/17

13

CHNG I

CC KHI NIM C BN CA L THUYT TH

(tun 1: Tng cng c 2 tit l thuyt v 2 tit hng dn bi tp/thc hnh)

1. NH NGHA TH

th l mt cu trc ri rc bao gm cc nh v cc cnh ni cc nh ny. Chng ta

phn bit cc loi th khc nhau bi kiuv s lng cnh ni hai nh no ca

th.

nh ngha 1.

n th v hng G = (V,E) bao gm V l tp cc nh, v E l tp cc cp khng cth t gm hai phn t khc nhau ca V gi l cc cnh.

nh ngha 2.

a th v hng G= (V, E) bao gm V l tp cc nh, v E l tp cc cp khng c th

t gm hai phn t khc nhau ca V gi l cc cnh. Hai cnh e 1v e2c gi l cnh

lp nu chng cng tng ng vi mt cp nh.

nh ngha 3.

Gi th v hng G = (V, E) bao gm V l tp cc nh v E l tp cc cp khng c

th t gm hai phn t (khng nht thit phi khc nhau) ca V gi l cnh. Cnh e c

gi l khuyn nu n c dng e = (u, u).

nh ngha 4.

n th c hng G = (V, E) bao gm V l tp cc nh v E l tp cc cp c th t

gm hai phn t khc nhau ca V gi l cc cung.

nh ngha 5.

a th c hng G = (V, E) bao gm V l tp cc nh v E l tp cc cp c th t

gm hai phn t khc nhau ca V gi l cc cung. Hai cung e1, e2tng ng vi cng mt

cp nh c gi l cung lp.

Trong cc phn tip theo ch yu chng ta s lm vic vi n th v hng v n

th c hng. V vy, cho ngn gn, ta s b qua tnh t n khi nhc n chng.

2. CC THUT NG C BN

nh ngha 1.

Hai nh u v v ca th v hng G c gi l k nhau nu (u,v) l cnh ca th G.

Nu e = (u, v) l cnh ca th ta ni cnh ny l lin thuc vi hai nh u v v, hoc

7/23/2019 LTDT- C1.pdf

2/17

14

cng ni l ni nh u v nh v, ng thi cc nh u v v s c gi l cc nh u ca

cnh (u, v).

c th bit c vao nhiu cnh lin thuc vi mt nh, ta a vo nh ngha sau

nh ngha 2.

Ta gi bc ca nh v trong th v hng l s cnh lin thuc vi n v s k hiu l

deg(v).

Th d 1.

Xt th cho trong hnh 1, ta c

deg(a) = 1, deg(b) = 4, deg(c) = 4, deg(f) = 3,

deg(d) = 1, deg(e) = 3, deg(g) = 0nh bc 0 gi l nh c lp. nh bc 1 c gi l nh treo. Trong v d trn nh g l

nh c lp, a v d l cc nh treo. Bc ca nh c tnh cht sau:

nh l 1.

Gi s G = (V, E) l th v hng vi m cnh. Khi tng bc ca tt c cc nh bng

hai ln s cnh.

Th d 2.

th vi n nh c bc l 6 c bao nhiu cnh?Gii: Theo nh l 1 ta c 2m = 6n. T suy ra tng cc cnhca th l 3n.

H qu.

Trong th v hng, s nh bc l (ngha l c bc l s l) l mt s chn.

nh ngha 3.

Nu e = (u, v) l cung ca th c hng G th ta ni hai nh u v v l k nhau, v ni

cung (u, v) ni nh u vi nh v hoc cng ni cung ny l i ra khi nh u v vo nh v.

nh u(v) s c g l nh u (cui) ca cung (u,v).

Tng t nh khi nim bc, i vi th c hng ta c khi nim bn bc ra v bn bc

vo ca mt nh.

7/23/2019 LTDT- C1.pdf

3/17

15

nh ngha 4.

Ta gi bn bc ra (bn bc vo) ca nh v trong th c hng l s cung ca th i

ra khi n (i vo n) v k hiu l deg+(v) (deg-(v))

Th d 3.

Xt th cho trong hnh 2. Ta c

deg-(a)=1, deg

-(b)=2, deg

-(c)=2, deg

-(d)=2, deg

-(e) = 2.

deg+(a)=3, deg

+(b)=1, deg

+(c)=1, deg

+(d)=2, deg

+(e)=2.

Do mi cung (u, v) s c tnh mt ln trong bn bc vo ca nh v v mt ln trong bn

bc ra ca nh u nn ta c:

nh l 2.

Gi s G = (V, E) l th c hng. Khi

Tng tt c cc bn bc ra bng tng tt c cc bn bc vo bng s cung.

th v hng thu c bng cch b qua hng trn cc cung c gi l th v

hng tng ngvi th c hng cho.

3. NG I. CHU TRNH. TH LIN THNG

nh ngha 1.

ng i di n t nh u n nh v, trong n ls nguyn dng, trn th v

hng G = (V, E) l dyx0, x1,, xn-1, xntrong u = x0, v = xn, (xi, xi+1) E, i = 0, 1, 2,, n-1.

ng i ni trn cn c th biu din di dng dy cc cnh:

(x0, x1), (x1, x2), , (xn-1, xn)

nh u gi l nh u, cn nh v gi l nh cui ca ng i. ng i c nh u

trng vi nh cui (tc l u = v) c gi l chu trnh. ng i hay chu trnh c gi

l nnu nh khng c cnh no b lp li.

Th d 1.

7/23/2019 LTDT- C1.pdf

4/17

16

Trn th v hng cho trong hnh 1: a, d, c, f, e l ng i n di 4. Cn d, e, c, a

khng l ng i, do (c,e) khng phi l cnh ca th. Dy b, c, f, e, b l chu trnh

di 4. ng i a, b, e, d, a, b c di l 5 khng phi l ng i n, do cnh (a, b) c

mt trong n 2 ln.

Khi nim ng i v chu trnh trn th c hng c nh ngha hon ton tng t

nh trong trng hp th v hng, ch khc l ta c ch n hng trn cc cung.

nh ngha 2.

ng i di n t nh u n nh v, trong , n l s nguyn dng, trn th c

hng G = (V, E) l dyx0, x1,, xn-1, xn

trong u = x0, v = xn, (xi, xi+1) E, i = 0, 1, 2,, n-1.

ng i ni trn cn c th biu din di dng dy cc cung:

(x0, x1), (x1, x2), , (xn-1, xn)

nh u gi l nh u, cn nh v gi l nh cui ca ng i. ng i c nh u

trng vi nh cui (tc l u = v) c gi l chu trnh. ng i hay chu trnh c gi

l nnu nh khng c cnh no b lp li.

Th d 2.

Trn th c hng cho trong hnh 1: a, d, c, f, el ng i n di 4. Cn d, e, c, a

khng l ng i, do (c,e) khng phi l cnh ca th. Dy b, c, f, e, b l chu trnh

di 4. ng i a, b, e, d, a, b c di l 5 khng phi l ng i n, do cnh (a, b) cmt trong n 2 ln.

nh ngha 3.

th v hng G = (V, E) c gi l lin thng nu lun tm c ng i gia hai

nh bt k ca n.

nh ngha 4.

Ta gi th con ca th G = (V, E) l th H = (W, F), trong WV v FE.

7/23/2019 LTDT- C1.pdf

5/17

17

Trong trng hp th l khng lin thng, n s r ra thnh mt s th con lin thng

i mt khng c nh chung. Nhng th con lin thng nh vy ta s gi l cc thnh

phn lin thngca th.

nh ngha 5.

nh v c gi l nh r nhnhnu vic loi b v cng vi cc cnh linthuc vi n

khi th lm tng s thnh phn lin thng ca th. Cnh e c gi l cunu vic

loi b n khi th lm tng s thnh phn lin thng ca th.

nh ngha 6.

th c hng G = (V,E) c gi l lin thng mnh nu lun tm c ng i gia

hai nh bt k ca n.

nh ngha 7.

th c hng G = (V, E) c gi l lin thng yu nu th v hng tng ng vi

n l v hng lin thng.

4. MT S DNG TH C BIT

th y .

th y n nh, k hiu bi Kn, l n th v hngm gia hai nh bt k ca

n lun c cnh ni.

Cc th K3, K4, K5cho trong hnh di y.

Hnh 1. th y K3, K4, K5 th y Knc tt c n(n-1)/2 cnh, n l n th c nhiu cnh nht.

th hai pha.

n th G=(V,E) c gi l hai pha nu nh tp nh V ca n c th phn hoch

thnh hai tp X v Y sao cho mi cnh ca th ch ni mt nh no trong X vi mt

nh no trong Y. Khi ta s s dng k hiu G=(XY, E) ch th hai pha vi

tp nh X Y.

nh l sau y cho php nhn bit mt n th c phi l hai pha hay khng.

7/23/2019 LTDT- C1.pdf

6/17

18

nh l 1.

n th l th hai pha khi v ch khi n khng cha chu trnh di l.

Hnh 2. th hai pha

th phng.

th c gi l th phng nu ta c th v n trn mt phng sao cho cc cnh ca n

khng ct nhau ngoi nh. Cch v nh vy s c gi l biu din phng ca th.

Th d th K4l phng, v c th v n trn mt phng sao cho cc cnh ca n khng

ct nhaungoi nh (xem hnh 6).

Hnh 3. th K4l th phng

Mt iu ng lu nu th l phng th lun c th v n trn mt phng vi cc cnh

ni l cc on thng khng ct nhau ngoi nh (v d xem cch v K4trong hnh 6).

nhn bit xem mt th c phi l th phng c th s dng nh l Kuratovski, m

pht biu n ta cn mt s khi nim sau: Ta gi mt php chia cnh (u,v) ca th l

vic loi b cnh ny khi th v thm vo th mt nh mi w cng vi hai cnh

(u,w), (w, u) . Hai th G(V,E) v H=(W,F) c gi l ng cu nu chng c th thu

c t cng mt th no nh php chia cnh.

nh l 2 (Kuratovski).

th l phng khi v ch khi n khng cha th con ng cu vi K3,3 hoc K5.

Trong trnghp ring, th K3,3 hoc K5khng phi l th phng. Bi ton v tnhphng ca th K3,3l bi ton ni ting v ba cn h v ba h thng cung cp nng

7/23/2019 LTDT- C1.pdf

7/17

19

lng cho chng: Cn xy dng h thng ng cung cp nng lng vi mi mt cn h

ni trn sao cho chng khng ct nhau.

th phng cn tm c nhng ng dng quan trng trong cng ngh ch to mch in.

Biu din phng ca th s chia mt phng ra thnh cc min, trong c th c c

min khng b chn. Th d, biu din phng ca th cho trong hnh 7 chia mt phng ra

thnh 6 min R1, R2,. . . .R6.

Hnh 4.Cc min tng ng vi biu din phng ca th

Euler chng minh c rng cc cch biu din phng khc nhau ca mt th u

chia mt phng ra thnh cng mt s min. chng minh iu , Euler tm c mi

lin h gia s min, s nh ca th v s cnh ca th phng sau y.

nh l 3 (Cng thc Euler).

Gi s G l th phng lin thng vi n nh, m cnh. Gi r l s min ca mt phng b

chia bi biudin phng ca G. Khi

r = m-n + 2

C th chng minh nh l bng qui np. Xt th d minh ho cho p dng cng thc

Euler.

Th d.:

Cho G l th phng lin thng vi 20 nh, mi nh u c bc l 3. Hi mt phng b

chia lm bao nhiu phn bi biu din phng ca th G?

Gii:

Do mi nh ca th u c bc l 3, nn tng bc ca cc nh l 3x20=60. T suy

ra s cnh ca th m=60/2=30.

V vy, theo cng thc Euler, s min cn tm l

r=30-20+2=12.

Bi ton t mu th

Cho n th v hng G. Hy tm cch gn mi nh ca th mt mu sao cho hai

nh k nhau khng b t bi cng mt mu. Mt php gn mu cho cc nh nh vy c

7/23/2019 LTDT- C1.pdf

8/17

20

gi l mt php t mu th. Bi ton t mu i hi tm php t mu vi s mu phi s

dng l t nht. S mu t nht cn dng t mu th c gi l sc s ca th.

Hy lp trnh cho bi ton ny.

Thut gii 1:

Dng mu th nht t cho tt c cc nh ca th m c th t c, sau dng mu

th hai t tt c cc nh ca th cn li c th t c v c nh th cho n khi t ht

cho tt c cc nh ca th.

m=1;

s nh c t=0;

mi nh u cha c t

do

{ for i=1 to n

if nh i l cha xt v c th t c bng mu m then

{ t nh i bng mu m

tng s nh c t ln 1 n v

}

m++

}

while (s nh c t

7/23/2019 LTDT- C1.pdf

9/17

21

Gi ci t cho thut gii 2

D liu vo c lu trn mt trn vung c[i][j].

Nu c[i][j]=1 th hai thnh ph i,j l k nhau. c[i][j]=0 th hai thnh ph i,j khng k nhau.

Thut ton

Tnh bc ca tt c cc nh

while (cn nh cha c t )

{

-Tm nh(cha c t) c bc ln nht; chng hn l nh i0.

-tm mu t nh i0; chng hn l mu j.

-Ngn cm vic t mu j cho cc nh k vi nh i0

-T mu nh i0 l j.

-Gn bc ca nh c t 0.

}

M gi:

+Danh sch bng mu cho cc nh c cho l 1 v bc ca cc nh cho l 0.

for (int i=1;i

7/23/2019 LTDT- C1.pdf

10/17

22

for (int j=1;jmaxtemp && dinh[j]==0)

{

maxtemp=bac[j];

i0=j;

}

//tm v t mu cho nh c bc cao nht (gi s l i0 ) v t mu cho nh ny (gi s

l mu j)

//Tm v t mu cho nh c bc cao nht mu m

j=1;

while (mau[i0][j]==0)

j++;

//bc ca cc nh k vi nh i0 th tr i 1 v ngn cm vic t mu j cc nh k vi

nh i0

for (int k=1;k

7/23/2019 LTDT- C1.pdf

11/17

23

chu trnh s cp. R rng rng mt ng i (t.. chu trnh) s cp l ng i (t.. chutrnh) n.

Th d

Trong n th trn, x, y, z, w, v, y l ng i n (khng s cp) di 5; x, w,v, z, y khng l ng i v (v, z) khng l cnh; y, z, w, x, v, u, y l chu trnh s cp di 6.

nh ngha:Mt th (v hng) c gi l lin thng nu c ng i gia micp nh phn bit ca th.

Mt th khng lin thng l hp ca hai hay nhiu th con lin thng, mi cpcc th con ny khng c nh chung. Cc th con lin thng ri nhau nh vyc gi l cc thnh phn lin thng ca th ang xt. Nh vy, mt th l linthng khi v ch khi n ch c mt thnh phn lin thng.

Th d

G G

th G l lin thng, nhng th G khng lin thng v c 3 thnh phn lin thng.

nh ngha:Mt nh trong th G m khi xo i n v tt c cc cnh lin thucvi n ta nhn c th con mi c nhiu thnh phn lin thng hn th G cgi l nh ct hay im khp. Vic xo nh ct khi mt th lin thng s to ramt th con khng lin thng. Hon ton tng t, mt cnh m khi ta b n i s tora mt th c nhiu thnh phn lin thng hn so vi th xut pht c gi lcnh ct hay l cu.

Th d

Trong th trn, cc nh ct l v, w, s v cc cu l (x,v), (w,s).

Mnh :Gia mi cp nh phn bit ca mt th lin thng lun c ng i scp.

Chng minh:Gi s u v v l hai nh phn bit ca mt th lin thng G. V G linthng nn c t nht mt ng i gia u v v. Gi x0, x1, ..., xn, vi x0=u v xn=v, ldy cc nh ca ng i c di ngn nht. y chnh l ng i s cp cn tm.Tht vy, gi s n khng l ng i n, khi x i=xjvi 0 i < j. iu ny c ngha

7/23/2019 LTDT- C1.pdf

12/17

24

l gia cc nh u v v c ng i ngn hn qua cc nh x0, x1, ..., xi-1, xj, ..., xnnhnc bng cch xo i cc cnh tng ng vi dy cc nh x i, ..., xj-1.

Mnh :Mi n th n nh (n 2) c tng bc ca hai nh tu khng nh hnn u l th lin thng.

Chng minh:Cho n th G=(V,E) c n nh (n 2) v tho mn yu cu ca biton. Gi s Gkhng lin thng, tc l tn ti hai nh u v v sao cho khng c ngi no ni u v v. Khi trong th G tn ti hai thnh phn lin thng l G1c n1nh v cha u, G2cha nh v v c n2nh. V G1, G2l hai trong s cc thnh phnlin thng ca G nn n1+n2n. ta c:

deg(u)+deg(v) (n11)+(n2 1) = n1+n22 n2

7/23/2019 LTDT- C1.pdf

13/17

25

B sung cnh vo G nhn c th G c m1cnh sao cho k thnh phn linthng l nhng th y . Ta c m m1nn ch cn chng minh

m12

)1)(( knkn.

Gi s Giv Gjl hai thnh phn lin thng ca G vi niv njnh v ninj>1 (*).Nu ta thay Giv Gjbng th y vi ni+1 v nj1 nh th tng s nh khngthay i nhng s cnh tng thm mt lng l:

12

)2)(1(

2

)1(

2

)1(

2

)1(

ji

jjjjiiii nnnnnnnnnn

.

Th tc ny c lp li khi hai thnh phn no c s nh tho (*). V vy m1l lnnht (n, k l c nh) khi th gm k-1 nh c lp v mt th y vi n-k+1nh. T suy ra bt ng thc cn tm.

nh ngha: th c hng G c gi l lin thng mnh nu vi hai nh phnbit bt k u v v ca G u c ng i t u ti v v ng i t v ti u.

th c hng G c gi l lin thng yu nu th v hng nn ca n llin thng.

th c hng G c gi l lin thng mt chiu nu vi hai nh phn bit btk u v v ca G u c ng i t u ti v hoc ng i t v ti u.

Th d

G G

th G l lin thng mnh nhng th G l lin thng yu (khng c ng it u ti x cng nh t x ti u).

Mnh :Cho G l mt th (v hng hoc c hng) vi ma trn lin k A theoth t cc nh v1, v2, ..., vn. Khi s cc ng i khc nhau di r t viti vjtrong r l mt s nguyn dng, bng gi tr ca phn t dng i ct j ca ma trn Ar.

Chng minh:Ta chng minh mnh bng quy np theo r. S cc ng i khc nhau di 1 t viti vjl s cc cnh (hoc cung) t vi ti vj, chnh l phn t dng i ct

j ca ma trn A; ngha l, mnh ng khi r=1.

Gi s mnh ng n r;ngha l, phn t dng i ct j ca Arl s cc ng ikhc nhau di r t viti vj. V A

r+1=A

r.A nn phn t dng i ct j ca Ar+1bng

bi1a1j+bi2a2j+ ... +binanj,

trong bikl phn t dng i ct k ca Ar. Theo gi thit quy np bikl s ng i

khc nhau di r t viti vk.

ng i di r+1 t viti vjs c to nn t ng i di r t viti nhtrung gian vkno v mt cnh (hoc cung) t vkti vj. Theo quy tc nhn s ccng i nh th l tch ca s ng i di r t viti vk, tc l bik, v s cc cnh

7/23/2019 LTDT- C1.pdf

14/17

26

(hoc cung) t vkti vj, tc l akj. Cng cc tch ny li theo tt c cc nh trung gianvkta c mnh ng n r+1.

Ghi ch v ti liu tham kho

Tan ri rc(Phn 2: L thuyt th) ca tc gi Nguyn c Ngha Nguyn T Thnh.

7/23/2019 LTDT- C1.pdf

15/17

27

Bi tp l thuyt

1-1.V th (nu tn ti)

a.V mt th c 4 nh vi bc cc nh l 3, 2, 2, 1.

b.V cc th m mi nh ca n u c bc l ln lt l k (1 k 5)

c.V cc th m mi nh ca n u c bc l 3 v c s nh ln lt l:4,5,6,8.

d.V mt th c 15 nh v mi nh ca n u c bc l 5.

1-2.a.Mt th phng lin thng c 8 nh, cc nh ln lt c bc l 2, 2, 3, 3, 3, 3, 4, 6. Hi

th c bao nhiu cnh ?

b.Mt n th phnglin thng c 10 mt, tt c cc nh u c bc 4. Tm s nh ca

th.

c.Xt mt th lin thng c 8 nh bc 3. Hi biu din phng ca th ny s chia mt

phng thnh my min.d.n th phng lin thng G c 9 nh, bc cc nh l 2,2,2,3,3,3,4,4,5. Tm s cnh v s

mt ca G.

1-3.a.Mt th c 19 cnh v mi nh u c bc 3, hi th ny c ti a bao nhiu nh

?

b.Cho mt th v hng c n nh. Hi th ny c th c ti a bao nhiu cnh. Trong

trng hp s cnh l ti a th mi nh s c bc l bao nhiu ?

c.Cho mt th v hng c n nh v 2n cnh. Chng minh rng trong th ny lun tn timt nh c bc khng nh hn 4.

d.Chng minh rng trong mt n th v hng nu khng cha chu trnh th s lun tn ti t

nht l hai nh treo.

e.Chng minh rng nu th G c cha mt chu trnh c di l th s mu ca G t nht phi

l 3.

1-4.a.Xt th v hng n c s nh n > 2 . Chng minh rng th c t nht 2 nh cng

bc vi nhau.

b.Cho 1 th G c cha ng 2 nh bc l (cc nh khc nu c phi bc chn) Chng minh

rng 2 nh ny lin thng vi nhau.

c.xt th v hng n c s nh n > 2. Gi s th khng c nh no c bc < (n-1)/2.

Chng minh rng th ny lin thng

d.Chng minh rng mt n th v hng l hai pha nu v ch nu s mu ca n l 2.

1-5.V th phnglin thng

a.c 6 cnh v 3 min.

b.c 4 nh v 5 min.

c.c 6 nh v 7 cnh.

7/23/2019 LTDT- C1.pdf

16/17

28

1-6.Gi s c 6 cuc mitting A,B,C,D,E,F cn c t chc. Mi cuc mitting c t chc

trong mt bui. Cc cuc mitting sau khng c din ra ng thi:BEF, CEF, ABE, CD, AD.

Hy b tr cc cuc mitting vo cc bui sao cho s bui din ra l t nht.

1-7.Chng minh rng mt th y c 5 nh khng l th phng.

1-8.Hy tm sc s ca thsau:

1-9.C ba nh gn ba ci ging, t mi nh c ng i thng n mi ging. C ln do bt

ha vi nhau, c ba ngi ny mun tm cch lm cc con ng khc n cc ging sao cho

cc ng ny khng ctnhau. Hi nh ny c thc hin c khng ? v sao ?

1-10.Tm s nh, cnh v min ca cc th sau:

1-11.Vi mi th sau y hy cho bit n c phi l th phng hay khng ? Nu c hy v

sao cho cc cnh ca th khng ct nhau ngoi nh.

C

D

L

G

F

E

H

KA

B

7/23/2019 LTDT- C1.pdf

17/17

29

a) b)

c) d)

e)f)

g) h) i)