ltdt- c1.pdf
TRANSCRIPT
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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
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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.
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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.
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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.
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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.
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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
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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
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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
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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
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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
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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
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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
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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
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(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.
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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.
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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
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a) b)
c) d)
e)f)
g) h) i)