28601026 tom tat bai giang toan roi rac nguyen ngoc trung

TRNG I HC S PHM TP.HCMKHOA TON TIN HC

TM TT BI GING

Mn

TON RI RC

Ging vin bin son: Nguyn Ngc Trung

TP.HCM 9.2006

MC LC

Chng 1. Mnh ........................................................................................ 31.1 Mnh - Tnh cht.......................................................................... 3

1.1.1 Mnh v cc php ton mnh ............................................ 31.1.2 Dng mnh ............................................................................ 51.1.3 Cc quy tc suy din .................................................................. 7

1.2 V t - Lng t.............................................................................. 111.3 Nguyn l quy np.......................................................................... 14

Chng 2. Php m .................................................................................... 152.1 Tp hp Tnh cht ........................................................................ 152.2 nh x ............................................................................................ 172.3 Gii tch t hp ............................................................................... 18

2.3.1 Cc nguyn l c bn ca php m: ....................................... 182.3.2 Gii tch t hp ........................................................................ 192.3.3 Nguyn l Dirichlet. (nguyn l chung b cu) ...................... 23

Chng 3. Quan h ...................................................................................... 243.1 Quan h .......................................................................................... 243.2 Quan h tng ng ..................................................................... 253.3 Quan h th t - Biu Hasse ...................................................... 26

Chng 4. i s Boole ............................................................................... 304.1 i s Boole: nh ngha Tnh cht ............................................. 304.2 Hm Boole Dng ni ri chnh tc ............................................... 364.3 Bi ton mch in Mng cc cng.............................................. 424.4 Tm cng thc a thc ti tiu Phng php Karnaugh ............... 44

TI LIU THAM KHO.......................................................................... 51

Tm tt bi ging Ton ri rc Trng HSP TP.HCM

Trang 3

1 Chng 1. Mnh

1.1 Mnh - Tnh cht 1.1.1 Mnh v cc php ton mnh

nh ngha. Mnh l cc khng nh c gi tr chn l xc nh (ng hoc sai, nhng khng th va ng, va sai). Cc mnh ng c ni l c chn tr ng, cc mnh sai c ni l c chn tr sai.

V d: - Cc khng nh sau l mnh :. 1 + 2 = 5 l mnh sai.. 10 l s chn l mnh ng.

- Cc khng nh sau khng phi l mnh :. Ti i hc. n l s nguyn t

K hiu: Ta thng k hiu cc mnh bng cc ch ci in hoa: P, Q, R, v chn tr ng (sai) c k hiu bi 1 (0).

Cc php ton mnh :

x Php ph nh: ph nh ca mnh P c hiu bi P (c l khng P hoc ph nh ca P. Chn tr ca P l 0 nu chn tr ca P l mt v ngc li.

VD. P = 3 l s nguyn t l mnh ng. Do mnh P = 3 khngl s nguyn t l mnh sai.Bng sau gi l bng chn tr ca php ph nh:

P P0 11 0

x Php ni lin: Mnh ni lin ca hai mnh P v Q c k hiu bi P Q (c l P v Q. Chn tr ca P Q l 1 nu c P ln Q u c chn tr l 1, trong cc trng hp khc P Q c chn tr l 0.

VD. P = Hm nay tri p v Q = Trn bng hp dn. Khi ta c mnh ni lin ca P v Q l: P Q = Hm nay tri p v trn bng hp dn. Mnh ni lin ny s ng nu nh c hai mnh P v Q u


Trang 4

ng. Ngc li nu c mt trong hai mnh trn sai hoc c hai cng sai th mnh ni lin s l sai.

Bng chn tr ca php ni lin:

P Q P Q0 0 00 1 01 0 01 1 1

x Php ni ri: Mnh ni ri ca hai mnh P v Q c k hiu bi P Q (c l P hay Q. Chn tr ca P Q l 0 nu c P ln Q u c chn tr l 0, trong cc trng hp khc P Q c chn tr l 0.

VD. P = An l ca s, P = An l gio vin. Khi ta c mnh ni ri ca P v Q l P Q = An l ca s hay An l gio vin. Mnh ni lin ny s ng nu nh mt trong hai mnh trn l ng hoc c hai mnh trn u ng. Nu c hai mnh P v Q u sai th P Q s sai.

Bng chn tr ca php ni ri:

P Q P Q0 0 00 1 11 0 11 1 1

x Php ko theo: Mnh P ko theo Q c k hiu l P Qo . xc nh chn tr ca mnh P ko theo Q ta xt v d sau: P = An trng s, Q = An mua xe my, khi mnh P kp theo Q s l Nu An trng s thAn s mua xe my. Ta c cc trng hp sau y: An trng s v anh ta mua xe my: hin nhin mnh P Qo l

ng. An trng s nhng anh ta khng mua xe my: r rng mnh

P Qo l sai. An khng trng s nhng anh ta vn mua xe my: mnh P Qo vn

ng. An khng trng s v anh ta khng mua xe my: mnh P Qo ng.


Trang 5

Bng chn tr ca php ko theo:

P Q P Qo0 0 10 1 11 0 01 1 1

x Php ko theo hai chiu: Mnh P ko theo Q v ngc li c k hiu bi P Ql l mnh c chn tr ng khi P v Q c chn tr ging nhau (cng ng hoc cng sai) v c chn tr sai khi P v Q c chn tr khc nhau.

VD. P = An hc gii, Q = An c im cao. Khi mnh P Ql = Nu An hc gii th An s c im cao v ngc li.

Bng chn tr ca php ko theo hai chiu nh sau:

P Q P Ql0 0 10 1 01 0 01 1 1

1.1.2 Dng mnh

nh ngha. Dng mnh c xy dng t:- Cc mnh (l cc hng mnh )- Cc bin mnh p, q, r, c th ly gi tr l cc mnh no .- Cc php ton trn mnh , v cc du ngoc ( ).

V d. ( , , )E p q r p q r p o l mt dng mnh trong p, q, r l cc bin mnh .

rng ta c th xy dng nhiu dng mnh phc tp t cc dng mnh n gin hn bng cch s dng cc php ton mnh kt hp chng li. Chng hn nh dng mnh E(p,q,r) trn c kt ni t hai dng mnh

1( , , )E p q r p q v 2 ( , , )E p q r r p o bng php ton ni ri ( ).

Mi dng mnh s c s c mt bng chn tr xc nh trong mi dng cho bit chn tr ca dng mnh theo cc chn tr c th ca cc bin.


Trang 6

V d. ( , , )E p q r p q r p o c bng chn tr nh sau:

p q r r p q r p o E(p,q,r)0 0 0 1 0 0 00 0 1 0 0 1 10 1 0 1 1 0 10 1 1 0 1 1 11 0 0 1 1 1 11 0 1 0 1 1 11 1 0 1 1 1 11 1 1 0 1 1 1

nh ngha. Hai dng mnh E v F c ni l tng ng logic nu chng c cng bng chn tr. Khi y ta vit E F .

Ch rng nu E v F tng ng logic th dng mnh P Ql lun ly gi tr l 1 d cc bin c ly bt c gi tr no.

nh ngha. i. Mt dng mnh c gi l mt hng ng nu n lun lun ly chn tr

1ii. Mt dng mnh c gi l mt hng sai nu n lun ly chn tr 0.

Mnh . Hai dng mnh E v F tng ng logic khi v ch khi P Ql l mt hng ng.

nh ngha. Dng mnh F c ni l h qu logic ca dng mnh E nu E Fo l mt hng ng. Khi y ta vit E F .

Cc quy lut logic:

nh l. Vi p, q, r l cc bin mnh , 1 l hng ng, 0 l hng sai, ta c cc tng ng logic:

i. Ph nh ca ph nh:p p

ii. Quy tc De Morgan: p q p q

v p q p q iii. Lut ko theo:

p q p qo iv. Lut giao hon:

p q q p


Trang 7

v p q q p v. Lut phn phi:

p q r p q p r v p q r p q p r

vi. Lut kt hp: p q r p q r

v p q r p q r vii. Lut ly ng:

p p p v p p p

viii. Lut trung ha:1p p

v 0p p ix. Phn t b:

0p p v 1p p

x. Lut thng tr:0 0p

v 1 1p xi. Lut hp th:

p p q p v p p q p

V d. s dng cc quy lut logic chng minh rng dng mnh ( , )E p q p p q q o o l hng ng.

Gii. E(p,q) p p q q o p p p q 0 p q q o p q q o p q q 1p 1

1.1.3 Cc quy tc suy din

Trong chng minh ton hc, xut pht t mt s khng nh ng (mnh ng) gi l tin , ta s p dng cc quy tc suy din suy ra chn l ca mt khng nh q m ta gi l kt lun. Ni cch khc, ta s phi tm cch chng minh dng mnh 1 2 np p p q o! l mt hng ng, trong 1 2, ,..., ,np p p q l cc dng mnh theo mt s bin logic no .


Trang 8

V d. Gi s ta c cc tin :p1: Nu An hc chm th An t im caop2: Nu An khng hay i chi th An hc chmp3: An khng c im cao

Ta mun dng cc quy tc suy din suy ra kt lun: q = An hay i chi. Mun vy, ta phi tru tng ha cc mnh nguyn thy: An hc chm, An hay i chi v An c im cao thnh cc bin mnh p, q, r. Nh vy cc tin by gi tr thnh cc dng mnh :

1p p r o2p q p o3p r

Ta phi chng minh dng mnh sau l mt hng ng: p r q p r qo o o

Ta c th chng minh iu ny bng cch lp bng chn tr ca dng mnh trn. Tuy nhin cch ny s gp rt nhiu kh khn khi cc bin mnh ln (s dng ca bng chn tr bng 2n, vi n l s bin mnh ). Mt phng php khc l s dng cc quy tc suy din chia bi ton ra thnh nhiu bc nh, ngha l t cc tin ta suy ra mt s kt lun trung gian trc khi p dng cc quy tc suy din suy ra kt lun. tin ta m hnh ha php suy din thnh s nh sau:

1

2

n

pp

p

q?

#

Sau y l mt s quy tc suy din thng dng m chn l ca n c th c kim tra d dng bng cch lp bng chn tr.

Quy tc Modus Ponens (Phng php khng nh):

Quy tc ny c th hin bi hng ng: p q p qo o hoc di dng s :

p qp

q

o

?V d. Nu An hc chm th An s c im cao, m An hc chm. Suy ra An c im cao.


Trang 9

Tam on lun (Syllogism).

Quy tc ny c th hin bng hng ng: p q q r p ro o o o hay di dng m hnh:

p qq r

p r

oo

? o

V d. Nu An khng hay i chi th An hc chm v nu An hc chm th An s c im cao. Suy ra nu An khng hay i chi th An s c im cao.

Quy tc Modus Tollens (phng php ph nh)

Quy tc ny c th hin bi hng ng: p q q po o hay di dng m hnh:

p qq

p

o

?

V d. Nu tri ma th ng t m ng khng t. Suy ra tri khng ma.

Quy tc mu thun (chng minh bng phn chng)Quy tc ny da trn tng ng logic sau:

1 2 1 2 0n np p p q p p p q o o ! !

V d. Hy s dng phng php phn chng cho chng minh sau:p r

p qq s

r s

o oo

? o- Trc ht, ta ly ph nh ca kt lun: r s r s r s o - Sau ta s thm vo cc tin hai gi thit ph r v s tm cch

chng minh suy lun sau l ng:


Trang 10

0

p rp q

q srs

o oo

?

Cc bc suy lun s c thc hin nh sau:

p q oq so

p s? o (Tam on lun)m s

p? (PP ph nh)m p ro

r? (PP khng nh)Kt lun r cng vi gi thit ph r cho ta 0r r . Do theo

phng php phn chng, chng minh ban u l ng.

Quy tc chng minh theo trng hp:Quy tc ny c th hin bng hng ng sau:

p r q r p q ro o o o ngha ca quy tc ny l nu mt gi thit c th tc ra thnh hai trng hp p ng hay q ng, v ta chng minh c ring r cho tng trng hp l kt lun r ng, khi y r cng ng trong c hai trng hp.

V d. chng minh rng f(n) = n(n+1) lun chia ht cho 2 vi mi s t nhin n, ta xt hai trng hp l n chn, n l v thy rng trong c hai trng hp f(n) lun chia ht cho 2. Vy ta rt ra kt lun cn chng minh l f(n) lun chia ht cho 2 vi mi s t nhin n.

Trn y l mt s quy tc suy din ta thng dng trong cc qu trnh suy lun. Sau y ta s xt mt v d c th c s dng kt hp nhiu quy tc suy din:

V d. Kim tra suy lun sau ng hay sai: Nu ngh s Vn Ba khng trnh din hay s v bn ra t hn 50 v th m din s b hy b v ng bu s rt bun. Nu m din b hy b th phi tr li v cho ngi xem. Nhng tin v khng c tr li cho ngi xem. Vy ngh s Vn Ba trnh din.

kim tra suy lun trn, ta thay cc mnh nguyn thy bng cc bin mnh :

p: ngh s Vn Ba trnh din


Trang 11

q: s v bn ra t hn 50 vr: m din b hy bs: ng Bu rt bunt: tr tin v li cho ngi xem

T , suy lun ban u c th m hnh nh sau: p q r s

r tt

o o

p?

Suy lun trn c th c thc hin theo cc bc sau: p q r s o ( tin ) r s r o (hng ng)r to (tin ) p q t? o (tam on lun m rng)

m t (tin )p q (phng php ph nh v lut De

Morgan)m p q p o (hng ng)

p? (phng php khng nh)Vy suy lun ban u l chnh xc.

1.2 V t - Lng t

nh ngha. Mt v t l mt khng nh p(x,y, ) trong c cha mt s bin x,y, ly gi tr trong nhng tp hp cho trc A, B, sao cho:

i. Bn thn p(x,y,) khng phi l mnh ii. Nu thay x, y, bng nhng a A , b B , ta s c mt mnh

p(a,b,) ngha l chn tr ca n hon ton xc nh. Cc bin x, y, c ni l bin t do ca v t.

V d. a. p(n) = n l s nguyn t l mt v t theo bin t do n` , vi n = 2, 11,

13 ta c cc mnh ng p(2), p(11), p(13) cn vi n = 4, 10, 20 ta c cc mnh sai p(4), p(10), p(20).

b. q(x,y) = x + y l s l l v t theo 2 bin t do ,x y] , chng hn q(2,5) l mt mnh ng, q(-3, -7) l mt mnh sai,


Trang 12

nh ngha. Cho trc cc v t p(x), q(x) theo mt bin x A . Khi :i. Ph nh ca p, k hiu l p l v t theo bin x m khi thay x bng mt

phn t a c nh ca A th ta c mnh p a .ii. Pho ni lin (tng ng ni ri, ko theo, ) ca p v q, k hiu bi p q

( p q , p qo , ) l v t theo bin x m khi thay x bng mt phn t a c nh ca A th ta c mnh p a q a ( ) ( ), ( ) ( ),...p a q a p a q a o

V d: p(x) = x l s nguyn t, q(x) = x l s chn. Khi ta c: Ph nh ca p: ( )p x = x khng l s nguyn t Php ni lin ca p v q: ( )p q x = x l s nguyn t va x l s

chn . . .

nh ngha. Cc mnh , ( )x A p x v , ( )x A p x c gi l lng t ha ca v t p(x) bi lng t ph dng ( ) v lng t tn ti ( ).

Ch :a. Trong cc mnh lng t ha, bin x khng cn l bin t do na, ta ni

n b buc bi cc lng t hay .b. Mnh , ( )x A p x s ng nu thay x bng gi tr a bt k nhng c nh

trong A, p(a) lun l mnh ng.c. Mnh , ( )x A p x s ng nu c mt gi tr a trong A sao cho p(a) l

mnh ng.

Xt mt v t p(x,y) theo hai bin ,x A y B . Khi y nu thay x bng mt phn t c nh nhng ty a A , ta s c mt v t p(a,y) theo bin y. Khi ta c th lng t ha n theo bin y v c hai mnh sau: , ( , )y B p a y v , ( , )y B p a y . Do x c thay bng mt phn t c nh nhng ty a ca A, nn ta c hai v t sau y l hai v t theo bin x A : , ( , )y B p x y v , ( , )y B p x y . Tip tc lng t ha hai v t trn theo bin x, ta c 4 mnh sau y:

, , ( , )x A y B p x y , , ( , )x A y B p x y , , ( , )x A y B p x y , , ( , )x A y B p x y

Tng t ta cng c th lng t ha p(x,y) theo bin x trc ri bin y sau c 4 mnh na:

, , ( , )y B x A p x y , , ( , )y B x A p x y , , ( , )y B x A p x y , , ( , )y B x A p x y


Trang 13

Cu hi t ra lc ny l liu th t lng t ha c quan trng hay khng? Ni cch khc, cc mnh tng ng c tng ng logic vi nhau khng? nh l sau s cp n vn ny

nh l. Nu p(x,y) l mt v t theo 2 bin x, y th ta c cc iu sau:i. , , ( , )x A y B p x y , , ( , )y B x A p x y ii. , , ( , )x A y B p x y , , ( , )y B x A p x y iii. , , ( , )x A y B p x y , , ( , )y B x A p x y

Chng minh. Xem ti liu tham kho [1].

Cng tng t cch lm nh trn, ta c th m rng d dng cho cc v t theo nhiu bin t do. c bit, ta c:

Mnh . Trong mt mnh lng t ha t mt v t theo nhiu bin c lp, nu ta hon v hai lng t ng cnh nhau th:

i. Mnh mi vn cn tng ng logic vi mnh c nu hai lng t ny cng loi.

ii. Mnh mi s l h qu logic ca mnh c nu hai lng t trc khi hon v c dng

nh l sau s cho chng ta bit cch ly ph nh ca mt mnh lng t ha:

nh l. Ph nh ca mt mnh lng t ha t v t p(x,y, ) c c bng cch thay lng t bi lng t , thay lng t bng lng t v thay v t p(x,y,) bng ph nh p(x,y,).

V d. Ph nh ca mnh ,x y ` ` , x + y l s chn l mnh , ,x y ` ` x+y khng l s chn.

Quy tc c bit ha ph dng:Nu mt mnh ng c dng lng t ha, trong mt bin x A b buc bi lng t ph dng , khi nu thay x bng a A ta s c mt mnh ng.

V d. Ta c mnh ng sau: , ( 1) 2n n n ` # , khi nu thay n bng s t nhin bt k, chng hn n = 5, khng cn kim tra li, ta cng chc chn rng mnh 5*(5+1)#2 l mnh ng.


Trang 14

1.3 Nguyn l quy np

Nguyn l quy np: Mnh , ( )n p n ` l h qu logic ca > @(0) , ( ) ( 1)p n p n p n o `

Nguyn l quy np c c th ha thnh phng php chng minh quy np nh sau: Gi s ta phi chng minh mnh : , ( )n p n ` , khi ta s thc hin cc bc sau:

Bc 1. Kim tra p(0) l mnh ng (trong thc t, ta c th bt u bng gi tr nh nht c th c ca n, khng nht thit phi bt u t 0)

Bc 2. Vi n bt k, gi s p(n) l mnh ng, ta s phi chng minh p(n+1) cng l mnh ng.

V d. chng minh n ` , ( 1)0 1 ...2

n nn ,ta xt v t p(n) =

( 1)0 1 ...2

n nn . Ta c:

p(0) = 0.102

, y r rng l mt mnh ng.

Xt n l s t nhin bt k, gi s p(n) l mnh ng, ngha l ta c:( 1)0 1 ...

2n nn

ta s chng minh p(n+1) cng ng. Tht vy, ta c: > @1 ( 1) 1( 1)0 1 ... ( 1) ( 1)

2 2n nn nn n n

suy ra p(n+1) l mnh ng.

Theo nguyn l quy np ta c iu phi chng minh.


Trang 15

2 Chng 2. Php m

2.1 Tp hp Tnh cht

Trong chng trc, chng ta mt vi ln s dng khi nim tp hp trong mt s v d, c bit l trong nh ngha ca cc lng t. Trong chng ny ta s ni r hn v khi nim ny.

Trong ton hc, tp hp l mt khi nim c bn, khng th nh ngha thng qua cc i tng khc c. Mt cch trc quan, tp hp l mt khi nim ch cc i tng c nhm li theo mt tnh cht no . Nu a l i tng ca tp hp A, ta vit a A , nu khng, ta vit a A .

rng t tnh cht c hiu theo mt ngha rng ri. Thng thng n s c biu hin bi mt v t p(x) theo mt bin x U . Khi y tp hp c k hiu bi:

^ `( )A x U p x U c gi l tp v tr.

V d. 1. ^ `2A x x ` # , y l tp hp cc s t nhin chn.2. ^ `2 5A x x ] , y l tp hp cc s nguyn c bnh phng nh hn 5.

Tp hp cn c th c biu din bng cch lit k cc phn t ca n. Chng hn nh tp hp trong mc 2 ca v d trn c th c vit l: A = {-2,-1,0,1,2}.

Tp hp khng cha bt c phn t no gi l tp hp rng v c k hiu l .

Xt hai tp hp A v B trong tp v tr U. Ta ni A l tp hp con ca B (hay A cha trong B) nu: ,x U x A x B . Ta c th hiu n gin l bt c phn t no nm trong A cng nm trong B.

Ta c th s dng cc php ton trn mnh v v t nh ngha cc php ton trn tp hp nh php hp ( ), php giao ( ) v phn b theo nh ngha sau y:

nh ngha. Gi s A, B l hai tp hp con ca tp hp v tr U. Khi y:^ `^ `

^ `

( ) ( )

( ) ( )

A B x U x A x B

A B x U x A x B

A x U x A


Trang 16

nh l sau s gii thiu cc tnh cht ca cc php ton trn tp hp:

nh l. A, B, C l cc tp con ty ca U, ta c:i. Tnh giao hon:

=A B B AA B B A

ii. Tnh kt hp:

A B C A B C

A B C A B C

iii. Lut De Morgan:A B A B

A B A B

iv. Tnh phn phi:

A B C A B A C

A B C A B A C

v. Phn t trung ha:A AA U A

vi. Phn b:A A UA A

vii. Tnh thng tr:A U UA

viii. Tnh ly ng:A A AA A A

ix. Lut hp th:

A A B A

A A B A

Chng minh. Cc tnh cht trn c suy ra t nh ngha v cc quy lut logic.

Ch . Ta c th ly hp hoc giao ca nhiu tp hp v k hiu nh sau:

1 21

...n

i ni

A A A A

*v 1 2

1

...n

i ni

A A A A


Trang 17

2.2 nh x

nh ngha.i. Mt nh x f t tp hp A vo tp hp B l php tng ng lin kt mi phn

t x ca A vi mt phn t duy nht y ca B m ta k kiu l f(x) v gi l nh ca x bi f. Ta vit:

:( )

f A Bx f xo6

ii. Hai nh x f v g t A vo B c ni l bng nhau nu: , ( ) ( )x A f x g x

nh ngha. i. Nu E l mt tp hp con ca A th nh ca E bi f l tp hp:

^ ` ^ `( ) , ( ) ( )f E y B x A y f x f x x E ii. Nu F l mt tp hp con ca B th nh ngc ca F qua nh x f l tp

hp:^ `1( ) ( )f F x A f x F

Ch :1. Nu y B ta vit ^ ` 1 1( )f y f y 2. Nu 1( )f y th y khng nm trong nh f(A) ca A.3. Nu 1( ) {x}f y th x l phn t duy nht c nh l y.

nh ngha. Gi f l mt nh x t tp A v tp B. Khi y ta ni:i. f l ton nh nu f(A)=Bii. f l n nh nu hai phn t khc nhau bt k ca A c nh khc nhau,

ngha l:, ' , ' ( ) ( ')x x A x x f x f x z z

iii. f l song nh nu n va n nh va ton nh.

Xt f l mt song nh t A ln B, khi vi mi y B ty , s c duy nht mt phn t x A sao cho f(x)=y. Nh th, tng ng y x6 l mt nh x t B vo A m ta k hiu l 1f v gi l nh x ngc ca nh x f. Ta c:

1( ) ( )f y x f x y .

Ta c tnh cht ca nh x ngc nh sau:1

1

( ( )) ,( ( )) ,

f f y y y Bf f x x x A

nh ngha. Cho hai nh x: :f A Bo v : 'g B Co , trong 'B B . nh x hp h l nh x t A vo C c xc nh bi:


Trang 18

:( ) ( ( ))

h A Cx h x g f xo

6Ta k hiu: h f g D .

Cc tnh cht ca nh x:

nh l. Cho f l mt nh x t tp A vo tp B, E1 v E2 l hai tp con ty ca A, F1 v F2 l hai tp con ty ca B. Ta c:

i. 1 2 1 2( ) ( ) ( )f E E f E f E ii. 1 2 1 2( ) ( ) ( )f E E f E f E iii. 1 11 2 1 2( ) ( ) ( )f F F f F f F iv. 1 11 2 1 2( ) ( ) ( )f F F f F f F

2.3 Gii tch t hp

Php m cc phn t ca mt tp hp A chnh l tm mt song nh gia A v mt tp con {1,2,,n} ca N. Nu A hu hn th n chnh l s phn t ca n. nh ngha sau y s cp n vn ny:

nh ngha. i. Mt tp hp A c ni l hu hn v c n phn t nu tn ti mt song nh

gia A v tp hp con {1,2,...,n} ca N. Khi ta vit A n .ii. Nu A khng hu hn, ta ni A v hn.

2.3.1 Cc nguyn l c bn ca php m:

Nguyn l cng: Nu mt cng vic c th c thc hin bng mt trong n cch loi tr ln nhau: k1, k2, , kn. Trong thc hin theo cch ki li c ti phng n khc nhau (i=1..n). Khi tng s phng n thc hin cng vic ban u l: t1 + t2 + + tn.

V d: i t Tp.HCM ra H Ni c 3 cch: i t, i tu ha hoc i my bay. i bng t c 3 cch: i taxi, i xe , thu xe ring. i bng tu ha c 2 cch: i bng tu nhanh v i bng tu bnh thng. i bng my bay cng c hai cch: i bng Vietnam airline hoc i bng Pacific airline. Nh vy tng cng c 3 + 2 + 2 = 7 cch i t Tp.HCM ra H Ni.


Trang 19

Nguyn l cng m rng:i. Cho A v B l hai tp hp bt k, khi ta c:

A B A B A B ii. Cho A1, A2, , An l n tp hp bt k. Ta c:

11 2 1 2... ... ( 1)

nn nA A A N N N

trong Nk l tng s phn t ca cc tt c cc phn giao ca k tp bt k trong s n tp ban u.

V d. Vi 3 tp A, B, C (n=3) theo nguyn l cng tng qut ta c cng thc sau: A B C A B C A B B C C A A B C

Nguyn l nhn: Nu mt cng vic phi thc hin c theo n giai on lin tip nhau: k1, k2, , kn. Mi giai on ki c ti cch thc hin. Nh vy s c t1.t2tncch khc nhau thc hin cng vic ban u.

V d. Mt ngi c 3 ci o s-mi khc nhau, 2 ci qun di khc nhau v 4 i giy khc nhau. Nh vy c tt c 3.2.4 = 24 b trang phc khc nhau ngi ny la chn khi mc.

nh ngha. Tch -cc ca 2 tp hp A, B k hiu l A Bu l tp hp tt c cc cp th t (a,b) vi a A v b B trong hai cp (a,b) v (a,b) c ni l bng nhau nu v ch nu a=a v b=b.

nh l. Nu A v B l hai tp hp hu hn th A Bu cng l tp hp hu hn v ta c:

.A B A Bu

2.3.2 Gii tch t hp

Hon v.

nh ngha. Cho n vt khc nhau. Mt hon v ca n vt ny l mt cch sp xp chng theo mt cch no .

Mnh . S cc hon v ca n vt khc nhau l: ! 1.2.3...nP n n .

Chng minh. Ta chng minh cng thc ny da trn nguyn l nhn. Xt cng vic xy dng mt hon v ca n vt ban u. Cng vic ny c chia thnh cc bc sau:

- Bc 1: Chn vt ng u: c n cch chn (n vt u c th ng u)- Bc 2: Chn vt ng th hai: c n-1 cch chn (do chn vt ng u

nn by gi ta ch cn n-1 vt )-


Trang 20

- Bc n: Chn vt cn li cui cng: ch c 1 cch duy nht.Nh vy theo nguyn l nhn, s cch xy dng hon v, cng chnh l s cc hon v ca n vt ban u l n.(n-1)2.1 = n!.

V d. S cc s t nhin c 5 ch s khc nhau c to t cc ch s 1, 2, 3, 4, 5 l 5! = 120 cch.

Hon v lp

Hon v lp cng mang ngha nh hon v, iu khc bit y l trong n vt ban u c nhiu vt ging nhau. Mnh sau y s cho chng ta bit cng thc tnh hon v lp.

Mnh . Cho n vt, trong c t1 vt loi 1 ging nhau, t2 vt loi 2 ging nhau, , tk vt loi k ging nhau. Khi , s cc hon v khc nhau ca n vt ban u l

1 2

!! !... !k

nt t t

.

Chng minh. u tin, nu xem nh n vt l khc nhau, ta c n! hon v. Tuy nhin do c t1 vt loi 1 ging nhau, nn ng vi mt hon v ban u, nu ta hon v t1 phn t ny (c t1! hon v nh vy) ta vn c hon v . Chnh v vy thc cht t1! hon v kiu ny ch l mt hon v. Do , s hon v thc s khc nhau nu c t1 vt loi 1 ging nhau ch cn l:

1

!!

nt

. Tip theo ta li c t2 phn t loi 2

ging nhau, nn s hon v thc s khc nhau by gi s l 1 2

!! !n

t t. C tip tc nh

vy cho n khi kt thc, ta s c cng thc cn chng minh trong mnh .

V d: S cc s t nhin c 6 ch s, trong bt buc phi c 3 ch s 1, 2 ch s 2 v 1 ch s 3 l: 6! 60

3!2!1! .

Chnh hp

nh ngha. Cho n vt khc nhau. Mt chnh hp chp k ca n vt ny l mt cch chn k vt khc nhau c k n th t trong s n vt .

Mnh . S chnh hp chp k ca n vt khc nhau l: !( )!

kn

nAn k

Chng minh. Dng nguyn l nhn.


Trang 21

V d. Trong mt lp hc c 20 thnh vin. S cch chn ra mt ban cn s lp gm 3 ngi, trong c mt lp trng, mt lp ph v mt th qu l:

320

20! 6840(20 3)!

A

.

Chnh hp lp

Trong phn trn, ta xt s cch chn k vt t n vt khc nhau ban u. Nu ta c php chn trong s k phn t , c nhiu phn t ging nhau (ngha l mt vt c th c chn nhiu ln) th kt qu s thay i nh th no? Mnh sau y s cp n yu t ny.

Mnh . S cch chn k vt t n vt khc nhau ban u trong mt vt c th c chn nhiu ln (lp) l: nk.

Chng minh. S dng nguyn l nhn.

V d. C bao nhiu s t nhin c 5 ch s c xy dng t 3 ch s 1, 2, 3, trong cc ch s c th xut hin nhiu ln trong cng mt s? Cu tr li trong trng hp ny l 35 = 243 s t nhin nh vy.

T hp

nh ngha. Cho n vt khc nhau. Mt t hp chp k ca n vt ny l mt cch chn k vt khc nhau khng k n th t trong s n vt .

Mnh . S t hp chp k ca n vt khc nhau l: !!( )!

kn

nCk n k

Chng minh. D thy rng s khc nhau gia t hp v chnh hp ch l vn c xt n hay khng th t ca k vt c chn. i vi t hp, ta khng xt n yu t th t, iu c ngha l nu hon v k vt c chn mt cch ty , t hp

ca chng ta ban u cng khng thay i. Do , ta c cng thc: !

kk nn

ACk

. V ta

d dng c c cng thc ca s t hp.

V d. Trong mt lp hc c 20 hc sinh. C bao nhiu cch chn ra mt i vn ngh gm 5 ngi? Cu tr li l: 520

20! 155045!15!

C .

T hp lp

Xt bi ton sau y: C bao nhiu cch cho 4 ci bnh ging nhau vo trong 3 ci hp khc nhau m khng c rng buc no.


Trang 22

i vi bi ton ny, iu gy cho chng ta kh khn chnh l vic nhng ci bnh ging nhau. N lm cho cng vic m ca chng ta s phc tp nu ta m theo kiu thng thng v s c rt nhiu trng hp trng nhau.

gii bi ton ny, chng ta s lm nh sau: - Biu din mi ci bnh l 1 du +, v dng du | th hin vch ngn gia cc

hp. C 3 ci hp th ch cn 2 vch ngn l . Khi ta s c mt dy k hiu gm 4 du + v 2 du | nh sau: + + + + | | .

- By gi ta s hon v dy k hiu ny bt k, mt hon v nhn c s tng ng 1-1 vi mt cch cho bnh vo hp. S du + ng trc du | u tin s l s bnh nm trong hp th nht, s du + ng gia hai du | s l s bnh trong hp th hai v s du + nm sau du | th hai s l s bnh trong hp th ba. Chng hn nh hon v: + + | | + + s tng ng vi cch cho 2 bnh vo hp th nht, 0 bnh vo hp th hai v 2 bnh vo hp th ba.

- Nh vy s cch chia 4 ci bnh ging nhau vo trong 3 ci hp khc nhau chnh l s hon v ca mt dy gm 4 du + v 2 du |. Theo cng thc ca hon v lp, s hon v ny l: (2 4)!

4!2! . mt cht th cng thc ny cng

chnh l cng thc ca t hp 4 24 2 4 2C C .

Tng qut, ta c mnh sau:

Mnh . S cch cho n vt ging nhau vo trong k hp khc nhau l: 11 1n kn k n kC C

Dng t hp lp c rt nhiu ng dng, mi ng dng i hi nhng cch vn dng khc nhau ca cng thc. Chnh v th khi gii, chng ta phi rt linh hot a v bi ton gc chia bnh m chng ta c cng thc.

V d.

a. Tm s nghim nguyn khng m ca phng trnh sau: x1 + x2 + x3 = 10. i vi bi ton ny, nu t x1 l s bnh c cho vo hp 1, x2 l s bnh c cho vo hp 2 v x3 l s bnh c cho vo hp 3 th bi ton trn tr thnh c bao nhiu cch cho 10 ci bnh ging nhau vo trong 3 ci hp khc nhau v kt qu cn tm l: 10 1010 3 1 12C C .

b. Mt ngi M c 4 a con. Mt ngy n, ngi M c 20 ci ko v mun chia cho 4 a con sao cho mi a c t nht 2 ci ko. Hi c bao nhiu cch chia nh vy?. R rng l cch t vn y cng ging nh vic ta cho 20 ci bnh ging nhau vo trong 4 ci hp khc nhau sao cho mi ci hp c t nht 2 ci bnh. R rng l ta khng th p dng ngay cng thc bit v nu nh vy th s c rt nhiu trng hp s c hp c t hn 2 bnh (thm ch l c th khng c ci bnh no). gii quyt trng hp ny, ta s cho vo mi hp hai ci bnh trc, sau mi chia 12 ci bnh cn li


Trang 23

mt cch t do vao 4 ci hp, nh vy s cch chia tha mn yu cu ban u l: 12 1212 4 1 15C C .

2.3.3 Nguyn l Dirichlet. (nguyn l chung b cu)

Nguyn l. Nu chung b cu c t ca hn s b cu th c t nht hai con chim b cu chung trong mt ca.

V d.a. Trong mt lp c 50 ngi s c t nht 2 ngi c ngy sinh nht trong

cng mt thng.b. Trong 5 s t nhin bt k lun tm c 2 s sao cho hiu ca chng chia

ht cho 4.

Nhn thy rng trong phn a. ca v d trn, vic chc chn l c 2 ngi c cng ngy sinh nht trong mt thng no l chnh xc nhng dng nh vn cha lm chng ta tha mn. iu ny cng d hiu v tht ra chng ta cn c th c c kt qu mnh hn. Sau y l nguyn l Dirichlet tng qut.

Nguyn l Dirichlet tng qut.

Nu nht n con chim b cu vo trong mt chung c k ca th chc chn s c mt ca cha khng t hn n

kcon chim b cu.

V d: Trong mt lp c 50 ngi th chc chn s c t nht 5 ngi c ngy sinh nht trong cng mt thng.


Trang 24

3 Chng 3. Quan h

3.1 Quan h

nh ngha. Mt quan h gia tp hp A v tp hp B l mt tp con ca A Bu . Nu ( , )a b , ta vit a b . c bit, mt quan h gia A v A c gi l mt quan h hai ngi trn A.

V d. 1. A = {1, 2, 3, 4}, B = {4, 5}, ={(1,4),(1,5),(3,5)(4,4)} l mt quan h gia

A v B. Quan h c th c biu din bi s sau:

2. Quan h = trn mt tp hp A bt k: a b a b 3. Quan h d trn N, Z hay R: a b a b d4. Quan h ng d trn Z: (mod 7)a b a b {

Cc tnh cht ca quan h:

nh ngha. Cho l quan h hai ngi trn tp hp A. Ta ni:i. c tnh phn x nu ,x A x x ii. c tnh i xng nu , ,x y A x y y x iii. c tnh phn xng nu , ,x y A x y y x x y iv. c tnh bc cu nu , , ,x y z A x y y z x z

Nhn xt. 1. Mt quan h hai ngi trn tp A c tnh cht phn x nu mi phn t ca

A u quan h vi chnh n. Trong trng hp ny, nu ta biu din quan h trn h trc th n s cha cc im nm trn ng cho chnh.

2. Nu c tnh cht i xng th khi biu din n trn th, ta s thy cc im c xc nh s i xng qua ng cho chnh.

1 2 3 4


Trang 25

3. Hai tnh cht i xng v phn xng khng phi l ngc nhau. Mt quan h khng c tnh cht i xng khng c ngha l n c tnh cht phn xng v ngc li. S c nhng quan h va i xng va phn xng v cng s c nhng quan h khng i xng v cng khng phn xng.V d. Xt A = {1,2,3,4}, v ={(1,1),(3,3),(4,4)}, khi l quan h va i xng va phn xng. Cn nu xt = {(1,2),(2,1),(3,2)} th khng i xng (v (3,2) nhng (2,3) ) cng khng phn xng (v (1,2) 'v (2,1) ' nhng 1 2z ).

V d. Xt quan h hai ngi trn tp Z c nh ngha nh sau:22,, babaZba

Chng minh rng c cc tnh cht phn x, i xng v bc cu.Gii.

phn x. Za , ta c a2 = a2, do , theo nh ngha, ta c a a hay c tnh phn x

i xng. Xt a, b bt k thuc Z, gi s a b , ta s chng minh b a . Tht vy: 2 2 2 2a b a b b a b a , suy ra c tnh i xng.

bc cu. Xt a, b, c bt k thuc Z, gi s a b v b c ta s chng minh a c . Tht vy: 2 2 2 2 2 2a b b c a b b c a c a c , suy ra c tnh bc cu.

3.2 Quan h tng ng

nh ngha. Mt quan h hai ngi trn tp hp A c gi l quan h tng ng nu n c cc tnh cht phn x, i xng v bc cu.

V d. 1. Quan h nh nh ngha trong phn trc, 22,, babaZba ,

chnh l mt quan h tng ng.2. Cho trc mt nh x :f A Bo , ta nh ngha quan h trn A nh sau:

, , ( ) ( )x y A x y f x f y khi , ta c th kim chng c y l mt quan h tng ng.

3. Cho trc mt s t nhin n. Xt quan h trn Z c nh ngha nh sau:)(mod,, nbabaZba

y cng l mt quan h tng ng trn Z.

nh ngha. Cho l mt quan h tng ng trn A v x A . Khi y lp tng ng cha x, k hiu l x hay [x], l tp hp con ca A sau y:

^ `x y A y x

V d. Xt A = {1,2,3,..10}. Xt quan h trn A: (mod 3)a b a b { . y l mt quan h tng ng trn A. V ta c:

- Lp tng ng cha 1: ^ `1 1,4,7,10


Trang 26

- Lp tng ng cha 5: ^ `5 2,5,8 - rng: 1 4 7 10 , 2 5 8 v 3 6 9

nh l. Cho l mt quan h tng ng trn tp hp A. Khi y:i. ,x A x x ii. , ,x y A x y x y iii. Hai lp tng ng x v y sao cho x y z th trng nhau.

Chng minh. i. Do c tnh cht phn x nn ta c ,x A x x . Theo nh ngha ca lp

tng ng, ta suy ra x x .ii. Xt x v y l hai phn t bt k ca A. Gi s x y , ta s chng minh x y .

Xt z l mt phn t bt k trong x . T nh ngha ca lp tng ng, ta suy ra z x . Mt khc, do c tnh cht bc cu nn kt hp vi gi thit ban u l x y , ta suy ra z y . iu ny cng c ngha l z y . T , ta c x y . Bng cch tng t ta cng chng minh c y x .

iii. Gi s x y z . Khi tn ti phn t z x y , ngha l z x v z y . T ta suy ra z x v z y , do c tinh i xng v bc cu nn ta suy ra x y . Theo phn ii) ta c x y .

T cc tnh cht trn ca cc lp tng ng, ta c th ni rng cc lp tng ng to thnh mt phn hoch ca tp A. Ngha l hp ca cc lp tng ng s chnh bng A v cc lp tng ng hoc trng nhau, hoc tch ri hn nhau.

V d.1. Xt quan h trn Z: 2 2m n m n . Ta kim chng c y l mt

quan h tng ng. Cc lp tng ng to thnh phn hoch ca ] l:{0}, {1,-1}, {2,-2}, , {k,-k},

v ta ni Z c phn hoch thnh v s lp tng ng hu hn.2. Xt quan h ng d theo modulo n trn tp Z. y cng l mt quan h

tng ng v ta c Z s c phn hoch thnh n lp tng ng: 0, 1,..., 1n

mi lp tng ng l mt tp con v hn ca Z, chng hn nh 0 l tp hp tt c cc s nguyn chia ht cho n.

3.3 Quan h th t - Biu Hasse


Trang 27

nh ngha. Mt quan h hai ngi trn tp hp A c ni l mt quan h th t nu n c tnh cht phn x, phn xng v bc cu. Khi y ta ni A l tp hp c th t hay A l tp hp c sp.

K hiu. Thng thng, ta s k hiu mt quan h th t l E v k hiu cp ,A E l cp c th t.

V d. 1. d,Z l mt tp hp c th t.2. Trn tp hp P(E) ta c quan h: A B A B E . Khi E l mt quan h

th t trn P(E).3. Xt n l mt s nguyn dng. t

^ `naZaU n | k hiu |a n ch a l c s ca n (hay n chia ht cho a). nU chnh l tp hp cc c s ca n. Trn nU ta nh ngha mt quan h:

|x y x yETa s kim chng rng ,nU E l mt tp hp c th t. Tht vy d thy rng E c tnh phn x v bc cu. Mt khc gi s a bE v b aE , ngha l a l c ca b v b l c ca a. iu ny ch c th xy ra khi v ch khi a = b. Vy E c tnh phn xng. Suy ra E l quan h th t v tp ,nU E l mt tp hp c th t.

biu din quan h th t, chng ta c th s dng hai phng php l lit k hoc dng th. Tuy nhin c hai phng php ny u khng th hin c mt cch trc quan v quan h th t. Chnh v th, chng ta s phi dng mt cch khc biu din: l biu Hasse. Trc ht, ta xt nh ngha sau:

nh ngha. Cho ,A E l tp c th t v x, y l hai phn t bt k trong A.a. Nu x E y, ta ni y l tri ca x hay x c tri bi y.b. y l tri trc tip ca x nu y l tri ca x v khng tn ti mt phn

t z A no sao cho x z yE E v x y zz z .

V d. Xt tp 12 ,U E , d dng nhn thy rng:- Tri ca 2 l 4, 6, 12- Tri trc tip ca 2 l 4, 6.

nh ngha. Cho ,A E l tp c th t hu hn. Biu Hasse ca ,A E bao gm:


Trang 28

a. Mt tp hp cc im trong mt phng tng ng 1 1 vi A, gi l cc nh

b. Mt tp hp cc cung c hng ni mt s cp nh: hai nh x v y c ni bng mt cung c hng (t x sang y) nu v ch nu y l tri trc tip ca x.

V d. Xt ^ `12 1,2,3,4,6,12U a. Biu Hasse ca 12 ,U d l:

b. Biu Hasse ca 12 ,|U l:

c. Cho tp E = {1,2,3}. Xt tp P(E) tp tt c cc tp con ca E. Trn P(E) ta nh ngha quan h E nh sau:

, ( ),A B E A B A B 5 EKhi biu Hasse ca ( ),P E E nh sau:

nh ngha. Tp ,A E c ni l c th t ton phn nu v ch nu hai phn t bt k u so snh c, ngha l mnh sau l ng:

1 2 3 4 6 12

13

6

12

2

4

{1}

{2}

{3}

{1,3}

{1,2}

{2,3}

{1,2,3}


Trang 29

, ,x y A x y y x E E

V d. Tp N, Z, Q, R vi th t ,d t thng thng l cc tp c th t ton phn.

Mnh . Biu Hasse ca ,A E l mt dy chuyn khi v ch khi ,A E l tp c th t ton phn.

nh ngha. Cho ,A E l mt tp c th t. Khi ta ni:a. Mt phn t m ca A c ni l ti tiu (tng ng l ti i) nu m

khng l tri thc s ca bt c phn t no (m khng c tri thc s bi bt c phn t no) ca A.

b. Mt phn t M ca A c ni l cc tiu (tng ng l cc i) nu M c tri bi mi phn t ca A (M l tri ca mi phn t trong A).

V d.a. Xt tp 12 ,U d , ta c:

- Phn t ti tiu l 1 y cng l phn t cc tiu- Phn t ti i l 12 y cng l phn t cc i

b. Xt tp {1,2,3,4,5,6},| , ta c:- Phn t ti tiu l 1 y cng l phn t cc tiu- Phn t ti i l: 4, 5, 6 khng c phn t cc i


Trang 30

4 Chng 4. i s Boole

4.1 i s Boole: nh ngha Tnh cht

nh ngha. Mt i s Boole l tp hp A z cng vi hai php tnh hai ngi v tha mn cc tnh cht sau:

i. Tnh kt hp: vi mi , ,x y z A :x ( ) (x )y z y z

v x ( ) (x )y z y z ii. Tnh giao hon: vi mi ,x y A :

x y y x v x y y x

iii. Tnh phn phi: vi mi , ,x y z A :x ( ) (x ) ( )y z y x z

v x ( ) (x ) ( )y z y x z iv. Phn t trung ha: tn ti hai phn t trung ha 1, 0 i vi hai php ton

, sao cho vi mi x A , ta c:x x 0

v x x 1v. Phn t b: vi mi x A , tn ti x A sao cho:

x x 1v x x 0

V d. a. Xt tp hp {0,1}B . Trn B ta nh ngha hai php ton sau:

, , ..

x y B x y x yx y x y x y

By gi ta s kim chng rng tp hp B vi hai php ton trn s l mt i s Boole. Tht vy, vi mi x, y, z B ta c:

- Tnh kt hp: ( )x y z = ( )x y z yz

= ( ) ( )x y z yz x y z yz = x y z xy yz xz xyz = ( ) ( )x y xy z x y xy z = ( )x y xy z = ( )x y z

( )x y z = ( ) ( ) ( )x yz xy z x y z Vy tnh cht kt hp c tha


Trang 31

- Tnh giao hon: Tnh cht ny hin nhin c tha v trong nh ngha ca cc php ton, vai tr ca x v y l nh nhau.

- Tnh phn phi: Ta c:o ( )x y z = ( )x yz x yz xyz (1)

Mt khc ta c:( ) ( )x y x z = ( )( )x y xy x z xz

= 2 2 2 2x xz x z xy yz xyz x y xyz x yz rng {0,1}x B , do ta c x2 = x. T , ta suy ra:( ) ( )x y x z = x yz xyz (2)T (1), (2) ta suy ra ( )x y z = ( ) ( )x y x z .

o ( )x y z = ( )x y z yz = xy xz xyz (3)

Mt khc, ta c:( ) ( )x y x z = 2 2( )xy yz xy yz x yz xy yz xyz do x x (4)T (3), (4), ta suy ra x ( ) (x ) ( )y z y x z Vy tnh cht phn phi c tha.

- Phn t trung ha:

Ta c: vi mi x B , 0 0 .01 .1

x x x xx x x

. Do , 1=1 l phn t trung

ha ca php v 0 =0 l phn t trung ha ca php .- Phn t b: Vi mi x B , ta c phn t b ca x l 1x x . Tht vy, ta

c:2(1 ) 0

(1 ) (1 ) 1x x x x x xx x x x x x

01

Vy B={0,1} vi hai php ton nh ngha trn l mt i s Boole.

b. Xt tp ^ `30 1,2,3,5,6,10,15,30U l tp cc c s ca 30. Trn tp ny, ta nh ngha hai php ton nh sau:

, , ( , )( , )

x y B x y UCLN x yx y BCNN x y

D dng nhn thy rng do cc tnh cht ca php ly c chung ln nht v bi chung nh nht nn hai php ton , ny tha mn cc tnh cht kt hp, giao hon v phn phi. By gi ta xt xem liu c tn ti 2 phn t trung ha cho 2 php ton v c tn ti phn t b cho mi phn t ca U30 hay khng.

Trc ht, ta nhn thy rng UCLN(30,x) = x v BCNN(1,x) = x vi mi 30x U . Do , 30 v 1 ln lt chnh l hai phn t trung ha ca php v php

. Ta c: 30, 1 1 0 .Xt mt phn t x bt k trong U30. Ta s chng minh phn t b ca x chnh

l 30xx

:


Trang 32

- Nhn thy rng nu x l mt c s ca 30 th 30x

cng l mt c s ca

30, ngha l 3030x Ux

.

- Ta nhn thy rng: 30 = 2.3.5. iu c ngha l nu mt s nguyn t 2, 3, hoc 5 l c s ca x th n khng th no l c s ca 30x

x . Suy

ra, gia x v x khng c chung mt c s nguyn t no. Suy ra: 30( , ) 1x x UCLN xx

0

- Tng t ta cng c: 30, 30x x BCNN xx

1

Vy U30 l mt i s Boole vi hai php ton k trn.

Ch . U12 vi 2 php ton k trn khng th l mt i s Boole v phn t x = 2 khng c phn t b tha mn cc tnh cht nh trong nh ngha (phn kim chng xem nh bi tp).

nh l. Cho i s Boole A. Trn A, ta nh ngha:, ,x y A x y x y x E

khi , E s l mt quan h th t trn A. Hn na, 1 chnh l phn t cc i v O chnh l phn t cc tiu trong ,A E .

Chng minh.x E l mt quan h th t trn A.

o Phn x:,x A ta c: x x = ( )x x 0 (t/c phn t trung ha)

= ( ) ( )x x x x (t/c phn t b)= ( )x x x (t/c phn phi)= x 1 (t/c phn t b)= x (t/c phn t trung ha)

Suy ra x xE hay E c tnh cht phn xo Phn xng:

, ,x y x y x

x y A x yy x y x y

EE

(do c tnh cht giao hon)

Vy, E c tnh cht phn xng.o Bc cu:

, , ,x y x y x

x y z Ay z y z z

EE

. Xt x z , ta c:

x z = ( )x y z


Trang 33

= ( )x y z = x y= x

Suy ra x zE hay E c tnh cht bc cu.

Vy E l quan h th t trn Ax 1 l cc i, 0 l cc tiu trong ,A E

Vi mi x A , ta c x x x E1 1. Suy ra 1 l phn t cc i trong ,A E . Mt khc, ta c:

( ) ( ) ( ) ( )x x x x x x x x x 0 0 0 0 0 0 suy ra ,x x A E0 , hay 0 l phn t cc tiu trong ,A E .

Nh vy, trn i s Boole lun lun tn ti mt quan h i s Boole tr thnh tp c sp. lm r hn kha cnh ny, chng ta xt cc v d sau y:

V d. a. Xt i s Boole ^ 1`,0 B , theo nh l trn, trn B chng ta c quan h th

t c nh ngha nh sau:

(*).,,

xyxxyxyxByx

E

rng do, x, y ch c th nhn gi tr 0 hay 1 nn ta ch c 4 trng hp khc nhau ca x v y. Trong 4 trng hp , tha (*), c 3 trng hp:

TH1: x = 1, y = 1.TH2: x = 0, y = 0.TH3: x = 0, y = 1.

Trng hp cn li (x=1, y=0) th khng tha (*). Nhn li 3 trng hp trn, chng ta thy rng: chng u c mt c im chung l yx d . V vy, ta c th vit:

yxxyx

xyxyxByx

d

(*).,, E

V nh vy th chng ta nhn thy rng: mc d quan h th t E c nh ngha bng cc php ton trong i s Boole tng nh xa l li chnh l quan h d m ta rt quen thuc. Hn na, theo quan h ny, 0=0 chnh l phn t cc tiu v 1=1 chnh l phn t cc i, ng theo nh l trn.

b. Xt i s Boole U30, theo nh l trn, trn U20 chng ta c quan h th t c nh ngha nh sau:


Trang 34

(**)),(,,

xyxUCLNxyxyxByx

E

Nhn thy rng, iu kin (**) ch tha khi v ch khi x l c s ca y. V vy, ta c th vit:

yxxyxUCLN

xyxyxByx

|(**)),(

,,

E

V nh vy th chng ta nhn thy rng: mc d quan h th t E c nh ngha bng cc php ton trong i s Boole li chnh l quan h l c s ca m ta rt quen thuc. Hn na, theo biu Hasse ca (U30,|) di y, 0=1 chnh l phn t cc tiu v 1=30 chnh l phn t cc i, ng theo nh l trn.

c. Cho tp E = {1,2,3}. Xt tp P(E) tp tt c cc tp con ca E. Trn P(E) ta c quan h E nh sau:

*)*(*),(,

ABABABAEPBA

E

Nhn thy rng iu kin (***) c tha khi v ch khi BA . V nh vy ta c th vit:

BAABA

BABAEPBA

*)*(*

),(, E

Quan h c nh ngha trong nh l trn chnh l quan h l tp con ca m ta kho st trong cc phn trc.

Theo biu Hasse trn, d dng nhn thy 0= l phn t cc tiu v 1={1,2,3} l phn t cc i trong i s Boole P(E).

1

3

2

5

15

6

10

30

{1}

{2}

{3}

{1,3}

{1,2}

{2,3}

{1,2,3}


Trang 35

Ch : trn mt i s Boole c th c rt nhiu quan h th t khc. Mc du vy, t y tr v sau khi ta ni n quan h th t trong i s Boole m khng ni g thm th ta hiu rng y l quan h th t nh nh ngha trong nh l trn.

nh ngha. Cho i s Boole A. Mt tri trc tip ca phn t 0 trong A c gi l mt nguyn t ca A.

V d. a. Trong i s Boole U30, 2, 3 v 5 chnh l cc nguyn t v chng l

tri trc tip ca 0=1.b. Trong i s Boole P(E), vi E l tp {1,2,3}, cc tp hp {1}, {2},

{3} chnh l cc nguyn t.

Khi ni n nguyn t, chng ta hiu rng l nhng i tng khng th chia nh c na v chng thng c dng cu thnh nn nhng i tng khc. y, trong i s Boole, khi nim nguyn t vn c gi nguyn vn ngha nh vy. Mnh sau y cp n ngha ny.

Mnh . Trong i s Boole A, mi phn t x khc 0 u c th c biu din di dng:

kmmmx ...21vi m1, m2, ..., mk l cc nguyn t no ca A

Mnh trn c ngha rt ln. N khng nh rng tp hp cc nguyn t ca mt i s Boole A c th c dng biu din tt c cc phn t khc 0 trong i s Boole .

V d.a. Xt i s Boole U30. Nhc li rng, 2 php ton c nh ngha trn

i s Boole ny l 2 php ton UCLN v BCNN, v cc nguyn t ca i s Boole ny chnh l: 2, 3, v 5. Theo mnh trn, bt k mt phn t no khc 0(=1) ca U30 cng u c th biu din bng mt php mt s nguyn t no ca n. Chng hn nh:

)3,2(326 BCNN hay )3,5(3515 BCNN hay )5,3,2(53230 BCNN

b. Xt i s Boole P(E), vi E l tp {1,2,3}. Nhc li rng, 2 php ton c nh ngha trn i s Boole ny l 2 php ton v , v cc nguyn t ca i s Boole ny chnh l cc tp hp: {1}, {2}, {3}. Theo mnh trn, bt k mt phn t no khc 0(=) ca P(E) cng u c th biu din bng mt php mt s nguyn t no ca n. Chng hn nh:

}2{}1{}2{}1{}2,1{ hay }3{}2{}3{}2{}3,2{


Trang 36

hay }3{}2{}1{}3{}2{}1{}3,2,1{

4.2 Hm Boole Dng ni ri chnh tc

nh ngha. Mt hm Boole f theo n bin l mt nh x:

),...,(),...,(:

11 nn

n

xxfxxBBf

6o

Nhn xt. a. Nu ng nht tp B={0,1} vi tp cc mnh {ng, Sai} th hm Boole

cng khng phi l mt khi nim qu xa l: mt hm Boole chnh l mt dng mnh (cng thc logic xem li chng 1). N nhn cc bin x1, x2, ,xn chnh l cc bin mnh v kt qu thu c cng l cng l mt mnh (ng hoc sai).

b. Tng qut, mt hm Boole theo n bin s l mt hm s theo n bin s, mi bin s s nhn gi tr 0 hoc 1 v kt qu ca hm s cng l gi tr 0 hoc 1. Mc d khng ni ra nhng ta vn phi hiu rng cc php ton biu din trong cng thc ca hm Boole vn chnh l cc php ton trong i s Boole B={0,1}.

V d.a. yxyxf ),( l mt hm Boole theo 2 bin: f(0,0) = 1, f(0,1)=0, b. zyxzyxf )(),,( l mt hm Boole theo 3 bin: f(0,0,1)=0,

f(1,0,1)=1,

Bng chn tr ca hm Boole.Tng t nh mt dng mnh , chng ta c th lp bng chn tr ca hm Boole xc nh gi tr ca hm Boole khi gi tr ca cc bin thay i.

V d. Xt hm Boole: zyxzyxf )(),,( . Bng chn tr ca hm Boole ny c xc nh nh sau:

x y z y yx z f(x,y,z)0 0 0 1 0 1 10 0 1 1 0 0 00 1 0 0 0 1 10 1 1 0 0 0 01 0 0 1 1 1 11 0 1 1 1 0 11 1 0 0 0 1 11 1 1 0 0 0 0

Nh vy, mi hm Boole s tng ng vi mt v ch mt bng chn tr. Ni cch khc, hai hm Boole c cng bng chn tr s l hai hm Boole bng nhau (thc cht ch c coi nh l mt hm Boole duy nht).


Trang 37

Nhn thy rng, vic xy dng bng chn tr ca mt hm Boole hon ton ging nh xy dng bng chn tr ca mt dng mnh (xem li chng 1). Cu hi t ra l: T mt bng chn tr, lm th no ta xy dng li c cng thc tng minh ca hm Boole?. y l mt cu hi c t ra rt thc t, chng ta phi thng xuyn xy dng cng thc ca hm Boole khi bit bng chn tr ca n. thc hin c iu ny, chng ta phi quay li kin thc ca i s Boole hc trong phn trc.

Xt Fn l tp hp tt c cc hm Boole theo n bin. Trn Fn, ta nh ngha 2 php ton , nh sau: Vi mi nFgf , , gf v gf c xc nh bi:

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

xgxfxgxfxgfxgxfxgfBx n

R rng vi gf v gf c xc nh nh trn th gf v gf u l nhng hm Boole theo n bin, do 2 php ton , l 2 php ton trn Fn. By gi, ta s kim tra xem n c tha mn cc tnh cht ca i s Boole hay khng.

D thy rng cc php ton trn c nh ngha tng t nh trong i s BooleB={0,1} (xem li VD trong mc 4.1). Chnh v th, khng kh c th kim tra hai php ton ny tha mn cc tnh cht giao hon, kt hp v phn phi.

K tip, chng ta s xc nh 2 phn t trong Fn ng vai tr l 0 v 1. Thc cht y l 2 hm Boole theo n bin. Gi s nFt l hm ng vai tr phn t 1. Khi , ta c:

(*))()().(,,

)())((,,

,

xfxtxfBxFfxfxtfBxFf

ftfFf

nn

nn

n

Theo , do (*) phi tha vi f v x c ly bt k, iu ny ch c th xy ra khi t(x)=1 vi mi nBx . Nh vy hm Boole ng vai tr l phn t trung ha ca php chnh l mt hm hng: hm ny lun nhn gi tr l 1 vi mi gi tr ca cc bin. Ni cch khc, bng chn tr ca hm ny bng 1 tt c cc dng.

Tng t, hm ng vai tr l phn t 0 trong Fn, phn t trung ha ca php , chnh l hm m gi tr ca n lun bng 0 vi bt k gi tr ca cc bin.

By gi vi hm nFf bt k, liu c lun tn ti mt hm nFf tha mn iu kin: ff 0 v ff 1? Cu tr li l c, tht vy, vi nFf bt k, ta t

nBxxfxf ),(1)( . Khi ta c:

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

)}1,0{)((0))(1).(()().())((,

xfxfxfxfxfxfxfxfxfxfxff

BxfdoxfxfxfxfxffBx n


Trang 38

Nh vy hm f chnh l phn t b ca hm f.

T nhng iu kho st trn, ta thy rng Fn vi hai php ton nh ngha nh trn s l mt i s Boole. y l mt iu cng ht sc th v khi cc hm Boole c nh ngha thng qua i s Boole B={0,1} v by gi, tp hp nhng hm Boole theo n bin cng s li l mt i s Boole.

Cng trong mc 4.1, chng ta thy rng mi i s Boole u l tp hp c th t theo mt quan h th t no . V Fn cng khng phi ngoi l, trn Fn, ta c quan h th t sau:

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

)()()(,

,,

xgxfBxxfxgxfBx

xfxgxfBxggfgfFgf

n

n

nn

d

E

rng, thng qua nh ngha trn, mt hm Boole f c ni l b hn mt hm Boole g nu v ch nu ti cng mt dng trn bng chn tr, gi tr ca f(x)lun b hn hay bng gi tr ca g(x).

V d. Cho 2 hm Boole f v g c xc nh trong bng chn tr sau:

x y z f g0 0 0 1 10 0 1 1 10 1 0 0 10 1 1 0 01 0 0 1 11 0 1 1 11 1 0 0 01 1 1 0 1

Khi , ta thy rng trn mt dng bt k, gi tr ca f lun nh hn hay bng gi tr ca g. Do , ta c: gf E .

By gi, ta s kho st cc nguyn t ca Fn. Theo nh ngha, nguyn t ca mt i s Boole s l cc tri trc tip ca phn t 0 theo quan h c xc nh. Nh vy nhng nguyn t ca i s Boole chnh l nhng hm Boole m bng chn tr ca n ch khc 0 ti mt dng duy nht (v nu c 2 dng khc 0 th y khng cn l tri trc tip ca 0 na). Bng chn tr ca mt hm Boole theo n bin s c 2n dng, chnh v vy, Fn s c tt c l 2n nguyn t.

Trn y, ta xc nh c cc nguyn t ca Fn theo bng chn tr, tuy nhin chng ta cng cn bit cng thc tng minh ca cc nguyn t ny. Mnh sau y s cho chng ta r hn v iu ny.


Trang 39

Mnh . Mi nguyn t ca Fn u c dng:nbbb ...21

trong bi s l xi hoc ix (i=1..n, xi l cc bin ca trong cc hm Boole).

Vic chng minh mnh trn khng kh. Tuy nhin chng ta s tp trung hn vo vic xc nh cng thc cho tng nguyn t c th.

rng cc hm Boole c xc nh theo cng thc trn ch khc 0 ti mt dng duy nht. Dng duy nht ny c xc nh bi cc thnh phn bi trong cng thc. Ti dng duy nht , nu xi = 1 th bi = xi, cn nu xi=0 th bi = ix . c th hn, ra xt v d sau y:

V d. Xt i s Boole F3. Ta c 8 nguyn t (gi l f0,,f7) v bng chn tr ca chng c xc nh nh sau:

x y z f0 f1 f2 f3 f4 f5 f6 f70 0 0 1 0 0 0 0 0 0 00 0 1 0 1 0 0 0 0 0 00 1 0 0 0 1 0 0 0 0 00 1 1 0 0 0 1 0 0 0 01 0 0 0 0 0 0 1 0 0 01 0 1 0 0 0 0 0 1 0 01 1 0 0 0 0 0 0 0 1 01 1 1 0 0 0 0 0 0 0 1

xc nh cng thc ca f0, ta thy rng f0 ch khc 0 ti mt dng duy nht ng vi x = 0, y = 0, v z = 0. Nh vy theo mnh trn, ta c zyxf 0 . Tng t, ta c cng thc ca 7 nguyn t cn li nh sau:

zyxfzyxfzyxfzyxfzyxfzyxfzyxf

7

6

5

4

3

2

1

Ch . cho tin, t y v sau, ta s s dng du chm (.) thay cho k hiu , ngha l ta s vit yx. thay cho yx .

n y ta xc nh c c th cc nguyn t ca Fn. Ta quay tr li cu hi t ra ban u: Lm th no xc nh c cng thc tng minh ca mt hm


Trang 40

Boole khi bit bng chn tr ca n?. Phn sau y s gii thiu c th cc xc nh ny.

Cng thc dng ni ri chnh tc ca hm Boole:

Cn nhc li rng chng ta c mt mnh rng mi phn t khc 0 trong i s Boole u c th biu din c di dng t hp mt s nguyn t c ni vi nhau bi php hp ( ). C th, mt hm Boole f z 0 bt k u c th c biu din di dng:

kmmmf ...21vi m1, m2, , mk l mt s nguyn t no ca Fn. Vn y l nhng

nguyn t no c chn thit lp cng thc ca hm f.

rng cc nguyn t ca Fn u khng ng thi bng 1 vi mt b gi tr no ca cc bin. Ni cch khc vi nu mt nguyn t no c gi tr 1 vi mt b gi tr ca cc bin th cc nguyn t cn li u c gi tr bng 0 vi chnh b gi tr ca cc bin . Tn dng tnh cht ny, ta c th thit lp cng thc ca mt hm Boole t bng chn tr ca n theo cc bc sau y:

- Chn tt c cc nguyn t ca Fn ng vi cc dng bng 1 trong bng chn tr ca f.

- Ni tt c cc nguyn t c chn trn bng php ton hp ( ) ta c cng thc biu din ca hm f.

V d. Cho hm Boole f(x,y) c bng chn tr nh sau:

x y f0 0 10 1 11 0 01 1 0

Hy thit lp cng thc biu din ca hm f.

Gii.

thit lp cng thc ny, ta tun th 2 bc nh nu trn.


Trang 41

u tin ta xc nh cc nguyn t tng ng vi cc dng m ti gi tr ca f bng 1. Do f c gi tr bng 1 ti 2 dng u tin nn ta chn 2 nguyn t tng ng vi 2 dng : yxf .0 v yxf .1 .

K tip, ni chng li bng php hp, ta c cng thc ca hm f . Ta c :

yxyxyxf ..),(

l cng thc cn tm.

V d. Cho hm Boole f(x,y,z) c bng chn tr nh sau:

x y z f0 0 0 10 0 1 10 1 0 00 1 1 01 0 0 01 0 1 11 1 0 01 1 1 1

Hy thit lp cng thc biu din ca hm f.

Gii.

u tin ta xc nh cc nguyn t tng ng vi cc dng m ti gi tr ca f bng 1. Do f c gi tr bng 1 ti 4 dng nn ta chn 4 nguyn t tng ng vi 4 dng : zyxf ..0 , zyxf ..1 , zyxf ..5 , v zyxf ..7

K tip, ni chng li bng php hp, ta c cng thc ca hm f . Ta c :

zyxzyxzyxzyxzyxf ........),,(

l cng thc cn tm.

Cng thc biu din hm Boole c xc nh theo quy tc trn c gi l dng ni ri chnh tc ca n. S d ta ni y l dng ni ri v n c ni bi php ni ri php hp v ta ni l chnh tc v trong tt c cc s hng (cc nguyn t) u lun c s xut hin ca tt c cc bin.

Ch :


Trang 42

- Dng ni ri chnh tc i khi cn c gi l dng tng cc tch v trong cng thc ny ta ly tng (php ) ca cc s hng l tch (php ) ca cc bin.

- Dng ni ri chnh tc c u im l d thit lp, mc d vy cng thc ny thng s rt di.

4.3 Bi ton mch in Mng cc cng

Cc hm Boole c mt s tng ng rt cht ch vi cc mch in t. Cc hm Boole nhn c cc bin nhn gi tr 0 hoc 1, v kt qu ca hm Boole cng l gi tr 0 hoc 1. Trong khi , cng tng t, mch in t thng c thit k vi nhiu u vo v mt hoc nhiu u ra, cc u vo l cc ng truyn tn hiu ch nhn cc gi tr nh phn (0 hoc 1), cc u ra cng l cc ng tn hiu nh phn l t hp cc u vo theo mt quy tc no . Nh vy, ta vy ta c th coi mi u ra ca mch in t tng ng vi mt hm Boole v cc u vo chnh l cc bin ca hm Boole .

Cc cng in t c bn.

Trong hm Boole, chng ta s dng cc php ton trn i s Boole B={0,1} biu din. Nh ni trn, cc mch in, cc u ra s l mt t hp cc u vo. Cng tng t nh hm Boole cc bin c ni vi nhau bng cc php ton cc u vo cng c lin kt vi nhau thng qua cc cng c bn. V hn na, cc cng c bn ny cng hon ton tng ng vi cc php ton trn i s Boole B={0,1}.

MCH IN T...

.

.

.

x1

x2

xn

f1 (x1,x2,,xn)

f2(x1,x2,,xn)

fk(x1,x2,,xn)

U VO U RA

x yxyCng AND

xy x y

Cng OR

x

Cng NOTxx


Trang 43

D thy rng cng AND s tng ng vi php ton , cng OR s tng ng vi php ton , cn cng NOT s tng ng vi php ly phn t b.

T nhng s tng ng gia mch in t vi hm Boole trn, ta thy rng vic thit k cc mch in t cng s tng ng vi vic xy dng cc hm Boole. Chng hn, xt v d sau:

V d. Xy dng mch cng 2 bit. Mch cng 2 bit s thc hin php cng 2 bit x v y, kt qu thu c s l 2 gi tr nh phn S (tng) v C (nh). S d ta phi dng thm gi tr C v trong trng hp x=1, y=1 th S khng biu din kt qu (1+1 bng 0, nh 1). Nh vy mch cng 2 bit s c 2 u vo x, y v 2 u ra l S v C. Bng chn tr m t yu cu v quan h gia u ra vi cc u vo nh sau:

x y S C0 0 0 00 1 1 01 0 1 01 1 0 1

Hai u ra S v C s c biu din bng 2 hm Boole theo 2 bin x v y. p dng quy tc tm dng ni ri chnh tc ca hm Boole, ta c cng thc ca 2 hm Boole biu din S v C l:

yxyxyxfS ..),( yxyxgC .),(

T , m hnh mch cng 2 bit s nh sau:

V d. Trong mt trn thi u v thut c 3 trng ti s cho im cc n nh c thc hin trong mt trn u. Mi trng ti s c mt nt bm chm im cc n nh. Mi n nh s c tnh im nu c t 2 trng ti bm nt tr ln. Mch in dng cho vic chm im ny s gm c 3 u vo tng ng vi 3 nt bm ca cc trng ti. Nu trng ti no bm nt th tn hiu 1 t ng dy s c truyn vo mch, nu khng bm th tn hiu trn ng dy vn l 0. u ra ca mch s l mt tn hiu nh phn (0/1) th hin vic ng nh c c tnh im hay khng. (0 khng tnh im, 1 tnh im).

Gi f l hm Boole th hin tn hiu u ra, bng chn tr ca f theo m t trn s nh sau:

x y z f

xy

C

S

yx.

yx.


Trang 44

0 0 0 00 0 1 00 1 0 00 1 1 11 0 0 01 0 1 11 1 0 11 1 1 1

Theo quy tc xy dng dng ni ri chnh tc ca hm Boole, ta c cng thc ca hm Boole f l:

zyxzyxzyxzyxzyxf ........),,( T , mch in c thit k nh sau:

Thng qua cc v d trn, chng ta thy rng vic xy dng cng thc ca hm Boole lin quan mt thit n vic xy dng cc mch in t. Vi vic xy dng c cng thc dng ni ri chnh tc ca cc hm Boole, ta c mt c s l thuyt v mt cng c hu dng cho vic thit k cc mch in.

4.4 Tm cng thc a thc ti tiu Phng php Karnaugh

Nh ta thy trn m hnh cc mch in c thit k da theo cng thc ca hm Boole (cc cng u tng ng vi cc php ton). Mt khc, cng cn rng, i vi mt hm Boole, c th c nhiu dng cng thc biu din khc nhau. Cng thc dng ni ri chnh tc ca mt hm Boole, nh phn tch trn, rt d thit lp, tuy vy trong thc t, cng thc dng ni ri chnh tc khng phi l cng thc dng ngn nht ca mt hm Boole.

zyx ..

zyx ..

zyx ..

zyx ..

f

xyz


Trang 45

i vi mc ch thit k mch in, cng thc hm Boole cng ngn th mch in thit k s c chi ph cng thp (do s dng t cng in t). Chnh v th, ta cn phi tm cng thc hm Boole di dng ngn nht c th.

C nhiu phng php khc nhau nhm rt gn cng thc ca hm Boole nh: s dng cc tnh cht ca i s Boole bin i, rt gn mi phng php c nhng u khuyt ring v s hu dng cho mt s trng hp no .

Trong phn ny chng ta s gii thiu phng php Karnaugh, mt phng php hu hiu cho vic tm cng thc dng n gin nht ca hm Boole c 3 hoc 4 bin.

Trc ht, ta tm hiu khi nim biu Karnaugh ca mt hm Boole f.

nh ngha. Cho hm Boole f theo n bin. Biu Karnaugh ca hm f l mt hnh ch nht gm 2n sao cho:

- Mi s tng ng vi mt dng trong bng ca f.- Mt s c nh du nu v ch nu ti dng tng ng vi n trong

bng chn tr, gi tr ca f bng 1.- Cc c cho tng ng vi cc dng sao cho hai dng tng ng vi

hai cnh nhau lun sai khc nhau v gi tr ca ch mt bin duy nht.

C th, hnh v sau y l cch t chc biu Karnaugh cho mt hm Boole theo 3 bin.

Cc c xc nh tng ng vi cc dng da vo cch nh a ch ca cc bin nh trong hnh v. Chng hn nh (1) c a ch l zyx .. , do , n s tng ng vi dng {x=0,y=0,z=0} trong bng chn tr. Hoc, (2) c a ch l zyx .. v n tng ng vi dng {x=0,y=1,z=0}. R rng (1) v (2) l hai cnh nhau v n cng tng ng vi hai dng trong bng chn tr ch sai khc nhau gi tr ca bin y.

Ch :- C nhiu cch b tr v tr ca cc bin khc nhau, min sao phi m bo

c nhng yu cu ca mt biu Karnaugh.- Hai nm bin vn c coi l hai cnh nhau (tng tng rng biu

Karnaugh c cun li). Chng hn nh trong biu trn, (1) v c a ch zyx .. (gc trn bn phi) vn c coi l hai nm cnh nhau.

V d. Hm Boole f theo 3 bin c bng chn tr:

(1)

xx

yy

z

yz

(2)


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x y z f0 0 0 00 0 1 00 1 0 00 1 1 11 0 0 01 0 1 11 1 0 11 1 1 1

Khi , biu Karnaugh ca hm Boole f ny nh sau:

R rng, chng ta thy rng, trong bng chn tr c bao nhiu dng bng 1 th biu Karnaugh c by nhiu c nh du, v ngc li. Hn th na, bng chn tr v biu Karnaugh thc ra l bn sao ca nhau. Nu ta c bng chn tr th ta d dng xy dng c biu Karnaugh v tng t nu ta c biu Karnaugh th hon ton c th xy dng li bng chn tr.

nh ngha. Cho biu Karnaugh ca mt hm Boole f theo n bin. Ta nh ngha:

a. Mt t bo l mt hnh ch nht gm 2k ( nk dd0 ) c nh du lin nhau.

b. Mt t bo ln l mt t bo m khng b ph bi bt c t bo no khc.

V d. Xt biu Karnaugh trong v d trn, ta thy:- Cc t bo: zyx .. , zyx .. , zyx .. , zyx .. (t bo 1 ), xy, yz, xz (t bo 2 )- Cc t bo ln: xy, yz, xz.

Nh ni phn u ca mc ny, mc ch ca chng ta l tm cng thc a thc ti tiu (cng thc a thc dng n gin nht) ca hm Boole. Sau y chng ta s kho st phng php Karnaugh thc hin iu ny.

Phng php Karnaugh tm cng thc a thc ti tiu ca hm Boole.

xx

yy

z

yz


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Cho hm Boole f (di dng cng thc hoc bng chn tr), tm cng thc a thc ti tiu ca f , ta thc hin theo cc bc sau:

Bc 1. Xy dng biu Karnaugh ca fBc 2. Xc nh tt c cc t bo ln trong biu Karnaugh va xy dngBc 3. Tm mt s lng t nht cc t bo ln ph kn cc nh du trong

biu Karnaugh, ghp chng li vi nhau bng php , ta s c cng thc a thc ti tiu cn tm.

V d: Cho hm Boole f c bng chn tr di y. Hy tm cng thc a thc ti tiu ca f.

x y z f0 0 0 00 0 1 00 1 0 00 1 1 11 0 0 01 0 1 11 1 0 11 1 1 1

Bc 1. Xy dng biu Karnaugh ca hm f:

Bc 2. Xc nh tt c cc t bo ln: xy, yz, xz

Bc 3. Tm mt s lng t nht cc t bo ln ph cc c nh du.

Tng tng rng chng ta c th tch ring cc t bo ln, khi hnh nh v cc t bo ln nh sau:

xx

yy

z

y

xx

yy

z

y

xx

yy

z

y

xx

yy

z

yz

T bo ln xy T bo ln yz T bo ln xz

z z


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Nh vy thc cht vic chn cc t bo ln ph kn cc c nh du trong biu Karnaugh chnh l vic chn cc t bo ln ri xp chng chng ln nhau sao cho hnh thu c ging biu Karnaugh ban u.

Trong v d trn, r rng l trong trng hp ny, ta phi chn c 3 t bo ln. Nh vy, cng thc a thc ti tiu ca hm f ny s l:

xzyzxyf

V d: Cho hm Boole zyxyxzyxzyxf .....),,( . Hy tm cng thc a thc ti tiu ca f.

Bc 1. Xy dng biu Karnaugh ca hm f.Trong trng hp ny, ta c cung cp bng chn tr ca hm f. Chng

ta c th lp bng chn tr ca f ri thc hin nh lm v d trc. Mc d vy, mt cch nhanh chng hn, ta c th xy dng biu Karnaugh trc tip t cng thc trn. D thy rng, cc thnh phn ca cng thc trn tng ng vi 3 t bo:

t chng cc t bo ny ln nhau, ta s c biu Karnaugh ca hm f ban u:

Bc 2. Xc nh tt c cc t bo ln: zyyxzx .,.,.

Bc 3. Tm mt s lng t nht cc t bo ln ph cc c nh du.

Tng t nh v d trc, d hnh dung, ta tch ring cc t bo ln:

xx

yy

z

yz

xx

yy

z

yz

xx

yy

z

yz

xx

yy

z

yz

T bo ln zx. T bo ln yx. T bo ln zy.

xx

yy

z

yz

xx

yy

z

yz

xx

yy

z

yz

T bo zyx .. T bo yx. T bo zyx ..


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Mc d c 3 t bo ln, nhng trong trng hp ny, ph kn cc nh du trong bn Karnaugh, ta ch cn chn 2 t bo ln: zx. v zy. . Nh vy, cng thc a thc ti tiu ca hm f ny s l:

zyzxf ..

i vi hm Boole theo 4 bin, biu Karnaugh s bao gm 16 c sp xp nh di y:

v phng php tin hnh tm cng thc a thc ti tiu cng tng t nh i vi hm 3 bin.

Ch : i vi biu Karnaugh ca hm Boole 4 bin, hai trn cng v di cng (ca cng mt ct) vn c coi l cnh nhau, 4 4 gc vn c coi l cnh nhau. iu ny l bi v biu Karnaugh c th c gp li theo c 2 chiu ngang v dc.

V d: Tm cng thc a thc ti tiu ca hm Boole sau:tzyxtzyxtzyxtzyxtzyxtzyxtzyxtzyxf .....................),,,(

Bc 1. Tng t nh hm Boole vi 3 bin, ta c biu Karnaugh ca hm Boole trn nh sau:

Bc 2. Xc nh cc t bo ln: c 3 t bo ln:

x x

y y y

z

z

t

t

t

x x

y y y

z

z

t

t

t

x x

y y y

z

z

t

t

t

x x

y y y

z

z

t

t

t

x x

y y y

z

z

t

t

t

T bo ln zx. T bo ln zy. T bo ln tyx ..


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Bc 3. Tm mt s lng t nht cc t bo ln ph cc c nh du. y, d thy l ta khng th b t bo ln no. Nh vy, cng thc a thc ti tiu ca hm Boole ban u l:

tyxzyzxtzyxf ....),,,(

n y, chng ta bit cch xy dng cng thc a thc ti tiu da trn phng php biu Karnaugh. y l mt phng php cc k hu hiu i vi cc hm Boole 3 hoc 4 bin. i vi cc hm nhiu bin hn, phng php ny kh thc hin v khi biu Karnaugh phi m rng ra khng gian v iu ny s lm cho ta kh nhn bit cc t bo v t bo ln.

Khi p dng phng php biu Karnaugh, cn lu cc im sau:

- V biu Karnaugh phi tuyt i chnh xc. Ch nhm ln vic nh du 1 s dn ti kt qu hon ton sai lch trong nhng bc tip theo.

- Vic xc nh cc t bo ln phi thn trng, nu xc nh khng chnh xc cc t bo ln th cng thc thu c cui cng c th khng phi l cng thc dng n gin nht. (Do cc t bo ln c nhiu hn, nhng s bin biu din li t hn).

- Khi chn cc t bo ln ph cc c nh du cn u tin chn nhng t bo ln bt buc (khng th khng chn) trc. Vi nhng c nh du ch thuc mt t bo ln duy nht th t bo ln ny bt buc phi c chn. Bn cnh khi c hai t bo ln cng ph qua mt th ta u tin chn t bo ln c nhiu hn ph.


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TI LIU THAM KHO

[1]. Nguyn c Ngha Nguyn T Thnh, Ton ri rc, NXB i hc Quc Gia H Ni, 2003.

[2]. Kenneth H.Rosen Ton ri rc ng dng trong Tin hc (Bn dch),

NXB Khoa hc v K thut, 2000.

[3]. Nguyn Hu Anh, Ton ri rc, NXB Gio dc, 2000.[4]. Hong Chng , i cng v Ton hc hu hn, NXB Gio Dc,

1999.

28601026 tom tat bai giang toan roi rac nguyen ngoc trung

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