Giao Trinh Toan Roi Rac - Chuong 6

8/6/2019 Giao Trinh Toan Roi Rac - Chuong 6

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CHNG VI

CY

Mt th lin thng v khng c chu trnh c gi l cy. Cy c dng

t nm 1857, khi nh ton hc Anh tn l Arthur Cayley dng cy xc nh nhngdng khc nhau ca hp cht ho hc. T cy c dng gii nhiu bi ton

trong nhiu lnh vc khc nhau. Cy rt hay c s dng trong tin hc. Chng hn,

ngi ta dng cy xy dng cc thut ton rt c hiu qu nh v cc phn ttrong mt danh sch. Cy cng dng xy dng cc mng my tnh vi chi ph r nht

cho cc ng in thoi ni cc my phn tn. Cy cng c dng to ra cc m

c hiu qu lu tr v truyn d liu. Dng cy c th m hnh cc th tc m thihnh n cn dng mt dy cc quyt nh. V vy cy c bit c gi tr khi nghin cu

cc thut ton sp xp.

6.1. NH NGHA V CC TNH CHT C BN.6.1.1. nh ngha: Cy l mt th v hng lin thng, khng cha chu trnh v ct nht hai nh.

Mt th v hng khng cha chu trnh v c t nht hai nh gi l mt rng.

Trong mt rng, mi thnh phn lin thng l mt cy.

Th d 1: Rng sau c 3 cy:

6.1.2. Mnh : Nu T l mt cy c n nh th T c t nht hai nh treo.Chng minh: Ly mt cnh (a,b) tu ca cy T. Trong tp hp cc ng i s cp

cha cnh (a,b), ta ly ng i t u n v di nht. V T l mt cy nn u { v. Mt

khc, u v v phi l hai nh treo, v nu mt nh, u chng hn, khng phi l nh treoth u phi l u mt ca mt cnh (u,x), vi x l nh khng thuc ng i t u n v.

Do , ng i s cp t x n v, cha cnh (a,b), di hn ng i t u n v, tri vi

tnh cht ng i t u n v chn.6.1.3. nh l: Cho T l mt th c n u 2 nh. Cc iu sau l tng ng:1) T l mt cy.

2) T lin thng v c n1 cnh.

3) T khng cha chu trnh v c n1 cnh.

4) T lin thng v mi cnh l cu.

5) Gia hai nh phn bit bt k ca T lun c duy nht mt ng i s cp.

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6) T khng cha chu trnh nhng khi thm mt cnh mi th c c mt chu trnh duy

nht.Chng minh:1)2) Ch cn chng minh rng mt cy c n nh th c n1 cnh. Ta

chng minh bng quy np. iu ny hin nhin khi n=2. Gi s cy c k nh th c k1cnh, ta chng minh rng cy T c k+1 nh th c k cnh. Tht vy, trong T nu ta xo

mt nh treo v cnh treo tng ng th th nhn c l mt cy k nh, cy ny ck1 cnh, theo gi thit quy np. Vy cy T c k cnh.

2)3) Nu T c chu trnh th b i mt cnh trong chu trnh ny th T vn lin thng.

Lm li nh th cho n khi trong T khng cn chu trnh no m vn lin thng, lc

ta c mt cy c n nh nhng c t hn n1 cnh, tri vi 2).

3)4) Nu T c k thnh phn lin thng T1, ..., Tk ln lt c s nh l n1, ..., nk (vi

n1+n2+ ~ +nk=n) th mi Ti l mt cy nn n c s cnh l ni1. Vy ta c

n1=(n11)+(n21)+ ... +(nk1)=(n1+n2+ ~ +nk)k=nk.

Do k=1 hay T lin thng. Hn na, khi b i mt cnh th T ht lin thng, v nucn lin thng th T l mt cy n nh vi n2 cnh, tri vi iu chng minh trn.

4)5) V T lin thng nn gia hai nh phn bit bt k ca T lun c mt ng i scp, nhng khng th c ni bi hai ng i s cp v nu th, hai ng s to

ra mt chu trnh v khi b mt cnh thuc chu trnh ny, T vn lin thng, tri vi gi

thit.

5)6) Nu T cha mt chu trnh th hai nh bt k trn chu trnh ny s c ni bi

hai ng i s cp. Ngoi ra, khi thm mt cnh mi (u,v), cnh ny s to nn ving i s cp duy nht ni u v v mt chu trnh duy nht.

6)1) Nu T khng lin thng th thm mt cnh ni hai nh hai thnh phn linthng khc nhau ta khng nhn c mt chu trnh no. Vy T lin thng, do n lmt cy.

6.2. CY KHUNG V BI TON TM CY KHUNG NH NHT.6.2.1. nh ngha: Trong th lin thng G, nu ta loi b cnh nm trn chu trnhno th ta s c th vn l lin thng. Nu c loi b cc cnh cc chu trnh

khc cho n khi no th khng cn chu trnh (vn lin thng) th ta thu c mt cy

ni cc nh ca G. Cy gi l cy khung hay cy bao trm ca th G.

Tng qut, nu G l th c n nh, m cnh v k thnh phn lin thng th pdng th tc va m t i vi mi thnh phn lin thng ca G, ta thu c th gi

l rng khung ca G. S cnh b loi b trong th tc ny bng mn+k, s ny k hiul R(G) v gi l chu s ca th G.

6.2.2. Bi ton tm cy khung nh nht: Bi ton tm cy khung nh nht ca th l mt trong s nhng bi ton ti u trn th tm c ng dng trong nhiu lnh


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vc khc nhau ca i sng. Trong phn ny ta s c hai thut ton c bn gii bi

ton ny. Trc ht, ni dung ca bi ton c pht biu nh sau.Cho G=(V,E) l th v hng lin thng c trng s, mi cnh eE c trng

s m(e)u0. Gi s T=(VT,ET) l cy khung ca th G (VT=V). Ta gi di m(T) cacy khung T l tng trng s ca cc cnh ca n:

m(T)= TE

)(e

em .

Bi ton t ra l trong s tt c cc cy khung ca th G, hy tm cy khung c

di nh nht. Cy khung nh vy c gi l cy khung nh nht ca th v bi ton

t ra c gi l bi ton tm cy khung nh nht. minh ho cho nhng ng dng ca bi ton cy khung nh nht, di y l

hai m hnh thc t tiu biu cho n.

Bi ton xy dng h thng ng st: Gi s ta mun xy dng mt h thng ng

st ni n thnh ph sao cho hnh khch c th i t bt c mt thnh ph no n bt kmt trong s cc thnh ph cn li. Mt khc, trn quan im kinh t i hi l chi phv xy dng h thng ng phi l nh nht. R rng l th m nh l cc thnh

ph cn cc cnh l cc tuyn ng st ni cc thnh ph tng ng, vi phng n

xy dng ti u phi l cy. V vy, bi ton t ra dn v bi ton tm cy khung nhnht trn th y n nh, mi nh tng ng vi mt thnh ph vi di trn

cc cnh chnh l chi ph xy dng h thng ng st ni hai thnh ph.

Bi ton ni mng my tnh: Cn ni mng mt h thng gm n my tnh nh s t 1n n. Bit chi ph ni my i vi my j l m(i,j) (thng thng chi ph ny ph thuc vo

di cp ni cn s dng). Hy tm cch ni mng sao cho tng chi ph l nh nht.Bi ton ny cng dn v bi ton tm cy khung nh nht.

Bi ton tm cy khung nh nht c nhng thut ton rt hiu qu gii

chng. Ta s xt hai trong s nhng thut ton nh vy: thut ton Kruskal v thut ton

Prim.

6.2.3. Thut ton Kruskal:Thut ton s xy dng tp cnh ET ca cy khung nhnht T=(VT, ET) theo tng bc. Trc ht sp xp cc cnh ca th G theo th t

khng gim ca trng s. Bt u t ET=, mi bc ta s ln lt duyt trong danh

sch cnh sp xp, t cnh c di nh n cnh c di ln hn, tm ra cnhm vic b sung n vo tp ET khng to thnh chu trnh trong tp ny. Thut ton skt thc khi ta thu c tp ET gm n1 cnh. C th c th m t nh sau:

1. Bt u t th rng T c n nh.

2. Sp xp cc cnh ca G theo th t khng gim ca trng s.

3. Bt u t cnh u tin ca dy ny, ta c thm dn cc cnh ca dy c xp

vo T theo nguyn tc cnh thm vo khng c to thnh chu trnh trong T.


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4. Lp li Bc 3 cho n khi no s cnh trong T bng n1, ta thu c cy khung nh

nht cn tm.Th d 2: Tm cy khung nh nht ca th cho trong hnh di y:

Bt u t th rng T c 6 nh.Sp xp cc cnh ca th theo th t khng gim ca trng s:

{(v3, v5), (v4, v6), (v4, v5), (v5, v6), (v3, v4), (v1, v3), (v2, v3), (v2, v4), (v1, v2)}.

Thm vo th T cnh (v3, v5).

Do s cnh ca T l 1


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6.2.4. Thut ton Prim: Thut ton Kruskal lm vic km hiu qu i vi nhng th dy ( th c s cnh m } n(n1)/2). Trong trng hp , thut ton Prim t rahiu qu hn. Thut ton Prim cn c gi l phng php ln cn gn nht.

1. VT:={v*}, trong v* l nh tu ca th G.ET:=.

2. Vi mi nh vjVT, tm nh wjVT sao chom(wj,vj) = min m(xi, vj)=:Fj

xiVTv gn cho nh vj nhn [wj, Fj]. Nu khng tm uc wj nh vy (tc l khi vj khng k

vi bt c nh no trong VT) th gn cho vj nhn [0, g].

3. Chn nh vj* sao cho

Fj* = min FjvjVT

VT := VT {vj*},

ET := ET {(wj*, vj*)}.

Nu |VT| = n th thut ton dng v (VT, ET) l cy khung nh nht.Nu |VT| < n th chuyn sang Bc 4.

4. i vi tt c cc nh vjVT m k vi vj*, ta thay i nhn ca chng nh sau:Nu Fj > m(vj*, vj) th t Fj:=m(vj*, vj) v nhn ca vj l [vj*, Fj]. Ngc li, ta

gi nguyn nhn ca vj. Sau quay li Bc 3.

Th d 3: Tm cy khung nh nht bng thut ton Prim ca th gm cc nh A, B,C, D, E, F, H, I c cho bi ma trn trng s sau.

g

g

g

g

g

g

g

g

1418211119121814172321202032

18173430211920

21233422293423

11213022131319

19202129133316

12201934133315

18322023191615

.

Yu cu vit cc kt qu trung gian trong tng bc lp, kt qu cui cng cn a ra

tp cnh v di ca cy khung nh nht.

V.lp A B C D E F H I VT ET

A B C D E F H

A

I

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C

D

E

F

H

I


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K.to [A,15] [A,16] [A,19] [A,23] [A,20] [A,32] [A,18] A 1 [A,16] [B,13] [A,23] [B,19] [B,20] [B,12] A, B (A,B)2 [A,16] [I,11] [I,21] [I,18] [I,14] A, B, I (A,B), (B,I)3 [D,13] [I,21] [I,18] [I,14] A, B, I, D (A,B), (B,I), (I,D)

4 [I,21] [I,18] [I,14] A, B, I, D, C (A,B), (B,I), (I,D),(D,C)

5 [I,21] [H,17] A, B, I, D, C,H (A,B), (B,I), (I,D),(D,C), (I,H)6 [I,21] A, B, I, D, C,

H, F(A,B), (B,I), (I,D),(D,C), (I,H), (H,F)

7 A, B, I, D, C,H, F, E

(A,B), (B,I), (I,D),(D,C), (I,H), (H,F),

(I,E)

Vy di cy khung nh nht l:15 + 12 + 11 + 13 + 14 + 17 + 21 = 103.

Tnh ng n ca thut ton: chng minh thut ton Prim l ng, ta chng minh

bng quy np rng T(k) (k=1, 2, ...,n), th nhn c trong vng lp th k, l mt th con ca cy khung nh nht ca G, do T(n) chnh l mt cy khung nh nht ca

G.

T(1) ch gm nh v* ca G, do T(1) l th con ca mi cy khung ca G.Gi s T(i) (1ei


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6.3.1. nh ngha: Cy c hng l th c hng m th v hng nn ca n lmt cy.

Cy c gc l mt cy c hng, trong c mt nh c bit, gi l gc, t gc

c ng i n mi nh khc ca cy.Th d 4:

Trong cy c gc th gc r c bc vo bng 0, cn tt c cc nh khc u c bc

vo bng 1.Mt cy c gc thng c v vi gc r trn cng v cy pht trin t trn

xung, gc r gi l nh mc 0. Cc nh k vi r c xp pha di v gi l nh

mc 1. nh ngay di nh mc 1 l nh mc 2, ...Tng qut, trong mt cy c gc th v l nh mc k khi v ch khi ng i t r

n v c di bng k.Mc ln nht ca mt nh bt k trong cy gi l chiu cao ca cy.

Cy c gc hnh trn thng c v nh trong hnh di y lm r mcca cc nh.

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Trong cy c gc, mi cung u c hng t trn xung, v vy v mi tn ch

hng i l khng cn thit; do , ngi ta thng v cc cy c gc nh l cy nnca n.

6.3.2. nh ngha: Cho cy T c gc r=v0. Gi s v0, v1, ..., vn-1, vn l mt ng itrong T. Ta gi:

vi+1 l con ca vi v vi l cha ca vi+1. v0, v1, ..., vn-1 l cc t tin ca vn v vn l dng di ca v0, v1, ..., vn-1.

nh treo vn l nh khng c con; nh treo cng gi l l hay nh ngoi; mt nh

khng phi l l mt nh trong.

6.3.3. nh ngha: Mt cy c gc T c gi l cy m-phn nu mi nh ca T cnhiu nht l m con. Vi m=2, ta c mt cy nh phn.

Trong mt cy nh phn, mi con c ch r l con bn tri hay con bn phi;

con bn tri (t.. phi) c v pha di v bn tri (t.. phi) ca cha.

Cy c gc T c gi l mt cy m-phn y nu mi nh trong ca T uc m con.

6.3.4. Mnh : Mt cy m-phn y c i nh trong th c mi+1 nh v c(m1)i+1 l.

Chng minh: Mi nh trong ca cy m-phn y u c bc ra l m, cn l c bc

ra l 0, vy s cung ca cy ny l mi v do s nh ca cy l mi+1. Gi l l s l thta c l+i=mi+1, nn l=(m1)i+1.

6.3.5. Mnh :1) Mt cy m-phn c chiu cao h th c nhiu nht l mh l.2) Mt cy m-phn c l l th c chiu cao h u [logml].

Chng minh:1) Mnh c chng minh bng quy np theo h. Mnh hin nhinng khi h=1. Gi s mi cy c chiu cao ke h1 u c nhiu nht mk-1 l (vi hu2).Xt cy T c chiu cao h. B gc khi cy ta c mt rng gm khng qu m cy con,

mi cy con ny c chiu cao e h1. Do gi thit quy np, mi cy con ny c nhiu

nht l mh-1 l. Do l ca nhng cy con ny cng l l ca T, nn T c nhiu nht lm.mh-1=mh l.

2)le mh h u [logml].

6.4. DUYT CY NH PHN.

6.4.1. nh ngha: Trong nhiu trng hp, ta cn phi im danh hay thm mtcch c h thng mi nh ca mt cy nh phn, mi nh ch mt ln. Ta gi l vicduyt cy nh phn hay c cy nh phn.

C nhiu thut ton duyt cy nh phn, cc thut ton khc nhau ch yu

th t thm cc nh.

Cy nh phn T c gc r c k hiu l T(r). Gi s r c con bn tri l u, conbn phi l v. Cy c gc u v cc nh khc l mi dng di ca u trong T gi l cy


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con bn tri ca T, k hiu T(u). Tng t, ta c cy con bn phi T(v) ca T. Mt cy

T(r) c th khng c cy con bn tri hay bn phi.Sau y l ba trong cc thut ton duyt cy nh phn T(r). Cc thut ton u

c trnh by quy. Ch rng khi cy T(x) ch l mt nh x th duyt T(x) cngha l thm nh x.

Th d 5:

6.4.2. Cc thut ton duyt cy nh phn:1) Thut ton tin th t:

1. Thm gc r.

2. Duyt cy con bn tri ca T(r) theo tin th t.

3. Duyt cy con bn phi ca T(r) theo tin th t.

Duyt cy nh phn T(a) trong hnh trn theo tin th t:1. Thm a

2. Duyt T(b)2.1. Thm b2.2. Duyt T(d)

2.2.1. Thm d

2.2.2. Duyt T(g)2.2.2.1. Thm g

2.2.2.3. Duyt T(l): Thm l

2.2.3. Duyt T(h): Thm h

2.3. Duyt T(e)2.3.1. Thm e

2.3.2. Duyt T(i)2.3.2.1. Thm i

2.3.2.2. Duyt T(m): Thm m

2.3.2.3. Duyt T(n): Thm n3. Duyt T(c)

a

b c

d e f

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3.1. Thm c

3.3. Duyt T(f)3.3.1.Thm f

3.3.2. Duyt T(j)3.3.2.1. Thm j

3.3.2.2. Duyt T(o): Thm o3.3.2.3. Duyt T(p): Thm p

3.3.3. Duyt T(k)

3.3.3.1. Thm k

3.3.3.2. Duyt T(q): Thm q3.3.3.3. Duyt T(s): Thm s

Kt qu duyt cy T(a) theo tin th t l:

a, b, d, g, l, h, e, i, m, n, c, f, j, o, p, k, q, s.

2) Thut ton trung th t:1. Duyt cy con bn tri ca T(r) theo trung th t.

2. Thm gc r.

3. Duyt cy con bn phi ca T(r) theo trung th t.

Duyt cy nh phn T(a) trong hnh trn theo trung th t:

1. Duyt T(b)1.1. Duyt T(d)

1.1.1. Duyt T(g)1.1.1.2. Thm g

1.1.1.3. Duyt T(l): thm l1.1.2. Thm d1.1.3. Duyt T(h): Thm h

1.2. Thm b

1.3. Duyt T(e)1.3.1. Duyt T(i)


1.3.1.2. Thm i

1.3.1.3. Duyt T(n): Thm n1.3.2. Thm e

2. Thm a3. Duyt T(c)

3.2. Thm c

3.3. Duyt T(f)3.3.1. Duyt T(j)


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3.3.1.1. Duyt T(o): Thm o

3.3.1.2. Thm j3.3.1.3. Duyt T(p): Thm p

3.3.2. Thm f3.3.3. Duyt T(k)

3.3.3.1. Duyt T(q): Thm q3.3.3.2. Thm k3.3.3.3. Duyt T(s): Thm s

Kt qu duyt cy T(a) theo trung th t l:

g, l, d, h, b, m, i, n, e, a, c, o, j, p, f, q, k, s.

3) Thut ton hu th t:

1. Duyt cy con bn tri ca T(r) theo hu th t.

2. Duyt cy con bn phi ca T(r) theo hu th t.

3. Thm gc r.Duyt cy nh phn T(a) trong hnh trn theo hu th t:

1. Duyt T(b)1.1. Duyt T(d)

1.1.1. Duyt T(g)

1.1.1.2. Duyt T(l): thm l1.1.1.3. Thm g

1.1.2. Duyt T(h): thm h1.1.3. Thm d

1.2. Duyt T(e)1.2.1. Duyt T(i)


1.2.1.2. Duyt T(n): Thm n

1.2.1.3. Thm i1.2.3. Thm e

1.3. Thm b

2. Duyt T(c)

2.2. Duyt T(f)2.2.1. Duyt T(j)

2.2.1.1. Duyt T(o): Thm o2.2.1.2. Duyt T(p): Thm p

2.2.1.3. Thm j

2.2.2. Duyt T(k)2.2.2.1. Duyt T(q): Thm q


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2.2.2.2. Duyt T(s): Thm s

2.2.2.3. Thm k2.2.3. Thm f

2.3. Thm c3. Thm a

Kt qu duyt cy T(a) theo trung th t l:l, g, h, d, m, n, i, e, b, o, p, j, q, s, k, f, c, a.

6.4.3. K php Ba Lan:Xt biu thc i s sau y:

(a+b)(c2

d) (1)

Ta v mt cy nh phn nh hnh di y, trong mi nh trong mang du

ca mt php tnh trong (1), gc ca cy mang php tnh sau cng trong (1), y l

du nhn, k hiu l

, mi l mang mt s hoc mt ch i din cho s.

Duyt cy nh phn trong hnh trn theo trung th t l:a + b c d / 2 (2)

v y l biu thc (1) b i cc du ngoc.

Ta ni rng biu thc (1) c biu din bng cy nh phn T( ) trong hnh trn,hay cy nh phn T( ) ny tng ng vi biu thc (1). Ta cng ni: cch vit (k php)

quen thuc trong i s hc nh cch vit biu thc (1) l k php trung th t km theo

cc du ngoc.

Ta bit rng cc du ngoc trong (1) l rt cn thit, v (2) c th hiu theo nhiucch khc (1), chng hn l(a + b c) d / 2 (3)

hoc l a + (b c d) / 2 (4)

Cc biu thc (3) v (4) c th biu din bng cy nh phn trong cc hnh sau.Hai cy nh phn tng ng l khc nhau, nhng u c duyt theo trung th t l

(2).

+

a b c /

2d


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i vi cy trong hnh th nht, nu duyt theo tin th t, ta c

+ a b c / d 2 (5)

v nu duyt theo hu th t, ta c:a b + c d 2 / (6)

C th chng minh c rng (5) hoc (6) xc nh duy nht cy nh phn trong

hnh th nht, do xc nh duy nht biu thc (1) m khng cn du ngoc. Chng

hn cy nh phn trn hnh th hai c duyt theo tin th t l

+ a b c / d 2 khc vi (5).

v c duyt theo hu th t l

a b c + d 2 / khc vi (6).

V vy, nu ta vit cc biu thc trong i s, trong lgic bng cch duyt cytng ng theo tin th t hoc hu th t th ta khng cn dng cc du ngoc m

khng s hiu nhm.Ngi ta gi cch vit biu thc theo tin th t l k php Ba Lan, cn cch vit

theo hu th t l k php Ba Lan o, ghi nh ng gp ca nh ton hc v lgic

hc Ba Lan Lukasiewicz (1878-1956) trong vn ny.

+ /

a d 2

cb

a /

d

2

cb

+


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Vic chuyn mt biu thc vit theo k php quen thuc (c du ngoc) sang

dng k php Ba Lan hay k php Ba Lan o hoc ngc li, c th thc hin bngcch v cy nh phn tng ng, nh lm i vi biu thc (1). Nhng thay v v cy

nh phn, ta c th xem xt xc nh dn cc cng thc b phn ca cng thc cho. Chng hn cho biu thc vit theo k php Ba Lan l

o / a b 5 c 2 3 o c d 2 a c d / o b 3 d 3 5

Trc ht, ch rng cc php ton +, , *, /, o u l cc php ton hai ngi, v

vy trong cy nh phn tng ng, cc nh mang du cc php ton u l nh trongv c hai con. Cc ch v s u t l. Theo k php Ba Lan (t.. Ba Lan o) th

T a b (t.. a b T) c ngha l a T b, vi T l mt trong cc php ton +, , *, /, o.

] ] 533*/*2325*/* 35 dcadccba dbdcadccba o

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dbdcadccba

ooo

53)3(/)(*)(32)5(/*3)3(

2

2

5

dbcba

dbdcadccba

oo

5

)3(

32

2

53

3

5)3(/)(*)(32

5*

dbcba

dbdcadccba

o

5

)3()(

3

)(2

5

23

3

23

5

)3()(*)(

2

5*

dbdcadc

cba

dbdcadc

cba

5

)3()()(

2

5 323

dbdcadc

cba


15/17

101

BI TP CHNG VI:

1. V tt c cc cy (khng ng cu) c:

a) 4 nh b) 5 nh c) 6 nh

2. Mt cy c n2 nh bc 2, n3 nh bc 3, , nk nh bc k. Hi c bao nhiu nh bc

1?3. Tm s ti a cc nh ca mt cy m-phn c chiu cao h.

4. C th tm c mt cy c 8 nh v tho iu kin di y hay khng? Nu c, v

cy ra, nu khng, gii thch ti sao:

a) Mi nh u c bc 1.

b) Mi nh u c bc 2.

c) C 6 nh bc 2 v 2 nh bc 1.

d) C nh bc 7 v 7 nh bc 1.

5. Chng minh hoc bc b cc mnh sau y.a) Trong mt cy, nh no cng l nh ct.

b) Mt cy c s nh khng nh hn 3 th c nhiu nh ct hn l cu.

6. C bn i bng A, B, C, D lt vo vng bn kt trong gii cc i mnh khu vc.

C my d on xp hng nh sau:

a) i B v ch, i D nh.

b) i B nh, i C ba.

c) i A nh, i C t.Bit rng mi d on trn ng v mt i. Hy cho bit kt qu xp hng ca

cc i.7. Cy Fibonacci c gc Tn uc dnh ngha bng hi quy nh sau. T1 v T2 u l cyc gc ch gm mt nh v vi n=3,4, cy c gc Tn c xy dng t gc vi Tn-1

nh l cy con bn tri v Tn-2 nh l cy con bn phi.

a) Hy v 7 cy Fibonacci c gc u tin.

b)Cy Fibonacci Tn c bao nhiu nh, l v bao nhiu nh trong. Chiu cao

ca n bng bao nhiu?

8. Hy tm cy khung ca th sau bng cch xo i cc cnh trong cc chu trnh n.

a) a b c

d e f g

h i j


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102

b)

9. Hy tm cy khung cho mi th sau.

a) K5 b) K4,4 c) K1,6

d) Q3 e) C5 f) W5.

10. th Kn vi n=3, 4, 5 c bao nhiu cy khung khng ng cu?11. Tm cy khung nh nht ca th sau theo thut ton Kruskal v Prim.

12. Tm cy khung nh nht bng thut ton Prim ca th gm cc nh A, B, C, D,E, F, H, I c cho bi ma trn trng s sau.

g

g

g

g

g

g

g

g

18142112191120

18172321201932

14173430212018

21233422292419

12213022133323

19202129131315

11192024331316

20321819231516

.

Yu cu vit cc kt qu trung gian trong tng bc lp, kt qu cui cng cn a ra

tp cnh v di ca cy khung nh nht.

a b

c d e

h

fi

k

g

j

l

a b

c d e f

g

h

42

14104

3 1 11

3

155

7

20 9

A B C D E F G

A

H

B

C

D

E

F

G

H


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103

13. Duyt cc cy sau y ln lt bng cc thut ton tin th t, trung th t v hu

th t.a) b)

14. Vit cc biu thc sau y theo k php Ba Lan v k php Ba Lan o.

a)BDC

BDA

DCBA

DCBA

2

2

)(

))((.

b)5

)243(

35

3)(

3424 dbadad

cba

.

15. Vit cc biu thc sau y theo k php quen thuc.

a) x y + 2 x y 2 x y * /.

b) o / a b 3 c 2 4 o c d 5 a c d / o b 2 d 4 3.

a

cb

d e

g

f

h

i j

ed gf

b c

a

h i

m n

j k

p q

l

o

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