bai hoc cua bo

8/8/2019 bai hoc cua bo

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BI TP TON RI RC

***

Full

TH TH

Ging vin : Nguyn Mu HnSinh vin thc hin : Nguyn Th Diu Hng

Lp : Tin K30D

* Bi 1:

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Cho G l mt th c v nh v e cnh.M v m tng ng l bc ln nhtv nh nht ca cc nh ca G.Chng minh rng:

m 2.e/v M

Li gii:Theo ra ta c:M: bc ln nht ca nh ca G.m: bc nh nht ca nh ca G.

Nh vy:m deg(vi) M (vi deg(vi): bc ca nh vi)v.m deg(vi) v.M

v.m 2.e v.Mm 2.e M

Vy ta c iu phi chng minh.

* Bi 2:Chng minh rng nu G l n th phn i c v nh v ecnh, khi

e v2/4.Li gii :

Ta c:G=(V,E) l n th phn i.V=V1 U V2, V1 V2 =, V1 , V2 .

Gi n1 v n2 ln lt l s phn t ca V1 v V2.n1 + n2 = vG l th phn i nn e t gi tr max khi G l th phn i y

.Khi :e = n1.n2

C ngha l trong trng hp tng qut th:e n1.n2

Nh vy, chng minh e v2/4ch cn chng minh:n1.n2 v2/4

Tht vy:n1.n2 v2/4

n1.n2 (n1+ n2)2/44.n1.n2 n12 + n22 + 2.n1.n2n12 + n22 - 2.n1.n2 0 v2/4(n1- n2)2 0 (hin nhin ng)

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P(3,1)

P(2,1)

P(2,2)P(2,3)

P(3,2)P(3,3)

Suy ra:e n1.n2 v2/4

Vy ta c iu phi chng minh.

* Bi 3:Trong mt phng n mng kiu li kt ni n=m2 b x l song song, b

x l P(i,j) c kt ni vi 4 b x l (P(i 1) mod m, j), P(i, (j 1) modm), sao cho cc kt ni bao xung quanh cc cnh ca li. Hy v mngkiu li c 16 b x l theo phng n ny.

Li gii:

* Bi 4:Hy v cc th v hng c biu din bi ma trn lin k sau:

3

P(0,0)

P(0,3)

P(1,0)

P(0,2)

P(1,1)

P(1,3) P(1,2)

P(0,1)

P(2,0)

P(3,0)


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a) b)1 2 3 1 2 0 12 0 4 2 0 3 0

3 4 0 0 3 1 11 0 1 0

c)0 1 3 0 41 2 1 3 03 1 1 0 10 3 0 0 24 0 1 2 3

Li gii:

a) b)

c)

*Bi 5:

V

1

V

3

V

2

4

V

4

V

3

V1

V2

V

1

V

2

V

5

V

3

V

4


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Nu ngha ca tng cc phn t trn mt hng (tng ng ct) ca mtma trn lin k i vi mt th v hng ? i vi th c hng ?

Li gii:Cho th G=(V,E).V= {v1,v2,...,vn }

Ma trn lin k ca th G=(V,E) l ma trn:A=( aij ) vi 1i,jna11 a12 ... a1na21 a22 ... a2n

A= .........an1 an2 ... ann

*Nu G l thv hng:aij l s cnh ni nh vi v vj-Tng hng i ca ma trn A:

n

aij chnh lbc ca nh vij=1

-Tng ct j ca ma trn A:naij chnh lbc ca nh vji=1

*Nu G l thc hng:aij l s cung ni vi v vj m vj l nh cui-Tng hng i ca ma trn A:

n

aij chnh lbc ra ca nh vij=1

-Tng ct j ca ma trn A:naij chnh lbc ra ca nh vji=1

*Bi 6:Tm ma trn lin k cho cc ma trn sau:

a) Kn b) Cn c) Wn d) Km,n e) QnLi gii:

a) Ma trn lin k ca th y Kn:ai1 ai2 ... aij ... ain

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a1j 0 1 ... 1 ... 1a2j 1 0 ... 1 ... 1... ... ... ... ... ... ...aij 1 1 ... 0 ... 1... ... ... ... ... ... ...anj 1 1 ... 1 ... 0

Hay vit cch khc:Ma trn lin k ca th y Kn l:

0 nu i = jA = (aij), trong aij =

1 nu i j

b) Ma trn lin k ca th vng Cn:

ai1 ai2 ai3 ... aij-1 aij aij+1 ... ain-1 ain

a1j 0 1 0 ... 0 0 0 ... 0 1a2j 1 0 1 ... 0 0 0 ... 0 0a3j 0 1 0 ... 0 0 0 ... 0 0... ... ... ... ... ... ... ... ... ... ...aij 0 0 0 ... 1 0 1 ... 0 0... ... ... ... ... ... ... ... ... ... ...anj 1 0 0 ... 0 0 0 ... 1 0

Vit cch khc:Ma trn lin k ca th vng Cn l:

A = (aij), trong :1 nu j=2 hoc j=n

- Vi i=1: aij=0 nu j2v jn

1 nu j=1 hoc j=n-1

- Vi i=n: aij=0 nu j1 v jn-1

-Vi i1 v in:

1 nu j=i+1, j=i-1

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aij =0 nu ji+1 v ji-1

c)Ma trn lin k A ca th bnh xe Wn:

ai1 ai2 ai3 ... aij-1 aij aij+1 ... ain-1 ain ain +1

a1j 0 1 0 ... 0 0 0 ... 0 1 1a2j 1 0 1 ... 0 0 0 ... 0 0 1... ... ... ... ... ... ... ... ... ... ... ...aij 0 0 0 ... 1 0 1 ... 0 0 1... ... ... ... ... ... ... ... ... ... ... ...anj 1 0 0 ... 0 0 0 ... 1 0 1

an+1j 1 1 1 ... 1 1 1 ... 1 1 0

d) Ma trn lin k ca th phn i y Km,n:

Cho G=(V,E)=Km,n, trong V=V1 U V2V1={v1,v2,...,vm}V2={v'1,v'2,...,v'n}

Ta c ma trn lin k ca Km,n nh sau:v1 v2 ... vm v'1 v'2 ... v'n

v1 0 0 ... 0 1 1 ... 1v2 0 0 ... 0 1 1 ... 1... ... ... ... ... ... ... ... ...

vm 0 0 ... 0 1 1 ... 1v'1 1 1 ... 1 0 0 ... 0v'2 1 1 ... 1 0 0 ... 0... ... ... ... ... ... ... ... ...

v'n 1 1 ... 1 0 0 ... 0

e) Ma trn lin k ca th lp phng Qn( 2n nh ng vi n xunh phn khc nhau cha bit 0, 1)

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00..00 00..01 00..10 00..11 ... 10..00 10..01 ... 11..11

00..00 0 1 1 0 ... 1 0 ... 000..01 1 0 0 1 ... 0 1 ... 000..10 1 0 0 1 ... 0 0 ... 000..11 0 1 1 0 ... 0 0 ... 0

... ...10..00 1 0 0 0 ... 0 1 ... 010..01 0 0 0 0 ... 1 0 ... 0

...11..11 0 0 0 0 ... 0 0 ... 0

*Bi 7:

Hai n th vi ma trn lin k sau y c l ng cu khng?

0 1 0 1 0 1 1 11 0 0 1 1 0 0 10 0 0 1 1 0 0 11 1 1 0 1 1 1 0

Ma trn 1 Ma trn 2

Li gii:

Cch 1:Da vo ma trn lin k, ta c th v c 2 th tng ng nh

sau:

G1 G2

G1=(V,E): th ng vi ma trn 1G2=(V',E'): th ng vi ma trn 2

V1

V4 V3

V2

V'4

V'1

V'3

V'2

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D dng nhn thy:- S cnh ca 2 th khc nhau: G1 c 4 cnh, G2 c 5 cnh- Ngoi ra:

G1 c 1 nh bc 1 (V3)2 nh bc 2 (V1,V2)1 nh bc 3 (V4)

G2 khng c nh bc 12 nh bc 2(V'2,V'3)2 nh bc 3(V'1,V'4)

Vy 2 th trn khng ng cu.

Cch 2:Tng cc phn t trong ma trn lin k ca n th bng tng s

bc ca cc nh v bng 2 ln s cnh ca th.

T 2 ma trn trn ta c:- th ng vi ma trn 1 c 8:2=4 cnh- th ng vi ma trn 2 c 10:2=5 cnh

Nh vy, 2 n th ng vi 2 ma trn lin k trn khng ng cu.

*Bi 8:Hai n th vi ma trn lin thuc sau c l ng cu khng?

1 0 0 0 0 0 1 0 0 11 0 1 0 1 0 1 1 1 00 0 0 1 1 1 0 0 1 00 1 1 1 0 1 0 1 0 1

Li gii:- Ma trn 1:

e1 e2 e3 e4 e5

u1 1 1 0 0 0u2 1 0 1 0 1

u3 0 0 0 1 1 ng vi th G=(U,E)u4 0 1 1 1 0

- Ma trn 2:

e'1 e'2 e'3 e'4 e'5

v1 0 1 0 0 1

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v2 0 1 1 1 0 ng vi th G'=(V,E')v3 1 0 0 1 0v4 1 0 1 0 1

e1 e'2

e2 e3 e5 e'5 e'3 e'4

e4 e'1

G=(U,E) G'=(V,E')Xt php ng cu f:

e1e'2e2e'5e3e'3e4e'1e5e'4

Lc ny, ta biu din li ma trn lin thuc ca th G' theo th tcc nh v1, v2, v3,v4 v th t cc cnh e'2, e'5, e'3, e'1, e'4 nh sau:

e'2 e'5 e'3 e'1 e'4

v1 1 1 0 0 0v2 1 0 1 0 1v3 0 0 0 1 1v4 0 1 1 1 0

Ma trn n y v ma trn lin thuc ca G bng nhau.Vy G v G' ng cu vi nhau.

* Bi 9: Cc th G v G sau c ng cu vi nhau khng?

10

U1

U4

U3

U2

V4

V1

V3

V2


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a)

b)

Li gii:a) Xt php ng cu f:

u1v2u2v3u3v6u4v5

u5v4u6v1Lc ny, ma trn lin k ca G (theo th t cc nh u6, u1, u2, u5, u4,

u3) vma trn lin k ca G' l bng nhau v bng:0 1 1 1 1 11 0 1 1 0 01 1 0 1 0 11 1 1 0 1 11 0 0 1 0 11 0 1 1 1 0

Vy G v G ng cu vi nhau.

b)Xt php ng cu f:u1v3u2v5

11

u1

u2

u3

u4

u5

u6

v2

v4

v5

v6

v3

v1

u2

u3

u4 u5 u6

v1

v4

v6

u1

v2

v5

v3


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u3v1 u4v2

u5v4u6v6

Lc ny, ma trn lin k ca G(theo th t cc nh v3, v4, v1, v5, v2,v6) v na trn lin k ca G bng nhau v bng:

0 1 0 0 0 10 0 0 0 1 00 1 0 1 0 01 0 0 0 0 00 0 0 1 0 00 0 1 0 1 0

Vy, hai th G v G ng cu vi nhau.

* Bi 10:Cho V={2,3,4,5,6,7,8} v E l tp hp cc cp phn t (u,v) ca V sao

cho u


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* Bi 11:Hy tm s ng i di n gia hai nh lin k (t.. khng lin

k) ty trong K3,3 vi mi gi tr ca n sau:a) n=2 b) n=3 c) n=4 d) n=5

Li gii:

K3,3 *Cch 1:

Tp cc nh ca K3,3 c chia lm 2 phn:- Phn 1 gm V1, V2, V3- Phn 2 gm V4, V5, V6

Trong , 2 nh thuc cng 1 phn th khng lin k2 nh thuc 2 phn khc nhau th lin k.Gi d l s ng i di n gia 2 nh thuc K3,3.* Nu n chn th im u v im cui ca ng i phi nm

trong cng 1 phn (chng khng lin k).*Nu n l th im u v im cui ca ng i phi nm 2

phn khc nhau (chng lin k vi nhau).M khi xut pht t 1 nh ta lun c 3 cch i(do mi phn gm 3

nh). p dng quy tc nhn ta c s ng i c di n gia 2 nh l:- Nu 2 nh lin k:

+ n chn: d=0+ n l : d=3n-1(do cnh cui cng ni vi nh cui ch c 1 cch)

- Nu 2 nh khng lin k:+ n chn : d=3n-1(do cnh cui cng ni vi nh cui ch c 1 cch)+ n l : d=0

p dng c th:

13

V

1

V

4

V

5

V

6

V

2

V

3


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dinh

n=2 n=3 n=4 n=5

Lin k d=0 d=9 d=0 d=81

Khng lin k d=3 d=0 d=27 d=0* Cch 2: th K3,3 c ma trn lin k theo th t cc nh V1, V2, V3, V4,

V5, V6 nh sau:0 0 0 1 1 10 0 0 1 1 1

A= 0 0 0 1 1 11 1 1 0 0 01 1 1 0 0 01 1 1 0 0 0

Ta c mnh : Cho G l mt th (v hng hoc c hng) vima trn lin k A theo th t cc nh v1, v2, ..., vn. Khi s cc ng ikhc nhau di r t vi ti vj trong r l mt s nguyn dng, bng gi trca phn t dng i ct j ca ma trn Ar.

nTa c:

An = A.A...A.A

3n-1 3n-1 3n-1 0 0 03n-1 3n-1 3n-1 0 0 0

A= 3n-1 3n-1 3n-1 0 0 0 , nu n chn0 0 0 3n-1 3n-1 3n-1

0 0 0 3n-1 3n-1 3n-1

0 0 0 3n-1 3n-1 3n-1

0 0 0 3n-1 3n-1 3n-1

0 0 0 3n-1 3n-1 3n-1

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0 0 0 3n-1 3n-1 3n-1

A= 3n-1 3n-1 3n-1 0 0 0 , nu n l3n-1 3n-1 3n-1 0 0 03n-1 3n-1 3n-1 0 0 0

Nh vy, theo mnh trn, p dng vo cc trng hp c th bi cho ta c kt qu nh cch 1.

* Bi 12:Mt cuc hp c t nht 3 i biu n d.Mi ngi quen t nht 2i biu khc.Chng minh rng c th sp xp mt s i biu ngixung quanh mt bn trn mi ngi ngi gia 2 ngi m i

biu quen.Li gii:

* Ta c th biu din mi quan h ca cc i biu n tham dcuc hp bng n th G=(V,E).

G c n nh (n3, n l s i biu) v e cnh.Mi nh ca th ng vi 1 i biu, gia 2 nh ng vi 2 i

biu quen nhau tn ti 1 cnh.Gi Vi (i=1,2,...,n): nh ca th (ng vi 1 i biu)Do mi ngi quen t nht 2 i biu khc nndeg(Vi) 2 n

deg(Vi) 2ni=1

S cnh ca th: e n (1)* Mt khc, theo ra ta c: cc i biu ngi xung quanh 1 bn

trn.V vy, th biu din cch sp xp ch ngi ca cc i biu tha

yu cu l th vng Cn.Trong th vng Cn c n (cnh), n cnh ny c ly t e cnh ca

G(do n biu th mi quan h gia cc i biu) (2)* Tp nh ca G v Cn bng nhau v bng n. (3)T (1), (2) v (3) cho thy, Cn l th con bao hm ca G.(Cn c

to ra bng cch b i mt s cnh thch hp ca G)Vy, da trn mi quan h gia cc i biu nh trn ta c th sp

xp cc i biu ngi quanh bn trn sao cho mi ngi ngi gia 2 ngim h quen.( pcm)

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*Bi 13:Mtlp hc c t nht 4 sinh vin. Mi sinh vin thn vi t nht 3

sinh vin khc. Chng minh rng c th xp mt s chn sinh vin ngiquanh mt ci bn trn mi sinh vin ngi gia 2 sinh vin m h thn.

Li gii:* Mi quan h gia cc sinh vin trong lp c th biu din bng 1

n th G=(V,E) n nh(n4, n: s sinh vin), e cnh.Hai nh ng vi 2 sinh vin thn nhau lin k vi nhau.Gi Vi (i=1,2,...,n): nh ca th ng vi 1 sinh vin.Mi sinh vin thn vi t nht 3 ngi

deg(Vi) 3 n

deg(Vi) 3n i=1

Tng s cnh ca G l: e 3n/2 (1)* Mt khc, theo ra ta c: cch sp xp ch ngi ca cc sinh

vin c th biu din bng th vng Cn (do cc sinh vin ngi quanh bntrn).

Cn c n cnh (n cnh ny ly t e cnh ca G)M e phi l s nguyn suy ra n phi chia ht cho 2 (n chn)Tp nh ca Cn v G bng nhau v bng n.

T , ta thy Cn chnh l th con bao hm ca G.(Cn c th to ra t Gbng cch b i mt s cnh thch hp)Hay: c th sp xp mt s chn sinh vin ngi quanh mt ci bn

trn sao cho mi ngi ngi gia 2 ngi m h thn.( pcm)

* Bi 14:Trong mt cuc hp c ng 2 i biu khng quen nhau v mi i

biu ny c mt s l ngi quen n d.Chng minh rng lun lun c thxp mt s i biu ngi chen gia 2 i biu ni trn, mi ngi ngi

gia 2 ngi m i biu quen.Li gii:

Mi quan h giacc i biu n tham d cuc hp c th biudin bng 1 n th G=(V,E).Trong mi nh l mt i biu, gia 2nh ng vi 2 i biu quen nhau tn ti 1 cnh.

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Trong cuc hp c ng 2 i biu khng quen nhau v c s lngi quen n tham d.Vy G c ng 2 nh khng lin k v 2 nh nyc bc l.

T mnh : Nu mt th c ng hai nh bc l th hai nhny phi lin thng, tc l c mt ng i ni chng ta suy ra c th tm ramt s i biu ngi chen vo gia 2 i biu ny sao cho mi i biu ngigia 2 ngi m i biu quen.(do 2 nh ng vi 2 ngi ny khng linthng, 2 ngi khng ngi st nhau v h quen vi n-2 ngi cn li)

*Bi 15:Mt thnh ph c n (n 2) nt giao thng v hai nt giao thng bt

k u c s u mi ng ngm ti mt trong cc nt giao thng ny ukhng nh hn n. Chng minh rng t mt nt giao thng tu ta c th in mt nt giao thng bt k khc bng ng ngm.

Li gii:- Ta c th xem h thng ng ngm ca thnh ph l mt n th

c cc nh l cc nt giao thng.S nh ca th chnh l s nt giao thng: n (n2)Cnh ca th l ng ngm ni 2 nt giao thng.Theo ra ta c:

Hai nt giao thng bt k u c s u mi ng ngm timt trong cc nt giao thng u khng nh hn n.

- Ta c mnh :Mi n th n nh (n2) c tng bc ca 2 nh ty khng

nh hn n u l th lin thng.Vy, theo nh l trn, h thng ng ngm ca thnh ph l th

lin thng.Suy ra, tmt nt giao thng tu ta c th i n mt nt giao thng

bt k khc bng ng ngm.(pcm).

*Bi 16:C bao nhiu n th ng cu vi n nh khi:a) n=2 b) n=3 c) n=4

Li gii:a) Vi n=2, c 2 n th khng ng cu nh sau:

v

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b) Vi n=3, c 4 n th khng ng cu:

c) Vi n=4 c 11 n th khng ng cu:

BI TP TON RI RC***

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CHNG 3:

TH EULER TH EULER

V THV THHAMILTONHAMILTON

Ging vin : Nguyn Mu HnSinh vin thc hin : Nguyn Th Diu HngLp : Tin K30D

*Bi 1:Vi gi tr no ca n th cc th sau c chu trnh Euler?

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a) Kn b) Cn c) Wn d) QnLi gii:

Kn, Cn, Wn, Qn u l th lin thng. th lin thng cha chu trnh Euler l th Euler. Ta c th hiu

bi ton l tm gi tr ca n cc th trn l th Euler.Ta c nh l:

th(v hng) lin thng G l th Euler khi v ch khimi nh ca G u c bc chn.

a) KnMi nh ca Kn u c bc l n-1. Do , Kn l th Euler th n-1

phi l s chn.Hay n l s l: n=2k+1 (kZ*)

b) Cn (n3)Mi nh ca Cn u c bc 2(chn). Vy, Cn lun l th Euler.

c) WnWn c n+1 nh.Trong , c 1 nh bc n v n nh bc 3.Nh vy,

Wn khng th l th Euler.d) QnTrong Qn c 2n nh, mi nh c bc l n. Vy, Qn l th Euler

th nphi chn.

*Bi 2:

Vi gi tr no ca m, n th cc th phn i y Km,n c:a) Chu trnh Euler b) ng i Euler Li gii:

a) th phn i y Km,n c chu trnh Euler th cc nh caKm,nphi c bc chn.

M cc nh ca Km,n c bc m hoc n.Vy mun Km,n c chu trnh Euler th m, n phi l s chn.

b) th phn i y c ng i Euler th trong Km,n phic ng 2 nh bc l.Vi n=m=1 th th phn i khng phi l th cng i Euler.

Hay mt trong hai gi tr m hoc n phi bng 2 v gi tr cn li phil s l.

* Bi 3:

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Vi gi tr no ca m v n th th phn i y Km,n c chu trnhHamilton.

Li gii:Cch 1: ( theo nh l Dirac)

nh l Dirac pht biu nh sau: Nu G l mt n th c nnh v mi nh ca G u c bc khng nh hn n/2 th G l thHamilton.

Suy ra: Km,n c chu trnh Hamilton th mi nh ca Km,n phi cbc khng nh hn n/2:

deg(Vi) (n+m)/2 (1)M trong Km,n, deg(Vi)={m,n}

T (1) ta c:n (m+n)/2 (n-m)/2 0m (m+n)/2 (m-n)/2 0

(n-m)/2 0(n-m)/2 0

n-m= 0n=m

Vy vi n=m th Km,n c chu trnh Hamilton.

Cch 2: (theo nh l Ore)nh l Ore c pht biu nh sau: Nu G l mt n th c n

nh v bt k 2 nh khng k nhau cng c tng s bc khng nh hn nth G l th Hamilton.

Hai nh khng lin k ca Km,n nm cng mt phn, bt k 2 nhkhng lin k no u c tng bc l n+m.

Km,n c chu trnh Hamilton, theo nh l Ore th:n+n n+m n-m 0 n=mm+m n+m m-n 0

Vy vi n=m th Km,n c chu trnh Hamilton.

Cch 3:Ta c nh l: Nu G l d th phn i vi 2 tp nh l V1 v V2 c

s nh cng bng n v bc ca mi nh ln hn n/2 th G l thHamilton.

Vy vi n=m th Km,n c chu trnh Hamilton.*Bi 4:

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Chng minh rng th lp phng Qn l mt th Hamilton.Vcy lit k tt c cc chu trnh Hamilton ca th lp phng Q3.Li gii:

Theo nh l Dirac: Nu G l n th c n nh v mi nh caG u c bc khng nh hn n/2 th G l th Hamilton.

M trong th lp phng Qn, mi nh u c bc n.Vy, th lp phng Qn l th Hamilton (pcm)

* V cy chu trnh Hamilton ca th lp phng Q3:

* Bi 5:Trong mt cuc hp c 15 ngi mi ngy ngi vi nhau chung mt

bn trn mt ln. Hi c bao nhiu cch sp xp sao cho mi ln ngi hp,mi ngi c 2 ngi ngi bn cnh l bn mi, v sp xp nh th no?

Li gii:

22

000

100010001

011 101 110 011 110 101

001

010

110

100

101

111 010100 111 001100111100111010111

000

111

110

101

100

000 000

100

110

010

110 110

011

111

010

000

011

001

101

100

000

101

111

001

011

000 000

100

111

110

101

000

001

101

111

110

000

010

011

001

101 011

111

101

001

000

110

010

011

001

000

011

111

110

100

000


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Xt n th gm n=15 nh, mi nh ng vi mt i biu thamgia cuc hp, hai nh k nhau khi hai i biu mun lm quen vi nhau.Vyta c n th y K15.

y l th Hamilton, mi chu trnh Hamilton chnh l mt cch spxp ch ngi cho cc i biu tha mn yu cu bi.

Theo nh l, trong Kn vi n l, n3 c ng (n-1)/2 chu trnhHamilton phn bit.Vy c (15-1)/2 = 7 cch sp xp ch ngi nh trn.

Mi cch sp xp l mt chu trnh Hamilton ca K15.

* Bi 6:Hiu trng mi 2n(n2) sinh vin gii n d tic.Mi sinh vin gii

quen vi t nht n sinh vin gii khc n d tic.Chng minh rng lunlun c th xp tt c cc sinh vin gii ngi xung quanh mt bn trn

mi ngi ngi gia hai ngi m sinh vin quen.Li gii:

Cho th G=(V,E), mi nh ca G l mt sinh vin, gia 2 sinh vinquen nhau tn ti mt cnh.G l n th c 2n nh.

Do mi sinh vin n d tic quen vi t nht n sinh vin khc nn bcca mi nh ca th G deg(Vi) n (2n/2)

Theo nh l Dirac th G l th Hamilton.Suy ra, tn ti chu trnhHamilton trong G.Mi chu trnh Hamilton l mt cch sp xp ch ngi chocc sinh vin xung quanh bn trn sao cho mi ngi ngi gia 2 ngi hquen.Vy ta c iu phi chng minh.

* Bi 7: th trong hnh sau gi l th Peterson

a) Tm mt ng i Hamilton trong Gb) Chng minh P\{v}, vi v l nh bt kca P, l mt th Hamilton.

Li gii:a) Mt ng i Hamilton trong G:

23

e

k i

bg

f h

d c

a


24/70

abcdefhkgib) Chng minh P\{v} l mt th Hamilton(vi v l mt nh bt

k ca P):R rng trong P, cc nh c chia lm 2 phn

- Phn 1: gm a, b, c, d, e- Phn 2: gm f, g, h, i, k

Trong mi phn, cc nh c vai tr nh nhau.Xt 2 trng hp:

* Trng hp 1: v l nh thuc phn 1Do vai tr cc nh nh nhau, gi s ta b nh a.Lc ny, trong P\

{a} tn ti chu trnh Hamilton: i f e d c b h k g i.Suy ra, P\{a} l th Hamilton.

* Trng hp 2: v l nh thuc phn 2

Cng nh trng hp trn, vai tr cc nh nh nhau, gi s ta bnh f. Trong P\{f} tn ti chu trnh Hamilton:

hk d e a g i c b hSuy ra, P\{f} l th Hamilton

Nh vy, trng hp tng qut, vi v l nh bt k ta lun c P\{v} l th Hamilton(pcm)

* Bi 8:Gii bi ton ngi pht th Trung Hoa vi th cho trong hnh

sau:

Li gii:

Trc tin, ta gn nhn cho th:

24


25/70

Cc nh bc l:V0(G)={B, E,I, L}

Tp cc phn hoch cp:P={P1, P2, P3}

Trong :

P1={B,E),(I,L)}d(P1) = d(B,E) + d(I,L) = 2 + 2 = 4P2={(B,I),(E,L)}d(P2) = d(B,I) + d(E,L) = 3 + 1 = 4P3={(B,L),(E,I)}d(P3) = d(B,L) +d (E,I) = 2 + 3 = 5

m(G) = min{d(P1), d(P2), d(P3)} = 4Vy, GT c c bng cch thm vo th G 4 cnh (A,B), (A,J),

(I,J), (E,L).GT l th Euler. ng i ngn nht l chu trnh Euler trongGT, l:

H, I, G, K, F, L, K, J, I, J, A, J, C, B, A, D, E, L, D, E, L, E, F, G, H

*Bi 9:Chng minh rng th G cho trong hnh sau c ng i Hamilton

(t s n r) nhng khng c chu trnh Hamilton.

Li gii:Trong G tn ti ng i Hamilton (t s n r).

25

J C

I K DL

F EGH

A B

s

ec

b f

a

g

d

h

r


26/70

Tht vy,trong G tn ti ng i:sabcefgdhr lng i Hamilton.

Ta gi s trong G tn ti chu trnh Hamilton.Theo hnh v ta thy, i ti s th phi i qua a hoc c.Mt khc, i ti b cng phi i qua a hocc.Nh vy, trong chu trnh ny, nh a hoc c s xut hin 2 ln.V l, v yl chu trnh Hamilton, mi nh ch xut hin 1 ln(ngoi tr nh u vnh cui trng nhau).

Vy, khng tn ti chu trnh Hamilton trong th G.

* Bi 10:Cho v d v:a) th c mt chu trnh va l chu trnh Euler, va l chu trnh

Hamilton.

b) th c mt chu trnh Euler v mt chu trnh Hamilton, nhng haichu trnh ny khng trng nhau.

c) th c 6 nh, l th Hamilton nhng khng phi l thEuler.

d) th c 6 nh, l th Euler nhng khng phi l thHamilton.Li gii:

a) th c mt chu trnh va l chu trnh Euler, va l chu trnhHamilton:

b) th c mt chu trnh Euler, mt chu trnh Hamilton nhng 2 chutrnh ny khng trng nhau:

b1)

26

A

DE

B

CG


27/70

Trong th trn c cha chu trnh Hamilton v chu trnh Euler.* Chu trnh Hamilton:

A, B, C, D, E, G, A* Chu trnh Euler:

A, B, C, D, G, B, D, E, G, AHai chu trnh ny khng trng nhau.

b2)

th trn c:* Chu trnh Hamilton:

A, B, C, D, E, A* Chu trnh Euler:

A, B, C, D, E, B, D, A, C, E, A

c) th c 6 nh, l th Hamilton m khng phi l th Euler:

th trn l th Hamilton, do c cha chu trnh Hamilton:A, B, C, D, E, G

Nhng th trn khng phi l th Euler, do mi nh ca thu c bc l( bc 3)

27

A

BE

D C

A B

G

E D

C


28/70

d) th c 6 nh, l th Euler nhng khng phi l thHamilton:

th trn cha chu trnh Euler: A, B, C, D, E, B, G, A nn n l

th Euler.Ta thy rng, E v G khng lin k.Mun i t E qua G hay ngc li

th cn thng qua B.Vy, trong bt k chu trnh no cha tt c cc nh thc nh B xut hin hn 1 ln.Vy khng tn ti chu trnh Hamilton, hay th trn khng phi l th Hamilton.

---&0&---

28

G

E D

A

C

B


29/70

BI TP TON RI RC---&0&--

CHNG 4:

MT S BIMT S BITON TI UTON TI UTRN THTRN TH

Ging vin : Nguyn Mu Hn Sinh vin thc hin : Nguyn Th Diu Hng

Lp : Tin K30D

29


30/70

* Bi 1:Dng thut ton Dijkstra tm ng i ngn nht t nh a n cc

nh khc trong th sau:

Li gii:

L(a) L(b) L(c) L(d) L(e) L(g) L(h) L(k)

0 a- 4 11 2 b

- 4 7 9 - 3 c

- 4 5 7 15 9 - - d

- - 5 7 15 9 - - e

- - - 7 15 8 - - g

- - - - 14 8 - - h

- - - - 13 - - - k

b

a

k

h

c

e

g

d

4

32

4

2

1

11

5

7

12

2

7

34

5

30


31/70

* Bi 2:Dng thut ton Dijkstra tm ng i ngn nht t nh a n cc

nh trong th sau:

Li gii:L(a) L(b) L(c) L(d) L(e) L(f) L(g) L(h) L(i) L(k)

0 a

- 1 10 6 3 b

- - 10 6 3 5 c

- - 10 5 - 5 9 11 d

- - 9 - - 5 8 11 e

- - 6 - - - 7 7 11 10 f

- - 6 - - - - 7 11 9 g

- - 6 - - - - - 11 9 h

- - - - - - - - 10 9 i

10 - k

* Bi 3:

a

c

b

g

d

i

f

h

e

k

1

10

6

3 2

4

10

4

1

4

1

3

6

8

5

3

25

2

8

5

31


32/70

Cho th c trng s nh hnh di y.Hy tm ng i ngn nhtt nh A n nh N.

Li gii:

L(A) L(B)

L(C) L(D) L(E)

L(F) L(G) L(H) L(I) L(J) L(K) L(L)

L(M) L(N)

0 A

- 7 4 1 B

- 7 3 - 3 C

- 6 - - 3 D

- 6 - 7 - - 12 E

- - 9 - 7 - - 12 F

- - 9 - - - - 12 G

- - - 17 - - 15 - - 11 H

- - - 17 - - 14 - - - 16 I

- - - 16 - - - - - - 16 J

- - - - 19 - - - 18 - - - 16 K

- - - - 19 - - - 18 - - - - 23 L

- - - - 19 - - - - - - - - 20 M

- - - - - - - - - - - - - 20 N

* Bi 4:

A

F

J

G H I

K

B

L

C

M

D

N

E7 3 8 3

2 9 5 7

1 4 2 2 3

4 3 2 2 5 2 2 5

2 3 4 5 3 4 3 2

32


33/70


34/70

A B C D E F GA 3 6B 3 2 4C 6 2 1 4 2D 4 1 2 4E 4 2 2 1F 2 2 4G 4 1 4

W=W0p dng thut ton Floyd ta c:

A B C D E F GA 3 6B 3 6 2 4C 6 2 12 1 4 2

D 4 1 2 4E 4 2 2 1F 2 2 4G 4 1 4

W1

A B C D E F GA 6 3 5 7B 3 6 2 4C 5 2 4 1 4 2D 7 4 1 8 2 4E 4 2 2 1F 2 2 4G 4 1 4

W2A B C D E F G

A 6 3 5 6 9 7B 3 4 2 3 6 4C 5 2 4 1 4 2

D 6 3 1 2 2 3 4E 9 6 4 2 8 2 1F 7 4 2 3 2 4 4G 4 1 4

W3

A B C D E F G

34


35/70

A 6 3 5 6 8 7 10B 3 4 2 3 5 4 7C 5 2 2 1 3 2 5D 6 3 1 2 2 3 4E 8 5 3 2 4 2 1F 7 4 2 3 2 4 4G 10 7 7 4 1 4 8

W4

A B C D E F GA 6 3 5 6 8 7 9B 3 4 2 3 5 4 6C 5 2 2 1 3 2 4

D 6 3 1 2 2 3 3E 8 5 3 2 4 2 1F 7 4 2 3 2 4 3G 9 6 4 3 1 3 2

W5=W6

A B C D E F GA 6 3 5 6 8 7 9B 3 4 2 3 5 4 6C 5 2 2 1 3 2 4D 6 3 1 2 2 3 3E 8 5 3 2 2 2 1F 7 4 2 3 2 4 3G 9 6 4 3 1 3 2

W7=W*

* Bi 5:Tm W* bng cch p dng thut ton Floyd vo th sau:

35


36/70

Li gii:Ta c ma trn trng s ca th trn l: (nhng trng l )

A B C D E FA 3 1 20B 8 2C 6 8DE 4 3F 5 13

Wp dng thut ton Floyd ta c:

A B C D E FA 3 1 20B 8 2C 6 8D

E 4 3F 5 13

W0=W1=W

A

E

F

B C

D

3

1

2

0

8

6

13

43

5

8

2

36


37/70

A B C D E FA 3 11 1 5B 8 2C 6 8DE 4 3F 5 13

W2

A B C D E FA 3 11 17 1 5B 8 14 16 2C 6 8D

E 4 3F 5 11 13

W3

A B C D E FA 3 11 5 1 4B 8 14 16 2C 6 8 11DE 4 3F 5 11 13 14

W4=W5

A B C D E FA 3 9 5 1 4B 7 13 15 2C 16 6 8 11DE 8 4 16 3

F 5 11 13 14W6=W*

* Bi 6:

37


38/70

Gii bi ton mng vn ti sau bng thut ton Ford-Fulkerson vilung vn ti khi u bng 0:

( Kh nng thng qua ca cc cung l cc s mu xanh)Li gii:

Lung c ng i (V0, V1), (V1, V5), (V5, V7) gm cc cung chabo ha. Nng lung ln 4 n v c lung 1:

1

Xt xch =(V0, V1, V4, V2, V6, V7). Nng lung 1 ln 2 n v ta cc lung 2:

V

3

V

4

V

0

V

2

V

6

V

7

V

1

V

5

8

4

2

6

24

4

6

8

2

4

3

4

V

3

V

4

V

0

V

2

V

6

V

7

V

5

8

4

2

6

24

4

6

8

2

4

3

4

V

1

+0 +1

+5

4

4

4

0

38


39/70

2

Xt xch =(V0, V1, V5, V3, V4, V7). Nng lung 2

ln 2 n v ta clung 3:

3

V

3

V

4

V

0

V

2

V

6

V

7

V

5

8

4

2

6

24

4

6

8

2

4

3

4

V

1

+0

+1

+4

6

4

4

+2

0

2

2

2

2

V

1

V

4

V

2V

6

V

0

V

7

Xch

0

+0

+1 +4

+2

+6

0+2 0+20+2

0+24+2

V

3

V

4

V

0

V

2

V

6

V

7

V

5

8

4

2

6

24

4

6

8

2

4

3

4

V

1

+0

+1

+4

8

6

4

+2

0

2

2

2

2

2

2

2

39


40/70

Xt xch =(V0, V3, V6, V7). Nng lung 3 ln 2 n v ta c lung4:

4

Xt xch =(V0, V2, V6, V7). Nng lung 4 ln 4 n v ta c lung 5

V

1

V

5

V

3 V

4

V

0

V

7

0

+0

+1+5

+3

+446+2

4+2

0+2 0+2

0+2

Xch

V

3

V

4

V

0

V

2

V

6

V

7

V5

8

4

2

6

24

4

6

8

2

4

3

4

V1

+0

+1

+4

8

6

4

+2

0

2

2

2

2

2

2

4

2

2

V

0

V

3

V

6

V

7

Xch

0 +0 +3 +4

0+2 0+2 2+2

40


41/70

5

Tip theo ta ch c th nh du c V0 nn qu trnh nng lung kt

thc v ta c gi tr ca lung cc i l:5=8+4+2=14

* Bi 7:Gii bi ton mng vn ti sau bng thut ton Ford-Fulkerson vi

lung vn ti khi u c cho km theo:

(gi tr thng qua l cc s mu )

V

3

V

4

V

0

V

2

V

6

V

7

V

5

8

4

2

6

24

4

6

8

2

4

3

4

V

1

+0

+1

+4

8

6

4

+2

0

2

2

2

2

2

6

8

2

2

4

V

1

V10

V11

V

9

V8

V

7

V

6

V4

V3

V

5

V0

V

2

10

4

6

10

5

8

28

15

20

10

306

15

1510 8

12

3

1

420

2

6

7

8

8

6

6

8

20

16

2

16

8

0 10

2

00

2

0

3

25

7

200

2

0

41


42/70

Li gii:Lung c ng i (V0, V1), (V1, V3), (V3, V7), (V7, V10), (V10, V9),

(V9, V11) bao gm cc cung cha bo ha. Ta c th nng lung ln 2 nv c lung 1:

1

Xt xch =(V0, V5, V8, V9, V11). Nng lung 1 ln 2 n v ta clung 2:

V

1

V10

V11

V

9

V

8

V

7

V

6

V

4

V

3

V

5

V

0

V

2

10

4

6

10

5

8

28

15

20

10

306

15

1510 8

12

3

1

420

2

6

7

8

10

6

6

8

4 0

16

2

16

8

0 10

2

00

2

2

325

7

2

20

0

2

0

+0

+1

+3

+7

-10

+9

V

0

V

5

V

8V

9

V11

0

+0

+5

+8

+9

Xch

2+2

2+22+2

2+2

42


43/70

2

Xt xch =(V0, V6, V7, V3, V4, V11).Nng lung 2 ln 2 n v tac 3:

V

1

V10

V11

V

9

V8

V

7

V

6

V

4

V

3

V

5

V

0

V

2

10

4

6

10

5

8

28

15

20

10

306

15

1510 8

12

3

1

420

2

6

7

8

10

6

6

8

4 0

16

4

16

8

0 10

4

00

4

2

3

25

7

2

20

0

4

0

+0

+8

+9

+5

V

0

V

6

V

7 V

3

V

4

V11

0

+0

+6

-7

+3

+4Xch

16+2

10+22-2

16+2

25+2

43


44/70

3

Tip theo ta ch c th nh du V0 nn qu trnh nmg lung kt thcv ta c gi tr lung cc i l:

3=27+2+7+4=40

* Bi 8:Hy gii bi ton ngi du lch vi 6 thnh ph c s liu cho trong

ma trn trng s sau:

25 45 14 32 249 16 2 34 2322 11 33 7 0

23 14 27 20 2114 44 29 46 325 3 4 7 8

V

1

V10

V11

V

9

V8

V

7

V

6

V

4

V

3

V

5

V

0

V

2

10

4

6

10

5

8

28

15

20

10

306

15

1510 8

12

3

1

420

2

6

7

8

10

6

6

8

4 0

18

4

18

8

0 12

4

00

4

0

3

27

7

2

20

0

4

0 +3-7

+4

+6+0

44


45/70

Li gii:Ma trn trng s M:

25 45 14 32 249 16 2 34 23

22 11 33 7 023 14 27 20 2114 44 29 46 325 3 4 7 8

* Ma trn rt gn M:

1 2 3 4 5 61 11 30 0 13 102 0 13 0 27 213

15 11 33 2 04 2 0 12 1 75 4 41 25 43 06 15 0 0 4 0

Tng hng s rt gn s=49Trong M c m14=m21=m24=m36=m42=m56=m62=m63=m65=0

14=10 21=2 24=0 36=2 42=1 56=462=0 63=12 65=1Ta thy 63=12l ln nht. Vy chn (6,3) phn nhnh. Cn

di ca nhnh khng cha (6,3) l s+63=49+12=61.Xa dng 6 ct 3 ri

t m'36=.Tp cc hnh trnh

Cn di =49

Hnh trnh cha cnh (6,3) Hnh trnh khng cha (6,3)

Cn di=49 Cn di=61

1 2 4 5 6

1 11 0 13 102 0 0 27 213 15 11 33 2 4 2 0 1 75 4 41 43 0

45


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*Ma trn rt gn M'':

1 2 4 5 6

1 11 0 13 102 0 0 27 213 13 9 31 0 4 2 0 1 75 4 41 43 0

Tng hng s rt gn s''=2.m''14=m''21=m''24=m''35=m''42=m''56=014=10 21=2 24=0 35=1042=10 56=11

Ta thy 56=11 l ln nht. Vy chn (5,6) phn nhnh.Cndi ca nhnh cha (5,6) l 49+2=51.Cn di ca nhnh khng cha cnh(5,6) l 51+11=62.Xa dng 5 ct 6, sau t m''35=.

Hnh trnh cha (6,3) Cn di=49

Hnh trnh cha (5,6) Hnh trnh khng cha (5,6)Cn di=51 Cn di =62

1 2 4 51 11 0 132 0 0 273 13 9 31 4 2 0 1

* Ma trn rt gn M''': Tng hng s rt gn s'''=10

1 2 4 5

1 11 0 122 0 0 263 4 0 22 4 2 0 0

Trong ma trn trn c m'''14=m'''21=m'''24=m'''42=014=11 21=2 24=0 42=032=4 45=12

46


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Ta thy 45=12 l ln nht. Vy chn (4,5) phn nhnh.Cndi ca nhnh cha (4,5) l 51+10=61.Cn di ca nhnh khng chacnh (4,5) l 61+12=73. Xa dng 4 ct 5, sau t m'''34 = .

Hnh trnh cha (5,6)Cn di=51

Hnh trnh cha (1,4) Hnh trnh khng cha (1,4)Cn di=61 Cn di=73

1 2 4

1 11 02 0 03 4 0

* Ma trn rt gn M'''':

1 2 4

1 11 02 0 03 4 0

Trong ma trn trn c:Tng hng s rt gn s''''=0

m''''21=m''''32 =m''''14 =m''''24 =0

21=4 14=11 24=0 32=15Ta thy 32=15 l ln nht nn ta chn (3,2) tip tc phn

nhnh.Cn di ca nhnh cha (3,2) l 61. Xa dng 3 ct 2 , sau tm''''24=. Cn di ca nhnh khng cha (3,2) l 61+15=76.

Hnh trnh cha (1,4)Cn di=51

Hnh trnh cha (2,1) Hnh trnh khng cha (2,1)Cn di=61 Cn di=76

1 41 02 0

* Ma trn rt gn M''''': Tng hng s rt gn s'''''=0

47


48/70

1 4

1 02 0

Hai cnh cn li ca chu trnh khng phi chn na m c avo chu trnh.

y ta c cc cnh (6,3), (5,6), (4,5), (3,2), (1,4), (2,1).Vy ta c chu trnh: 6, 3, 2, 1, 4, 5, 6 vi chi ph 61 l ti u.

48


49/70

BI TP TON RI RC---&0&---

CHNG 5:

CYCY

Ging vin : Nguyn Mu HnSinh vin thc hin : Nguyn Th Diu Hng

Lp : TinK30D

49


50/70

* Bi 1:V tt c cc cy ( khng ng cu ) c:a) 4 nh b) 5 nh c) 6 nh

Li gii:a)

b)

* Bi 2:Mt cy c n2 nh bc 2, n3 nh bc 3, ..., nk nh bc k. Hi c bao

nhiu nh bc 1?Li gii:

50


51/70

Gi n1: s nh bc 1 ca cy.Ta c:

- S nh ca cy: n1+n2+n3+...+nk- S cnh ca cy: n1+n2+n3+...+nk-1- S bc ca cy: n1+2.n2+3.n3+...+k.nk

Cy l mt n th, do vy, n s c tng bc bng 2 ln scnh. Tc l:

n1+2.n2+3.n3+...+k.nk=2(n1+n2+n3+...+nk-1)n1= n3+2n4+...+(k-2)nk+2

* Bi 3:Tm s ti a cc nh ca mt cy m-phn c chiu cao h.

Li gii:Cy m-phn c chiu cao h. Vy trong cy ny, cc nh s c xp

vo h+1mc(t mc 0n mc h).S nh ca cy chnh l tng cc nhtrong cc mc ny .

Ta c: Gi tr max ca:- S nh thuc mc 0: 1 =m0 (do ch c 1 nh gc)- S nh thuc mc 1: m = m1- S nh thuc mc 2: m.m= m2

- S nh thuc mc 3: m.m.m= m3

- ...- S nh thuc mc h: mh

Suy ra:S nh ti a ca cy m-phn c chiu cao h l:

h1+m1+m2+m3+...+mh = mi

i=0

* Bi 4:C th tm c mt cy c 8 nh v tha iu kin di y hay

khng? Nu c, v 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.

Li gii:a) Khng c. Gii thch: Cy l mt n th lin thng, khng cha chu trnh

v c t nht 2 nh. Trong n th lin thng, gia 2 nh bt k lun tnti ng i gia chng nn trong cy s cha mt s nh c bc khc 1.

51


52/70

b) Khng c. Gii thch: Ta c: Nu T l mt cy c n nh th T c t nht 2

nh treo( c bc 1). Vy, khng th tm ra mt cy c mi nh u c bc2.

c) C.V d:

d) C. V d:

* Bi 5:Chng minh hoc bc b cc mnh sau: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 lcu.

Li gii:a) Trong mt cy, nh no cng l nh ct.Mnh trn sai.Trong mt cy lun c t nht 2 nh treo v nh treo khng phi l

nh ct(do khi xa n v cnh lin k vi n th khng to ra nhiu thnhphn ln thng hn).

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

Mnh trn sai.Cho mt cy T c n nh. Trong mt cy, mi cnh u l cu. Nh

vy, s cu ca T l n-1( do trong T c n-1 cnh).Mt khc, trong mt cy, ngoi cc nh treo th tt c cc nh cn li

u l nh ct.Mt cy cha t nht 2 nh treo, do s nh ct ln nhtc th c l n-2

S cu = n-1

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S nh ct n-2Vy, trong mt cy, cu c nhiu hn nh ct.

* Bi 6:C 4 i bng A, B, C, D lt vo vng bn kt gii i mnh khu

vc. C my d on xp hng nh sau:- i B v ch, i D nh.- i B nh, i C ba.- i A nh, i C t.Bit rng mi d on trn ng v mt i. Hy cho bit kt qu xphng ca cc i.

Li gii:Theo ra th mi d on ng v mt i.Gi s d on u tin: B v ch, D nh th d on v D

ng.Vy D nh. d on 2: B nh, C ba.Do D nh nn d on B nh khng

ng C ba ng. d on 3: A nh, C t.Do C ng th ba nn d on C t

sai A nh( v l, do d on D nh l ng)Vy d on u tin, d on B v ch ng.B v ch nn d

on 2, d on i C ba ng, v d on th 3 i A nh ng, v cuicng, i D ng th t.

Xp hng cc i l:V ch : B

Nh : ABa : CT : D

* Bi 7:Cy Fibonacci c gc Tn c nh ngha bng hi quy nh sau: T1v T2 u l cy c gc ch gm 1 nh; vi n=2, 3, 4... th cy c gc Tn

D on

B v ch

D nh

C ba

C ba

A nh

A nh

V l

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c xy dng t gc vi Tn-1 nh l cy con bn tri v Tn-2 nh cy conbn phi.

a) Hy v 7 cy Fibonacci c gc u tinb) Cy Fibonacci Tn c bao nhiu nh, l v bao nhiu nh trong.

Chiu cao ca n bng bao nhiu?Li gii:

a) 7 cy Fibonacci c gc u tin:

T1 T2 T3 T4

T5

T6

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b)Gi li, ni v ti ln lt l s l, nh v nh trong ca cy Fibonacci

Ti(i3), hi l chiu cao.Ta c:li = li-1 + li-2ni = ni-1 + ni-2 + 1ti = ni - lihi = hi-1 + 1Trng hp ring:

l1 = l2 = 1n1 = n2 = 1h1 = h2 = 0

* Bi 8:Hy tm cy khung ca th sau bng cch xa i cc cnh trong

cc chu trnh n:a)

T7

d

h i j

e f g

a b c

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Li gii:C nhiu cy khung c th c to ra bng cch xa i cc cnh

trong cc chu trnh n. Sau y l mt s v d:a)

a b

c

dfghh

e

h

f

k

i j

g

l

d

h i j

e f g

a b c

d

h i j

e f g

a b c

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b)

d

h i j

e f g

a b c

a b

c

dfgh

h

e

h

f

k

i j

g

l

a b

c

dfghh

e

h

f

k

i j

g

l

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* Bi 9:Hy tm cy khung cho mi th sau:

a) K5 b) K4,4 c) K1,6d) Q3 e) C5 f) W5

Li gii:a) K5

b) K4,4

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c) K1,6

Bn thn K1,6 l mt cy khung.d) Q3

e) C5

K1,6

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f) W5

* Bi 10: th Kn vi n=3, 4, 5 c bao nhiu cy khung khng ng cu

Li gii:K3: c 1 cy khung

K4: c 2 cy khung( khng ng cu )

K5:

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* Bi 11:Tm cy khung nh nht ca th sau theo thut ton Kruskal v

Prim:

Li gii:

+ Gii theo thut ton Kruskal:Th t ca cc cnh sp xp theo th t khng gim:(d,e), (b,f), (c,d), (a,c), (d,g), (g,h), (h,f), (a,e), (e,f), (b,e), (c,g), (d,h), (a,b)

Bt u t th rng T c 6 nh.- Thm vo T cnh (d,e) s cnh ca T l 1


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+ Gii theo thut ton Prim:Ta c ma trn trng s ca th trn l:

a b c d e f g ha 42 4 10

b 42 14 3 c 4 3 15 d 3 1 5 20e 10 14 1 11 f 3 11 9g 15 5 7h 20 9 7

V.lp a b c d e f g h VT ETK.To - [a,42]

[a,4]

[a,10]

a

1 - [a,42]

- [c,3] [a,10]

[c,15] a, c (a,c)

2 - [a,42]

- - [d,1] [d,5] [d,20] a,c,d (a,c),(c,d)

3 - [e,14] - - - [e,11] [d,5] [d,20] a,c,d,e (a,c),(c,d)(d,e)

4 - [e,14] - - - [e,11] - [g,7] a,c,d,e,g (a,c),(c,d)(d,e),(d,g)

5 - [e,14] - - - [h,9] - - a,c,d,e,g,h (a,c),(c,d)(d,e),(d,g),(g,h)

6 - [f,3] - - - - - - a,c,d,e,g,h,f (a,c),(c,d)(d,e),(d,g),(g,h),(h,f)

7 - - - - - - - - a,c,d,e,g,h,f,b (a,c),(c,d)(d,e),(d,g),(g,h),(h,f),

(f,b)

Vy cy khung nh nht ca th trn c di l:4+3+1+5+7+9+3=32

Cy khung vi tp cnh:E={(a,c),(c,d)(d,e),(d,g),(g,h),(h,f),(f,b)}

* Bi 12:

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

A B C D E F H IA 16 15 23 19 18 32 20B 16 13 33 24 20 19 11C 15 13 13 29 21 20 19D 23 33 13 22 30 21 12E 19 24 29 22 34 23 21F 18 20 21 30 34 17 14H 32 19 20 21 23 17 18I 20 11 19 12 21 14 18

Yu cu vit kt qu trung gian trong tng bc lp, kt qu cui

cng cn a ra tp cnh v di ca cy khung nh nht.Li gii:

A B C D E F H I VT ET- [A,16] [A,15] [A,23] [A,19] [A,18] [A,32] [A,20] A - [C,13] - [C,13] [A,19] [A,18] [C,20] [C,19] A,C (A,C)- - - [C,13] [A,19] [A,18] [B,19] [B,11] A,C,B (A,C),(C,B)- - - [I,12] [A,19] [I,14] [I,18] - A,C,B,I (A,C),(C,B)

(B,I)- - - - [A,19] [I,14] [I,18] - A,C,B,I,D (A,C),(C,B)

(B,I),(I,D)- - - - [A,19] - [F,17] - A,C,B,I,D,F (A,C),(C,B)

(B,I),(I,D),(I,F)- - - - [A,19] - - - A,C,B,I,D,F,H (A,C),(C,B)

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

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

(A,E)

di cy khung nh nht tm c:15+13+11+12+14+17+19=101

Tp cnh ca cy khung nh nht:

E={(A,C),(C,B)(B,I),(I,D),(I,F),(F,H),(A,E)}

* Bi 13:Duyt cc cy sau y ln lt bng cc thut ton tin th t, trung

th t v hu th t:

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a)

b)

Li gii:a)*Duyt theo tin th t:1. Thm a.

2. Duyt T(b):2.1. Thm b.2.2. Duyt T(d): Thm d.2.3. Duyt T(e):

2.3.1. Duyt T(g): Thm g.3. Duyt c:

3.1. Thm c

g

ed

ji

h

f

b c

a

ed

cb

a

h i

m n

p q

f g

lj k

o

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3.2. Duyt T(f):3.2.1. Thm f3.2.2. Duyt T(h):

3.2.2.1. Thm h3.2.2.2. Duyt T(i): Thm i3.2.2.3. Duyt T(j): Thm j

Kt qu duyt cy theo tin th t l:a, b, d, e, g, c, f, h, i, j

* Duyt theo trung th t:1. Duyt T(b):

1.1. Duyt T(d): Thm d1.2. Thm b1.3. Duyt T(e):

1.3.1. Duyt T(g): Thm g1.3.2. Thm e

2.Thm a3. Duyt T(c):

3.1. Thm c3.2 Duyt T(f):

3.1.1. Duyt T(h):3.1.1.1. Duyt T(i): Thm i3.1.1.2. Thm h3.1.1.3. Duyt T(j): Thm j

Kt qu duyt cy theo trung th t:d, b, g, e, a, c, i, h, j, f

* Duyt theo hu th t:1. Duyt T(b):

1.1. Duyt T(d): Thm d1.2. Duyt T(e):

1.2.1. Duyt T(g): Thm g1.2.2. Thn e

1.3. Thm b2. Duyt T(c):

2.1.Duyt T(f):2.1.1. Duyt T(h):

2.1.1.1. Duyt T(i): Thm i2.1.1.2. Duyt T(j): Thm j2.1.1.3.Thm h

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2.1.2.Thm f2.2. Thm c

3. Thm aKt qu duyt cy theo hu th t:

d,g, e, b, i, j, h, f, c, a

b)* Duyt theo tin th t:1. Thm a2. Duyt T(b):

2.1. Thm b2.2. Duyt T(d): Thm d2.3. Duyt T(e):

2.3.1. Thm e2.3.2. Duyt T(g): Thm g

3. Duyt T(c):3.1. Thm c3.2. Duyt T(f):

3.2.1. Thm f3.2.2. Duyt T(h):

3.2.2.1. Thm h3.2.2.2. Duyt T(i): Thm i3.2.2.3. Duyt T(j): Thm j

Kt qu duyt cy theo tin th t:a, b, d, e, g, c, f, h, i, j

* Duyt cy theo trung th t:1. Duyt T(b):

1.1. Duyt T(d):1.1.1. Duyt T(h): Thm h1.1.2. Thm d1.1.3. Duyt T(i):

1.1.3.1. Duyt T(m): Thm m

1.1.3.2. Thm i1.1.3.3. Duyt T(n):

1.1.3.3.1. Duyt T(p): Thm p1.1.3.3.2. Thm n1.1.3.3.3. Duyt T(q): Thm q

1.2. Thm b1.3. Duyt T(e): Thm e

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2.Thm a3. Duyt T(c):

3.1. Duyt T(f):3.1.1. Duyt T(j): Thm j3.1.2. Thn f3.1.3. Duyt T(k):

3.1.3.1. Duyt T(o): Thm o3.1.3.2. Thm k

3.2. Thm c3.3. Duyt T(g):

3.3.1. Thm g3.3.2. Duyt T(l): Thm l

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

* Duyt cy theo hu th t:1. Duyt T(b):

1.1. Duyt T(d):1.1.1. Duyt T(h): Thm h1.1.2. Duyt T(i):

1.1.2.1. Duyt T(m): Thm m1.1.2.2. Duyt T(n):

1.1.2.2.1. Duyt T(p): Thm p1.1.2.2.2. Duyt T(q): Thm q1.1.2.2.3.Thm n

1.1.2.3. Thm i1.1.3. Thm d

1.2. Duyt T(e): Thm e1.3. Thm b

2. Duyt T(c):2.1. Duyt T(f):

2.1.1. Duyt T(j): Thm j2.1.2. Duyt T(k):

2.1.2.1. Duyt T(o): Thm o2.1.2.2. Thm k

2.1.3. Thm f2.2. Duyt T(g):

2.2.1. Duyt T(l): Thm l2.2.2. Thm g

2.3. Thm c

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3. Thm a.Kt qu duyt cy theo hu th t:

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

* Bi 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)(

342

4 dbadad

cba

+

+

Li gii:

a) BDCBDA

DCBA

DCBA

++

+

++

2

2

)(

))((

Ta c th v cy nh phn tng ng vi biu thc trn nh sau:

Kt qu duyt cy nh phn trn theo tin th t s cho ta cch vitca biu thc trn theo k php Ba Lan, nu duyt theo hu th t th s cho

ta cch vit theo k php Ba Lan o.Vy ta c biu thc trn c biu din theo:

- K php Ba Lan:+ / * + A B + C D + * - A B C D / + A 2 * B D - C 2 * B D

- K php Ba Lan o:A B + C D + * A B - C * D + / A 2 B D * + C 2 B D * - / +

/

+

/

* +

+ +

A B C D

* D

- C

BA

-+

A 2

*

B D

C 2

*

B D

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b)5

)243(

35

3)(

342

4 dbadad

cba

+

+

Cy nh phn tng ng:

Kt qu duyt cy nh phn trn theo tin th t s cho ta cch vitca biu thc trn theo k php Ba Lan, nu duyt theo hu th t th s chota cch vit theo k php Ba Lan o.

Vy ta c biu thc trn c biu din theo:- K php Ba Lan:

+ - - a b 4 + / c 3 * 5 d 2 * / - a d 3 4 / - + * 3 a * 4 b * 2 d 3 5- K php Ba Lan o:

a b - 4 c 3 / 5 d * + - 2 a d - 3 / 4 3 a * 4 b * + 2 d * - 3 5 / * +

* Bi 15:Vit cc biu thc sau y theo k php quen thuc:a) x y + 2 x y 2 - x y * /

b) - * / - - a b * 3 c 2 4 - c d 5 * - - a c d / - b * 2 d 4 3Li gii:

+

-

a b

4

+

-

2

/

c 3

*

5 d

*

/ 4

- 3

a d

/

5

- 3

+ *

* *

2 d

3 a 4 b

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

(x+y) 2 (x-y) 2 - xy /(x+y)2 (x-y)2 - xy /

[ (x+y)2 (x-y)2 ] xy /

[(x+y)2 (x-y)2 ]__________________________

xy

b) - * / - - a b * 3 c 2 4 - c d 5 * - - a c d / - b * 2 d 4 3

- * / - (a-b) 3c 2 4 (c-d) 5 * - (a-c) d / - b 2d 4 3

- * / (a-b-3c) 2 4 (c-d)5 * (a-c-d) / (b-2d)4 3

a-b-3c (b-2d)4

- * ________ 4 (c-d)5 * (a-c-d) _______

2 3

a-b-3c 4 (b-2d)4

- * ________ (c-d)5 (a-c-d) _______

2 3

a-b-3c 4 (b-2d)4

- ________ (c-d)5 (a-c-d) _______

2 3

4 4

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