Toán rời rạc 2015

TS. NGUYN VIT NG Hng dn, p s BI TP TON RI RC

January 18, 2015

1

LI GII TM TT HOC P S

1) t u (1)

r (s t) (2)

(p q ) r (3)

(s u ) (4) ______________

p

s u ( Do tin (4) v lut i ngu ) (5) u (Do (5) v lut n gin ni lin) (6) t ( Do (1),(6) v lut ph nh) (7) s (Do (5) v lut n gin ni lin) (8)

t s ( Do (7), (8) v php tan ni lin) (9)

(t s) ( Do (9) v lut i ngu) (10) r (Do (2), (10) v lut ph nh) (11)

( p q) (Do (3), (11) v lut ph nh) ( 12) p q ( Do (12) v lut i ngu) (13) p (Do (13) v lut n gin ni lin).

2) a) C > 0, m , n ,( n m xn < C n ).

b) C > 0, m , n ,( n m xn C n ).

3) a)C > 0, d , m , n ( n m T(n) < C nd).

b) C > 0, d , m , n ( n m T(n) C nd).

4)

1) ( )q s (tin )

2) q s (lut De Morgan)

3) q v s (lut n gin)

4) p q (tin )

5) p (PP ph nh)

6) p s (T 3, 5 v nh ngha php ni lin) 7) ( )p s (lut De Morgan)


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8) r (p s) (tin )

9) r (PP ph nh)

10) (t p) r (tin )

11) (t p) (PP ph nh)

12) t p (lut De Morgan) 13) t (lut n gin) Vy suy lun cho l ng. 5)

1) ))()()(( xRxQxPRx (Tin )

2) )()()( aRaQaP (Qui tc c bit ph dng vi a bt k)

3) )()()( aRaQaP (Lut ko theo)

4) )()()( aRaPaQ (Lut ko theo)

5) )()(( xQxPRx (Tin )

6) )()( aQaP (Qui tc c bit ha ph dng vi a bt k)

7) )()( aQaP (Lut ko theo)

8) )()()( aRaPaP (T 4 v 7, Tam an lun)

9) )()()( aRaPaP (Lut ko theo)

10) )()( aRaP (Lut ly ng)

11) )()( aPaR (Lut ko theo)

12) ))()(( xPxRRx (Qui tc tng qut ha ph dng).

6) .


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a) T q ta c [ t q ] (1). T (t p) ta c [ t p ] (2).

T (2) v (1) ta c [ t p ] [ t q ], ngha l [ t (p q) ] (3)

T (3) ta c [ t (p q) ] (4). T (4) v [ (p q) s ], ta suy ra (t s)

b) Gn chn tr p = 1, q = 0, s = 0 v t = 0, ta thy 3 dng mnh pha trn ng v dng

mnh di sai nn suy lun l sai

.

7)

a) Ta c p q m nn q.

M r nn r.

Mt khc s , do .

b) Ta c ko theo x > 0. Do P ng.

Ph nh ca P l .

8)

a) A ng v (9) Z, q Q, q2 6q = (q 3)2 9 9.

A = k Z, q Q, (0,25 ) q2 6q < k

b) B y z y x z

(y z) (y x ) z

[ (y z) z ] (y x ) [ (y z ) (z z ) ] (y x )

[ ( z y) 1 ] (y x ) z [ y (y x ) ]

z y

9) a) S n thc c c l s nghim nguyn 0 ca phng trnh m + n + p + q = 16, ngha l

bng 16 34 19 969K C

b) S n thc c c l s nghim nguyn ca phng trnh m + n + p + q = 16 (m 2, n

0,

p = 3 v 1 q 4). Ta c phng trnh tng ng

m + n + q = 10 (m = (m 2) 0, n 0, 0 q = (q 1) 3).

Phng trnh m + n + q = 10 (m 0, n 0, q 0) c s nghim nguyn l 10 23 12 66K C .

Phng trnh m + n + q = 10 (m 0, n 0, q 4) m + n + q = 6 (m 0, n 0,


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q = (q 4) 0) c s nghim nguyn l 6 23 8 28K C .

p s cn tm l 66 28 = 38.

c) (x y + 4z 3t)16 = *16 (8,5,1,2)P x

8(y)5(4z)1( 3t)2 + ... (php hon v lp trn 16 phn t)

= 16!

8!5!1!2!4

13

2(x

8y

5z

1 t2) + ... = 77.837.760 x8y5z t2 + ...

10) a) phn x v (x,y) F, x x v y y nn (x,y) (x,y).

phn xng v (x,y), (z,t) F, [ (x,y) (z,t) v (z,t) (x,y) ] ( x z x v

y t y )

(x,y) = (z,t).

truyn v (x,y),(z,t),(u,v) F, [ (x,y) (z,t) v (z,t) (u,v) ] ( x z u v y

t v )

( x u v y v ) (x,y) (u,v) ).

Vy l mt quan h th t trn F.

b) Do | D | = 50 v | E | = 60 nn | F | = | D x E | = | D |.| E | = 50 x 60 = 3000.

min( F, ) = (11, 39) v max(F, ) = (60, 20).

11)

a) A = x Q, y R, (0,25 ), 4x2 + 8x 2y .

A sai (x c nh nn 4x2 + 8x c nh v cho y + th 2y + , ngha l khng th xy

ra 4x2 + 8x 2y ) v suy ra A ng .

b) B [ p (p r) ] [ (q r) q ]

[ ( p p) ( p r) ] [ (q q ) ( r q ) ]

[ 1 ( p r) ] [ O ( r q ) ] ( p r) ( r q )

p [ r ( r q ) ] ( p r ) = C .


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12) a) S dy s c th c : 10!

1!2!3!4! (0,5 ) = 12.600

b) Ch s cui l 2 hoc 4 : 9!

2!3!4! +

9!

1!2!3!3! (0,5 ) = 1260 + 5040 = 6300

c) Xem hai ch s 7 lin nhau nh l mt ch s : 9!

1!1!3!4! (0,25 ) = 2520

13)

Ta c ao = 5, a1 = 17 (*) v an + 2 = 4an + (20n + 67)3n n 0 (**).

Xt h thc an + 2 = 4an n 0 (*) vi a thc tng ng f(x) = x2 4 = (x 2)(x + 2) .

(*) c nghim tng qut bn = p.2n + q(2)n n 0 (p, q ) .

Do f(3) = 5 0 nn (**) c mt nghim c dng cn = (rn + s)3n n 0 (r, s ) .

Thay cn = (rn + s)3n n 0 vo (**), [ r(n + 2) + s ]3n + 2 = 4(rn + s)3n + (20n + 67)3n n 0.

Suy ra 9(rn + 2r + s) = (4rn + 4s + 20n + 67) n 0, ngha l 5r = 20 v 5s + 18r = 67 v do

r = 4, s = 1, cn = (4n 1)3n n 0 . Ta c (**) c nghim tng qut

an = bn + cn = p.2n + q(2)n + (4n 1)3n n 0 (p, q ) . T (*), ta c

(p + q = 6 v p q = 4), ngha l (p = 5, q = 1) v an = 5.2n + (2)n + (4n 1)3n n 0 .

14)

a) t f(x) = x3 + 2x2 x S. phn x [ x S, f(x) = f(x) nn xx ] .

i xng [ x,y S, xy f(x) = f(y) f(y) = f(x) yx ] .

truyn [ x,y S, (xy & yz f(x) = f(y) = f(z) f(x) = f(z) xz ] .

b) a0 f(a) = f(0) a3 + 2a2 = 0 a2(a + 2) = 0 (a = 0 hoc a = 2) .

b1 f(b) = f(1) b3 + 2b2 = 3 (b 1)(b2 + 3b + 3) = 0 b = 1 .

c(1) f(c) = f(1) c3 + 2c2 = 1 (c + 1)(c2 + c 1) = 0 c = 1 hay c = 1 5

2


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15) 4 4

12 8. .1

3!

C C= 5775.

16) t t = 2 (x + y + z) th t v ta c phng trnh x + y + z + t = 2

t x = (x + 20) , y = (y + 7) v z = (z 3) , ta c phng trnh mi tng

ng vi bt phng trnh cho :

x + y + z + t = 22 vi x, y, z, t v z < 7

Ta c 2 bi ton :

BT (1) : Tm s nghim ca phng trnh x + y + z + t = 22 vi x, y, z, t

BT (2) : Tm s nghim ca phng trnh x + y + z + t = 22 vi x, y, z, t v z 7

BT (1) c kt qu l 325C = 2300

t z = (z 7) th BT (2) c cng kt qu vi bi ton (3)

BT (3) : Tm s nghim ca phng trnh x + y + z + t = 15 vi x, y, z, t

BT (3) c kt qu l 318C = 816

S nghim nguyn ca bt phng trnh cho l 2300 816 = 1484

17) thi 2013.

a) Gi x, y, z l s ko cn chia cho mi ngi.

Theo gi thit ta c x, y, z {3,4,,10} v x + y + z = 20. (1)

Ta c (1) (x 3) + ( y 3 )+(z 3) = 11.

Do s nghim nguyn ca (1) tha x, y, z 3 l =78.

Ta li c (1) (x 11) + ( y 3 )+(z 3) = 3.

Do s nghim nguyn ca (1) tha x 11, y,z 3 l =10.


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Tng t cho trng hp y 11, x,z 3 v z 11, y,x 3. Do s nghim ca (1) tha mn yu

cu ca bi tan l 78 3.10 = 48 nghim , ngha l c 48 cch chia.

b) S quan h trn A l s tp con ca A A, nn bng 216 = 65536.

S quan h tng ng trn A bng s cch phn hach A thnh cc tp con ri nhau( lp tng

ng).

- Phn hoch thnh 4 lp tng ng; 1 cch.

- Phn hach thnh 3 lp tng ng : = 6 cch.

- Phn hach thnh 2 lp tng ng :

= 7 cch.

- Phn hach thnh 1 lp tng ng : 1 cch.

Vy, s cc quan h tng ng l 1+6+7+1 = 15 quan h.

18) a) 3281 b) 29615.

19) a)5461512; b) 486000; c) 1959552; d) 1958040

20) a)2118760; b) 1050000 21) (366 266) + (367 267 ) + (368 268) = 2684483063360.

22) X = {x : 1 x 20} vi quan h thng thng.

a) min(A) = 8 v |A| 10 A = B {8} vi B Y = {9, 10,, 20}, |B| 9.

S tp con A tha min(A) = 8 v |A| 10

= S tp con B Y = {9, 10,, 20}, |B| 9

9 10 11 12

12 12 12 12C C C C 299

b) min(A) = 6 v max(A) = 18 A = {6, 18} C vi C Z = {7, 8,, 17}.

S tp con A tha min(A) = 6 v max(A) = 18= S tp con C Z = {7, 8,, 17}.

= 211

= 2048.

23) t aj l s trn m i bng chi cho n ht ngy th j trong thng. Ta c a1, a2, , an l mt dy tng gm cc s nguyn dng khc nhau tng i v aj 45. Hn na, a1+14 , a2 + 14, , a30 + 14 cng l mt dy s tng gm cc s nguyn dng khc nhau vi 15 aj +14 59.

Ta thy rng 60 s nguyn dng a1, a2, , a30, a1 +14, a2 +14, , a30 + 14 u nh hn hoc bng 59. p dng nguyn l Dirichlet ta thy c t nht hai trong 60 s nguyn dng ni trn phi


January 18, 2015

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bng nhau. Nh th phi c t nht hai ch s i v j sao cho ai = aj +14. Do ng 14 trn c i bng chi t ngy th j + 1 n ngy th i.

24) a) V H l n th v hng nn mi nh ca H khng c vng v ch c th ni vi

cc nh khc khng qu mt cnh, ngha l mi nh ca H c bc ti a l (n 1).

Suy ra H c ti a l n(n 1) / 2 cnh.

b) Nu H c nh c lp th bc ca cc nh ca H thuc tp hp {0,1,, n 2}. Nn theo

nguyn l Dirichlet, H phi c t nht hai nh cng bc.

Nu H khng c nh c lp th bc ca cc nh ca H thuc tp hp {1, 2, , n 1 }. Nn

theo nguyn l Dirichlet, H phi c t nht hai nh cng bc.

25)

a) s2 < s3 < s1

s3 = aba < ab * * * * < s1 = ac

b) Mi v tr * c 3 cch chn. Do c 3* 3 *3 *3 = 81 chui.

26) S: 42580.

27) Ta c h thc qui cp 1 l s1 = 1.2.21 = 4 v sn = sn 1 + n(n + 1)2

n n 2 (1)

H thc qui thun nht sn = sn 1 n 2 (2) c nghim tng qut un = t n 1.

H thc qui (1) c mt nghim ring dng vn = (an2 + bn + c)2

n n 1

Th vn vo (1) v n gin cho 2n 1

, ta c

2(an2 + bn + c) = [a(n 1)2 + b(n 1)+ c ] + 2n(n + 1) n 2 .

Th n = 0, 1 v 2, ta c 3 phng trnh (a b c = 0, 2a + 2b + c = 4, 7a + 3b + c = 12)

Gii h, ta c ( a = 2, b = 2 v c = 4 ), ngha l un = (n2 n + 2)2n + 1 n 1

(1) c nghim tng qut sn = un + vn = t + (n2 n + 2)2n + 1 n 1

Do s1 = 4 nn t = 4 .Vy sn = (n2 n + 2)2n + 1 4 n 1.

28) PT c trng ca h thc qui thun nht l : k2 3k + 2 = 0 c nghim l k =1 , k= 2.

Do nghim tng qut ca h thc qui tuyn tnh thun nht l xn = A + 2nB.

V phi ca PT c dng P(n)rn vi P(n) l a thc bc nht, r = 1 v r l nghim n ca PT c

trng , nn nghim ring ca h thc qui tuyn tnh khng thun nht c dng

xn = n( Cn+ D). Thay vo h thc qui cho, ta c


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(n+1)[C(n+1) +D] 3n(Cn + D) +2(n -1)[C(n-1) +D] = n.

Ln lt cho n = 1 v n= 2 ta c C D = 1 v C D = 2, suy ra C = -1/2 v D = - 3/2.

Nghim tng qut ca h thc qui cho l xn = A + 2nB -

.

T x0 = 1 v x1 = 2 ta c A+ B = 1 v A + 2B = 4, suy ra A = - 2 v B = 3. Do nghim

ring tha u kin u cho l xn = -2 + 3. 2n -

29)

a) 1 2 2 0.n n nx x x

Phng trnh c trng

2 2 = 0

C hai ngim thc l 1 = 1, 1 = 2. Nghim tng qut l

xn = C1(1)n

+ C22n

b) 11 2 2 (6 5)2n

n n nx x x n (1) tha iu kin u x0 = 7, x1 = 4.

V 2 l nghim n ca phng trnh c trng nn (1) c mt nghim ring dng

xn = n(an + b)2n

Th vo (1) ta c

n(an + b)2n (n 1)[a(n 1) + b]2n 1 2(n 2)[a(n 2) + b]2n 2 = (6n 5)2

n 1

12an 10a + 6b = 12n 10

12a = 12, 10a + 6b = 10

a = 1, b = 0.

Vy mt nghim ring ca (1) l

xn = n22

n.

Nghim tng qut ca (1) l

xn = C1(1)n

+ C22n

+ n22

n

Th iu kin x0 = 7, x1 = 4 ta c


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1 2 1

1 2 2

7 4

2 2 4 3

C C C

C C C

Vy xn = 4(1)n

+ 3.2n + n

22

n

30)

a) S: an = c 3n +dn 3

n .

b)S: an = (2 +n )3n +

(n + 3) 3n

31)

a) an = ( A + n B) 3 n + (n 2) n 2 3 n

b) Tm s cc chui nh phn chiu di n cha chui con 00.

Gi an l s chui nh phn chiu di n cha chui con 00.

Ta c a0 = 0, a1 = 0.

Ta tnh an:

- TH1 : Nu bit u tin l bit 1 th c an 1 cch chn n 1 bit cn li. - TH2 : Nu bit u tin l bit 0 th c hai TH xy ra:

Bit th 2 l bit 1 : c an 2 cch chn n 2 bit cn li

Bit th 2 l bit 0 : c 2n 2 cch chn n 2 bit cn li ( cc bit ny chn 0 hay 1 u c)

Vy an = an 1 + an 2 + 2n 2

( n 2) (1).

H thc qui TTTN : an = an 1 + an 2 (2)

PTT : x2 x 1 = 0 c 2 nghim n l

Nghim tng qut ca (2) l an = A (

)

(

)

Ta tm mt nghim ring ca (1) di dng an = C2n. Thay vo (1) :

C2n = C2

n 1 + C2

n 2 + 2

n 2 4C = 2C + C + 1 C = 1.

Nghim TQ ca (1) l an = A (

)

(

)

+ 2n.

S dng K u : A + B + 1 = 0

A(

) (

) +2 = 0.


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A = -

, B = -

an =

(

)

(

)

+ 2n

32)

a) an = an-1 + 6an-2 c vit li an - an-1 - 6an-2 = 0 (1)

Phng trnh c trng ca (1) l x2 - x - 6 = 0 c 2 nghim l x = -2 v x = 3.

Nn nghim tng qut ca (1) l an = C1(-2)n + C23

n.

b) t fn = 10n(-2)n - 3(-2)

n-1 = (-2)

n(10n + 3/2).V -2 l 1 nghim ca phng trnh c trng nn

nghim ring c dng n(-2)n(An + B). (3)

Th (3) vo h thc ban u ta c:

n(-2)n(An + B) = (n-1)(-2)

n-1(A(n-1) + B) + 6(n-2)(-2)

n-2(A(n-2) + B) + (-2)

n(10n + 3/2) (4).

Th n = 2 vo (4), ta c:

2(-2)2(2A + B) = (-2)(A + B) + (-2)

2(10.2 + 3/2)

16A + 8B = -2A - 2B + 86 18A + 10B = 86 9A + 5B = 43 (5)

Th n = 1 vo (4), ta c:

(-2)(A + B) = 6(-1)(-2)-1

(B - A) + (-2)(10 + 3/2)

-2A - 2B = 3B - 3A - 23 A - 5B = -23 (6)

T (5) v (6) ta c h phng trnh

ny ta c: A = 2 v B = 5

Nh vy nghim tng qut ca h thc l:


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an = C1(-2)n + C23

n + n(-2)

n(2n + 5) (7)

Th iu kin u vo (7), ta c:

a0 = 8 = C1(-2)0 + C23

0 + 0(-2)

0(2.0 + 5)

C1 + C2 = 8 (8)

a1 = 5 = C1(-2)1 + C23

1 + 1(-2)

1(2.1 + 5)

-2C1 + 3C2 = 19 (9)

T (8) v (9) ta c h phng trnh:

C1 + C2 = 8

-2C1 + 3C2 = 19

Gii h phng trnh trn ta c C1 = 1 v C2 = 7.

Vy nghim ca h thc qui (1) l:

an = (-2)n + 7.3

n + n(-2)

n(2n + 5).

33)

a)Gi Pn l tng s tin c trong ti khon vo cui nm th n. Nh vy: - S tin c trong ti khon vo ngy u ca nm th nht s l: P0 = 100 triu - S tin c trong ti khon vo ngy cui nm ca nm th nht l: P1 = P0 + Li 1 + Li 2

Trong : Li 1 = 20% tng s tin c trong ti khon c nm = 0.2 * P0 Li 2 = 45% tng s tin c trong ti khon ca nm trc = 0.45 *0

Vy : P1 = P0 + 0.2 P0

- S tin c trong ti khon vo ngy cui nm th hai s l: P2= P1 + 0.2*P1 + 0.45*P0 Tng s tin c trong ti khon vo cui nm th n s l: Pn= Pn-1 + 0.2*Pn-1 + 0.45*Pn-2 = 0.45*Pn-2 +1.02*Pn-1

b)Gii h thc qui tuyn tnh thun nht vi P0 =100, P1 = 120 ta c

Pn = 250 3 50 3

3 2 3 10

n n

.


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34)

a)Gi Ln l s cch xp s xe my, xe p cho y n l - s cch xp cho y n l vi v tr u tin l xe p l Ln-1 - s cch xp cho y n l vi v tr u tin l xe my l Ln-2 Vy:

Ln = Ln-1 + Ln-2

b)Gii h thc qui vi iu kin u: L0 = 1, L1 = 1 ta c

1 1

1 1 5 1 1 5

2 25 5

n n

nL

35)

Gi s n-1 ng thng chia mt phng thnh xn-1 min.

ng thng th n ct n-1 ng thng cho trc ti n-1 giao im .

Trong :

- c n-2 an thng hu hn - c 2 an c mt u v hn Mi an thng ny phn min mt phng n i qua thnh 2 min .

Do vy s tng thm n min.

Vy: xn = xn-1 + n

Gii h thc qui tuyn tnh khng thun nht vi iu kin u x0 = 1, x1 = 2 ta c

( 1)1

2n

n nx

.

36) a) Kar(f) = Kar( x y z t) Kar x z t ) Kar( x y z t) Kar(x z t ) Kar( x y z t )

Kar(x y t)


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S = Kar(f)

S c 6 t bo ln T1 = x t , T2 = x y , T3 = y z t, T4 = x z t, T5 = x y z v T6 = y z t

b) Cc t bo ln phi chn l T1 v T2 .

S c 3 php ph nh sau (tt c u l cc php ph ti tiu) :

T5

T1 T2 T4 T6

T5 T3

T4

S = T1 T2 T4 T5 = T1 T2 T4 T6 = T1 T2 T5 T3

f c cc cng thc a thc ti tiu (u n gin ngang nhau) nh sau :

f(x,y,z,t) = x t x y x z t x y z = x t x y x z t y z t = x t x y x y z y z t


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37) 2013( t 2).

a)

x x x x

z t

z t

z t

z t

y y y y

S c 4 t bo ln x t, x y, x z t

b) S c 2 php ph ti tiu, suy ra f c hai cng thc a thc ti tiu l :

( , , , ) .

( , , , )

f x y z t xy x y t xzt

f x y z t xy x y t yzt

c)T Kar(f) suy ra dng ni ri chnh tc ca f l

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

38)

x x x x

z t

z t

z t

z t

y y y y


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S c 6 t bo ln T1 = x z, T2 = x t, T3 = z t, T4 = x y t , T5 = y z t v T6 = x y z

b) S c 3 php ph nh sau :

T1 T3 T4 (php ph ti tiu)

T5 T4 (php ph cha ti tiu) : loi

T6 (php ph ti tiu)

Suy ra f(x,y,z,t) = x z z t x y t ( y l cng thc a thc ti tiu ca f )

= x z z t y z t x y z ( b loi v phc tp hn cng thc trn )

39) Tm tt c cc cng thc a thc ti tiu ca hm Bool 4 bin sau:

( , , , ) ( ) ( ). f x y z t xyz y xz zt xy z t z

( , , , ) .f x y z t xyz x yz y zt xyz t xy z

V kar(f):

x x x x

z 1 t

z 2 3 t

z 4 5 t

z 6 7 8 t

y y y y

Cc t bo ln: x yz , x yt , yzt , xyt , xz , y zt .


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x x x x x x x x

z 1 t z 1 t

z 2 3 t z 2 3 t

z 4 5 t z 4 5 t

z 6 7 8 t z 6 7 8 t

y y y y y y y y

x yz x yt

x x x x x x x x

z 1 t z 1 t

z 2 3 t z 2 3 t

z 4 5 t z 4 5 t

z 6 7 8 t z 6 7 8 t

y y y y y y y y

yzt xyt

x x x x x x x x

z 1 t z 1 t

z 2 3 t z 2 3 t

z 4 5 t z 4 5 t

z 6 7 8 t z 6 7 8 t

y y y y y y y y

xz y zt

T bo ln nht thit phi chn: xz .

Cc cng thc a thc tng ng vi cc ph ti tiu gm cc t bo ln:


January 18, 2015

18

1

2

3

4

f xz yzt x yt (F )

f xz yzt xyz yzt (F )

f xz xyt x yz x yt (F )

f xz xyt x yz y zt (F )

So snh ta thy cng thc (F1) thc s n gin hn cc cng thc khc. Suy ra f c mt cng thc

a thc i tiu l

1

f xz yzt x yt (F )

40)

a)Biu Karnaugh ca f gm cc gch cho

Suy ra biu Karnaugh ca gm cc trng.

b)Dng ni ri chnh tc ( dng tuyn chun tc) ca .

t x x y z y z y z t y t

41)

a) Cc t bo ln: , , ,, , ,y t z y x zt x yt x y z x z t

b) S : C ba cng thc a thc ti tiu l ,

,

y t y z x zt x y z

y t y z x yt x y z

y t y z x yt x z t

42) G ng cu vi G.


January 18, 2015

19

f(u1) = v6, f(u2) =v3, f(u3) = v4, f(u4) = v5, f(u5) = v1, f(u6) = v2

1 2 3 4 5 6

1

2

3

4

5

6

0 1 0 1 0 0

1 0 1 0 0 1

0 1 0 1 0 0

1 0 1 0 1 0

0 0 0 1 0 1

0 1 0 0 1 0

G

u u u u u u

u

u

M u

u

u

u

6 3 4 5 1 2

6

3

' 4

5

1

2

0 1 0 1 0 0

1 0 1 0 0 1

0 1 0 1 0 0

1 0 1 0 1 0

0 0 0 1 0 1

0 1 0 0 1 0

G

v v v v v v

v

v

M v

v

v

v

MG = MG

43)


January 18, 2015

20

v 1 v 2 v 3 v 4 v 5 v 6 v 7 v 8 v9 v10 0 ( ,-) ( ,-) ( ,-) ( ,-) ( ,-) ( ,-) ( ,-) ( ,-) ( ,-)

- (1,v1)* ( ,-) ( ,-) (10,

v1)

( ,-) ( ,-) (6,v1) (3,v1) ( ,-)

- -

(5, v2) ( ,-) (10,

v1)

( ,-) ( ,-) (6,v1) (3,v1)* ( ,-)

- - (5,

v2)*

( ,-) (10,

v1)

( ,-) (9,v9) (5,v9) - (11,v9)

- - - (10,

v3)

(6,v3) (7,v3) (9,v9) (5,v9)* - (11,v9)

- - - (10,

v3)

(6,v3)* (7,v3) (9,v9) - - (11,v9)

- - - (10,

v3)

- (7,v3)* (7,v5)

- - (11,v9)

- - - (9,v6) - - (7,v5)* - - (11,v9)

- - - (9,v6)* - - - - - (10, v7)

- - - - - - - - - (10,

v7)*

44)

a b c d e f g z 0 ( ,-) ( ,-) ( ,-) ( ,-) ( ,-) ( ,-) ( ,-)

0* (4,a) (3,a) ( ,-) ( ,-) ( ,-) ( ,-) ( ,-)

- (4,a)

(3,a)* ( ,-) (10, c) ( ,-) ( ,-) ( ,-)

- (4,a)* - (9,b) (10, c)

( ,-) ( ,-) ( ,-)

- - - (9,b)*

(10, c) (14,d) (11,d) ( ,-)

- - - - (10, c) (14,d) (11,d) ( ,-)

- - - - -

(12,g) (11,d)*

(15,g)

- - - - - (12,g)* - (15,g)

- - - - - - - (15,g)*

ng i ngn nht t a n z l abdgz vi chiu di 15.

45)


January 18, 2015

21

s x y z t

0* (, - ) (, - ) (, - ) (, - )

- (10, s ) (5, s )* (, - ) (, - )

- (8, y ) - (14, y ) (7, y )*

- (8, y )* - (13, t ) -

- - - (9, x )* -

d(s, x) = 8. ng i : syx ; d(s,y) = 5. ng i: sy; d(s,z) = 9. ng i: syxz

d(s, t) = 7. ng i: syt.Tng t ta c bng sau vi th mi

s x y z t

0* (, - ) (, - ) (, - ) (, - )

- (10, s ) (5, s )* (, - ) (, - )

- (2, y )* - (14, y ) (7, y )

- - - (3, x )* (7, y )

- - - - (7, y )*

Tuy nhin kt qu by gi khng phi l ng i ngn nht. Chng hn trong ct y ta c

ng i sy vi chiu di 5. Tuy nhin ng i syxy c chiu di l

5 3+2 = 4 < 5.

46) a) Dng thut ton Dijkstra ta tm c ng i ngn nht t a n cc nh e ( di 6):

a b c d e f g h

0* (, ) (, ) (, ) (, ) (, ) (, ) (, )

(4,a) (, ) (, ) (, ) (, ) (6,a) (2,a)*

(3,h)* (, ) (,) (, ) (10, h) (4,h)

(8,b) (9,b) (, ) (5,b) (4,h)*

(8,b) (9,b) (, ) (5,b)*

(8,b) (9,b) (6,f)*


January 18, 2015

22

a b

h

f e

1

2

2

1

b) ng i ngn nht t nh a n nh d nhng phi i qua nh e gm cc ng i ngn nht

t a n e v ng i ngn nht t e n d. Dng thut ton Dijkstra tm c ng i ngn nht

t e n d ( di 4) nh sau:

e a b c d f g h

0* (, ) (, ) (, ) (, ) (, ) (, ) (, )

(, ) (, ) (1,e)* (5,e) (1, e) (, ) (, )

(, ) (6,c) (4,c) (1,e)* (, ) (,)

(, ) (3,f)* (4,c) (10,f) (9,f)

(7,b) (4,c)* (17,f) (4,h)

d

c

e

1

3

Suy ra ng i cn tm nh sau (c di l 6 + 4 = 10):


January 18, 2015

23

a b

h

f

d

c

e

1

2

2

1

1

3

47) a) G khng c ng Euler v khng c chu trnh Euler v G c 4 nh bc l l c( bc 5

),

e( bc 5 ), f ( bc 5 ) v g( bc 7 ) .

b) d, f, df , g, dg , c, dc , b, cb , e, be , a, ea ( trng s 2 ), h, gh ( trng s 5 ), i, hi ( trng s 0 ).

Trng s ca T l 9 + 1 + 2 + 3 + 1 + 2 + 5 + 0 = 23

c) a, b, c, d, e, f, g, h, i, hi ( trng s 0 ), dg , cd , ae ( trng s 2 ), bc , ce , af , gh ( trng s 5 ).


January 18, 2015

24

Trng s ca Z l 0 + 1 + 2 + 2 + 3 + 4 + 5 + 5 = 22

48) a) Ma trn khang cch

0 3 2 5

3 0 1 4

2 0 2 1

5 1 2 0 3

4 3 0 2

1 2 0

D

b)Bng sau y lu cc bc chy ca thut tan

1 2 3 4 5 6

0 ( ( ( ( ( - (3, 1) (2,1)* (5,1) ( ( - (3,1)* - (4,3) ( (3,3) - - - (4,3) (7,2) (3,3)*

- - - (4,3)* (5,6) -

- - - - (5,6)* -

- (3,1) (2,1) (4,3) (5,6) (3,3)


January 18, 2015

25

th biu din ca cy bao trm l

2

3 2 1

2

Trng lng ca cy T l 3+2+1+2+2 = 8.

49) a) G c t nht mt chu trnh Hamilton l abcpqvursta .

b) qr (1), cq (8), st (8), bc (7), bv (6), tu (6), ab (5), at (4) v cp (4).

Trng s ca T l 1 + 8 + 8 + 7 + 6 + 6 + 5 + 4 + 4 = 49

c) b, u, bu (9), r, ru (2), s, rs (2), a, au (3), v, av (2), c, cv (2), p, cp (4), q, pq (3), t v at

(4).

1 3

6 4

2 5


January 18, 2015

26

Trng s ca Z l 9 + 2 + 2 + 3 + 2 + 2 + 4 + 3 + 4 = 31 .

50) a) G c 4 nh bc l l a, d, h (u bc 3) v e (bc 5) nn G khng c chu trnh

Euler v khng c ng Euler .

b)

V b c d e f g h T

a (1,a) (,) (,) (,) (4,a) (,) (8,a)

b (7,b) (,) (4,b) (3,b)* (,) (8,a) ab

f (7,b) (,) (4,b)* (10,f) (6,f ) bf

e (5,e)* (9,e) (10,f) (6,f ) be

c (8,c) (10,f) (6,f)* ec

h (8,c)* (8,h) hf

d (8,h)* cd

g (1,a) (5,e) (8,c) (4,b) (3,b) (8,h) (6,f ) gh


January 18, 2015

27

Trng s ca T l 1 + 2 + 3 + 2 + 3 + 1 + 3 = 15 .

51) Trong tp hp cc hm Boole ca 5 bin c 25 = 32 t ti tiu.

S cch chn 6 t ti tiu trong 32 t ti tiu l = 906102

52) th Kn c n(n 1 )/ 2 cnh.

Do G c n(n 1 )/ 4 cnh. Suy ra n chia ht cho 4 hoc n 1 chia ht cho 4.


January 18, 2015

28

53)

K4< K5


January 18, 2015

29

Ta CM qui np theo chiu cao h BT l 2h

- R rng bt ng thc ng khi h = 1( lc ny l = 2).

- Gi s BT ng vi mi cy c chiu cao h 1 . Xt T l cy c chiu cao h. Gi l1, l2 l s nt l ca cy con T1, T2 l cy con bn tri v bn phi ca nt gc. rng T1 v

T2 l cc cy c chiu cao h 1 nn theo gi thit qui np ta c

l =l1 +l2 2. 2h-1

= 2h.

Vy h log2l h log2(N + 1) h .

b) Do cy l cy cn bng nn cc nt mc h 2 u l nt trong. V vy tng s nt mc h 1 l 2h -1 . Tng s nt l mc h bng 2 ln s nt trong mc h 1 nn

2h 1

< N + 1 2h h 1 < log2(N + 1) h h =

Gii thch :

N + 1 = l= lh + lh 1 = 2Nh 1 + lh 1 = Nh 1 +Nh 1 + lh 1 = Nh 1+ 2h 1

> 2h 1

.

55) .

R khng phi l th t ton phn v (1, 2) v (2, 1) khng so snh c vi nhau.

nh ngha quan h R trn 2 bi ( a, b) R (c, d) khi v ch khi a < c hoc a = c v

b d. R rng R l th t ton phn trn 2 .

Gi s A l mt tp con khc rng ca 2 . Khi y tp cc thnh phn th nht ca nhng phn t

trong A l mt tp con khc rng ca nn c phn t b nht l m. Khi tp con cc thnh

phn th hai ca nhng cp trong A vi thnh phn th nht l m s c phn t b nht l n . R

rng (m ,n ) l phn t b nht ca A.

56) V 2310 = 2*3*5*7*11 nn mi c 1 ca 2310 l tch ca cc s nguyn t thuc mt tp

con ca S ={2, 3, 5, 7, 11}. Ta c th ng nht c ny vi dy s nguyn a1a2 am trong

2 a1 < a2


January 18, 2015

30

59)

7 5

7 6

4 1 11

D

2 3

3 4

1 2 3

Q

1

7 5

7 6

4 1 9

D

1

2 3

3 4

1 2 1

Q

2

7 5 13

7 6

4 1 8 7

D

2

2 3 2

3 4

1 2 2 2

Q

3

7 5 13

7 6

4 1 8 7

D

3

2 3 2

3 4

1 2 2 2

Q

4

17 7 5 13

10 7 7 6

4 1 8 7

D

4

2 2 3 2

4 4 3 4

1 2 2 2

Q

60)

a) Ta CM bng phn chng. Gi s G khng lin thng. Khi G c t nht hai thnh phn lin thng, trong phi tn ti thnh phn lin thng H vi < n/2 nh. Trong H bc ca

mi nh <

, tri gi thit.

b) Theo cu a) th G lin thng. Gi G l th thu c t G bng cch b i mt nh. Nu

G khng lin thng th tn ti mt thnh phn lin thng H c

nh. Trong H mi

nh P c bc

. Khi trong G nh P c bc

. Tri gi thit.

61)

R rng ta ch cn CM cho G lin thng l . Ta CM bng phn chng.

Gi s G lin thng v G - e c t nht 3 thnh phn lin thng. Tr li cnh e cho G. Ta thy e ch

c th ni nhiu lm l 2 trong 3 thnh phn lin thng ca G e vi nhau, v do G c t nht

hai thnh phn lin thng. Tri gi thit G lin thng.

62) Ta CM qui np theo s cnh m ca G.

Vi m = 0 th khng nh hin nhin ng ( Mi nh l mt thnh phn lin thng).


January 18, 2015

31

Gi s kt lun bi tan ng cho m = k cnh. Xt G ty c k+ 1 cnh. B mt cnh ra khi G

ta thu c G c k cnh. Trong G c t nht n k thnh phn lin thng. Theo Cu 51 s thnh

phn lin thng trong G khng vt qu 1 so vi G. Do s thnh phn lin thng trong G

khng t hn n k 1 = n ( k + 1). Vy kt lun ng cho m = k+1.

63)

a) (a + b)2 + c : + + a b 2 c

a b + 2 c +

a + b2 + c : + + a b 2 c

a b 2 + c +

(a + b)2 = a

2 + b

2 + 2ab : + a b 2 = + + a 2 b 2 * 2 * ab

a b + 2 = a 2 b 2 + 2 a b * * +

b) (a + b)2/ (c d ) [(x + y)

2 (x y )2]/(x*y)

64) a)V cy biu din ca biu thc

b)duyt cy theo tin th t ta c biu thc theo k php Ba Lan:

+ + + x^ x 2 ^ x 3 ^ x 4 .

65) a) 3 / 2 / ^ x y z x y z

+

+

+ ^

^

^ x x

x

x

2

3

4


January 18, 2015

32

+

^

*

y

y zx3

/

2

z/

x

b) Biu thc trn c vit theo k php thng thng nh sau: z

y x y3x .

z 2

Toán rời rạc 2015

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