Bai Tap Toan Roi Rac

2011

BI TP

Ton ri rc

Ti liu hc tp theo hc ch tn ch

dnh cho sinh vin Khoa CNTT

Phin bn 1.0

1

MC LC

CHNG 1: C S LOGIC ...................................................................................... 2 CHNG 2: PHNG PHP M ........................................................................ 5 CHNG 3: QUAN H .............................................................................................. 6 CHNG 4: I S BOOLE .................................................................................... 7 NI DUNG B SUNG: PHNG PHP HM SINH ........................................... 8 THI TRC NGHIM GIA K (THAM KHO) ......................................... 24 THI KT THC MN HC (THAM KHO) ................................................ 25 MT S THI THAM KHO TRNG KHC ........................................ 26

Bi tp ton ri rc

2

CHNG 1: C S LOGIC

1. Lp bng chn tr v ch ra cc mnh lun ng:

a. ( )x y z x

b. [( ) ( )] ( )a b c b a c

c. [ ( )] ( )m n p m p

d. [ ( )] [( ) ( )]p r p p q r p

2. p dng cc lut logic rt gn cc mnh v ch ra u l cng thc hng:

a. ( ) ( ) ( )a c b c a b

b. [( ) ( )] ( )x y y z x z

c. ( ) ( )m n p m n p

d. [[( ) ] [( ) ]]m n p m p p n

3. Mt x th bn cung vo mt mc tiu, kt qu c th hin theo cc mnh sau:

Pk = { Pht th k trng ch} k = 1, 2, 3

Hy gii thch cc mnh phc hp sau:

a. 1 2 3P P P

b. 1 2 3P P P

c. 11 2 3 1 3 1 2 3( ) ( ) ( )P P P P P P P P P

d. 1 2 1 2 3 1 2 3( ) ( ) ( )P P P P P P P P

4. Cho 4 th sinh A, B, C, D tham gia thi u xp hng. Kt qu xp hng nh th no

nu:

- Ngi th 1 d on: B hng nh, C hng ba.

- Ngi th 2 d on: A hng nh, C hng t.

- Ngi th 3 d on: B hng nht, D hng nh.

- c bit l mi ngi c phn ng phn sai.

5. Trong mt chatroom, c tng cng 5 ngi An, Bnh, Chinh, Dung, Yn ang tho

lun v ti logic ton vi nhau trn mng.

Bit rng:

- Hoc An, hoc Bnh hoc l c 2 ang tho lun.

- Hoc Chinh, hoc Dung, nhng khng phi c 2 cng tho lun.

- Nu Yn ang tho lun th Chinh cng vy.

Bi tp ton ri rc

3

- Dung v An, hoc c 2 cng tho lun, hoc khng ai tho lun.

- Nu Bnh ang tho lun th Yn v An cng vy.

Hy gii thch xem nu tt c khng nh trn u ng th hin ti ai ang tho lun?

6. Sau khi thu gn, hy tm cc b nghim (x, y, z, t ) cc cng thc hm sau t gi

l 0:

a. [( ) ( )] [ ( )]F yt xz xt y z x yt y z xt

b. [( ) ( )] [( ) ( )]zF x y zt yz xt xz yt y xt

c. )[( ) ( )] [( ]z y xz yF x y yt xt z yt x z

7. Chuyn tt c cc cng thc sau thnh dng chun tuyn hon ton v chun hi hon

ton:

a. ( ) ( )m n n p

b. [ ( )] ( )m n p n p

c. ( ( ))n m m n p

d. [( ) ( )] ( )m n n m m n

e. ( ) ( )m n p m n p

f. [ ( )] [( ) ( )]m n p m p n p

8. Tm chn tr ca cc v t ng vi cc gi tr tng ng sau:

m(x): x > 2

n(x): x-1 l

p(x): x < 0

a. m(2) b. n(210)

c. ( 1) (3)m n

d. (0) (4)m n

e. [ (7) (1)] (1)m n p

f. [ ( 1) ( 1)] ( 1)m n p

g. [ (3) (3)] (3)m n p

h. (1) [ (1) (1)]m n p

9. Xc nh chn tr ca v t sau:

p(x,y) : x l c ca y

a. p(1,5) b. p(3,4)

c. , ( , )x p x x

d. , ( , )y p y y

Bi tp ton ri rc

4

e. , ( , )x y p x y

f. , ( , )y x p x y

g. ,( ( , ) ( , )) ( )x y p x y p y x x y

h. ,( ( , ) ( , )) ( , )x y z p x y p y z p x z

10. Xc nh chn tr ca cc mnh sau :

a. 2,[( 6 5 0) ( 5 0)]x R x x x

b. , ,[( 2 1) ( 3)]x R y R x y x y

c. 2 2 2, , ,[( ) ( )]x R y R z R x y z x z

d. 2, , ,[( 2 ) (2 0)]x R y R z R x y z x y z

e. 2 2 2, , ,[( ) ( )]x R y R z R x y z x z

f. , , ,[( 2 ) (2 0)]x y z x y z x y z

g. 2 2 2, , ,[( ) ( )]x R y R z R x y z x z

11. Xc nh cc suy lun ng v cho bit qui tc suy din c p dng:

a. Nu Bnh i chi th Bnh khng hc logic ton. Nu Bnh khng hc bi th s thi trt mn logic ton. M Bnh li i chi nn Bnh thi trt mn logic ton.

b. Nu l sinh vin CNTT ca trng i hc S Phm th phi hc ton ri rc. Hng khng hc Ton ri rc nn Hng khng phi l sinh vin ngnh cng CNTT ca trng i hc S Phm.

c. Mi sinh vin nghim tc u khng np bi cha lm xong. Minh khng np bi cha lm xong. Vy Minh l sinh vin nghim tc.

d. Mi sinh vin li hc u khng chu n lp hc thng xuyn.

Tn n lp hc thng xuyn.

V th Vn l sinh vin khng li hc.

12. Dng cc qui tc suy din chng minh cc kt lun sau:

a. {p q , (s q) , (r s) , p u }

r, u

b. {(x y) z , z(w u) , (t w) , w t }

x

13. Trong 1 trn thi u i khng v thut Vovinam, th sinh s c tnh l 1 im nu

nh c t nht 2 trong s 3 trng ti pht c. Ngi ta thit k 1 my chm im cho

cc trn u vi th thc thi u nh vy.

a. Tm cng thc logic tng ng vi my chm im ny. b. Hy v mch in t tng ng vi cng thc ny (khng rt gn cng thc).

Bi tp ton ri rc

5

CHNG 2: PHNG PHP M

1. Cho tp X={5, 6, ..., 200} a. C bao nhiu s chn, l. b. C bao nhiu s chia ht cho 5. c. C bao nhiu s gm nhng ch s phn bit. d. C bao nhiu s khng cha s 0. e. C bao nhiu s ln hn 101 v khng cha s 6. f. C bao nhiu s c cc ch s c sp tng thc s.

g. C bao nhiu s c dng xyz trong 0 ,x y y z

2. C 10 cun sch khc nhau, trong c 5 cun sch thuc lnh vc tin hc, 3 cun sch thuc lnh vc ton hc, 2 cun sch thuc lnh vc vn hc.

a. C bao nhiu cch ly ra 4 cun sch bt k? b. C bao nhiu cch ly ra 4 cun sch trong c t nht 2 cun sch tin hc? c. C bao nhiu cch ly ra 4 cun sch vi c 3 loi sch? d. C bao nhiu cch xp cc cun sch ny trn 1 gi sch? e. C bao nhiu cch xp cc cun sch ny trn 1 gi sch sao cho ng theo th

t lit k? f. C bao nhiu cch xp cc cun sch ny trn 1 gi sch sao cho 3 cun sch

ton hc khng c xp cnh nhau? 3. Mt ngi lm vic trn my tnh qun mt mt khu ng nhp vo ti

khon s dng, anh ta ch cn nh mt khu c dng XXXX-YYY trong XXXX l cc ch s khc nhau ly t 10 s, v YYY l cc ch ci khc nhau ly t 26 ch ci. Hi trong trng hp xu nht phi th bao nhiu ln c c mt khu ban u?

4. C 6 ngi cng mt lc ng k th ti tin hc l A, B, C, D, E, F: a. C bao nhiu cch xp th t thi u DEF ng cnh nhau? b. C bao nhiu cch xp th t thi u lun bt u bi A v kt thc bi F 5. C bao nhiu chui 8 bit bt u bng 1100? 6. C bao nhiu chui 8 bit trong bit th 2 v bit th 4 l 1? 7. C bao nhiu chui 8 bit c xui hay ngc u ging nhau? 8. C bao nhiu xu nh phn c di 8 bt u bi 110 v 101? 9. C bao nhiu chui 8 bit hoc bt u bng 100 hoc c bit th 4 bng 1? 10. C bao nhiu chui 8 bit hoc bt u bng 10 hoc kt thc bi 01? 11. C bao nhiu xu nh phn di 16 m trong c ng 4 s 1? 12. Cc k th ABCDEF dng to thnh cc chui c di l 3?

a. C bao nhiu chui nh vy nu cho php lp. b. C bao nhiu chui nh vy nu khng cho php lp. c. C bao nhiu chui bt u bng A nu cho php lp. d. C bao nhiu chui bt u bng A nu khng cho php lp. e. C bao nhiu chui khng cha A nu cho php lp. f. C bao nhiu chui khng cha A nu khng cho php lp.

13. C bao nhiu xu k t c th c to t cc ch ci Telecommunication? 14. Phng trnh 1 2 3 4 23x x x x

C bao nhiu b nghim t nhin ( 1 2 3 4, , ,x x x x ) sao cho 12 5x

15. Phng trnh 1 2 10... 100x x x

C bao nhiu b nghim t nhin ( 1 2 10, ,...,x x x ) sao cho 1 2 101, 2,..., 10x x x

Bi tp ton ri rc

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CHNG 3: QUAN H

1. Cho 2 tp hp X={a, b, c} v Y = {b, c, d, e}

a. Tnh X x Y

b. Tm s quan h 2 ngi trn Y

c. Tm s quan h gia X v Y cha (b, c), (b, d)

d. Hy tm 1 quan h trn X c tnh phn x v bc cu nhng khng i xng

e. Hy tm 1 quan h trn Y c tnh phn x v i xng nhng khng bc cu.

2. R l mt quan h trn A = {1, 2, 3, 4, 5} vi:

R = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 4), (2, 5), (3, 1), (3, 3) , (4, 1), (4, 2), (4,

4), (5, 2), (5, 5)}

R c phi l 1 quan h tng ng hay khng?

3. Cho R l 1 quan h trn tp hp cc s t nhin vi R = {(x, y): x+y chn}

Chng minh R l 1 quan h tng ng

4. Cho R l 1 quan h trn A x A vi A= {1, 3, 5, 7, 8, 9} sao cho :

(a, b) R (c, d) b = d

a. Chng minh R l 1 quan h tng ng

b. Tm lp tng ng cha (1, 3)

c. Phn hoch A x A thnh cc lp tng ng tch bit phn hoch trn R

5. R l mt quan h trn A x A vi A = {1, 2, 3, 4, 5}:

R = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (2, 4), (2, 5), (3, 3), (3, 5) , (4, 4), (4, 5), (5, 5)}

R c phi l 1 quan h th t hay khng?

6. Cho l 1 quan h trn X={ 2, 3, 4, 5, 12, 15, 60} xc nh bi:

, , ,x y X x y y kx k Z

a. Chng minh l 1 quan h th t

b. Tm cc phn t ti i, ti tiu, ln nht, nh nht xc nh bi quan h trn

c. V biu Hasse tng ng

7. Cho l 1 quan h trn A x A vi A ={2, 4, 6, 12, 24} xc nh bi:

( , ),( , ) ,( , ) ( , )x y z t A x y z t x y z t

c phi l 1 quan h th t hay khng? Nu l 1 quan h th t th xc nh cc phn t ti i, ti tiu, ln nht, nh nht xc nh bi quan h trn

Bi tp ton ri rc

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CHNG 4: I S BOOLE

1. Cho (A, , ) l 1 i s Boole hy chng minh:

a. ,x A x x x

b. ,x A x x x

c. ,x A x x

d. , 1x A x x

e. , ,x y A x y x y

f. , ,x y A x y x y

g. , , ( )x y A x x y x

h. , , ( )x y A x x y x

2. Cho U30 = (1, 2, 3, 5, 6, 10, 15, 30)

Trn U30 ta nh ngha 2 php ton v nh sau:

30, , ( , )x y U x y UCLN x y

30, , ( , )x y U x y BCLN x y

Chng minh (U30, , ) l 1 i s Boole

3. n gin cc hm Bool sau v v mch ca hm n gin:

a. ( , , )f x y z x y z x y z x y z x y z x y z

b. ( , , , )f x y z t x y zt x y z t x y z t x y z t x y zt x y zt x y zt x y z t x y z t

c. ( , , , )f x y z t x y z t x y z t x y z t x y z t x y zt x y zt x y zt x y z t x y z t

d. ( , , , )f x y z t x y z x y z x y z t x y z t x y zt y zt x y zt x y z t x y t

e. ( , , , )f x y z t x y z t x y z t x y z t x y z x y zt y zt x y z y z t

Bi tp ton ri rc

8

PHNG PHP HM SINH

Hm sinh l mt trong nhng sng to thn tnh, bt ng, nhiu ng dng ca ton ri rc. Ni mt cch nm na, hm sinh chuyn nhng bi ton v dy s thnh nhng bi ton v hm s. iu ny l rt tuyt vi v chng ta c trong tay c mt c my ln lm vic vi cc hm s. Nh vo hm sinh, chng ta c th p dng c my ny vo cc bi ton dy s. Bng cch ny, chng ta c th s dng hm sinh trong vic gii tt c cc dng ton v php m. C c mt ngnh ton hc ln nghin cu v hm sinh, v th, trong bi ny, chng ta ch tm hiu nhng vn cn bn nht v ch ny.

Trong bi vit ny, cc dy s s c trong ngoc < > phn bit vi cc i tng ton hc khc.

1. Hm sinh

Hm sinh thng ca dy s v hng l chui lu tha hnh thc

G(x) = g0 + g1x + g2x2 + g3x

3

Ta gi lm sinh l chui hnh thc bi v thng thng ta s ch coi x l mt k hiu thay th thay v mt s. Ch trong mt vi trng hp ta s cho x nhn cc gi tr thc, v th ta gn nh cng khng n s hi t ca cc chui. C mt s loi hm sinh khc nhng trong bi ny, ta s ch xt n hm sinh thng.

Trong bi ny, ta s k hiu s tng ng gia mt dy s v hm sinh bng du mi tn hai chiu nh sau

g0 + g1x + g2x2 + g3x

3 +

V d, di y l mt s dy s v hm sinh ca chng

0 + 0.x + 0.x2 + 0.x3 + = 0

1 + 0.x + 0.x2 + 0.x3 + = 1

3 + 2x + x2 + 0.x3 + = x2 + 2x + 3

Quy tc y rt n gin: S hng th i ca dy s (nh s t 0) l h s ca xi trong hm sinh.

Nhc li cng thc tnh tng ca cc s nhn li v hn l

.1

1...1 32

zzzz

ng thc ny khng ng vi |z| 1, nhng mt ln na ta khng quan tm n vn hi t. Cng thc ny cho chng ta cng thc tng minh cho hm sinh ca hng lot cc dy s

1 + x + x2 + x3 + = 1/(1-x)

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9

1 - x + x2 - x3 + = 1/(1+x)

1 + ax + a2x2 + a3x3 + = 1/(1-ax)

1 + x2 + x4 + = 1/(1-x2)

Cc php ton trn hm sinh

Php mu ca hm sinh nm ch ta c th chuyn cc php ton thc hin trn dy s thnh cc php ton thc hin trn cc hm sinh tng ng ca chng. Chng ta cng xem xt cc php ton v cc tc ng ca chng trong thut ng dy s.

2.1. Nhn vi hng s

Khi nhn hm sinh vi mt hng s th trong dy s tng ng, cc s hng s c nhn vi hng s . V d

1 + x2 + x4 + = 1/(1-x2)

Nhn hm sinh vi 2, ta c

2/(1-x2) = 2 + 2x

2 + 2x

4 +

l hm sinh ca dy s

Quy tc 1. (Quy tc nhn vi hng s)

Nu F(x) th cF(x)

Chng minh.

cf0 + (cf1)x + (cf2)x2 + (cf3)x

3 +

= c(f0 + f1x+f2x2 + f3x

3 + )

= cF(x).

2.2. Cng

Cng hai hm sinh tng ng vi vic cng cc s hng ca dy s theo ng ch s. V d, ta cng hai dy s trc

1/(1-x)

+ 1/(1+x)

1/(1-x) + 1/(1+x)

By gi ta thu c hai biu thc khc nhau cng sinh ra dy (2, 0, 2, 0, ). Nhng iu ny khng c g ngc nhin v thc ra chng bng nhau:

1/(1-x) + 1/(1+x) = [(1+x) + (1-x)]/(1-x)(1+x) = 2/(1-x2)

Quy tc 2. (Quy tc cng)

Nu F(x), G(x)

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10

th F(x) + G(x)

Chng minh.

f0+g0+ (f1+g1)x + (f2+g2)x2 +

= (f0 + f1x + f2x2 + ) + (g0 + g1x + g2x

2 + )

= F(x) + G(x)

2.3. Dch chuyn sang phi

Ta bt u t mt dy s n gin v hm sinh ca n

1/(1-x)

By gi ta dch chuyn dy s sang phi bng cch thm k s 0 vo u

xk + xk+1 + xk+2 +

= xk(1+x+x2 + )

= xk/(1-x)

Nh vy, thm k s 0 vo u dy s tng ng vi vic nhn hm sinh vi xk. iu ny cng ng trong trng hp tng qut.

Quy tc 3. (Quy tc dch chuyn phi)

Nu F(x)

th xk.F(x) (c k s 0)

Chng minh.

f0xk + f1x

k+1 + f2x

k+2 +

= xk(f0 + f1x + f2x

2 + )

= xkF(x)

2.4. o hm

iu g s xy ra nu ta ly o hm ca hm sinh? Chng ta hy bt u t vic ly o hm ca mt hm sinh tr nn quen thuc ca dy s ton 1:

2

2

32

432

)1(

1,...4,3,2,1

)1(

1...4321

)1

1(...)1(

x

xxxx

xdx

dxxxx

dx

d

Ta tm c hm sinh cho dy s !

Tng qut, vic ly o hm ca hm sinh c hai tc ng ln dy s tng ng: cc s hng c nhn vi ch s v ton b dy s c dch chuyn tri sang 1 v tr.

Quy tc 4. (Quy tc o hm)

Nu F(x)

th dF(x)/dx

Bi tp ton ri rc

11

Chng minh.

f1 + 2f2x + 3f3x2 +

= (d/dx)(f0 + f1x + f2x2 + f3x

3 + )

= dF(x)/dx

Quy tc o hm l mt quy tc rt hu hiu. Trong thc t, ta thng xuyn cn n mt trong hai tc ng ca php o hm, nhn s hng vi ch s v dch chuyn sang tri. Mt cch in hnh, ta ch mun c mt tc ng v tm cch v hiu ho tc ng cn li. V d, ta th tm hm sinh cho dy s . Nu ta c th bt u t dy th bng cch nhn vi ch s 2 ln, ta s c kt qu mong mun

=

Vn l ch php o hm khng ch nhn s hng dy s vi ch s m cn dch chuyn sang tri 1 v tr. Th nhng, quy tc 3 dch chuyn phi cho chng ta cch v hiu ho tc ng ny: nhn hm sinh thu c cho x.

Nh vy cch lm ca chng ta l bt u t dy s , ly o hm, nhn vi x, ly o hm ri li nhn vi x.

1/(1-x)

(d/dx)(1/(1-x)) = 1/(1-x)2

x/(1-x)2

(d/dx)( x/(1-x)2) = (1+x)/(1-x)3

x(1+x)/(1-x)3

Nh vy hm sinh cho dy cc bnh phng l x(1+x)/(1-x)3.

3. Dy s Fibonacci

i khi chng ta c th tm c hm sinh cho cc dy s phc tp hn. Chng hn di y l hm sinh cho dy s Fibonacci:

x/(1-x-x2)

Chng ta thy dy s Fibonacci bin i kh kh chu, nhng hm sinh ca n th rt n gin.

Chng ta s thit lp cng thc tnh hm sinh ny v qua , tm c cng thc tng minh tnh cc s hng tng qut ca dy s Fibonacci. D nhin, chng ta bit cng thc tnh s hng tng qut ca dy Fibonacci trong phn phng php gii cc phng trnh sai phn tuyn tnh h s hng. Nhng iu ny khng ngn cn chng ta mt ln na tm cch gii thch s xut hin ca phng trnh c trng, cch x l trng hp nghim kp thng qua cng c hm sinh. Hn na, phng php hm sinh cn gip chng ta gii quyt hng lot cc bi ton v dy s quy khc na, trong c nhng phng trnh m chng ta hon ton b tay vi cc phng php khc.

3.1. Tm hm sinh

Ta bt u bng cch nhc li nh ngha ca dy Fibonacci:

f0 = 0

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12

f1 = 1

fn = fn-1 + fn-2 vi n 2

Ta c th khai trin ng thc cui cng thnh dy v hn cc ng thc. Nh th dy s Fibonacci xc nh bi

f0 = 0

f1 = 1

f2 = f1 + f0

f3 = f2 + f1

f4 = f3 + f2

By gi, cch lm tng qut ca chng ta l: nh ngha F(x) l hm sinh ca dy s bn tri ca cc ng thc, chnh l cc s Fibonacci. Sau chng ta tm c hm sinh cho dy s v phi. Cho hai v bng nhau v ta gii ra c hm sinh F(x). Ta hy cng lm iu ny. u tin ta nh ngha

F(x) = f0 + f1x + f2x2 + f3x

3 + f4x

4 +

By gi, ta cn tm hm sinh cho dy s

Mt trong nhng cch tip cn l tch cc dy s ca chng ta thnh cch dy m chng ta bit hm sinh, sau p dng Quy tc cng

x

xF(x)

+ x2F(x)

x + xF(x) + x2F(x)

Dy s ny gn nh l dy s nm v phi ca dy Fibonacci, ch c 1 khc bit duy nht l 1+f0 v tr th hai. Nhng do f0 = 0 nn iu ny khng c ngha g.

Nh vy, ta c F(x) = x + xF(x) + x2F(x) v t F(x) = x/(1-x-x2), chnh l cng thc m chng ta ni n phn u.

3.2. Tm cng thc tng minh

Ti sao chng ta li phi tm hm sinh ca mt dy s? C mt vi cu tr li cho cu hi ny, nhng di y l mt trong nhng cu tr li : nu ta tm c hm sinh cho mt dy s, trong nhiu trng hp, ta c th tm c cng thc tng minh cho cc s hng ca dy s , v y l iu rt cn thit. V d cng thc tng minh cho h s ca xn trong khai trin ca x/(1-x-x2) chnh l cng thc tng minh cho s hng th n ca dy s Fibonacci.

Nh vy cng vic tip theo ca chng ta l tm cc h s t hm sinh. C mt vi cch tip cn cho bi ton ny. i vi cc hm phn thc, l t s ca cc a thc, chng ta c th s dng phng php phn tch thnh cc phn thc s cp m chng ta bit phn tch

Bi tp ton ri rc

13

phn cc hm hu t. Ta c th tm c d dng cc h s cho cc phn thc s cp, t tm c cc h s cn tm.

Ta s th lm cho hm sinh ca dy s Fibonacci. u tin, ta phn tch mu s ra tha s

x/(1-x-x2) = x/(1-1x)(1-2x)

trong 2

51,

2

5121

. Tip theo, ta tm cc hng s A1, A2 sao cho

x

A

x

A

xx

x

2

2

1

1

2 111

Ta c th lm iu ny bng phng php h s bt nh hoc thay x cc gi tr khc nhau thu c cc phng trnh tuyn tnh i vi A1, A2. Ta c th tm c A1, A2 t cc phng trnh ny. Thc hin iu ny, ta c

.5

11,

5

11

21

1

21

1

AA

Thay vo ng thc ni trn, ta c khai trin ca F(x) thnh cc phn thc s cp

xxxx

x

21

2 1

1

1

1

5

1

1

Mi mt s hng trong khai trin c chui lu tha n gin cho bi cng thc

...11

1

...11

1

22

22

2

22

11

1

xxx

xxx

Thay cc cng thc ny vo, ta c chui lu tha cho hm sinh

...))1(...)1((5

1

1

1

1

1

5

1)(

22

22

22

11

21

xxxx

xxxF

T suy ra

.2

51

2

51

5

1

5

21

nnnn

nf

y chnh l cng thc m ta cng tm c trong phn gii cc phng trnh sai phn tuyn tnh h s hng. Cch tip cn mi ny lm sng t thm mt s vn ca phng php cp ti. Ni ring, quy tc tm nghim ca phng trnh sai phn trong trng hp phng trnh c trng c nghim kp l h qu ca quy tc tm khai trin phn thc s cp!

4. m bng hm sinh

Hm sinh c th c p dng trong cc bi ton m. Ni ring, cc bi ton chn cc phn t t mt tp hp thng thng s dn n cc hm sinh. Khi hm sinh c p dng theo cch ny, h s ca xn chnh l s cch chn n phn t.

Bi tp ton ri rc

14

4.1. Chn cc phn t khc nhau

Hm sinh cho dy cc h s nh thc c suy ra trc tip t nh l nh thc

kkk

kkk

k

kkkk xxCxCCCCCC )1(...,...0,0,,...,,,10210

Nh vy h s ca xn trong (1+x)k bng s cch chn n phn t phn bit t mt tp hp gm k phn t. V d, h s ca x2 l C

2k, s cch chn 2 phn t t tp hp k phn t.

Tng t, h s ca xk+1 l s cch chn k+1 phn t t tp hp k phn t v nh th, bng 0.

4.2. Xy dng cc hm sinh m

Thng thng ta c th dch m t ca bi ton m thng sang ngn ng hm sinh gii. V d, ta c th chng t rng (1+x)k s sinh ra s cc cch chn n phn t phn bit t tp hp k phn t m khng cn dng n nh l nh thc hay cc h s nh thc!

Ta lm nh sau. u tin, ta hy xt tp hp c mt phn t {a1}. Hm sinh cho s cch chn n phn t t tp hp ny n gin l 1 + x. Ta c 1 cch chn khng phn t no, 1 cch chn 1 phn t v 0 cch chn hai phn t tr ln. Tng t, s cch chn n phn t t tp hp {a2} cng cho bi hm sinh 1 + x. S khc bit ca cc phn t trong hai trng hp trn l khng quan trng.

V by gi l tng chnh: hm sinh cho s cch chn cc phn t t hp ca hai tp hp bng tch cc hm sinh cho s cch chn cc phn t t mi tp hp. Chng ta s gii thch cht ch iu ny, nhng trc ht, hy xem xt mt v d. Theo nguyn l ny, hm sinh cho s cch chn cc phn t t tp hp {a1, a2} l

(1+x). (1+x) = (1+x)2 = 1 + 2x + x

2

C th kim chng rng i vi tp hp {a1, a2} ta c 1 cch chn 0 phn t, 2 cch chn 1 phn t, 1 cch chn 2 phn t v 0 cch chn 3 phn t tr ln.

Tip tc p dng quy tc ny, ta s c hm sinh cho s cch chn n phn t t tp hp k phn t

(1+x).(1+x)(1+x) = (1+x)k

y chnh l cng thc hm sinh m ta nhn c bng cch s dng nh l nh thc. Nhng ln ny, chng ta i thng t bi ton m n hm sinh.

Chng ta c th m rng iu ny thnh mt nguyn l tng qut.

Quy tc 5 (Quy tc xon). Gi A(x) l hm sinh cho cch chn cc phn t t tp hp A v B(x) l hm sinh cho cch chn cc phn t t tp hp B. Nu A v B l ri nhau th hm

sinh cho cch chn cc phn t t A B l A(x).B(x).

Quy tc ny l kh a ngha, v cn hiu chnh xc cch chn cc phn t t mt tp hp c ngha l th no? Rt ng ch l Quy tc xon vn ng cho nhiu cch hiu khc nhau ca t cch chn. V d, ta c th i hi chn cc phn t phn bit, cng c th cho php c chn mt s ln c gii hn no , hoc cho chn lp li tu . Mt cch nm na, gii hn duy nht l (1) th t chn cc phn t khng quan trng (2) nhng gii hn p dng

cho vic chn cc phn t ca A v B cng p dng cho vic chn cc phn t ca A B

(Cht ch hn, cn c mt song nh gia cc cch chn n phn t t A B vi b sp th t cc cch chn t A v B cha tng th n phn t)

Bi tp ton ri rc

15

Chng minh. nh ngha

000

.)().()(,)(,)(n

n

n

n

n

n

n

n

n xcxBxAxCxbxBxaxA

u tin ta hy tnh tch A(x).B(x) v biu din h s cn thng qua cc h s a v b. Ta c th

sp xp cc s hng ny thnh dng bng

b0 b1x b2x2 b3x

3

a0 a0b0 a0b1x a0b2x2 a0b3x

3

a1x a1b0x a1b1x2 a1b2x

3

a2x2 a2b0x

2 a2b1x

3

a3x3 a3b0x

3

Ch rng cc s hng c cng lu tha ca x xp trn cc ng cho /. Nhm tt c cc s hng ny li, ta thy rng h s ca xn trong tch bng

cn = a0bn + a1bn-1 + + anb0

By gi ta chng minh rng y cng chnh l s cch chn n phn t t A B. Mt cch

tng qut, ta c th chn n phn t t A B bng cch chn j phn t t A v n-j phn t t B, trong j l mt s t 0 n n. iu ny c th c thc hin bng ajbn-j cch. Ly tng t 0 n n, ta c

a0bn + a1bn-1 + + anb0

cch chn n phn t t A B. chnh xc l gi tr cn c tnh trn.

Biu thc cn = a0bn + a1bn-1 + + anb0 c bit n trong mn x l tn hiu s; dy l xon (convolution) ca hai dy v .

4.3. Chn cc phn t c lp

Xt bi ton: C bao nhiu cch chn 12 cy ko t 5 loi ko? Bi ton ny c th tng qut ho nh sau: C bao nhiu cch chn ra k phn t t tp hp c n phn t, trong ta cho php mt phn t c th c chn nhiu ln? Trong thut ng ny, bi ton chn ko c th pht biu c bao nhiu cch chn 12 cy ko t tp hp

{ko sa, ko s-c-la, ko chanh, ko du, ko c-ph}

nu ta cho php ly nhiu vin ko cng loi. Ta s tip cn li gii bi ton ny t gc nhn ca hm sinh.

Gi s ta chn n phn t (c lp) t tp hp ch c duy nht mt phn t. Khi c 1 cch chn 0 phn t, 1 cch chn 1 phn t, 1 cch chn 2 phn t Nh th, hm sinh ca cch chn c lp t tp hp c 1 phn t bng

1 + x + x2 + x3 + = 1/(1-x)

Bi tp ton ri rc

16

Quy tc xon ni rng hm sinh ca cch chn cc phn t t hp ca cc tp hp ri nhau bng tch ca cc hm sinh ca cch chn cc phn t t mi tp hp:

nxxxx )1(

1

1

1...

1

1.

1

1

Nh th, hm sinh ca cch chn cc phn t t tp hp n phn t c lp l 1/(1-x)n.

By gi ta cn tnh cc h s ca hm sinh ny. lm iu ny, ta s dng cng thc Taylor:

nh l 1 (nh l Taylor)

...!

)0(...

!3

)0('''

!2

)0(''

!1

)0(')0()(

)(32 k

k

xk

fx

fx

fx

ffxf

nh l ny ni rng h s ca xk trong 1/(1-x)n bng o hm bc k ca n ti im 0 chia cho k!. Tnh o hm bc k ca hm s ny khng kh. t

g(x) = 1/(1-x)n = (1-x)

-n

Ta c

g(x) = n(1-x)-n-1

g(x) = n(n+1)(1-x)-n-2

g(x) = n(n+1)(n+2)(1-x)-n-3

g(k)(x) = n(n+1)(n+k-1)(1-x)-n-k

T , h s ca xk trong hm sinh bng

k kn

k

Cnk

kn

k

knnn

k

g1

)(

)!1(!

)!1(

!

)1)...(1(

!

)0(

Nh vy s cch chn k phn t c lp t n phn t bng .1k

knC

5. Mt bi ton m bt kh thi

T u bi n gi ta thy nhng ng dng ca hm sinh. Tuy nhin, nhng iu ny ta cng c th lm c bng nhng cch khc. By gi ta xt mt bi ton m rt kh chu. C bao nhiu nhiu cch sp mt gi n tri cy tho mn cc iu kin rng buc sau:

S to phi chn

S chui phi chia ht cho 5

Ch c th c nhiu nht 4 qu cam

Ch c th c nhiu nht 1 qu o

V d, ta c 7 cch sp gi tri cy c 6 tri:

To 6 4 4 2 2 0 0

Chui 0 0 0 0 0 5 5

Cam 0 2 1 4 3 1 0

o 0 0 1 0 1 0 1

Cc iu kin rng buc ny qu phc tp v c cm gic nh vic i tm li gii l v vng. Nhng ta hy xem hm sinh s x l bi ton ny th no.

Bi tp ton ri rc

17

Trc ht, ta i tm hm sinh cho s cch chn to. C 1 cch chn 0 qu to, c 0 cch chn 1 qu to (v s to phi chn), c 1 cch chn 2 qu to, c 0 cch chn 3 qu to Nh th ta c

A(x) = 1 + x2 + x

4 + = 1/(1-x2)

Tng t, hm sinh cho s cch chn chui l

B(x) = 1 + x5 + x

10 + = 1/(1-x5)

By gi, ta c th chn 0 qu cam bng 1 cch, 1 qu cam bng 1 cch, Nhng ta khng th chn hn 4 qu cam, v th ta c

C(x) = 1 + x + x2 + x

3 + x

4 = (1-x

5)/(1-x)

V tng t, hm sinh cho s cch chn o l

D(x) = 1 + x = (1-x2)/(1-x)

Theo quy tc xon, hm sinh cho cch chn t c 4 loi tri cy bng

...4321)1(

1

1

1.

1

1.

1

1.

1

1)().().().( 32

2

25

52

xxx

xx

x

x

x

xxxDxCxBxA

Gn nh tt c c gin c vi nhau! Ch cn li 1/(1-x)2 m ta tm c chui lu tha t trc . Nh th s cch sp gi tri cy gm n tri cy n gin bng n+1. iu ny ph hp vi kt qu m ta tm ra trc , v c 7 cch sp cho gi 6 tri cy. Tht l th v!

6. Cc hm sinh thng gp

6.1. nh l nh thc m rng.

Vi u l mt s thc v k l s nguyn khng m. Lc h s nh thc m rng

k

u c

nh ngha nh sau

0,1

0,!/)1)...(1(

k

kkkuuu

k

u

nh l 2. Cho x l s thc vi |x| < 1 v u l mt s thc. Lc

0

.)1(k

ku xk

ux

nh l ny c th c chng minh kh d dng bng cch s dng nh l Taylor.

V d. Tm khai trin lu tha ca cc hm sinh (1+x)-n v (1-x)-n

Gii: Theo nh l nh thc m rng, c th suy ra

0

.)1(k

kn xk

nx

Theo nh ngha

k knkk C

k

knnn

k

knnn

k

n1)1(

!

)1)...(1()1(

!

)1)...(1)((

Bi tp ton ri rc

18

T

0

1 .)1()1(k

kk

kn

kn xCx

Thay x bng x, ta c

0

1 .)1(k

kk

kn

n xCx

V d. Tm khai trin lu tha ca (1-x)-1/2

Gii: Theo nh l nh thc m rng, ta c

0

2/1 .2/1

)1(k

kxk

x

Theo nh ngha

k

k

kk

k

C

k

k

k

k

k 4)1(

!

2

12...

2

3

2

1)1(

!

12

1...1

2

1.

2

1

2/12

T

0

22/1 .4

)1()1(k

k

k

k

kk xC

x

Thay x bng x, ta c

0

22/1 .4

)1(k

k

k

k

k xC

x

6.2. Bng cc hm sinh thng gp

Hm s Khai trin lu tha ak

1/(1-x) 1 + x + x2 + x

3 + 1

1/(1+x) 1 x + x2 x3 + (-1)k

1/(1-ax) 1 + ax + a2x

2 + a

3x

3 + ak

(1-xn+1

)/(1-x) 1 + x + x2 + + xn 1 nu k n, 0 nu k >

n

(1+x)n nn

nnn xCxCxC ...1221 knC

1/(1-x)n

...!3

)2)(1(

2

)1(1 32

x

nnnx

nnnx

k

knC 1

1/(1-x)2 1 + 2x + 3x

2 + 4x

3 + k+1

1/(1-ax)2 1 + 2ax + 3a

2x

2 + 4a

3x

3 + (k+1)ak

1/(1-xr) 1 + x

r + x

2r + x

3r + 1 nu r | k v 0 trong

Bi tp ton ri rc

19

trng hp ngc li

1/(1+xr) 1 - x

r + x

2r - x

3r + (-1)s nu k=sr v 0

trong trng hp ngc li

ln(1+x) x x2/2 + x3/3 x4/4 + 0 khi k = 0 v (-1)k/k

ln(1-x) - x x2/2 x3/3 x4/4 0 khi k = 0 v -1/k

arctgx x + x3/3 + x

5/5 + 0 vi k chn v

1/k vi k l

7. Cc v d c li gii

7.1. Cp s nhn cng

Ta th tm li cng thc tnh s hng tng qut cho cp s nhn cng, tc l dy s xc nh bi a0 = a, an = axn-1 + d vi mi n = 1, 2, 3,

t F(x) = a0 + a1x + a2x2 + a3x

3 +

Ta c F(x) = a0 + a1x + a2x2 + a3x

3 +

= a0 + (qa0 + d)x + (qa1+d)x2 + (qa2+d)x

3 +

= a0 + qx(a0+a1x+a2x2+) + dx(1+x+x2+)

= a + qxF(x) + dx/(1-x)

T suy ra

)1)(1(

)()(

qxx

xadaxF

Ta tm phn tch dng

qx

B

x

A

qxx

xada

11)1)(1(

)(

Quy ng mu s chung, ta c a + (d-a)x = A + B (B+qA)x. T

A + B = a, B + qA = a d

Suy ra A = d/(1-q) v B = a d/(1-q).

Cui cng, p dng cc cng thc khai trin quen thuc, ta c

q

dq

q

daa nn

11.

7.2. Phng trnh sai phn khng thun nht

Tip theo, ta xem hm sinh lm vic th no i vi cc phng trnh sai phn khng thun nht.

Xt bi ton: Tm cng thc tng qut ca dy s cho bi a0 = 1, an = 2an-1 + 3n vi n = 1, 2,

3, ..

Theo ng s trn, ta xt F(x) = a0 + a1x + a2x2 + a3x

3 +

Bi tp ton ri rc

20

V thc hin vic khai trin v phi:

F(x) = a0 + a1x + a2x2 + a3x

3 +

= a0 + (2a0+3)x + (2a1+32)x

2 + (2a

2+3

3)x

3 +

= (1 + 3x + (3x)2 + (3x)

3 + ) + 2x(a0+a1x+a2x2+)

= 1/(1-3x) + 2xF(x)

T suy ra

xxxx

xF21

2

31

3

)21)(31(

1)(

p dng cng thc khai trin lu tha cho cc hm s thng gp, ta tm c

an = 3n+1

2n+1.

Ta xem xt mt v d khc: Tm cng thc tng qut ca dy s cho bi a0 = 1, an = 2an-1 + n.3

n.

t F(x) = a0 + a1x + a2x2 + a3x

3 + Xt

F(x) = a0 + a1x + a2x2 + a3x

3 +

= a0 + (2a0+1.3)x + (2a1+2.32)x

2 + (2a

2+3.3

3)x

3 +

= 1 + 3x(1 + 2.(3x) + 3(3x)2 + ) + 2x(a0+a1x+a2x2+)

= 1 + 3x/(1-3x)2 + 2xF(x)

T suy ra

)21()31(

139)(

2

2

xx

xxxF

Dng phng php h s bt nh, ta tm c phn tch

2)31(

3

21

7

31

9)(

xxxxF

V t

an = 3(n+1)3n 9.3n + 7.2n = (n-2)3n+1 + 7.2n.

7.3. Trng hp phng trnh c trng c nghim kp

Tip theo, ta xem xt hm sinh x l trng hp phng trnh c trng c nghim kp nh thn no.

Xt bi ton: Tm cng thc tng qut ca dy s xc nh bi a0 = 1, a1 = 4, an = 4(an-1-an-2) vi mi n = 2, 3, 4,

Theo s chung, ta xt F(x) = a0 + a1x + a2x2 + a3x

3 + a4x

4 +

B qua hai s hng u, cc s hng t a2 tr i c tnh theo cc s hng trc

F(x) = a0 + a1x + (4a1-4a0)x2 + (4a2-4a1)x

3 + (4a3-4a2)x

4 +

Bi tp ton ri rc

21

= 1 + 4x + 4x(a1x+a2x2+a3x

3+) 4x2(a0+a1x+a2x2+)

= 1 + 4x + 4x(F(x)-1) 4x2F(x)

T

F(x) = 1/(1-2x)2 = 1 + 2.2x + 3.2

2x

2 +

Suy ra an = (n+1)2n.

7.4. Mt ng dng ca quy tc xon

Quy tc xon cn c mt cch pht biu khc: Nu c hm sinh l F(x), c hm sinh l G(x) th xon ca chng, dy vi cn = a0bn + a1bn-1 + + anb0 c hm sinh l F(x).G(x).

Ta s dng quy tc ny gii mt dy s quy kh c bit: Cho dy s {an} xc nh bi a0 = 1, a0an + a1an-1 + + ana0 = 1 vi mi n = 1, 2, 3 Hy tm cng thc tng qut tnh an.

Ta tnh th cc s hng u tin ca dy s: Vi n =1, ta c a0a1+a1a0 = 1 suy ra a1 = 1/2, vi n = 2, a0a2 + a1a1 + a2a0 = 1 suy ra a2 = 3/8. Tng t, a3 = 5/16, a4 = 35/128 Quy lut kh l phc tp!

Ta th dng phng php hm sinh. Mi vic vn bt u t cch t F(x) = a0 + a1x + a2x2

+ a3x3 +

Tuy nhin, nu dng php th thng thng an = (1 (a1an-1+a2an-2++an-1a1))/a0 th chng ta s khng thu c phng trnh tm F(x) nh mong mun. Tuy nhin, h thc a0an + a1an-1 + + ana0 = 1 gi cho chng ta ngh s dng quy tc xon. C th nu c hm sinh l F(x) th theo quy tc xon, F2(x) s l hm sinh ca dy bnh phng xon ca tc l dy vi cn = a0an + a1an-1 + + ana0. Nhng, theo nh iu kin bi th cn = 1 vi mi n = 0, 1, 2, 3, c ngha l

F2(x) (1, 1, 1, 1, ) 1 + x + x2 + x3 + = 1/(1-x)

T F2(x) = (1-x)-1, suy ra F(x) = (1-x)-1/2.

p dng cng thc tm c trong phn 6.1, ta tm c n

n

nn

Ca

4

2 .

8. Bi tp

1. Xc nh hm sinh cho cc dy s sau

a)

b)

c)

d)

e)

f)

2. Tm khai trin lu tha cho cc hm s sau

Bi tp ton ri rc

22

)1ln()2

411)1)

)1()1(

1)

)1)(1(

1)

651

1)

22

222

xfx

xexd

xxc

xxb

xx

xa

3. Gi s

0

2.

21

1

n

n

n xaxx

Chng minh rng vi mi n 0, tn ti s nguyn m sao cho

a2

n + a2

n+1 = am.

3. Tm h s ca x10 trong chui lu tha ca cc hm sau y

a) (1 + x5 + x

10 + x

15 + )3

b) (x

4+x

5+x

6)(x

3+x

4+x

5+x

6+x

7)(1+x+x

2+x

3+x

4+)

c) (1+x2+x

4+x

6+)(1+x4+x8+x12+)(1+x6+x12+x18+)

d) 1/(1+x)2 e) x

4/(1-3x)

3 f) x

3/(1+4x)

2

4. S dng hm sinh, gii cc h thc quy sau

a) a0 = 1, an = 3an-1 + 2

b) a0 = 1, an = 3an-1 + 4n-1

c) a0 = 6, a1 = 30, an = 5an-1 6an-2

d) a0 = 4, a1 = 12, an = an-1 + 2an-2 + 2n

e) a0 = 2, a1 = 5, an = 4an-1 4an-2 + n2

f) a0 = 20, a1 = 60, an = 2an-1 + 3an-2 + 4n + 6.

6. Hi c bao nhiu cch trao 25 phn thng ging nhau cho bn s quan cnh st mi ngi nhn c t nht ba nhng khng qu by phn thng.

7. Tm hm sinh cho dy {ck}, trong ck l s cch i k l ra cc t 1$, 2$, 5$ v 10$.

9. a) Gi an l s cc xu tam phn (xu gm cc ch s 0, 1 hoc 2) di n c cha hai ch s lin tip ging nhau, v d a2 = 3 v c 3 xu nh vy l 00, 11, 22. Ngoi ra, a0 = a1 = 0 v cc xu nh vy phi c di t nht l 2. Hy tm mt cng thc truy hi cho an.

b) Chng minh rng

)31)(21(

1

21 xxx

x

l hm sinh cho dy

c) Tm cc hng s r v s sao cho

x

s

x

r

xx 3121)31)(21(

1

d) Tm cng thc tng qut tnh an.

10. Gi s c 4 loi ko: s-c-la, chanh, du v sa. Tm hm sinh cho s cch chn n vin ko tho mn cc iu kin khc nhau sau y

Bi tp ton ri rc

23

a) Mi mt loi ko xut kin s l ln.

b) S mi mt loi ko chia ht cho 3.

c) Khng c ko s-c-la v c nhiu nht mt vin ko chanh.

d) C 1, 3 hay 11 vin ko s-c-la, 2, 4 hoc 5 vin ko chanh.

e) Mi mt loi ko xut hin t nht 10 ln.

11. V hnh phc. Mt chic v xe but c nh s t 000000 n 999999. V xe but c gi l v hnh phc nu tng ba ch s u ca s c ghi trn v bng tng ca ba ch s cui. V d 000000, 999999, 123006 l nhng v hnh phc. Bi ton ca chng ta c mc ch i tm s nhng chic v hnh phc trn tng 106 v (t 000000 n 999999).

a) Chng minh rng s nhng chic v hnh phc bng s nghim ca phng trnh

a1 + a2 + a3 = a4 + a5 + a6, trong 0 ai 9.

b) Gi ck l s nghim ca phng trnh

a1 + a2 + a3 = k vi 0 ai 9.

Chng minh rng s nhng chic v hnh phc bng

27

0

2 .k

kcN

c) Tm hm sinh cho dy ck.

d) Chng minh rng ck = c27-k vi k=0, 1, , 27.

e) Chng minh rng s nghim ca phng trnh

a1 + a2 + a3 = a4 + a5 + a6 vi 0 ai 9.

bng s nghim ca phng trnh

a1 + a2 + a3 + a4 + a5 + a6 = 27 vi 0 ai 9.

f) T N l h s ca x27 trong khai trin ca (1+x+x2++x9)6. Suy ra gi tr ca N.

12. S Catalan. S Catalan l s c xc nh mt cch truy hi nh sau

C0 = 1, Cn = C0Cn-1 + C1Cn-2 + + Cn-1C0 vi n = 1, 2, 3,

S Catalan c nhiu nh ngha t hp khc nhau, chng hn, s Catalan l s cc cch ni 2n im trn ng trn bng n dy cung khng ct nhau, l s cy nh phn c gc c n+1 l, l s ng i ngn nht trn li nguyn t im (0, 0) n im (n, n) khng vt qua ng thng y = x

a) Gi F(x) l hm sinh ca dy Cn. Chng minh rng

F(x) = 1 + xF2(x)

T suy ra x

xxF

2

411)(

b) S dng kt qu bi tp 2e, suy ra cng thc tng qut tnh Cn l

nnn Cn

C 21

1

.

Bi tp ton ri rc

24

THI TRC NGHIM GIA K (THAM KHO)

MN: TON RI RC

Thi gian: 45 pht khng k thi gian pht

1. Pht biu no l mnh ? a. Hm ny tri tht p qu! b. Ngy mai l th 7, bn c bit khng? c. Mt con nga au c tu b c. d. Thi em hy v, qu hng ang ch em .

2. Chn tr ca mnh 2,[( 3 0) ( 4 3 0)]x R x x x l:

a. ng. b. Sai. 3. Chn tr ca mnh lng t ha , ,[( 2 5) (2 4)]x R y R x y x y l:

a. ng. b. Sai.

4. Ph nh ca mnh 2 2 2, , ,[( ) ( )]x R y R z R x y z x z l:

a. 2 2 2, , ,[( ) ( )]x R y R z R x y z x z

b. 2 2 2, , ,[( ) ( )]x R y R z R x y z x z

c. 2 2 2, , ,[( ) ( )]x R y R z R x y z x z

d. 2 2 2, , ,[( ) ( )]x R y R z R x y z x z

e. Tt c u sai

5. Ph nh ca biu thc mnh [( ) ]xz y z xy

a. [( ) ]xz y z xy

b. [( ) ]xz y z xy

c. [( ) ]xz y z xy

d. [( ) ]xz y z x y

e. Tt c u sai

6. F = [ z )]x y y y x , biu din F di dng chun hi hon ton l:

b. ( )( )( )f x y z x y z x y z

c. ( ) ( )f x y z x y z

d. ( ) ( )f x y z x y z

e. ( )( )( )f x y z x y z x y z

e. Tt c u sai

7. F = ( )xz yz y z , biu din F di dng chun tuyn hon ton l:

a. f x y z x y z x y z x y z x y z

b. f x y z x y z x y z x y z x y z x y z

c. f x y z x y z x y z x y z x y z x y z

d. f x y z x y z x y z x y z x y z x y z

Bi tp ton ri rc

25

e. Tt c u sai

8. Cng thc l [ ( )] [( ) ( )] ( )x y x x z xy x z

a. Cng thc hng ng. b. Cng thc hng sai c. Cng thc kh ng/sai

9. Cng thc l ( ) ( ) ( )x y xz xy y z x

a. Cng thc hng ng. b. Cng thc hng sai c. Cng thc kh ng/sai

10. Khng nh no l ng i vi tp hp A = { 0, {1}, {2}}

a. {{1}}A b. {{2}}A c. {{}}A d. 0A

11. Chn tp hp rng:

a. {xN/ 2x2-7=1} b. {xZ/ 3x

2-7=2} c. {xQ/4x

2+3=1} d. {xR/ x

2+3=5}

12. Chn nh x no l song nh:

a. 2: , ( ) 2f R R f x x b. 3: , ( )f R R f x x

c. 3: , ( ) 2 2f R Z f x x d. 2: , ( ) 2 2f Q Q f x x

13. C bao nhiu cch ghp cp ga 7 nam v 7 n?

a. 77 b. 7*7 c. 7 ! d.77 e. Kt qu khc

14. C bao nhiu xu nh phn c di 10 bt u bi 110 hoc ht thc bi 011?

a. 28 b. 2

6 c. 2

8-2

6 d. 610C e. Kt qu khc

15. C bao nhiu s cha s 0 trong khong (1911, 2008)

a. 16 b. 61 c. 17 d. Kt qu khc

16. C bao nhiu s c sp tng thc s trong on [111, 234]

a. 34 b. 36 c. 38 d. Kt qu khc

17. S sinh vin d thi mn ton ri rc khng th t hn l bao nhiu t nht 234 th sinh c cng s im nu nu ly thang im 100.

a. 23301 b. 23401 c. 22999 d. Kt qu khc

18. S b nghim nguyn dng (1 2 3, ,x x x ) ca phng trnh :

1 2 3 1 220, 3, 4x x x x x

a. 18.19.20 b. 2026C c. 5.3

20C d.12!

3 e. Kt qu khc

19. H s ca x3y7z trong khai trin (3x-7y+z)11

20. Tng tt c cc h s ca khai trin: (2x-2y+z)8

Bi tp ton ri rc

26

THI KT THC MN HC (THAM KHO)

MN: TON RI RC

Thi gian: 90 pht khng k thi gian pht

Cu 1 (2 im) Xt cc mt m c dng ABC-XYZ, trong A, B, C l cc ch ci khc nhau (c 26 ch ci trong bng ch ci ting Anh) cn X,Y,Z l cc ch s khc nhau (c 10 ch s thp phn t 0 n 9).

a. C tt c bao nhiu mt m nh vy? b. C tt c bao nhiu mt m m tng 3 s 3 v tr tn cng lun bng 13?

(VD: mt m RAS-319 l mt m hp l. Mt m RAS-271 l khng hp l)

Cu 2 (1 im) Tm phn t x7 y3 z2 trong khai trin (4x - 5y + 2z)12

Cu 3 (3 im) Cho l 1 quan h trn X={2, 4, 5, 8, 10, 15, 16, 30} xc nh bi:

, , ,x y X x y y kx k Z

d. Chng minh l 1 quan h th t

e. Tm cc phn t ti i, ti tiu, ln nht, nh nht xc nh bi quan h trn

f. V biu Hasse tng ng

Cu 4 (3 im) Tm a thc ti tiu v v mch logic tng ng vi kt qu tm c vi

( , , , )f x y z t yzt xzt xyzt xyzt xyt xyzt xyzt

Cu 5 (1 im) Cho ( , , )A l 1 i s Boole

Chng minh: ,x y A x y x x y y

--- Ht ---

MT S THI THAM KHO TRNG KHC

THI MN TON RI RC - Thi gian: 90 pht - (khng s dng ti liu)

1 :

Cu 1. Kim tra suy lun sau y bng 2 cch khc nhau:

p (q r)

q p

p

----------------

r

Cu 2.

a) Hy tnh s dy 6 bit khc nhau trong s bit 1 l mt s chn. b) Cho n l mt s nguyn dng. Tnh s dy n bit khc nhau trong s bit 1 l mt s

chn.

Bi tp ton ri rc

27

Cu 3. Cho X = a,b,c,d,e.

a) Tm mt quan h th t trn X sao cho a l phn t nh nht, d v e l 2 phn t ti i. b) Hi c bao nhiu quan h th t trn X tha iu kin c yu cu trong cu (a).

Cu 4.Tm cng thc dng chnh tc v cc cng thc a thc ti tiu ca hm Bool f(x,y,z,t) c bng gi tr nh sau:

x y z t f

1 1 0 0 0

0 0 0 1 1

0 0 1 0 1

0 0 1 1 1

0 1 0 0 1

0 1 0 1 1

0 1 1 0 0

0 1 1 1 0

1 0 0 0 0

1 0 0 1 1

1 0 1 0 1

1 0 1 1 1

1 1 0 0 0

1 1 0 1 0

1 1 1 0 1

1 1 1 1 1

Cu 5. Tnh s cc hm Bool theo 3 bin f(x,y,z) tha iu kin

f(x,y,z) = f(x,z,y) = f(y,x,z) vi mi x, y, z.

2 :

Cu 1. Cho biu thc logic E theo 4 bin p, q, r, s nh sau:

A = (p ( q r) s ) ( s r p )

Hy rt gn biu thc A v tm cc gi tr ca cc bin p, q, r, s cho A = 1.

Cu 2. Cho n l mt s nguyn dng v t Sn = 1, 2, . . ., n.

Bi tp ton ri rc

28

a) Tnh s tp hp con ca Sn cha t nht mt s chn trong trng hp n = 14 v trong trng hp n = 15.

b) Tnh s tp hp con ca Sn cha t nht mt s chn trong trng hp tng qut (n ty ).

Cu 3.

a) Nu ln nh ngha v quan h th t trn mt tp hp v cho mt v d.

b) Cho X = a, b, c, d, e. Tm tt c cc quan h th t trn X tha mn iu kin: a l phn t nh nht v e l phn t ln nht.

Cu 4. Tm cng thc dng chnh tc v cc cng thc a thc ti tiu ca hm Bool f(x,y,z,t) c bng gi tr nh sau:

x y z t f

0 0 0 0 1

0 0 0 1 1

0 0 1 0 1

0 0 1 1 0

0 1 0 0 1

0 1 0 1 0

0 1 1 0 0

0 1 1 1 0

1 0 0 0 0

1 0 0 1 1

1 0 1 0 1

1 0 1 1 1

1 1 0 0 1

1 1 0 1 0

1 1 1 0 1

1 1 1 1 1

Cu 5. Cho X = x1, x2, . . ., xn l mt tp hp hu hn c n phn t. Gi s R l mt quan h th t trn X. Hy vit mt thut ton tm tt c cc phn t ti i ca X theo quan h th t R.

3 :

Bi tp ton ri rc

29

Cu 1: Xt cc v t theo bin nguyn sau y:

p(x) : x2 5x + 6 = 0

q(x) : x2 4x 5 = 0

r(x) : x > 0

us(x, y) : x l c s ca y

Hy xc nh chn tr ca cc mnh sau y:

a) x : p(x) r(x).

b) x : q(x) r(x).

c) y, x : us(x,y).

d) y, x : us(x,y).

Cu 2: Cho m v n l cc s nguyn dng. Tnh s dy bit gm n bit tha iu kin sau y: Tng s bit 1 cc v tr chn t nht l bng m. Hy tnh s dy bit theo iu kin trn trong trng hp n = 32, m = 8.

Cu 3:

a) Nu ln nh ngha v biu Hasse ca mt tp hp X c th t (tc l c mt quan h

th t ang c xt trn X). V biu Hasse ca tp P(a,b,c) theo quan h th t ,

trong P(a,b,c) l tp hp gm tt c cc tp hp con ca a,b,c.

b) Nu trn tp hp P(a,b,c) ta xt quan h th t th biu Hasse c dng nh th no? Khi hy cho bit phn t ln nht v phn t nh nht l g?

Cu 4: Cho mt hm Bool f(x,y,z,t) theo 4 bin x, y, z, t. Gi s f c biu Karnaugh nh sau:

1 1 1

1 1 1

1 1 1

1 1

Hy tm cng thc dng ni ri chnh tc v cc cng thc a thc ti tu ca hm Bool f.

Cu 5: Cho X l mt tp hp c n phn t. Tnh s quan h 2 ngi trn X c tnh cht phn x v c tnh cht i xng.

4 :

Cu 1: Xt cc v t theo bin nguyn sau y:

p(x) : x2 5x + 6 = 0

q(x) : x2 4x 5 = 0

r(x) : x > 0

us(x, y) : x l c s ca y

Hy xc nh chn tr ca cc mnh sau y:

x : q(x) r(x).

Bi tp ton ri rc

30

x : p(x) r(x).

x, y : us(x,y).

x, y : us(x,y).

Cu 2: Mt lp hc c 12 hc sinh gii vn hay gii ton trong c 2 hc sinh gii c 2 mn (Vn v Ton) v c 8 hc sinh gii Ton. Hi c bao nhiu hc sinh ch gii Vn m khng gii Ton.

Gi s ta phi chn 2 t hc sinh i din i thi hc sinh gii: t hc sinh gii Vn v t hc sinh gii Ton, mi t gm c 4 ngi v mi hc sinh c chn i thi ch tham gia vo mt t m thi. Hi c bao nhiu cch chn 2 t hc sinh nh th?

Cu 3: Nu ln nh ngha v phn t nh nht v phn t ti tiu trong mt tp hp X theo mt quan h th t R. Cho mt v d minh ha.

Gi s X = x1, x2, . . ., xn l mt tp hp hu hn c n phn t v R l mt quan h th t trn X. Vit thut ton tm phn t ti tiu v phn t nh nht (nu c) ca X theo quan h th t R.


1 1 1

1 1

1 1 1

1 1


Cu 5: Cho X l mt tp hp c n phn t. Tnh s quan h 2 ngi trn X c tnh cht cht i xng nhng khng c tnh cht phn x.

5 :

Cu 1. Kim tra suy lun sau y bng 2 cch khc nhau:

p (q r)

p s

q

r

----------------

s

Cu 2. Mt lp hc c 14 hc sinh gii vn hay gii ton trong c 10 hc sinh gii Ton v c 8 hc sinh gii Vn. Hi c bao nhiu hc sinh ch gii Vn m khng gii Ton.

Bi tp ton ri rc

31

Gi s ta phi chn 2 t hc sinh i din i thi hc sinh gii: t hc sinh gii Vn v t hc sinh gii Ton, mi t gm c 4 ngi v mi hc sinh c chn i thi ch tham gia vo mt t m thi. Hi c bao nhiu cch chn 2 t hc sinh nh th?

Cu 3. Cho X = a, b, u, v.

Tm mt quan h th t trn X sao cho a v b l cc phn t ti tiu nhng khng ti i, u v v l cc phn t ti i nhng khng ti tiu.

Hi c bao nhiu quan h th t trn X tha iu kin c yu cu trong cu (a).


x y z t f

0 0 0 0 0

0 0 0 1 1

0 0 1 0 1

0 0 1 1 1

0 1 0 0 1

0 1 0 1 1

0 1 1 0 0

0 1 1 1 0

1 0 0 0 0

1 0 0 1 1

1 0 1 0 1

1 0 1 1 1

1 1 0 0 1

1 1 0 1 1

1 1 1 0 1

1 1 1 1 0

Cu 5. Cho X l mt tp hp hu hn v R l mt quan h th t trn X. Chng minh rng X c t nht mt phn t ti tiu. Hi c phi trong X lun lun c phn t nh nht khng?

6 :

Cu 1:

Pht biu mt nguyn l qui np dng chng minh mnh c dng:

Bi tp ton ri rc

32

n 1 : p(n) trong p(n) l mt v t theo bin s t nhin n.

Hy dng nguyn l qui np chng minh cng thc di y ng i vi mi s nguyn dng n:

Trong cng thc trn, k hiu C(n,k) l s t hp n chn k.

Cu 2: Cho m v n l cc s nguyn dng. Tnh s dy bit gm n bit tha iu kin sau y: Tng s bit 1 cc v tr chn l mt s chn. Hy tnh s dy bit theo iu kin trn trong trng hp n = 32, m = 8.

Cu 3: Trn tp hp s t nhin N ta xt mt quan h 2 ngi R c nh ngha nh sau :

a R b n N : b.n = a

Chng minh rng quan h R l mt quan h th t trn N.

t D = 1, 2, 3, 5, 6, 10, 15, 30. Trn tp hp D ta cng xt mt quan h R nh ngha nh trong phn (a). Hi R c phi l mt quan h th t trn D khng? Nu c th hy v biu Hasse v cho bit phn t nh nht v phn t ln nht trong D theo th t R.


1 1

1 1 1 1

1 1 1

1 1 1


Cu 5: Tm mt hm Bool f(x,y,z,t) tha iu kin: Nu x+y+z+t l mt s chn th f(x,y,z,t) = 0. Tnh s cc hm Bool tha mn iu kin trn.

7 :

Cu 1: Pht biu mt nguyn l qui np dng chng minh mnh c dng:

n 0 : p(n), trong p(n) l mt v t theo bin s t nhin n.

Hy dng nguyn l qui np chng minh pht biu p(n) sau y l ng i vi mi s t nhin n : p(n) : Nu tp hp X c n phn t th s cc tp hp con ca X l 2n

Cu 2: Cho m v n l cc s nguyn dng. Tnh s dy bit gm n bit tha iu kin sau y: Tng s bit 1 cc v tr chn l mt s l. Hy tnh s dy bit theo iu kin trn trong trng hp n = 32, m = 8.

Cu 3: Nu ln nh ngha v quan h 2 ngi trn mt tp hp v nh ngha ca cc tnh cht: phn x, i xng v bc cu. Cho mt v d v mt quan h 2 ngi tp hp cc s nguyn c cc tnh cht phn x, i xng v bc cu.

Bi tp ton ri rc

33

Gi s X l mt tp hp hu hn v R l mt quan h 2 ngi trn X c cho bng cch lit k. Vit mt thut ton kim tra xem R c tnh cht bc cu hay khng.


1 1

1 1 1

1 1 1

1 1 1


Cu 5: Tnh s cc hm Bool f(x,y,z,t,u,v) tha iu kin:

Nu s cc bit 1 trong dy bit (x,y,z,t,u,v) l mt s nguyn t th f(x,y,z,t,u,v) = 0.

8 :

Cu 1. Cho biu thc logic E theo 4 bin p, q, r, s nh sau:

E = ( ( q r) s p) ( s r p )

Hy rt gn biu thc E v tm cc gi tr ca cc bin p, q, r, s cho E = 1.

Cu 2. Cho n l mt s nguyn dng v t Sn = 1, 2, . . ., n.

Tnh s tp hp con ca Sn cha t nht mt s l trong trng hp n = 11 v trong trng hp n = 12.

Tnh s tp hp con ca Sn cha t nht mt s l trong trng hp tng qut (n ty ).

Cu 3. Nu ln nh ngha v quan h th t trn mt tp hp v cho mt v d.

Cho X = a, b, c, d, e. Tm tt c cc quan h th t trn X tha mn iu kin: tp hp cc

phn t ti tiu l a, b v tp hp cc phn t ti i l c, d, e.


x y z t f

0 0 0 0 1

0 0 0 1 1

0 0 1 0 1

0 0 1 1 1

0 1 0 0 1

0 1 0 1 0

Bi tp ton ri rc

34

0 1 1 0 0

0 1 1 1 0

1 0 0 0 0

1 0 0 1 1

1 0 1 0 1

1 0 1 1 1

1 1 0 0 0

1 1 0 1 0

1 1 1 0 1

1 1 1 1 1

Cu 5. Cho X l mt tp hp c n phn t. Tnh s quan h 2 ngi trn X c tnh cht cht i xng nhng khng c tnh cht phn x.

9 :

Cu 1: Kim tra suy lun sau y:

p q

p r q

r s t

s

------------

t

Cu 2: Trong mt lp c n (n > 5) hc sinh, ngi ta mun chia thnh 2 t m mi t phi c t nht l 3 hc sinh. Hi c bao nhiu cch phn chia? Tnh s cch phn chia trong trng hp n = 10.

Cu 3: Biu Hasse ca mt cu trc th t (hu hn) l g? Da vo biu Hasse hy

lit k ra mt quan h th t R trn X = 1, 2, 3, 4 tha iu kin: trong X c phn t nh nht v c phn t ln nht. C bao nhiu quan h th t trn X tha iu kin trn?

Cu 4: Tm cng thc dng chnh tc v cc cng thc a thc ti tiu ca hm Bool f(x, y, z, t) c bng Karnaugh nh sau:

1 1 1 1

1

1 1

Bi tp ton ri rc

35

1 1

10 :

Cu 1:

Cho biu thc Logic E theo 4 bin sau :

E= (p [ ( q V r) s]) [ s (r p) ]

Xc nh cc gi tr ca p, q, r, s E =1

Cu 2:

Chng minh cng thc sau :

Cu 3:

Lp tin hc c 21 sinh vin phi thc hin 3 bi thc hnh. Bit rng tt c cc hc sinh u thc hnh c t nht 1 bi, 5 hc sinh khng lm bi th nht, 7 hc sinh khng lm bi th hai, 6 hc sinh khng lm bi th ba, v c 9 hc sinh lm c 3 bi. Hi c bao nhiu hc sinh ch lm 1 bi.

Cu 4:

Cho tp X = { 1,2,3,4}. Xt quan h hai ngi R nh ngha nh sau:

R = { (1,1) , (1,4), (2,2) , (2,3), (3,2), (3,3), (4,1) , (4,4) }

Biu din quan h ny di dng ma trn v dng biu d Hasse

Quan h ny c cc tnh cht g ? ti sao ?

Kt lun g v quan h ny (tng ng hay th t ).

Cu 5:

Cho hm Bool f(a.b.c.d) tha :

f-1

(1) = { 0101 , 0110 , 1000 , 1011 ]

V biu Karnaugh ca f

Tm cc cng thc a thc ti tiu ca f

V mch t hp ca f theo kt qu cu b.

11 :

Cu 1:

Cho biu thc Logic E theo 4 bin p, q, r, s sau :

E= (( q V r) V s p ) ( s r p)

Hy rt gn biu thc E v tm cc gi tr ca p, q, r, s E=0

Cu 2:

Hy kim tra suy lun sau:

Bi tp ton ri rc

36

Nu A c ln chc v lm vic nhiu th A c tng lng.

Nu c tng lng A s mua xe mi.

M A khng mua xe mi

Vy A khng c ln chc hay A khng lm vic nhiu

Cu 3:

CM : / A B C/ = /A/ + /B/ + /C/ - /AB/ - /AC/ - /BC/ - /ABC/

Lp tin hc c 21 sinh vin phi thc hin 3 bi thc hnh. Bit rng tt c cc hc sinh u thc hnh c t nht 1 bi, 5 hc sinh khng lm bi th nht, 7 hc sinh khng lm bi th hai, 6 hc sinh khng lm bi th ba, v c 9 hc sinh lm c 3 bi. Hi c bao nhiu hc sinh ch lm 1 bi.

Cu 4:

Cho tp X = { 1,2,3,4}. Xt quan h hai ngi R nh ngha nh sau:

R = { (1,1) , (1,3), (2,2) , (2,4), (3,1) , (3,3), (4,2) , (4,4) }

Biu din quan h ny di dng ma trn v dng biu d Hasse

Quan h ny c cc tnh cht g ? ti sao ?

Kt lun g v quan h ny (tng ng hay th t ).

Cu 5:

Cho hm Bool f(a.b.c.d) tha :

f-1

(0) = { 0100 , 0000 , 0110 , 0010 ]

V biu Karnaugh ca f

Tm cc cng thc a thc ti tiu ca f

V mch t hp ca f theo kt qu cu b.

12 :

Cu 1: Cho biu thc Logic E theo 2 bin sau :

E= [ (p V q) V [( p q) V q ]] [ (p q)]

Hy rt gn biu thc E v tm cc gi tr ca p, q cho E =1

Cu 2: Hy kim tra suy lun sau:

Nu B i lm v mun th v anh rt gin d.

Nu A thng xuyn vng nh th v anh rt gin d.

Nu v A hay v B gin d th c H l bn bn hc ca A v B s nhn c li than phin.

M H khng nhn c li than phin

Vy B i lm v sm v A t khi vng nh.

Bi tp ton ri rc

37

Cu 3: N l 1 hng s cho trc. Xc nh gi tr ca bin C sau khi thc hin on chng trnh

C:=0 ;

For i:= 1 to N do

For j:= 1 to N do

For k:= j to N do C:= C+1;

Cu 4: Cho tp X = { 1,2,3,4}. Xt quan h hai ngi R nh ngha nh sau:

R = { (1,1) , (1,2), (2,1) (2,2) ,(2,4), (3,3), (4,2) , (4,4) }

(a) Biu din quan h ny di dng ma trn v dng biu d Hasse

(b) Quan h ny c cc tnh cht g ? ti sao ?

(c) Kt lun g v quan h ny (tng ng hay th t ).

Cu 5: Tm cng thc dng chnh tc v cc cng thc a ti tiu ca hm Bool f(x,y,z) c bng gi tr:

X y z t f

0 0 0 0 1

0 0 0 1 1

0 0 1 0 1

0 0 1 1 1

0 1 0 0 1

0 1 0 1 0

0 1 1 0 0

0 1 1 1 0

1 0 0 0 0

1 0 0 1 1

1 0 1 0 1

1 0 1 1 1

1 1 0 0 0

1 1 0 1 0

1 1 1 0 1

1 1 1 1 1

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Bai Tap Toan Roi Rac