Otomat và ngôn ngữ hình thức

7/31/2019 Otomat v ngn ng hnh thc

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tmt v ngn ng hnh thc

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MC LC

LI NI U .......................................................................................................... 5

Chng I:Mu ................................................................................................... 81.1 Tp hp v cc cu trc i s ......................................................................... 8

1.1.1 Tp hp v cc tp con ............................................................................. 8

1.1.2 Tp hp v cc php ton hai ngi .......................................................... 9

1.3 Quan h v quan h tng ng .................................................................. 12

1.4 Hm s .......................................................................................................... 15

1.5 Logic mnh v tn t .............................................................................. 16

1.5.1 Logic mnh ........................................................................................ 16

1.6.2 Cng thc mnh ................................................................................. 18

1.5.3 Dng chun ca cc cng thc ............................................................... 20

1.5.4 Cc qui tc suy din trong tnh ton mnh ........................................ 23

1.6 Tn t (v t) v cc lng t ........................................................................ 25

1.7 Cc phng php chng minh...................................................................... 28

Bi tp v logic v lp lun ................................................................................. 30

Chng II:L thuyt tmt ............................................................................... 35

2.1 tmt hu hn ............................................................................................. 35

2.1.1 Cc tnh cht ca hm chuyn trng thi ................................................ 38

2.1.2 Cc phng php biu din tmt ........................................................ 39

2.1.3 Ngn ng on nhn c ca tmt ................................................... 40

2.2 tmt hu hn khng n nh .................................................................. 41

2.3 S tng ng ca tmt n nh v khng n nh ........................... 43

2.4 Cc tiu ho tmt hu hn........................................................................ 47

Bi tp v tmt hu hn ................................................................................... 51

Chng III:Vn phm v ngn nghnh thc .................................................. 53

3.1 Bng ch ci v cc ngn ng ...................................................................... 53

3.2 Cc dn xut v v ngn ng sinh bi vn phm .......................................... 55

3.3 Phn loi cc ngn ng ca Chomsky........................................................... 62

3.4 Tnh qui v cc tp qui ......................................................................... 70

3.5 Cc php ton trn cc ngn ng ................................................................... 73

3.6Ngn ng v tmt ...................................................................................... 76


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Bi tp v vn phm v cc ngn ng sinh ......................................................... 77

Chng IV:Tp chnh qui v vn phm chnh qui ........................................... 80

4.1 Cc biu thc chnh qui ................................................................................. 80

4.2 S tng ng ca cc biu thc chnh qui ................................................ 82

4.3 tmt hu hn v biu thc chnh qui......................................................... 83

4.3.1 Cc h bin i v cc biu thc chnh qui ............................................ 85

4.3.2 Loi b cc - dchchuyn trong cc h thng bin i trng thi ...... 87

4.3.3 Chuyn cc h chuyn trng thi khng n nh v h thng n nh......................................................................................................................... 88

4.3.4 Phng php i s ng dng nh l Arden ......................................... 90

4.3.5 Thit lp tmt hu hn tng ng vi biu thc chnh qui ............ 934.3.6 S tng ng ca hai biu thc chnh qui......................................... 96

4.4 B Bm i vi cc tp chnh qui ............................................................ 97

4.5 ng dng ca b Bm (iu kin cn ca ngn ng chnh qui) ......... 98

4.6 Cc tnh cht ng ca cc tp chnh qui ...................................................... 99

4.7 Cc tp chnh qui v vn phm chnh qui.................................................... 101

4.7.1 Xy dng vn phm chnh qui tng ng vi T cho trc ........... 101

4.7.2 Xy dng T hu hn tng ng vi vn phm chnh qui G .......... 1024.8iu kin cn v ca ngn ng chnh qui............................................... 103

4.8.1Quan h tng ng bt bin phi ..................................................... 103

4.8.2 iu kin cn v : nh l Myhill - Nerode .................................... 103

Bi tp v biu thc chnh qui ........................................................................... 105

Chng V:Cc ngn ngphi ngcnh ............................................................ 109

5.1 Cc ngn ng phi ng cnh v cc cy dn xut......................................... 109

5.2 S nhp nhng trong vn phm phi ng cnh ............................................. 1145.3 Gin lc cc vn phm phi ng cnh ........................................................ 115

5.3.1 Lc gin cc vn phm phi ng cnh ................................................. 115

5.3.2 Loi b cc qui tc rng ....................................................................... 118

5.3.3 Loi b cc qui tc n v ..................................................................... 119

5.4 Cc dng chun ca vn phm phi ng cnh............................................... 120

5.4.1 Dng chun Chomsky........................................................................... 120

5.4.2 Dng chun Greibach ........................................................................... 1225.5 B Bm cho ngn ng phi ng cnh ...................................................... 124


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5.6 Thut ton quyt nh c i vi cc ngn ng phi ng cnh ................ 127

Bi tp v ngn ng phi cnh ............................................................................ 128

Chng VI:tmt y xung .......................................................................... 130

6.1 Cc nh ngha c s.................................................................................... 130

6.2Cc kt qu on nhn bi PDA .................................................................. 133

6.3 tmt y xung PDA v ngn ng phi ng cnh.................................... 136

6.4 Phn tch c php v tmt y xung ...................................................... 141

6.4.1 Phn tch c php trn / xung ............................................................. 141

6.4.2 Phn tch c php di / ln ................................................................. 144

Bi tp v tmt y xung ............................................................................. 147

Chng VII:Vn phm LR(k) ........................................................................... 149

7.1 Vn phm LR(k) .......................................................................................... 149

7.2 Mt s tnh cht ca vn phm LR(k) ......................................................... 152

7.3 Cc tnh cht ng ca cc ngn ng .......................................................... 153

Bi tp v vn phm LR(K) ............................................................................... 155

Chng VIII:My Turing, tmt gii ni v nhng bi ton P, NP ........... 156

8.1 M hnh my Turing .................................................................................... 156

8.2 Biu din my Turing .................................................................................. 157

8.2.1 Biu din bng cc m t hin thi ...................................................... 157

8.2.2 Biu din bng th chuyn trng thi ............................................... 159

8.2.3 Biu din bng bng chuyn trng thi ................................................ 160

8.3 Thit k my Turing .................................................................................... 161

8.4 Ngn ng on nhn v Hm tnh c .................................................. 164

8.4.1 My Turing nh l mt my tnh hm s nguyn ................................ 164

8.4.2 Cc chng trnh con............................................................................ 1668.5 My Turing khng n nh ........................................................................ 167

8.6 Lun Church............................................................................................ 167

8.7 S tng ng gia vn phm loi 0 v my Turing ................................ 168

8.8 tmt tuyn tnh gii ni v vn phm cm ng cnh .............................. 171

8.8.1 tmt tuyn tnh gii ni.................................................................... 171

8.8.2 Vn phm cm ng cnh (CSG) ........................................................... 172

8.8.3 S tng ng gia LBA v CSG ..................................................... 1748.9 Bi ton dng ca my Turing .................................................................... 176


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8.9.1 Nhng bi ton khng quyt nh c ............................................... 176

8.9.2 Bi ton v s tng ng ca Post ....................................................... 178

8.10 Lp cc bi ton NP y ................................................................... 179

8.10.1 Cc lp bi ton P v NP .................................................................... 179

8.10.2 NP-y ........................................................................................... 180

Bi tp v cc bi ton quyt nh c v lp bi ton P, NP-y ............ 183

Ph lc: Hng dn v li gii cc bi tp ....................................................... 185

Danh sch cc tvit tt v thut ng.............................................................. 202

Ti liu tham kho ............................................................................................... 206


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LI NI U

L thuyt tmt ra i xut pht t nhng nhu cu ca thc tin k thut, chyu l t nhng bi ton v cu trc ca cc h thng tnh ton v cc my bin ithng tin t ng. Ngy nay, l thuyt ny c mt c s ton hc vng chc vnhng kt qu ca n c nhiu ng dng trong nhiu lnh vc khc nhau.

Song song vi l thuyt tmt, cc ngn ng hnh thc cng c tp trungnghin cu nhiu t nhng nm 50 ca th k 20 , khi cc nh khoa hc my tnhmun s dng my tnh dch t mt ngn ng ny sang ngn ng khc. Ccngn ng hnh thc to thnh mt cng c m t i vi cc m hnh tnh ton ccho dng thng tin vo / ra, ln kiu cc thao tc.

tmt v vn phm sinh ra cc ngn ng hnh thc thc cht l mt lnh vc linngnh, gp phn rt quan trng trong vic m t cc dy tnh ton v iu khin t

ng, c pht sinh trong nhiu ngnh khoa hc khc nhau t cc h thng tnhton, ngn ng hc n sinh hc. Khi nghin cu vi t cch l cc i tng tonhc, l thuyt ny cng cp n nhng vn c bn ca khoa hc my tnh, vcc kt qu nghin cu c nhiu ng dng ngay i vi cc ngnh ton hc trutng.

Ngy nay, cc l thuyt v ngn ng hnh thc, tmt v l thuyt tnh ton chnh thc ho thnh cc m hnh ton hc tng ng cho cc ngn ng lp trnh,cho cc my tnh, cho cc qu trnh x l thng tin v cc qu trnh tnh ton nichung. tmt v ngn ng hnh thc c p dng rng ri trong nhiu lnh vc

khoa hc v ng dng nh: m hnh ho, m phng cc h thng tnh ton, cc kthut dch, thng dch, tr tu nhn to, cng ngh tri thc, ...

Nhm p ng nhu cu ging dy, hc tp v nghin cu ca cc sinh vin ngnhCng ngh thng tin, chng ti bin son gio trnh tmt v ngn ng hnhthc theo hng kt hp ba l thuyt chnh: l thuyt tmt, ngn ng hnh thcv l thuyt tnh ton vi nhiu v d minh ho phong ph.

Gio trnh ny gii thiu mt cch h thng nhng khi nim c bn v cc tnhcht chung ca tmt v ngn ng hnh thc.

Chng m u trnh by cc khi nim c bn, cc tnh cht quan trng ca cc

cu trc i s, logic mnh , logic tn t v cc phng php suy lun ton hclm c s cho cc chng sau. Chng II gii thiu v l thuyt tmt, nhngkhi nim c s v cc hot ng ca tmt. Vn phm v cc ngn ng hnhthc c cp n chng III. Nhng vn lin quan n tp chnh qui,ngn ng chnh qui v tmt hu hn c trnh by chi tit chng IV.Chng V, VI nghin cu cc khi nim c s, mi quan h gia cc lp ngn ngv cc tnh cht rt quan trng ca ngn ng phi ng cnh, tmt y xung. Lpcc ngn ng phi ng cnh loi LR(k) c nhiu ng dng trong chng trnh dch,chng trnh phn tch c php c trnh by trong chng VII. M hnh tnhton my Turing, mi quan h tng ng tnh ton gia cc lp ngn ng c

cp chng cui. Trong chng ta tm hiu v phc tp tnh ton ca cch thng tnh ton, nhng bi ton quyt nh c v nhng bi ton thuc lp


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NP-y . Sau mi chng c cc bi tp h thng ho li kin thc v thng quamn hc, hc vin nm bt c cc khi nim c bn, nng cao s hiu bit vtmt v ngn ng hnh thc, ng thi pht trin kh nng ng dng chng trongnghin cu v trin khai ng dng Cng ngh thng tin. c bit trong s cc bi

tp cui chng c nhng bi c nh du * l nhng bi kh, phn ln lc trch t cc thi tuyn sinh sau i hc (u vo cao hc) ngnh cng nghthng tin trong nhng nm gn y. Hu ht nhng bi tp kh u c li hngdn hoc gii thiu cch gii phn ph lc ngi c c th tham kho v tnh gi c kh nng nm bt, gii quyt vn ca mnh.

y cng l ti liu tham kho, hc tp cho sinh vin, hc vin cao hc v nghincu sinh cc ngnh Ton Tin hc,Tin hc ng dng v nhng ai quan tm ntmt, ngn ng hnh thc v cc m hnh tnh ton ni chung.

Mc d c nhiu c gng, nhng ti liu ny chc vn khng trnh khi nhng

thiu st. Chng ti rt mong nhn c cc kin, gp ca cc ngnghip vbn c hon thin hn cun gio trnh tmt v ngn ng hnh thc. Th gp xin gi v theo a ch ca tc gi: B mn Khoa hc my tnh, Khoa Cng nghthng tin, i hc Thi Nguyn.

Cc tc gi xin chn thnh cm n cc bn ng nghip trong Vin Cng nghthng tin, Vin KH & CN Vit Nam, B mn Khoa hc my tnh, Khoa CNTT,i hc Thi Nguyn.

H Ni 2007

Ch bin

on Vn Ban


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CHNG I

M u

Chng m u gii thiu khi qut v tm lc li cc khi nim c bn, cc tnhcht v cc k hiu c s dng trong tt c cc chng sau:

Tp hp v cc cu trc i s,

Cc quan h trn tphp v quan h tng ng,

Xu k t v cc tnh cht ca chng,

Logic mnh , tn t v cc php ton logic, tn t, Cc qui tc suy lun v phng php chng minh ton hc.

1.1 Tp hp v cc cu trc i s

1.1.1 Tp hp v cc tp con

Tp hpl skt tp nhng i tng c cng mt s thuc tnh ging nhau, v dtp tt c cc sinh vin ca Khoa CNTT. Mi i tng thnh vin c gi l

phn tca tp hp.

Cc ch in hoa A, B, C,c s dng k hiu cho cc tp hp; cc ch thng a, b, c,c s dng k hiu cho cc phn t ca mt tp hp xc nh. Phn t a thuc tp A, khiu l aA, ngc li, khi a khng phi l phn t ca A, k hiu l aA.

Tp hp c th c m t theo cc cch sau:

(i) m c cc phn t.Ta c th vit cc phn t ca tp hp theo th tbt k trong cp du ngoc {, }, v d tp cc s t nhin chia ht cho 7 vnh hn 50 c th vit l {7, 14, 21, 28, 35, 42, 49}.

(ii)M t tp hp theo tnh cht ca cc phn t. V d tp cc s t nhinchia ht cho 7 v nh hn 50 c th vit:

{n | nl s nguyn dng chia ht cho 7 v nh hn 50}.

(iii)nh ngha qui. Cc phn t ca tp hp c th nh ngha thng quacc qui tc tnh ton t cc phn t bit trc. V d tp cc s giai thaca nc th nh ngha:

{fn | f0 = 1; fn = (n-1) * fn-1, n = 1, 2, }.


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Tp con v cc php ton trn tp hp

TpAc gi l tp concaB(vit A B) nu mi phn t caAu l phn tcaB.

Hai tp A v B l bng nhau (vit A = B) nu chng c cng tp cc phn t.Thng thng, chng minhA = B, chng ta cn chng minh A B v B A.

Mt tp c bit khng cha phn t no c gi l tp trng (rng), k hiu l .

Trn cc tp hp xc nh mt s php ton: php hp , php giao , php tr -v tch Ccc nh ngha nh sau:

A B = {x | x A hoc x B}, gi l hp ca hai tp hp,

A B = { x | x A v x B}, gi l giao ca hai tp hp,

A - B = { x | x A v x B}, gi l hiu ca hai tp hp. C = U - A, vi U tp v tr tt c cc phn t ang xt, c gi l phn bca A.

Tp tt c cc tp con ca A c k hiu l 2A = {B | B A}, hoc (A).

A B = {(a, b) | a A v b B}, gi l tch Cc caAvBhay cn cgi l tch t nhin ca hai tp hp.

nh ngha 1.1 Gi s S l mt tp hp bt k. Mt h cc tp con {A1, A2, An} ca Sc gi l mtphn hochca Snu:

(i) Ai Aj = , i j , cc tp con khc nhau l ri nhau,(ii) S = A1 A2 An, hp ca cc tp con chnh bng S

V d 1.1 S = {1, 2, 3, 10} c th phn hoch thnh hai tp A1 = {1, 3, 5, 7, 9}-tp cc s l v tp cc s chn A2 = {2, 4, 6, 8, 10}.

1.1.2 Tp hp v cc php ton hai ngi

Trc tin chng ta xt cu trc bao gm mt tp hp v mt php ton hai ngi.Trn tp hp S(sau ny l cc bngch) c th xc nh mt php ton hai ngi,nh nguyn thc hin gn mt cp phn t bt k a, bca Svo mt phn t duy

nht k hiu l a * b. Php ton ny cn c gi l php * (php tonsao) trn tphp tha mn cc tin sau.

Tin 1 Tnh ng ca *.Nu a, b S th a * b S.

Tin 2 Tnh kt hp.Nu a, b, c S th (a * b) * c = a * (b * c).

Tin 3 Phn t n v. Tn ti duy nht mt phn t c gi l n v e Ssao cho vi mi x S, x * e = e * x = x.

Tin 4 Phn t ngc. Vi mi phn t x S, tn ti duy nht mt phn t khiu x sao cho x * x = x* x = e. Phn t ny c gi l phn t ngc ca x.

Tin 5 Tnh giao hon.Nu a, b S th a * b = a * b.


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Lu : C nhng php ton xc nh trn tp hp khngtha mn bt k tin no nutrn. V d, N = {1, 2, 3, }- tp cc s t nhin v php tr ( a * b = ab). D dngkim tra c tt c cc tin trn u khng tha mn.

Vn m chng ta quan tm l nhng tp tha mt s hoc tt c cc tin nu trn. nh ngha 1.2

(i) Tp S v php ton hai ngi * c gi l cu trc na nhm(semigroup), gi tt l na nhmnu tha mn tin 1 v 2.

(ii) Tp Sv php ton hai ngi * c gi l monoid nu tha mn cctin 1, 2 v 3.

(iii) Tp Sv php ton hai ngi * c gi l cu trc nhm (group) nutha mn tin 1, 2, 3 v 4.

(iv) Na nhm (monoid hoc nhm) c gi l na nhm (monoid hocnhm tng ng) giao hon hocAbeliannu chng tha mn thm tin 5.

Mi quan h gia cc nhm, monoid vnhm, k hiu l G = (S, *), c th hinthng qua cc tin tha mn nhtrn hnh H1-1.

V d 1.2

(i) Tp cc s nguynZvi php + to thnh cu trc nhm Abelian.

(ii) Zvi php * (nhn) to thnh cu trc monoidAbelian(n khng phi lnhm v tin 4 khng tha mn).

(iii) 2A tp tt c cc tp con ca A (A ) to thnh monoid giao hon.(phn t n v l tp ).

(iv) Tp tt c cc ma trn 2 2 vi php nhn l monoid nhng khng giao hon.

Hnh H1-1 Tp hp v mt php ton hai ngi

Na nhm (Semigroup)

Na nhm Abelian Monoid

Nhm (group)Monoid Abelian

Nhm Abelian

Tp hp (Set) Khng c php ton

Tin 1, 2

3

4

5

3

4 5

5


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Cc s trn cc cnh ca th trn l cc tin c tha mn, v d na nhmm tha mn tin 3 s to thnh monoid, cn nu tha mn tin 5 s thnhna nhm Abelian (nhm giao hon).

Trong ton hc cng nh trong khoa hc my tnh,nhiu khi chng ta phi giiquyt nhng vn lin quan n nhng cu trc gm mt tp hp v hai phpton nh nguyn (hai ngi). Gi s tp Sv hai php ton *, c th tha mn 11tin sau.

Tin 1 - 5 Php * tha mn 5 tin nu trn.

Tin 6 Tnh ng ca php .Nu a, b S th a b S.

Tin 7 Tnh kt hp ca .Nu a, b, c S th (a b) c = a (b c).

Tin 8 Phn t n v. Tn ti duy nht mt phn t c gi l n v e S

sao cho vi mi x S, x e = e x = x.Tin 9 Nu S v * tha mn cc tin 1-5 th vi mi x thuc S, x e, tnti mt phn t duy nhtxtrong S sao cho x x = x x= e, trong el n vca .

Tin 10 Tnh giao hon. Nu a, b S th a b = b a.

Tin 11 Tnh phn phi. Vi mi a, b, c S, a (b * c) = (a b)*( a c).

Tp hp xc nh vi mt hoc hai php ton hai ngi trn c gi l h i s.Nh trn cp, h i s c mt php ton to thnh nhm, na nhm v

monoid. Sau y chng ta xtnhng cu trc i s c hai php ton.nh ngha 1.3

(i) Mttp hp v hai php ton *, to thnh cu trc vnh(gi tt l vnh), nu

a/ L nhm giao hon vi php *,

b/ Php tha mn tin 6, 7, v 11.

(ii) Vnh c gi l vnh giao honnu tha mn tin giao hon (10)i vi php .

(iii) Vnh giao hon c n v nu l vnh giao honv tha mn tin n v (tin 8) i vi .

(iv) Tp hp vi hai php ton tha mn c 11 tin c gi l trng.

V d 1.3

Tp cc s nguynNvi php cng v php nhn (tng ng vi * v ) tothnh vnh giao hon c n v (0 l n v ca php cng, 1 - n v ca

php nhn).

Tp tt c cc s hu t (cc phn s dng p/q, p, q l cc s nguyn, q 0)vi 2 php cng, nhn to thnh trng.


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Tp 2A, A vi hai php , s tha mn cc tin 1, 2, 3, 5, 6, 7, 8,9, 10 v 11 nhng khng phi l nhm, khng phi l vnh v cng chng

phi l trng. N l monoid Abelian i vi c 2 php ton.

Quan h gia cc h i s c m t th nh trong hnh H1-2.

Hnh H1-2 Cc cu trc i s

1.3 Quan h v quan h tng ng

Quan h l khi nim cs quan trng trong khoa hc my tnh cng nh trongcuc sng. Khi nim quan h xut hin khi xem xt gia cc cp i tng v sosnh mt i tng ny vi i tng khc, v d, quan h anh em gia haingi. Chng ta c th biu din mi quan h gia av bbi cp c sp xp(a, b)th hin al anh ca b, chng hn. Trong khoa hc my tnh, khi nim quanh xuthin khi cn nghin cu cc cu trc ca d liu.

nh ngha 1.4 Quan h R trn tp Sl tp con cc cp phn t ca S, R S S.Khi xvyc quan hRvi nhau ta c th vitx R yhoc (x, y) R.

V d 1. 4 Trn tp Z nh ngha quan h R:x R ynu x = y + 5.nh ngha 1.5 Quan h trn tp Sc cc tnh cht sau:

(i) Phn x.Nu xRx vi mi x S.

(ii) i xng. Vi mi x, y S, nu xRy th yRx

(iii) Bc cu. Vi mi x, y, z S, nu xRy v yRz th xRz.

nh ngha 1.5 Quan h trn tp Sc gi l quan h tngngnu n ctnh phn x, i xng v bc cu.

8

Vnh (ring)

Vnh giao hon Vnh c n v

Vnh giao hon c n v

Trng

Nhm Abelian Tin 1-5

Tin 6, 7, 11

10

1-11

8

10


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V d 1.5

a/ Trn tp S, nh ngha xRy x = y. D kim tra c ba tnh cht trn quanh = u tha mn, vy = l quan h tng ng.

b/ Trong tp cc sinh vin ca Khoa CNTT ta c th thit lp quan hR: sinhvin aquan hRvi sinh vin bnu c hai u hc cng mt lp. Hin nhin quanh ny cng l quan h tng ng.

c/ Trn tp S, nh ngha xRy x > y. D kim tra tnh i xng khngc tha mn, do quan h > khng phi l quan h tng ng.

nh ngha 1.6 Gi s R l quan h tng ng trn tp Sv a S. Lp ccphn t tngngvi a, k hiu l Ca = {b | b S v a R b}. Lp Cac th khiu l [a]R.

nh l 1.1 Tp tt c cc lp tng ng ca quan hR trn tp Sto thnh mtphn hoch ca S.

Chng minh: Chng ta cn chng minh rng

(i) S = Sa

aC

(ii) Ca Cb = nu Ca v Cb khc nhau, Ca Cb.

Gi ssSvRl quan h tng ng nnsCs (sRs - tnh phn x). V CsSa

aC

nn S Sa

aC

. Mt khc, theo nh ngha ca Ca, ta lun c Ca S vi mi aca S.

Vy Sa

aC

S. T suy ra (i).

Trc khi chng minh (ii), chng ta ch rng

Ca = Cbnu aRb (1.1)

Bi v aRbth bRa doRi xng. Gi s d Ca, theo nh ngha ca Ca, aRd. Hnna bRav aRd, nn cn c vo tnh cht bc cu ca quan h tng ng, ta cbRd. Suy ra dCb. T chng ta c Ca Cb. Tng t chng ta c th chngminh ngc li Ca Cb. Kt hp c hai chng ta c (1.1).

Chng ta chng minh (ii) bng phn chng. Gi s Ca Cb, ngha l tn tiphn t d Ca Cb. Khi d Cath aRd. Tng t d Cbth bRd. V R i xngnn dRb. Kt hp aRd, dRb suy ra aRb. S dng (1.1) ta c Ca = Cb, vy mu thunvi gi thit.

Kt lun: Cc lp tng ng l trng nhau hoc ri nhau.

V d 1.6

a/ Trn tp s t nhin N, nRm khi mv nu chia ht cho 2 s c hai lptng ng l tp cc s chn v tp cc s l.

b/ Cc lp tng ng ca quan h tng ng trong v d 1.5 (b/) chnhl cc lp hc c xp trong mt Khoa.


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Vn tip theo c quan tm nhiu khi nghin cu cc hnh vi ca mt h thngtnh ton, c bit cc mi quan h gia cc ph thuc hm l bao ng ca mtquan h.

Cho trc quan h R khng c tnh phn x, khng bc cu; phi b sung baonhiu cp quan h vi nhau Rc cc tnh cht . V d R = {(1, 2), (2, 3), (1, 1),(3, 3)} trn tp {1, 2, 3}.Rkhng phn x v (2, 2) R.Rkhng c tnh bc cu v(1, 2), (2, 3) R, nhng (1, 3) R. Trong s cc quan h m rng Rv c tnh

phn x, bc cu, v d T = {(1, 2), (2, 3), (1, 3), (1, 1), (2, 2), (3, 3)}, chng taquan tm nht ti quan h nh nht chaRm c cc tnh cht .

nhngha 1.7 Gi s Rl mt quan h trn tp S.Bao ng bc cucaR, khiuR+l quan h nh nht chaRv c tnh bc cu.

nh ngha 1.8 Gi s R l mt quan h trn tp S. Bao ng phn x,bc cu

caR, k hiuR*l quan h nh nht chaR, ctnh phn x v bc cu. xy dngR+v R*, chng ta nh ngha quan h hp thnh (ghp) ca hai quanhR1, R2.

nh ngha 1.9 Gi sR1, R2l hai quan h trn tp S.Hp thnh ca R1, R2,k hiuR1 R2c nh ngha nh sau.

(i) R1 R2 = {(a, c) | b S, aR1b v bR2c}

(ii) R12

= R1 R1

(iii) R1n = R1

n-1 R1, vi n 2.

Da vo cc tnh cht ca tp hp v quan h, chng ta c nh l khng nh stn ti ca bao ng bc cu ca mi quan h.

nh l 1.2 Gi s Sl tp hu hn,Rl quan h trn S. Bao ng bc cu R+caRluntn ti v R+ = R1 R2 R3... Bao ng phn x, bc cu ca R l R* = R0 R+. Trong R0 = {(a, a) | aS} l quan h ng nht.

V d 1.7 R = {(1, 2), (2, 3), (2, 4)} l quan h trn {1, 2, 3, 4}. Tm R+

Chng ta tnh Ri. Lu , nu c (a, b) v (b, c) R th (a, c) R2.

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

R2 = R R = {(1, 2), (2, 3), (2, 4)} {(1, 2), (2, 3), (2, 4)}

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

R3 = R2 R = {(1, 3), (1, 4)} {(1, 2), (2, 3), (2, 4)}

= (khngc cp no trong R2c th ghp vi cc cp trong R).

R4

= R5

= . . . = .

Vy, R+ = R R2 = {(1, 2), (2, 3), (2, 4), (1, 4), (1, 3)}.

V d 1.8 R = {(a, b), (b, c), (c, a)} l quan h trn {a, b, c}. Tm R*.

Trc tin tm R+.


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R2 = {(a, b), (b, c), (c, a)} {(a, b), (b, c), (c, a)}

= {(a, c), (b, a), (c, b)}

R3

= {(a, c), (b, a), (c, b)} {(a, b), (b, c), (c, a)}

= {(a, a), (b, b), (c, c)}

R4 = {(a, a), (b, b), (c, c)} {(a, b), (b, c), (c, a)}

= {(a, b), (b, c), (c, a)} = R.

Do vy, R* = R0 R+ = R0 R R2 R3

= {(a, b), (b, c), (c, a), (a, c), (b, a), (c, b), (a, a), (b, b), (c, c)}

1.4 Hm s

Khi nim hm s xut hin khi chng ta mun xt mi quan h tng ng duynht ca i tng vi mt i tng cho trc.

nh ngha 1.10 Hm hoc nh x f() t tp X vo tp Y l qui tc gn tngng mi phn tx trongXbng mt phn t duy nht trong Y, k hiu lf(x). Phntf(x)c gi l nh caxthng qua f(). Hmf()c k hiu f: X Y.

Cc hm c th nh ngha bi:

(i) Tp cc nh ca cc phn t,

(ii) Qui tc tnh tng ngf(x)vx.

Cho trc f: X Y v A X. K hiu f(A) = {f(a) | a A} Y.V d 1.9

a/ f: {1, 2, 3, 4}{a, b, c} c th nh ngha f(1)= a, f(2) = b, f( 3) = f(4)= c.

b/ f: Z Z c xc nh bi cng thcf(x) = x2 + x.

nh ngha 1.11 Cho trc hm f: X Y.

(i) fc gi l n nh (one-to-one) nux y th f(x) f(y).

(ii)fc gi l ton nh (onto) nu mi phn t yca Yu l nh ca

mt phn t no caX, ngha lf(X) = Y.(iii)fc gi lsong nhnufva l n nh v l ton nh.

V d 1.10 f: Z Z c cho bi f(n) = 2*n l n nh nhng khng phi hmton nh v cc s nguyn l khng l nh ca s no c. f(X)l tp con thc s caY.

nh l sau gip chng ta phn bit c s khc nhau cbn gia tp hu hn vtp v hn.

nh l 1.3 (Nguyn l chung b cu) Gi s Sl tp hu hn. Hmf: SSl

n nh khi v ch khi l hm ton nh.


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Trong v d 1.10, min xc nh ca hm khng phi l hu hn, nn tn ti hmf()l n nh nhng khng phi l hm ton nh.

1.5 Logic mnh v tn t

1.5.1 Logic mnh

Mt mnh (logic, hay ton hc) l mt pht biu (mt cu) hoc ng hoc sai,nhng khng th l c hai. Khi n ng ta ni rng cng thc c gi tr chn l lng (Ttrue hoc 1), ngc li l sai (FFalse hoc 0).

V d 1.11Hy xt cc cu sau:

1. H ni l th ca nc Vit Nam.

2. Bnh phng ca hai l tm.

3. Logic ton l kh

4. Hy trt t!

Hai cu u l mnh v cu mt c gi tr T (hoc gi tr 1), cn cu th hai cgi tr lF (hoc gi tr 0). Cu th ba khng phi l mnh logic v logic tonc th l kh i vi mt s ngi, nhng li c th l khng kh i vi mt sngi khc. Cu cui l mt mnh lnh, khng th gn gi tr chn l cho n c.

Mnh n (mnh nguyn t) l mt mnh khng th tch nh hn c,ngha l khi b i bt k mt tno trong cu th n s khng cn l mnh .

Mnh phc hp (gi tt l mnh ) l mt mnh c to lp t cc mnh n v cc lin t: v (AND), hoc (OR), ph nh (NOT), suy ra, hay ko theo(IF THEN ), v php tng ng. Cc lin t trong cc mnh logic cnc gi l cc php ton mnh hay php ton logic. Ta k hiu Ml tp ttc cc mnh .

(i) Php ph nh (NOT)

NuPMth khng P(ph nh ca P), k hiu l P(hocP), l mt mnh nhn gi tr TnuPc gi trFv gi trFnuPc gi tr T.

Bng 1.1 gi tr ca php ph nh

P P

T F

F T

(ii) Php tuyn, php hoc (OR)

NuP, Q Mth tuyn ca P v Q (Phoc Q), k hiu lPQ, l mt mnh nhn gi trF khi v ch khi c Pv QlF.


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Bng 1.2 gi tr ca php tuyn

P Q P Q

T T T

T F T

F T T

F F F

(iii) Php hi, php v (AND)

NuP, Q Mth hi ca P v Q (Pv Q), k hiu l P Q, l mt mnh nhn gi tr T khi v ch khi c Pv Ql ng (T).

Bng 1.3 gi tr ca php hi

P Q P Q

T T T

T F F

F T F

F F F

(iv) Php ko theo (IF THEN )NuP, Q M, mnh P ko theo Q(NuPth Q), k hiu lPQ, l mtmnh nhn gi trF khi v ch khi P l T v QlF.

Bng 1.4 gi tr ca php ko theo

P Q P Q

T T T

T F F

F T T

F F T

(v) Php tng ng(tng ng)

Ngoi nhng php ton trn, ngi ta thng s dng php tng ng mnh (php khi v ch khi) c nh ngha nh sau:

P Q tng ng vi (P Q) (Q P) c cc gi tr c nh nghanh sau:


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Bng 1.5 gi tr ca php tng ng

P Q P QT T T

T F F

F T F

F F T

1.6.2 Cng thc mnh

nh ngha 1.12 Bin mnh l mt khiu biu din cho mt mnh .

Mt bin mnh khng phi l mt mnh nhng c th thay th bng mtmnh bt k.

Ta thng s dng cc ch vit hoa nh:P, Q, R, S cho cc bin mnh .

nh ngha 1.13Mt cng thc mnh (gi tt l cng thc) c nh ngha qui nh sau:

(i) NuPl mt bin mnh thPl mt cng thc.

(ii) Nu l cng thc th cng l cng thc.

(iii) Nu v l cng thc th (), (), (), () cng lcng thc.

(iv) Dy cc bin mnh l cng thc khi v ch khi n c thit lp theocc qui tc (i) - (iii).

Lu :

1. Mt cng thc cng nh bin mnh , n khng phi l mnh , nhngnu th cc mnh vo ch cc bin mnh tng ng trong cng thcth n s tr thnh mt mnh . Do vy, gi tr ca mt cng thc c thc tnh bng bng gi tr chn l khi thay cc gi tr ca cc mnh vo

ch cc bin mnh tng ng. Hin nhin l nu cng thc c n binth bng gi tr ca s c 2nb gi tr.

2. Trong cc ngn ng lp trnh, cng thc mnh thng c gi l biuthc logic. Gi tr ca biu thc logic l gi tr kiu Boolean (c gi tr ThocF) v c xc nh tng t nh i vi cng thc nh trn.

V d 1.12 Tnh gi tr ca cng thc = (P Q) (P Q ) ( Q P)


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Bng 1.6 gi tr ca cng thc

P Q P Q P Q (P Q) P Q Q P

T T T T T T TT F T F F T F

F T T T T F F

F F F T F T F

Ngoi phng php lp bng tnh gi tr nhtrn, ta cng c th suy lun l ch ng khi c ba cng thc P Q, P Q, Q P, ngha l khi c Pv Qng.

nh ngha 1.14 Mt cng thc l hng ng (mt tautology) l cng thc lun c

gi tr T (ng) vi mi trng hp gn gi tr cho cc bin mnh .

V d: PP, (P Q) P l hai mnh hng ng.

nh ngha 1.15 Hai cng thc , c xy dng t cc bin mnh P1, P2, Pnc gi l tng ng (logic), k hiu l , nu cng thc l hng ng.

Sau y l cc tnh cht c bn (cc lut c s) ca cc cng thc tng ngc s dng nhiu trong suy lun, chng minh cc h hnh thc logic ton hc.

Bng 1.7 Cc lut ng nht logic

I1 Lut lu ng

P P P, P P P

I2 Lut giao hon

P Q Q P, P Q Q P

I3 Lut kt hp

P (Q R) (P Q) R, P (Q R) (P Q) R

I4 Lut phn phi

P (Q R) (P Q) (P R), P (Q R) (P Q) (P R)I5 Lut hp th

P (P Q) P, P (P Q) P

I6 Lut De Morgan

(P Q) Q P, (P Q) Q P

I7 Lut ph nh kp

( P) P

I8 P P T, P P FI9 P TT, P F P, P T P, P FF,


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I10 (P Q) (P Q) P

I11 Lut nghch o

(P Q) Q P

I12 (P Q) ( P Q)

I13 (P Q) (P Q) (Q P)

Trong ,P, Q, Rl cc cng thc.

1.5.3 Dng chun ca cc cng thc

nh ngha 1.16 Mt tuyn s cpl mt cng thc ch gm tuyn ca cc hngthc s cp v mt hi s cpl mt cng thc ch gmhi ca cc hng thc scp. Trong hng thc s cpl bin mnh hoc ph nh ca bin mnh .

V d: P Q l tuyn s cp vP Q hi s cp.

nh ngha 1.17 Mt cng thc dng chun tuynnu n l tuyn ca cc his cp.

V d: (P Q) R l dng chun tuyn.

nh ngha 1.18 Mt cng thc dng chun hinu n l hi ca cc tuyn scp.

V d: (P Q) R l dng chun hi.nh l 1.4 Mi cng thc u tn ti dng chun tuyn (hoc hi) tng ng.

nh l ny d dng c chng minh thng qua thut ton sau.

Thut ton 1.1 Chuyn mt cng thc v dng chun tuyn (tng t i vi dngchun hi).

(i) Loi b cc php, (s dng cc lut ng nht nh trong bng 1.6)

(ii) S dng lut De Morgan loi b php ph nh ng trc cc hihoc tuyn ca cc cng thc. Trong kt qu, php ph nh ch c th xut

hin trc cc bin mnh .(iii)p dng lut phn phi (I6) loi b hi ca cc tuyn (hoc hi). Cngthc cui cng s l tuyn cacc hi s cp.

V d 1.13 Tm dng chun tuyn ca cng thc = (P (Q R)) (P Q)

Gii:

(P (Q R)) (P Q)

(P (Q R)) (P Q) (bc (i) p dng I12)

(P (Q R)) (P Q) (bc (ii) p dng I7)

(P Q) (P R) P Q (bc (iii) p dng I4,I3)


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Vy, dng chun tuyn ca l (P Q) (P R) P Q.

Lu : Mt cng thc c th c nhiu dng chun tuyn khc nhau. V d P Qbn thn l dng chun tuyn (P Q) v n cn cdng chun tuyn khc l

(P Q R) (P Q R)

Mt cu hi t ra l mt cng thc c th chuyn v mt loi dng chun no ml duy nht hay khng?. Cu tr li l c loi dng chun tuyn c gi l chuntuyn (hi) chnh tc hay chun tuyn (hi tng ng) y biu din duynht cho mi cng thc.

nh ngha 1.19 Mthi s cp cc tiutrn tp cc bin mnh P1, P2, Pnlmt cng thc dng Q1 Q2 Qn, trong Qil Pihoc l Pi. Mt cng thcdng chun tuyn chnh tc (chun tuyn y ) nu n l tuyn ca cc hi scp cc tiu.

nh l 1.5 Mi cng thc u chuyn tng ng c v dng chun tc tuyn.

nh l c chng minh thng qua thut ton sau.

Thut ton 1.2 Chuyn mt cng thc v dng chun tuyn chnh tc.

(i) p dng thut ton 1.1 a cng thc v dng chun tuyn.

(ii) Loi b i nhng hi s cp l hng ng (nh P P)

(iii) Nu Pihoc Pikhng c mt trong mt hi s cp th thay bng

( Pi) (Pi)

(iv) Lpli bc (iii) cho n khi tt c cc hi s cp u l cc tiu.

V d 1.14 Tm dng chun tuyn chnh tc ca cng thc = (P Q) (P R)

Gii:

(P Q) (P R)

(P Q) (P R)

((P Q R) (P Q R)) ((P R Q) (P RQ))

(P Q R) (P Q R) (P Q R) (P Q R).

Vy dng chun tuyn chnh tc ca l

(P Q R) (P Q R) (P Q R) (P Q R).

Ta khng nh, mi cng thc u tng ng vi mt dng chun tuyn chnhtc duy nht.

Mi hi s cp cc tiu Q1 Q2 Qnc th biu din thnh dy a1a2an,trong ai= 0 nu Qi = Pi, v ai= 1 nu Qi = Pi. Do vy, dng chun tuyn chnh tcca mt cng thc c th vit thnh tng ca cc xu ca {0, 1}. V d, dng chun tuynchnh tc ca cng thc v d trn c vit thnh 111 + 110 + 011 + 001.


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T dng biu din nh phn nu trn ta thy dng chun tuyn chnh tc v bnggi tr chn l ca mi cng thc c miquan h cht ch vi nhau. Cch xc nhdng chun tuyn chnh tc ca cng thc da vo bng gi tr c thc hin ngin nh sau:

Trong bng gi tr ca cng thc , nhng hng m c gi tr ng (T)s xcnh tng ng cc hi s cp cc tiu tng ng theo dng biu din nh nguynnu trn. Trong , T ng vi 1 v F ng vi 0 (sai). Chun tuyn chnh tc cacng thc chnh l tuyn ca nhng hi s cp cc tiu ng vi nhng hng .

V d 1.15 Tm dng chun tuyn chnh tc ca cng thc c cho nh bnggi tr sau:

Bng 1.8 Cc gi tr ca cng thc

P Q R

T T T T

T T F F

T F T F

T F F T

F T T T

F T F F

F F T F

F F F T

Gii:

Cng thc nhn gi tr ng trong cc hng th 1, 4, 5 v 8. Cc hi s cp cc tiu tngng vi nhng hng l (P Q R), (P QR), (PQR) v (PQR). Vydng chun tuyn chnh tc ca s l

(P Q R) (P QR) (PQR) (PQR)

Mt khc, ta d nhn thyi ngu ca dng chun tuyn s l dng chun hi.Tng t nh trn, ta c cc nh ngha ca cc dng i ngu nh sau:

nh ngha 1.20 Tuyn s cp cc i ca cc bin mnh P1, P2, Pnl mt cngthc dng Q1 Q2 Qn, trong Qil Pihoc l Pi. Mt cng thc dng chunhi chnh tc hay (chun hi u )nu n l hi ca cc tuyn s cp cc i.

Lu : D dng suy ra nu l dng chun tuyn chnh tc th s l dng chunhi chnh tc.

Tng t nh i vi dng chun tuyn chnh tc,ta cng khng nh c l mi cngthc u chuyn tng ng c v dng chun hi chnh tc v l dng chunduy nht.

V d 1.16 Tm dng chun hi chnh tc ca cng thc = P (Q R)


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Gii:Trc tin tm = (P (Q R))

(P (Q R))

P(Q R) P (QR) PQ R

Vy, dng chun hi chnh tc ca s l (PQ R) PQ R.

Mt khc, ta cng c th da vo mi lin quan gia bng gi tr chn l ca cngthc vi dng chun hi chnh tc ca cng thc thit lp cc tuyn s cpcc i.

Trong bng gi tr ca cng thc , nhng hng m c gi tr sai (F)s xcnh tng ng cc tuyn s cp cc i tng ng theo dng biu din nh phn.Trong , T ng vi 0 v F ng vi 1. Hi chnh tc ca cng thc chnh l hi canhng tuyn s cp cc i tng ng vi nhng hng .

V d 1.17 Tm dng chun hi chnh tc ca cng thc c cho trong bng gi tr sau:

Bng 1.9 Cc gi tr ca cng thc

P Q R

T T T TT T F FT F T FT F F TF T T TF T F FF F T FF F F T

Gii:Cng thc nhn gi tr sai trong cc hng th 2, 3, 6 v 7. Cc tuyn s cp cc i tngng vi nhng hng l (P Q R), (P QR), (PQ R) v (P Q R).Vy dng chun hi chnh tc ca s l

(P Q R) (P QR) (PQR) (P Q R)

1.5.4 Cc qui tc suy din trong tnh ton mnh

Trong lp lun logic, ta thng da vo mt s mnh c cng nhn (cc githuyt, tin ) l ng suy dn ra nhng mnh ng khc. Nhng mnh suy ra c gi l h qu, cc tnh chthay nh l.

Cc qui tc suy din chnh l cc cng thc hng ng (tautology) dng kotheo: PQ, trong Pl gi thit v Ql kt lun.

ph hp hn vi cc qui tc chng minh trong cc chng trnh, chng ta cth vit cc cng thc dng mnh Horn: P1 P2 PnQ di dng:


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P1

P2

Pn

Q

Di y l bng cc qui tc suy din c bn trong lp lun logic ton hc, c sdng trong lp trnh logic, trong suy din v chng minh ton hc, ...

Bng 1.10 Cc qui tc suy din

Cc qui tc Cng thc dng ko theo

RI1: Gia tng

P P (P Q)P QRI2: n gin ho

P Q (P Q) PP

RI3: Modus ponens

P

P Q (P (P Q)) Q

QRI4: Modus tollens

Q

P Q (Q (P Q)) PP

RI5: Tam on lunP

P Q (P (P Q)) Q

QRI6: Suy lun bc cuP QQ R ((P Q) (Q R))) (P R)P R

RI7: nh kin thit

(P Q) (R S)P R ((P Q) (R S) (PR)) (QS)

Q SRI8: nh i thit


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(P Q) (R S)Q S ((PQ)(RS)(QS)) (P R)

P R

V d 1.18 Hy chng minh rng

P QQ RP SR

S

Gii:Theo RI7ta c

P QQ R

P RMt khc, R(R), kt hp vi kt lun trn v p dng lut RI4 (Modus tollens)ta c

(R)

P R

PS dng kt lun ny cng vi gi thit P S v p dng RI5 suy ra iu phichng minh.

P

P S

S

1.6 Tn t (v t) v cc lng t

Trong suy lun, chng minh ton hc v trong qu trnh tnh ton, ta thng sdng nhng mnh c cc bin. V d "x > 5", "x = y + 10". Nhng mnh nyc th ng, sai tu thuc vo cc gi tr ca cc bin x, y. Mt khc, mnh "x > 5" c hai phn:

+xl bin, l ch im m mnh khng nh,

+ "ln hn 5" - ni v tnh cht caxv cn c gi l tn t(hoc v t).

Ta c th k hiu P(x) "x ln hn 5". Pht biu P(x) cn c gi l hm tn thay hm v t. Khi P(4) = F, P(8) = T.


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Tng qut ho vi nbin:x1, x2, . . . xn, P(x1, x2, . . . xn)l hm mnh vi bn-phn t, trong Pl tn t.

Cc hm mnh thng xc nh gi tr ng, sai trong mt tp cc gi tr ca cc

bin: vi tt c hay vi mt s phn tno . Ton t xc nh s lng cgi l cc lng t. C hai loi lng t:

+ Lng t tng qut: vi mi phn t (trong v tr),

+ Lng t tn ti: vi mt phn t no trong phm vi xc nh ca bin.

nh ngha 1.21 Cho P(x) l hm xc nh trn min gi tr D.

a/ Vi mi x thuc D, P(x) gi l mt pht biu tng qut v c k hiu:

x D, P(x), trong k hiu l lng t (lng t) vi mi.

Mnh :x D, P(x) ng nu P(x) ng vi mi x D.

b/ Vi x no thuc D, P(x) gi l pht biu tn ti v c k hiu:

x D, P(x), trong l lng t (lng t) tn ti.

Mnh : xD, P(x)ng nu P(x)ng vi t nht mt gi trx no trongD.

V d 1.19

a/ Vi mi s thc x, x20 l mnh hng ng, vit ngn gn

x R, x2 0, vi R- tp cc s thc.

b/ Vi mi s x, c mt s y, x + y = 0 c th vit x R, y R, x + y = 0.

Lu :

chng minh x D, P(x) l ng th phi ch ra rng vi mi x Dth P(x) u ng.

chng minh x D, P(x) l ng th ch cn ch ra rng P(x) ngvi t nht mt gi tr xno thuc D, ngha l tm c mt gi trx P(x) ng.

chng minh x D, P(x) l sai th ch cn tm ra mt gi tr xthucD m P(x) l sai.

chng minh x D, P(x) l sai th phi ch ra rng vi mi xthuc Dth P(x) u sai.

Khi min gi tr D c xc nh trc, ta c th vit ngn gn cc hm lngt nh sau: x P(x) hoc (x)P(x) i vi lng t tn ti v x P(x)hoc (x) P(x) i vi lng t vi mi.

Tng t nh cc cng thc mnh nu trn, cc hm tn t (cng thc tn t)c xy dng t cc bin mnh , cc php ton logic v hai php lng t , .i vi cc cng thc tn t chng ta c cc qui tc suy din sau.


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Bng 1.11 Cc qui tc suy din

I14 Phn phi ca i vi php

x (P(x) Q(x)) x P(x) x Q(x)

x (R Q(x)) Rx Q(x)

I15 Phn phi ca i vi php

x (P(x) Q(x)) x P(x) x Q(x)

x (RQ(x)) Rx Q(x)

I16 (x (P(x)) x (P(x))

I17 (x (P(x)) x (P(x))

I18

x (P(x) R) x P(x) R

I19 x (P(x) R) x P(x) R

RI9 x (P(x)) x (P(x))

RI10 x P(x) x Q(x) x (P(x) Q(x))

RI11 x (P(x) Q(x)) x P(x) x Q(x)

RI12 x P(x)

P(c) cl mt phn t P(c) l ng

RI13 P(c)x P(x)

Trong , P(x), Q(x) l cc hm tn t ca bin mnh xc gi tr thuc mtmin xc nh cho trc vRl mt cng thc c lp vi cc bin mnh x.

t li bin trong cc cng thc tn t v kt hp hai qui tc I 14v I15v chng tasuy ra:

x P(x) x Q(x) x P(x) y Q(y)

x (P(x) y Q(y)) v y Q(y) c lp vi x

x y (P(x) Q(y)) v P(x) c lp vi y.

x P(x) x Q(x) x P(x) y Q(y)

x (P(x) y Q(y)) v y Q(y) c lp vi bin x

x y (P(x) Q(y)) v P(x) c lp vi bin y.

Lu : Khi min xc nh D l hu hn, |D| = n, khng mt tnh tng qut ta c thvit D = {x1, x2, , xn}. Khi ,

x D, P(x) = P(x1) P(x2) P(xn)

x D, P(x) = P(x1) P(x2) P(xn)


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1.7 Cc phng php chng minh

H thng ton hc bao gm cc tin , nh ngha v nhng khi nim (thng lkhng nh ngha hnh thc c). Trong

+Tin l nhng mnh c tha nhn l lun ng v khng cn phichng minh.

+T cc tin v cc nh nghac th pht biu thnh cc nh l. nhl cn c chng minh da vo cc gi thit v cc lut suy din tng ng nutrn.

+B c th xem nhmt nh l thng c s dng chng minh ccnh l khc.

+H qu (kt lun)l mnh c suy ra t cc nh l, cc tnh cht.

Vic khngnh tnh ng/sai ca mt mnh (nh l) c gi l chng minhnh l.

V d 1.20 Trong hnh hc Euclidechng ta bit c nhng tin phi ctha nhn nh:

Tin : Cho trc hai im phn bit trn mt phng, c ng mt ng thng iqua hai im .

Khi nim im, ng thng, ... c nh ngha khng tng minh trong ccpht biu, nh ngha, tin .

T h tin v cc khi nim cs, ngi ta a ra cc nh ngha v nhng khinim mi nhtam gic ng dng, cc gc k, b nhau, ...

nh l:Nu hai cnh ca mt tam gic bng nhau th hai gc i chng cng bng nhau.

H qu:Tam gic c cc cnh bng nhau th cc gc bng nhau.

Tm li, cc nh l thng c dng:

Vi mi x1, x2, . . . xn, nu P(x1, x2, . . . xn) th Q(x1, x2, . . . xn) (*)

Hay vit P(x1, x2, . . . xn) Q(x1, x2, . . . xn), vi x1, x2, . . . xn D

chng minh nh l (*) chng ta c th thc hin mt trong cc cch sau:

Da vo tnh cht ca php ko theo trong logic mnh ta c: nuP(x1, x2, ...xn) = F (gi thit sai) th nh l (*) l lun ng. Do vy, ta ch cnxt cc trng hp P(x1, x2,... xn) = T (gi thit ng). Nu Q(x1, x2, ..., xn) = Tc suy ra t cc tin , nh l, hay cc nh l khc c chng minhth nh l trn c chng minh. Phng php ny c gi l chng minhtrc tip.

Phng php th hai l chng minh gin tip (hay phn chng). Gi sP(x1, x2, , xn) = T v Q(x1, x2, . . ., xn) = F. Lc xem P, Q nhlcc nh l, kt hp cc tin , nh l, hay cc nh l khc c

chng minh dn ra iu bt hp l (mu thun). nh l trn cchng minh v ta c:


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p q q p

Chng minh bng phng php lp lun, suy din - lp lun loi tr cctrng hp.

Chng minh bng phng php qui np ton hc. Gi thit mnh S(n)xc nh trn mi s nguyn dng. chng minh mnh S(n) ngth chng ta thc hin nh sau:

(i) [Bc cs] Kim tra xem S(1) = T?. Nu ng vi cc trnghp cs th thc hin bc tip theo.

(ii) [Gi thit qui np] Gi thit S(k) = T vi mi k < n.

(iii) [Bc qui np] Da vo gi thit qui np v cc nh ngha, cctnh cht c chng minh lin quan n mnh S(k + 1), nuchng ta khng nh c S(n + 1) = T th kt lun c mnh trn ng vi mi nnguyn dng.

V d 1.21 Hy chng minh rng:

1. Vi mi s thc d, d1, d2, x nu d = min{ d1, d2} l gi tr cc tiu cad1, d2v x d th x d1 v x d2.

Chng minh: T nh ngha ca hm minta c d d1 v d d2. Tip theo t x dv d d1 suy ra x d1v quan h c tnh bc cu. Tng t ta c x d2. y l

phng php chng minh trc tip.

2. Vi mi s thc x, y nu x + y 2 th hoc x 1, hoc y 1.

Chng minh:(Phn chng) Gi s kt lun l sai, ngha l x < 1 v y < 1. Khi x + y < 2, iu ny dn n nghch l (p p l mnh mu thun). Do vymnh trn l ng.

3. Xt bi ton sau: Chng trnh c li v lp trnh vin pht hin ra:

+ Li pht hin module 17 hoc module 24. + y l li v tnh ton s hc,+ Module 24 khng c li.

Vy module 17 c li v tnh ton s hc!

Chng minh: chng minh khng nh trn chng ta s dng phng php lplun v suy din dng:

Nu p1v p2v ... pnth q.

4. Chng minh rng n! 2n1vi n = 1, 2, . . . (**)

Chng minh qui np:

Bc th cs: Kim tra xem (**) c ng vi n = 1?. Bi v

1! = 1 1 = 21 - 1

nn bc th qui np l ng.


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Bc gi thit qui np:Gi thit rng n! 2n1, (***)

ta cn chng minh rng (**) ng vi n + 1, ngha l

(n+1)! 2n (****)

Tht vy: Theo nh ngha ca giai tha ta c

(n+1)! = (n+1) *(n !)

(n+1) 2n-1 theo gi thit qui np (***) 2 * 2n-1 bi v n+1 2= 2n

Vy (n+1)! 2n. y l iu cn chng minh.

Bi tp v logic v lp lun

1.1 Cho S = {a, b}*. Vi mi x, y S, nh ngha x y = xy (ghp 2 ch).(a) S c ng vi ?(b) Php c tnh kt hp, giao hon?(c) S c phn t n v i vi ?

1.2 Tm quan h l bao ng i xng ca R trn tp S.

1.3Nu X l tp hu hn th |2X| = 2|X|

1.4Nhng quan h R sau c phi l quan h tng ng hay khng

(a) Trn tp tt c cc ng thng, l1Rl2nu l1song song vi l2,(b) Trn tp cc s t nhin N, mRn nu m - n chia ht cho 3,(c) Trn tp cc s t nhin N, mRn nu m chia ht cho n,(d) Trn S = {1, 2, , 10}, aRb nu a + b = 10.

1.5 Cho f: {a, b}* {a, b}* xc nh bi f(x) = ax, vi x {a, b}*. Hi f() cnhng tnh cht g?

1.6Nu w {a, b}*tha mn abw = wab, chng minh rng |w| l s chn.

1.7Nhng mnh sau ng hay sai trong trng s thc?

a/ x, y , nu x < y th x2 < y2b/ x, y , nu x < y th x2 < y2c/ x, y , nu x < y th x2 < y2d/ x, y , nu x < y th x2 < y2

1.8 Chng minh cc mnh sau:

a/ 5n1 chia ht cho 4b/ 1 + 3 + 5 + . . . + (2n - 1) = n

2

c/ 2n n2, vi n = 4, 5, . . .

d/ 2 + 4 + 6 + . . . + 2n = n*(n + 1)

1.9* ( thi cao hc nm 2000, cu 1 mn thi c bn: Ton hc ri rc)1. Hy lp bng gi tr ca cng thc mnh sau: P((QR)S)


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2. Hy bin i tng ng a cng thc sau y v dng: khng c ccdu tng ng (), khng c cc du ko theo (); khng k cc dulng t th n l tuyn ca cc thnh phn m mi thnh phn ny li lhi ca cng thc khng cha cc du tuyn () v hi ():

x (P(x, y)y Q(y, x))1.10* ( thi cao hc nm 2001, cu 1 mn thi c bn: Ton hc ri rc)

1. Cho cng thc mnh A (x P(x) x Q(x)) (x R(x)x F(x))

Thc hin cc php bin i tng ng sau i vi A:

a/ Kh php ko theo

b/ a php ph nh v trc tip lin quan ti cc v t P, Q, R v F.

c/ a cc lng t ln trc cng thc. d/ a cng thc ng sau lng t v dng chun hi v dng chun tuyn.

2. (x!) P(x) l k hiu mnh Tn ti duy nht mt x sao cho P(x) lng.

a/ Cho trng gi tr ca bin x l tp cc s nguyn. Xc nh gi tr chn lca cc cng thc (x!) (x3= 1) v (x!) (x23x + 2 = 0).

b/ Biu din mnh (x!) P(x) qua cng thc tng ng cha lng tton th, lng t tn ti v cc php ton logic khc.

1.11* ( thi cao hc nm 2002, cu 1 mn thi c bn: Ton hc ri rc) ChoP(x) l v t mt bin trn trng M no . Khi y mnh (x!) P(x) c lTn ti duy nht mt x sao cho P(x) l ng.

1. Gi s P(x, y) l v t y = 2x trn trng s Z Z, Z l trng s nguyn.Hy cho bit gi tr chn l ca cc cng thc sau:

(x)(y!) (P(x, y) ((y!)(x) P(x, y))

(y!)(x) (P(x, y) ((x)(y!) P(x, y))

2. Cho mnh (x!)(x > 2). Tm trng ca x mnh trn ng (mnh trn sai).

3. Cho cng thc A (x P(x) (x Q(x)) (X (x P(x) x Q(x))).Dng php bin i tng ng logic a A v dng rt gn:

A (x)(y) ((P(x) Q(y)) X), trong P, Q l v t trn mtbin, cn X l bin mnh s cp.

1.12* ( thi cao hc nm 2002, cu 1 mn thi c bn: Ton hc ri rc)

1. Cho trc cng thc (x)(x) P(x, y) xc nh trn trng M M vi M ={1, 2, 3}. Hy bin i tng ng cng thc trn v dng khng cn cclng t , m ch cn cc php hi, tuyn v cc v t trn trng

cho.2. Cho trc cng thc


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A (x P(x)x Q(x)) x P(x) x Q(x) (((X Y) Y)X).

Thc hin cc php bin i tng ng sau y i vi A:

a/ Kh php keo theo .

b/ a du ph nh v trc tip lin quan nX, Y, P v Q.

c/ a cc lng t ,ln ng trc cc cng thc logic khc.

d/ Bin i cng thc ng sau lng t v dng chun hi v dng chuntuyn.


1. Cho trc cng thc

A (x P(x) x Q(x)) (x F(x) x (R(x) P(x))). Thc hin cc

php bin i tng ng sau y i vi A:a/ Kh php keo theo .

b/ a du ph nh v trc tip lin quan n P, Q, R v F.


d/ Tm dng chun hi v dng chun tuyn ca A.

2. Ch ra trn trng M = {a, b, c} ta lun c:

x P(x) x Q(x) x y (P(x)Q(y)

3. Suy lun di y c ng khng? Nhng qui tc suy din no c p dng? X1 X2X3X2X4X3( X4 X5) X5

X11.14* ( thi cao hc nm 2003, cu 1 mn thi c bn: Ton hc ri rc)

1. Cho cng thc

A (x (P(x) Q(x)) x R(x)) (x F(x) (X Y)). Thc hin cc phpbin i tng ng sau y i vi A:

a/ Kh php keo theo .

b/ a du ph nh v trc tip lin quan n P, Q, R v F.


d/ Tm dng chun hi v dng chun tuyn ca A. T vit dng chuntuyn v dng chun hi ca A.

2. a cng thc B (x)(y)P(x, y) (x)(y) P(x, y) v cng thctng ng trn trng M = {a, b} {c, d} khng cn cc lng t , ,


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ch cn php hi, php tuyn v php ph nh. Php ph nh ch lin quantrc tip ti tng v t c th trn M.

3. a/ Ch ra hnh suy din di y l ng

X1 X2X1 (X3X2)X3 ( X4 X5)X4

X5b/ Chuyn m hnh suy din trn v dng cng thc hng ng tng

ng.


1. Cho m hnh suy din trong logic v t

(x)( P(x) Q(x))(x) P(x)(x)(Q(x) R(x))(x)(S(x) R(x)) (*)

(x) S(x)

y P(x), Q(x), R(x), S(x) l cc bin v t xc nh trn trng M.

a/ Vit cng thc tng ng vi m hnh suy din (*) nu trn v n c phi lcng thc hng ng khng?.

b/ M hnh suy din trn c ng trn trng M khng?, nhng quy tc suy din noc p dng trong m hnh suy din .

3. Hy din t nh ngha gii hn0

limxx

f(x) = L di dng mt cng thc v t.

4. Ch ra rng cng thc ((x) P(x)) tng ng vi cng thc (x) P(x) trntrng M = {a1, a2, , an}.

1.16* ( thi cao hc nm 2005, cu 1 mn thi c bn: Ton hc ri rc) 1. Pht biu sau ng hay sai, ti sao?

Tt c mi ngi c bng c nhn th u tt nghip i hc.Lan c bng c nhn.Vy suy ra Lan tt nghip i hc.

2. Hai cng thc sau c tng ng logic vi nhau khng, ti sao?A = P (Q R) v B = (P Q) (P R)

3. Cho trc cng thc F = (P Q) ( P R)a/ Kh php ko theo v rt gn cng thc F.

b/ Tm dng chun hi chnh tc (chun hi y ) v chun tuyn chnh tcca F.


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1.17* ( thi cao hc nm 2006, cu 1 mn thi c bn: Ton hc ri rc) 1. Cho trc cng thc F = ( P Q) ( P R)

a/ Kh php ko theo v rt gn cng thc F.b/ Tm dng chun tuyn chnh tc (chun tuyn y ) ca F.

2. Hai cng thc sau c tng ng logic vi nhau khng, ti sao?A = x ( P(x) Q(x))B = x (P(x) x Q(x))

3. Suy lun di y c ng khng? Nhng lut logic, qui tc suy dinno c p dng?

P QP Q (Q R)S P

S


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CHNG II

L thuyt tmt

Chng hai gii thiu nhng khi nim c s v cc tnh cht ca l thuyt tmt.Ni dung bao gm:

Cc khi nim cbn v cc tnh cht ca tmt,

Cc tnh cht ca hm chuyn i trng thi,

S tngng ca tmt hu hn n nh v khng n nh,

Mt s vn lin quan n cc tiu ho tmt hu hn.

2.1 tmt hu hn

Chng ta s tm hiu v mt nh ngha tng qut nht ca tmt v sau thuhp cho ph hp vi cc ng dng ca khoa hc my tnh. Mt tmt c nhngha nh l mt h thng ([2], [3], [4], [5], [6]), trong nng lng, vt chthoc thng tin c bin i; c truyn i v c s dng thc hin mt schc nng no m khng cn c s tham gia trc tip ca con ngi. V d nhmy ph-t-copy t ng, my tr tin t ng ATM, ...

Trong khoa hc my tnh, thut ng tmt c ngha l my x l t ng trn dliu ri rc. Mi tmt c xem nh l mt c ch bin i thng tin gm mt

b iu khin, mt knh vo v mt knh ra.

Hnh H2-1 M hnh ca tmt ri rc

Trong c cc thnh phn:

1. u vo (input). mi thi khong (ri rc) t1, t2, , cc d liu vo I1, I2, ,l nhng s hu hn cc gi tr t bng ch ci iu vo.

2. u ra (Output). O1, O2, Om: cc kt qu x l ca h thng l cc shu hn cc gi tr xc nh kt qu u ra.

3. Trng thi. Ti mi thiim, tmt c th mt trong cc trng thi q1,q2, qn.

I1I2

Ik

O1O2

Om

tmt

q1, q2, qn


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4. Quan h gia cc trng thi. Trng thi tip theo ca tmt ph thuc vohin trng v u vo hin thi, ngha l c xc nh ph thuc vo mthay nhiu trng thi trc v d liu hin thi.

5.

Quan h kt qu. Kt qu ca tmt khng nhng ch ph thuc vo cchin trng m cn ph thuc vo cc u vo. Nhvy, kt qu (u ra,output) ca tmt c xc nh theo u vo v cc trng thi thc hinca n.

Lu [4]:

tmt m u ra ch ph thuc vo u vo c gi l tmt khngc b nh.

tmt m u ra ph thuc vo cc trng thi c gi l tmt huhn b nh.

tmt m u ra ch ph thuc vo cc trng thi c gi l myMoore.

tmt m u ra ph thuc vo u vov cc trng thi mi thiim c gi lMy Mealy.

V d 2.1 Xt thanh ghi dch chuyn nh sau

Error!

Hnh H2-2 Thanh ghi dch chuyn 4 bit s dng D-flip flaps

Thanh ghi dch chuyn trn cn c gi l my hu hn trng thi c 2 4 = 16trng thi (0000, 0001, , 1111), mt dy vo v mt dy ra, bng ch ci vo(tn hiu vo) = {0, 1}v bng ch ci u ra (tn hiu ra) O = {0, 1}. Thanhghi dch chuyn 4 bit trn c th c m t bi tmt sau:

O

Hnh H2-3. My hu hn trng thi thc hin thanh ghi dch chuyn 4 bit.

Nhn xt: Hnh vi ca mi my tun t (cc thao tc thc hin tun t) u c thbiu din c bng mt tmt.

Sau y chng ta xt nh ngha hnh thc v tmt hu hn trng thi gi tt l

tmt hu hn.

D Q D Q D Q D QInput Output

tmtq1, q2, q16


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Mt tmt hu hn lm vic theo thi gian ri rc nh tt c cc m hnh tnhton ch yu. Nh vy, ta c th ni v thi im k tip khi c t hot ngca mt tmt hu hn. Mt cch hnh thc hn, tmt c nh ngha nh sau.

nh ngha 2.1 tmthu hn n nh (gi tt l tmt hu hn), k hiu l T,c th biu din c bi b 5 thnh phn M= (Q,,, q0, F), trong :

(i) Q l tp hu hn khc rng cc trng thi,

(ii) l tp hu hn khc rng cc k hiu u vo (bng ch ci), vi githit Q = ,

(iii) l nh x t Q vo Q v c gi l hm chuyn trng thi.Hm ny m t s thay i trng thi ca tmt v thng c cho

bit di dng bng chuyn trng thi hay th chuyn trng thi.

(iv)

q0Q l trng thi khi u,(v) F Q l tp cc trng thi kt thc. Tng qut: | F| 1.

M hnh trn c th c m t nhtrong hnh H2- 4.

Xuc x l

Bng d liu vo

u c R

Hnh H2-4 S khi ca tmt hu hn

tmt hu hn trng thi gm cc thnh phn sau:

(i) Bng d liu vo. Bng d liu vo c chia thnh cc , mi chamt k t t bng ch ci , trong u c nh du cho s btu bng v cui dng $ nh du kt thc. Khi khng s dng

cc nh du th bng d liu vo s c xem nhc di vhn.

(ii) u c R. Mi ln u c ch xem xt mt v c th dch qua phihoc qua tri mt . n gin, chng ta gi thit u c ch dchqua phi sau mi ln x l.

(iii) B iu khin hu hn.B ny iu khin hot ng ca tmt mikhi c mt d liu vo. Khi c mt k hiu vo, v d a, trng thiq c th cho cc kt qu sau:

(a)u cRvn nguyn ti ch, khng dch chuyn,(b)Chuyn sang trng thi khc c xc nh theo (q, a).

$

B iu khinhu hn


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V d, h thng thang my ca mt to nh nhiu tng c th m hnh ha nh lmt tmt hu hn. Thng tin vo gm cc yu cu i ln, i xung ti mi thiim khi ngi s dng c nhu cu. Trng thi c xc nh bi thng tin v ccyu cu vn cha c p ng (v nhng tng cn i ti t bn trong thang my,

hay trn cc tng), v nhng v tr hin ti ca thang my (ln hay xung). Trngthi tip theo lun c xc nh bi trng thi hin ti v thng tin vo. T cctrng thi hin ti cng xc nh c thng tin ra m t hnh vi ca h thngthang my.

hiu c hot ng ca cc tmt, chng ta cn phi tm hiu cc tnh cht,c trng ca hm chuyn trng thi.

2.1.1 Cc tnh cht ca hm chuyn trng thi

Hm chuyn trng thi c xc nh trong nh ngha tmt hu hn l mt nh

x xc nh trn cc cp trng thi v k hiu vo, do vy khng th s dng trctip on nhn cc xu, m phi m rng thnh nh x

: Q *Q

nh cc tnh cht sau.

Tnh cht 2.1

(i) q Q, (q, ) = q, vi l t rng. Ngha l trng thi ca tmt chthay i khi c d liu vo. (2.1)

(ii) w*, a , (q, aw) = ((q, a), w) (2.2a)

(q, wa) = ((q, w), a) (2.2b)

Mnh 2.1 Vi mi hm chuyn trng thi v vi mi xu vox, y,

(q, xy) = ((q, x), y) (2.3)

Chng minh:Qui np theo |y|, ngha l theo s k t trong y.

Cs:Khi |y| = 1, y = a, th (q, xa) = ((q, x), a) ng theo tnh cht (2.2b).

Gi s rng (2.3) ng vi mi xuxv vi nhng xuyc nphn t, | y| = n.

Khi xuyc di l n+1, ta c th vity = y1a, vi | y

1| = n. Khi

(q, xy) = (q, xy1a) = (q, x1a), vi x1 = xy1= ((q, x1), a), Theo (2.2b)= ((q, xy1), a)

= (((q, x), y1), a), Theo gi thit qui np= ((q, x), y1a), Theo (2.2b)= ((q, x), y).

Do vy, tnh cht (2.3) ng vi mi xux, y.


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2.1.2 Cc phng php biu din tmt

Cho mt tmt thc cht l ch cn cho bit hm chuyn trng thi ca n. Hmchuyn trng thi c th cho di dng bng chuyn trng thi hoc di dng

th.A/ Phng php cho bng chuyn trng thi

Bng cho trc mt tmt c th cho di dng bng chuyn trng thi

Cc trngthi

K hiu vo

a1 a2 an

q0

q1

qf

qk

(q0, a1) (q0, a2) (q0, an)

(q1, a1) (q1, a2) (q1, an)

(qf, a1) (qf, a2) (qf, an)

(qk, a1) (qk, a2) (qk, an)

trong , Q = {q1, q2, qk}, = {a1, a2,, an} v F l tp tt c nhng trng thiqf (nhng trng thi nm trong hnh trn), q0 l trng thi bt u (c mi tn i

n).V d 2.2 Xt tmt hu hn c hm chuyn trng thi c xc nh theo bngsau

Bng B2.1 Bng chuyn trng thi ca M

Dy vo

Trng thi 0 1

q0 q2 q1

q1 q3 q0

q2 q0 q3

q3 q1 q2

Q = {q0, q1, q2, q3}, = {0, 1} v F = {q0}, q0va l trng thi bt u (c mi tn i n) va l trng thi kt thc (nm trong mt hnh trn).

B/ Phng php biu din bng th

Hm chuyn trng thi c th biu din di dng th c hng, trong mitrng thi l mt nh. Nu t k t vo a , v t trng thi qtmt chuynsang trng thip ((q, a) = p) th s c cung i t qnpvi nhn l a. nh ngvi trng thi bt u (q0) c mi tn n, nhng nh ng vi cc trng thi kt


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thc (thuc tp F) c khoanh trong vng trn kp, nhng nh cn li ckhoanh trong vng trn n.

V d 2.3 tmt hu hn c hm chuyn trng thi c xc nh theo bng v

d 2.2 c biudin bng th nh sau:

Hnh H2-5 Biu din th ca tmt

2.1.3 Ngn ng on nhn c ca tmt

nh ngha 2.2 Mt xuxc on nhn bi tmt hu hn M = (Q, , , q0, F)nu(q0, x) = q, vi q F, ngha l (q0, x) F.

nh ngha 2.3 Tp tt c cc xu (cn c gi l ngn ng) c on nhn bitmt hu hn M = (Q, , , q0, F), c k hiu l

T(M) = {x * | (q0, x) F}V d 2.4 Xt tmt hu hn cho trc v d 2.2. Ta d dng kim tra dy110101 l chp nhn c bi M. Tht vy,

(q0, 110101) = (q1, 10101) = (q0, 0101)

= (q2, 101) = (q3, 01)

= (q1, 1) = (q0, ) = q0.

V d 2.5 Xt tmt hu hn M c cho nh hnh sau

Hnh H2-6 Biu din th ca tmt hu hn M cho trc

Da vo cc tnh cht v cch biu din ca hm chuyn trng thi trn th nhhng, chng ta nhn thy:

0

q1

q2 q3

q0

1

1

1

1

00 0

b

q1

q3

q0 b

a

a

b


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Mt xu (mt t) on nhn c bi M l dy cc k t u vo (cc nhntrn cc cung) trn ng i t nh bt u n mt trng thi kt thc no.

Ngn ng on nhn c bi M l tp tt c cc xu c ghp t ccnhn trn tt c cc ng i t nh bt u n cc nh kt thc.

T nhn xt trn, ta suy ra ngn ng on nhn c bi M s l

T(M) = {a(ba)na, bm | n 0, m > 0}

2.2 tmt hu hn khng n nh

Chng ta nghin cu nhng tmt m t mt trng thi (cha kt thc) v mtk t vo th trng thi tip theo ca chng c xc nh duy nht. Tip theochng ta xt trng hp nhng tmt hu hn m trng thi tip theo nh trn

khng xc nh duy nht.Chng ta hy xt tmt c cho bi th chuyn trng thi nhtrong Hnh H2-7.

Hnh H2-7 H bin i biu din tmt hu hn khng n nh

Khi trng thi q0v d liu vo l 0, th tmt chuyn n trng thi no?. Theoh thng trn th n c th hoc chuyn n q1hoc quay vng tr li chnh q0.

Nhvy, trng thi tip theo ca tmt khng xc nh duy nht khi nhn d liuvo 0. Nhng tmt nhvy c gi l tmt hu hn khng n nh(khng

tt nh).nh ngha 2.4 Mt tmt hu hn khng n nh(TK), gi tt l tmtkhng n nh,l b 5 phn t M = (Q, , , q0, F), trong

(i) Q l tp hu hn khc rng cc trng thi,

(ii) l tp hu hn khc rng cck hiu vo (bng ch ci), vi Q =,

(iii)l nh x t Q vo 2Q(tp tt c cc tp con ca Q),

(iv) q0Q l trng thi khi u,

(v) F Q l tp cc trng thi kt thc. Tng qut th | F| 1.

0

q0

1

0

1

q2

q1

1


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Lu : Khi mt nh c nhiu hn mt vng khuyn (mt cung quay li chnh n)vi cc nhn khc nhau th c th ch cn v mt vng khuyn vi cc nhn cchnhau bng du phy , nh cc nh q0, q1, q4 hnh trn.

Dy cc trng thi xc nh bi dy vo 0100 c m t qua hm chuyn trngthi nh sau:

(q0, 0100) = ({q0}, 0100) Theo (2.4)

= (({q0}, 0), 100) Theo (2.7)

= ((q0, 0), 100) Theo (2.6)

= ({q0}, 100) = (({q0}, 1), 00)

= (({q0, q3}, 0), 0)

= ((q0, 0) (q

3, 0), 0) Theo (2.6)

= ({q4} (q3, 0), 0)

= (q4, 0) ( (q3, 0), 0)

= {q4} ( (q3, 0), 0)

V q4l trng thi kt thc nn dy 0100 l on nhn c bi tmt khng nnh trn.

Tng t, dy0100 cng chnh l dy ca cc nhn trn ng i t q0, q0, q3, q4,v k thc q4, nn n c on nhn bi M (hnh H2-8).

T , chng ta c tp tt c cc t on nhn bi mt tmt c nh ngha nh sau.

nh ngha 2.6 Tp cc xu (t) c M (n nh hoc khng n nh) onnhn l tp tt c cc xu d liu (dy cc k t) u vo c on nhn bi M;k hiu T(M).

T(M) = {w * |(q0, w) F }Tp cc xu c on nhn (hay on nhn c) bi mt tmt th hin khnng tnh ton ca h thng v m t hnh vi ca mt tmt. Tp on nhn c

bi mt tmt thng cn c gi l ngn ng c on nhn bi tmt.

2.3 S tng ng ca tmt n nh v khng n nh

trn chng ta phn bit hai loi tmt n nh (T)v khng n nh(TK). Vn quan trng l chng c quan h vi nhau nhth no?, ngha lngn ng on nhn bi TK v T c quan h vi nhau hay khng?

Mt cch trc quan chng ta nhn thy:

T c th m phng hnh vi ca TK bng cch gia tng s cc trng thi.Ni cch khc, T (Q, , , q0, F) c th xem nhTK (Q, , , q0, F)

bng cch nh ngha (q, a) = {(q, a)}, ni cch khc T l trng hpc bit ca TK.


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TK l tng qut hn, song c v nhkhng mnh hn, ngha l tpm n on nhn khng nhiu hn T.

iu th hai c khng nh bi nh l sau.

nh l 2.1 Vi mi TK M lun tn ti T M tng ng theo nghachng on nhn cng mt tp d liu, T(M) = T(M).

Chng minh: Gi s TK M = (Q, , , q0, F) on nhn L. Chng ta xydng mt tmt T M = (Q, , , q0, F) nh sau:

(i) Q = 2Q ,

(ii) q0 = {q0},

(iii) Fl tp tt c cc tp con ca Q c cha phn t ca F,

F = { S Q | S F }

Trc khi xc nh , chng ta hy kho st cc c im ca Q, q0v F. M khing t q0v vi mt d liu vo bt k, v d a,Mc th chuyn sang mt trongcc trng thi ca (q0, a). kho st hot ng caMkhi x l a, chng ta phixt tt c cc trng thi c th t c khi n x l a. V th cc trng thi caMs l cc tp con ca Q. Khi Mbt u vi q0, q0c nh ngha nhtrn s ltrng thi khi u caM.Mt khc xu w T(M) = L nuMt n trng thikt thc khi x l w. Do vy, trng thi kt thc caM(phn t ca F) l tp conca Q v c cha mt s trng thi kt thc ca M.

Tip theo chng ta nh ngha :(iv) ({q1, q2, qi}, a) = ( q1, a) ( q2, a) (qi, a)

Ngha l ({q1, q2, qi}, a) = {p1, p2, pk} khi v ch khi

({q1, q2, qi}, a) = {p1, p2, pk} vi i, k |Q|.

Trc khi chng minh L = T(M), chng ta chng minh kt qu b tr:

(q0, x) = {q1, q2, qi}, khi v ch khi

(q0, x) = {q1, q2, qi} vi mi x trong *. (2.8)

Chng ta chng minh iu kin cn ca (2.8) qui np theo |x|.Nu (q0, x) = {q1, q2, qi} th (q0, x) = {q1, q2, qi} (2.9)

Khi |x| = 0, hin nhin (q0, ) = {q0} v theo nh ngha , (q0, ) = q0= {q0},ngha l (2.9) ng vi |x| = 0.

Gi thit (2.9) ng vi miyv |y| n. Xt xuxc di n + 1. Ta c th vit x = ya,trong |y| = n v a . Gi s (q0, y) = {p1, p2, pk} v (q0, ya) = {r1, r2, rl}. Bi|y| n nn theo gi thit qui np ta c

(q0, y) = {p1, p2, pk} (2.10)


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Mt khc, {r1, r2, rl} = (q0, ya) = ((q0, y), a) = ({p1, p2, pk}, a). Theo nhngha ca ,

({p1, p2, pk}, a) = {r1, r2, rl}. (2.11)

Do vy, (q0, x) = (q0, ya) = ((q0, y), a) = ({p1, p2, pk}, a)

= {r1, r2, rl} theo (2.11).

Suy ra (2.9) c chng minh vi x = ya.

iu kin chng minh tng t.

T suy ra, x L = T(M) khi v ch khi (q, x) cha mt trng thi ca F. Bi v(q0, x) cha trng thi thuc F khi v ch khi (q0, x) nm trong F. iu khng nhxT(M) x T(M), ng ngha vi vic Mcng on nhn ng L.

V d 2.7 Xt tmt khng n nh M = ({q0, q1}, {0, 1}, , q0, {q0}) vi hmchuyn trng thi c xc nh theo bng B2.2.

Bng B2.2 Bng cc trng thi ca M

Trng thi \ 0 1

q0 q0 q1

q1 q1 q0, q1

Xydng tmt n nh tng ng M theo nh l 2.1, c cc thnh phn:(i) Q ={, { q01}, {q1},{q0, q1}},

(ii) q0 = { q0},

(iii) F = {{ q0}, { q0, q1}}, nhng tp con cc trng thi c cha q0,

(iv) c nh ngha nhtrong bng B2.3


Trng thi \ 0 1

{q0} {q0} {q1}{q1} {q1} {q0, q1}

{q0, q1} {q0, q1} {q0, q1}

Nhn xt:Khi M c ntrng thi, M tng ng s c th c 2ntrng thi. Tuy nhin,chng ta khng nht thit phi xy dng tmt c tt c 2ntrng thi m ch cn nhngtrng thi t n t trng thi khi u v i n c trng thi kt thc.

Trong bng chuyn trng thi B2.3 trng thi l khng t n c t trng thibt u {q0} nn c th loi b i. Ngoi ra, cho tin li chng ta c th k hiu


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li tp cc trng thi, v d s0 = {q0}, s1 = {q1}, s2 = {q0, q1} v Q = { s0, s1, s1}.Khi , tmt c hm chuyn trng thi nh trn s c vit thnh


Trng thi \ 0 1

s0 s0 s1

s1 s1 s2

s2 s2 s2

Tng qut ha qu trnh trn, chng ta c th xy dng thut ton thit lp M nhsau.

Thut ton 2.1 Xy dng hm chuyn trng thi ca M tng ng vi M (Php

n nh ha)Input: Cho trc tmt M = (Q, , , q0, F) khng n nh.

Output: Hm chuyn trng thi ca tmt n nh tng ng.

1. Xy dng bt u t {q0},

2. Xc nh trng thi t c t nhng trng thi (tp con cc trang thi caM) tng ng vi cc ct d liu vo xc nh t trc,

3. Lp li bc 2 cho n khi khng c mt trng thi mi xut hin trong ccct ng vi d liu vo.

V d 2.8 Tm tmt n nh tng ng vi

M = ({q0, q1, q2}, {a, b}, , q0, { q2})

v c cho trong bng B2.5.


Trng thi \ a b

q0 {q0, q1} q2

q1 q0 q1

q2 {q0, q1}

tmt n nh tng ng vi M l M = (2Q, {a, b}, , {q0}, F), trong

F = {{q2}, {q0, q2}, {q1, q2}, {q0, q1, q2}}. Thc hin theo thut ton 2.1 ta thu c nh bng B2.6.


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Chng minh: Chng ta chng minh bng phn chng. Gi s q1v q2khng phil (k+1)-tng ng. Khi tn ti w = aw1vi di k+1 sao cho (q1, aw1) ltrng thi kt thc nhng (q2, aw1) li khng phi l trng thi kt thc. Nhvy,(q1, aw1) = ((q1, a), w1) l trng thi kt thc v (q2, aw1) = ((q2, a), w1)khng phi l trng thi kt thc. T suy ra (q1,a) v (q2, a) khng phi l k-tng ng, mu thun vi gi thit (b).

Da vo cc tnh cht trn chng ta c th xy dng cc lp (k+1)-tng ng khibit cc lp k-tng ng trn tp cc trng thi v t xy dng c thutton cc tiu ho tmt (c s trng thi cc tiu) tng ng vi tmt chotrc.

Thut ton 2.2 Xy dng tmt cc tiu ho

Input: Cho trc tmt M = (Q, , , q0, F) thng l khng n nh.

Output: tmt n nh v cc tiu M = (Q, , , q0, F)

1. Thit lp phn hoch 0. Theo nh ngha 0-tng ng, 0 = {Q10, Q2

0},trong Q1

0= F (tp cc trng thi kt thc), Q20 = Q - Q1

0.

2. Xy dng k+1tk. Xy dng Qik, i = 1, 2, l cc tp con ca kv l cc

lp (k+1)-tng ng. q1 v q2 nm trong Qik nu chng l (k+1) tng

ng ngha l (q1, a) v (q2, a) l k-tng ng, vi mi a trong bng chvo. iu ny xy ra khi (q1, a) v (q2, a) nm trong cng lp tng ngca k. Do vy, Qi

kl lp (k+1)-tng ng. Thc hin nhtrn cho n khi

cc tp con Qik

to thnh mt phn hoch k+1ca Q mn hn k.3. Lp li bc 2 thit lp kvi k = 1, 2, cho n khi k= k+1.

4. Xy dngtmt cc tiu. Cc trng thi ca tmt cc tiu chnh l cc lp tngng c xc nh nh trong bc 3, l cc phn t ca k. Bng chuyn trngthi thu c bng cch thay trng thi q bng lp tng ng tng ng [q].

Lu :

1. Da vo bngchuyn trng thi cho trc ta d dng xy dng c cc lp0 = {Q1

0, Q20}, Q1

0 = F, Q20 = Q - F;

2. Gi s xy dng c k, k = 0, 1, . Vi q1, q2 Qik

, Qik

k, i = 1, 2, xt cc trng thi cc ct tng ng vi d liu vo a , nu(q1, a) v (q2, a)cng thuc mt tp con no ca kth l (k+1)-tng ng, ngha l q1, q2cngnm trong mt phn hoch ca k+1; ngc li s khng phi l (k+1)-tngng, ngha l q1, q2nm trong hai phn hoch khc nhau ca k+1.

3. Lp li bc 2 cho n khi k= k+1.

V d 2.9 Xy dng tmt cc tiu tng ng vi tmt c th chuyntrng thi nhtrong hnh H2-9.


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Hnh H2-9 th chuyn trng thi ca v d 2.9

T th chuyn trng thi hnh H2.9 chng ta c bng chuyn trng thi.


Trng thi \ a b

q0 q1 q0

q1 q0 q2

q2 q3 q1

q3 q3 q0

q4 q3 q5

q5 q6 q4

q6 q5 q6

q7 q6 q3

Theo thut ton trn ta c Q10 = {q3}, Q2

0 = { q0, q1, q2, q4, q5, q6, q7} v

0 = {{q3},{ q0, q1, q2, q4, q5, q6, q7}}.

Chng ta xt tip quan h 1-tng ng tnh 1. Trc tin ta cQ1

1 = {q3}. Da vo bng B2.7 chng ta d kim tra c q0 l 1-tngng vi q1, q5, q6v (q0, t) v (qi, t), i = 1, 5, 6 v t = a, b l cng ktthc hoc cng khng phi l trng thi kt thc. Vy Q 2

1 = { q0, q1, q5, q6}.

Tng t ta c th kim tra c q2 -tng ng vi q4 v Q31

= { q2, q4}.Hin nhin cn li Q4

1= {q7}. T chng ta c

b

q0b

q1

a

q3

q2

q4

q7

q5

q6

b

b

a

ba

b

b

a a

a

a

b


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1 = {{q3}, { q0, q1, q5, q6}, { q2, q4}, {q7}}.

Phn hoch tip 1 xy dng 2. Q12 = {q3} v v (q0, t) v (q6, t),

t = a, b u thuc cng lp 1-tng ng (lp {q0, q1, q5, q6}) nn q0v q6

l 2-tng ng. Vy Q22

= {q0, q6}. Nhng q0v q1 (q0v q5) khng phil 2-tng ng v (q0, b) = q0 ((q1, b) = q2, (q0, b) = q0, (q1, b) = q2)v chng li khng cng lp 1-tng ng. Kim tra ta thy q1 l 2-tngng vi q5, nn Q3

2 = {q1, q5}. Tng t q2 vn l 2-tng ng vi q4,v th Q4

2= { q2, q4} v cn li Q5

2= {q7}. Vy

2 = {{q3}, { q0, q6}, { q1, q5}, { q2, q4}, {q7}}

Tip tc xy dng 3. Tng t nhtrn, chng ta thy Q13= {q3}; Q2

3= {q0, q6}

v q0 l 3-tng ng vi q6; tng t Q33 = {q1, q5}; Q4

3 = {q2, q4} vQ5

3 = {q7}. Ngha l 3 = {{q3}, {q0, q6}, {q1, q5}, {q2, q4}, {q7}}.

V 2 = 3, nn cc lp tng ng trong 2s lm cs xy dng tmt cc tiuM= (Q, {a, b}, , q0, {q3}). Trong Q= 2= {{q3}, {q0, q6}, {q1, q5}, {q2, q4}, {q7}}.

Hm chuyn trng thi c xc nh nhtrong bng B2.8.

Bng B2.8 Bng cc trng thi ca tmt cc tiu

Trng thi \ a b

{q0,q6} {q1, q5} {q0, q6}{q1, q5} {q0,q6} {q2, q4}

{q2, q4} {q3} {q1, q5}

{q3} {q3} {q0,q6}

{q7} {q0, q6} {q3}

th chuyn trng thi ca tmt cc tiu thu c s l

Hnh H2-10 tmt cc tiu ca tmt hnh H2-9

[q3]

[q0,q6]

b

b

a

[q1,q5][q2,q4]

[q7]

b

bb

a

a

a

a


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Bi tp v tmt hu hn

2.1 Chng minh rng trong tmt hu hn, nu (q, x) = (q, y) th(q, xz) = (q, yz) vi mi z trong +.

2.2 Cho trc tmt hu hn M = (Q, , , q0, F). Gi s R l mt quan h trnQ c nh ngha nh sau q1R q2nu (q1, a) = (q2, a) vi a . Hi R c

phi l quan h tng ng hay khng?

2.3 Xy dng mt tmt khng n nh on nhn {ab, ba}, v s dng n tm tmt n nh on nhn cng tp .

2.4 Xy dng mt h bin i on nhn c cc xu c cha cat hoc rat tbng ch a, b, c,

2.5 Xy dng mt tmt khng n nh on nhn c tp tt c cc xu t tp{a,b} kt thc bng xu con aba.

2.6 Xy dng tmt n nh tng ng viM = ({q0, q1, q2, q3}, {0, 1}, , q0, {q3}) vi c cho trong bng

Bng B

Trng thi \ a b

q0 q0, q1 q0

q1 q2 q1

q2 q3 q3

q3 q2

2.7 M = ({q1, q2, q3}, {0, 1}, , q1, {q3}) l tmt khng n nh, trong c cho bi

(q1, 0) = {q2, q3}, (q1, 1) = {q1}

(q2, 0) = {q1, q2}, (q2, 1) =

(q3, 0) = {q2}, (q3, 1) = {q1, q2}Tm tmt n nh tng ng vi M.

2.8 Xy dng tmt n nh trn {0, 1} on nhn tt c cc xu cha chnln cc ch s 1.

2.9 Xy dng tmt n nh trn = {a1, a2, , an} on nhn cc ngn ngsau:

a) L1 = {x *| x c di l}

b) L2 = {x *| x c di chn}

c) L3 = {x *| x c di chn v ln hn 1}


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2.10* Cho hai tmt sau y:

M1: M2:

a/ Tm ngn ng T(M1) on nhn c bi M1v T(M2) on nhn c bi M2

b/ Xy dng tmt M t M1v M2 sao cho T(M) = T(M1)T(M2)

c/ Xy dng tmt M t M1v M2 sao cho T(M) = T(M1) T(M2)2.11* Xy dng tmt cc tiu tng ng vi tmt hu hn cho trc

q00

q4

q2

q6

q7

q3

1

1q1

q5

0

1

1

1 1

0

0

1

0

01

0

0

q4

q0 q1 q2

q3q5

0 1

11

101s0 s1 s2

s4

a

s3

c

bbc

bc


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CHNG III

Vn phm v ngn ng hnh thc

Chng ba trnh by nhng khi nim, cc tnh cht cbn ca vn phm v ngnng hnh thc. Ni dung bao gm:

Cc khi nim cbn ca vn phm v ngn ng hnh thc, Phn loi vn phm v ngn ng hnh thc ca Chomsky,

Cc php ton trn cc ngn ng v cc tnh cht ca chng,

Tnh qui ca vn phm cm ng cnh,

Cy dn xut i vi vn phm phi ng cnh.

3.1 Bng ch ci v cc ngn ng

L thuyt ngn ng hnh thc l mt lnh vc quan trng v c rt nhiu ng dngtrong khoa hc my tnh. Ngay t u nhng nm 50 ca th k 20, nhiu nh ngnng hc mong mun v tm mi cch nh ngha nhng vn phm hnh thc(m t cc qui tc ca vn phm mt cch chnh xc ca ton hc) m t ccngn ng t nhin (ngn ng s dng giao tip trong cuc sng hng ngy nhting Anh, ting Php, hoc ting Vit, ...). Theo truyn thng, l thuyt ngn ng

hnh thc lin quan n cc c t c php ca ngn ng nhiu hn l n nhngvn ng ngha. Nhim v chnh ca l thuyt ngn ng hnh thc l nghin cucc cch c t hu hn ca cc ngn ng v hn.

Khi hnh thc ho c cc vn phm ([2], [3], [4], [5]), hin nhin l vic xydng cc b chng trnh dch gia cc ngn ng sinh bi cc vn phm trnmy tnh s tr nn n gin hn. Mt ngi ni ting trong s cc nh ngn nghc hnh thc lNoam Chomsky, ngi a ra m hnh ton hc cho vn phmvo nm 1956, tuy cha m t c hon ton ngn ng t nhin (ting Anh),nhng to ra m hnh cho nhng ngn ng hnh thc thc hin d dng trn mytnh. Thc t, nhng dng k hiuBackus-Naur(Backus-Naur Form) c s dng

m t nhng ngn ng lp trnh nh Algol, Pascal, C, ... cng chnh l vn phm phing cnh Chomsky.

Trc khi nh ngha hnh thc vn phm, chng ta hy kho st mt s mu cuchnh trong cc ngn ng t nhin. Trong ngn ng t nhin, v d ting Vit, chai mu cu chnh. Nhng cu m chng ta tp trung kho st c dng hoc , v d Ba chy nhanh hocBa chy. Trong cu Ba chy nhanh c cc t Ba, chy v nhanh cvit theo th t xc nh. Nu chng ta thay cc danh t, ng tv trng t bngcc danh t, ng t v trng t khc tng ng th vn nhn c nhng cu

chnh xc v mt vn phm. V d, khi thay Ba bng Nam, thay chy bng


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bi v thay nhanh bng chm chng ta vn c nhng cu ng vn phm.Tng t i vi nhng cu dng th hai.

Lu : khng phi l cu m ch l m t mt

mu cu. Nu chng ta thay cc thnh phn , , bng cc t tng thch th s nhn c cc cu ng vn phm. Chng ta gi, , l cc bin(variable) hay cc k hiu chakt thc v cc t Ba, chy, nhanh tng ng c s dng to ra cc cuc th l cc t kt thc hoc k hiu cui. Nhvy, mi cu ca mt ngn ngchnh l mt xu c to ra bng cc t kt thc. Chng ta s dng S k hiucho mt cu, khi c th to cc qui tc sinh ra hai mu cu trn nh sau:

S S Ba

Nam chy bi nhanh chm,

Trong , ch ra rng t bn phi (v phi) thay th cho t bn tri (v tri).Chng ta k hiu P l tp tt c cc qui tc nh trn, c gi l cc qui tc dnxut (gi tt l qui tc); VNl tp cc thnh phn ca cu nh: , , )v bng t vng l bao gm nh: Ba, Nam, chy, bi,

nhanh, chm, ... T chng ta c th nh ngha hnh thc vn phm l mtb 4 thnh phn G = (VN, , P, S), trong

VN = {(, , } - cc k hiu khng kt thc,

= {Ba, Nam, chay, bi, nhanh, chm} - cc k hiu kt thc.

Pl tp cc qui tc nhtrn v S l k hiu c bit khi u to sinh mt cu.Cc cu c to ra nh sau:

(i) Bt u t S,

(ii) Ln lt thay th cc k hiu cha kt thc (bin) bng nhng k hiu

kt thc theo qui tc dn xut,(iii) Kt thc khi xu nhn c ch ton l nhng k hiu kt thc.

Tng qut ha qu trnh nu trn, chng ta nh ngha vn phm hnh thc nh sau.

nh ngha 3.1 Vn phm l mt b 4 thnh phn G = (VN, , P, S), trong

(i) VNl tp hu hn khc rng cc phn t c gi l cc bin, hay cck hiu khng kt thc,

(ii) l tp hu hn khc rng cc phn t c gi l cc khiu ktthc, hay bng ch ci,

(iii) VN = , tp cc bin v tp cc k hiu kt thc l ri nhau,


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Khi chngta bt u bng qui tc S Sc v lp li mt s ln th s nhn cSc

i. Sau p dng cc qui tc cn li nhtrn th chng ta c anbnci L(G).

Kt lun: L(G) = {anbnci | n 1, i 0 }.

V d 3.9 Cho G = ({S, A}, {0, 1, 2}, P, S), trong P gm S 0SA2, S 012,2A A2, 1A 11. Chng minh rng L(G) = {0n1n2n | n 1}.Khi p dng S 012 th S 012, do vy 012 L(G).

Nu p dng S 0SA2

S*

0n-1

S(A2)n-1

(p dng S 0SA2 n-1 ln)*

0n-1012(A2)n-1 (p dngS 012)*

0n1An-12n (p dng 2AA2)

*

0n1

n2

n(p dng 1A 11 n-1 ln).

Vy, 0n1n2n L(G), n 1.

chng minh L(G) {0n1n2n | n 1}, chng ta thc hin nhsau: Nu u tins dng S 012 th s nhn c 012. Ngc li s dng S 0SA2 mt s lnth s c 0n-1S(A2)n-1. loi c S th bt buc phi s dng S 012 v snhn c 0n12(A2)n-1. Mun loi tip A chng ta s dng 2A A2 hoc1A 11. Qui tc 2A A2 ch i ch ca A v 2. S dng 1A 11 l loi c

bin A. Do vy,0n12(A2)n-1 = 0n12A2 A2 A2, trong c n-1 ln cp A2.p dng n-1 ln 2A A2,

0n12(A2)n-1*

0n1An-1 2n

Sau p dng n-1 ln 1A 11 th c

0n12(A2)n-1*

0n1n2n

Suy ra L(G) {0n1n2n | n 1} v cui cng L(G) = {0n1n2n | n 1}.

V d 3.10 Xy dng vn phm G sinh ra L = {anbncn | n 1}.

Chng ta c th xy dng vn phm G sinh ra L bng phngphp qui. Tuynhin, nu xy dng vn phm sinh trc tip ra L l tng i kh. Vy, chng tac th thc hin theo hai bc.

(i) Xc nh cc qui tc dn xut ra ann,

(ii) Chuyn nv bncn.

c cc xu anndng (i), chng ta c th s dng nhng qui tc n gin nh S aS | a. Nu bc (ii) chng ta s dng ngay qui tc bc th s c(bc)n. Nhng n y khng th chuyn (bc)nv c bncnv v tri khng c binno c. Do vy ta c th s dng = BC, vi B v C l cc bin. Sau dn cc


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bin B v bn tri cn bin C v bn phi bng cch s dng qui tc CB BC. Tnhng lu trn, chng c th nh ngha G = ({S, B, C}, {a, b, c}, P, S), trong

P = { S aSBC | aBC, CB BC, aB ab, bB bb, bC bc, cC cc}

Tip theo chng ta chng minh rng L(G) = {anbncn | n 1}.V S aBC abC abc, nn abc L(G).

S*

an-1S(BC)n-1 (p dng n-1 ln SaSBC)

*

an-1aBC(BC)n-1 (p dng qui tc SaBC)*

anBnCn (p dng n-1 ln qui tc CB BC)

*

an-1abBn-1Cn (p dng 1 ln qui tc aB ab)

*

anbnCn (p dng n-1 lnqui tc bB bb)*

anbn-1bcCn-1 (p dng 1 ln qui tc bC bc)*

anbncn (p dng n-1 ln qui tc cC cc)

T chng ta c L(G) {anbncn | n 1}.Chiu ngc li s c khng nh nu mi xu anbncn, n 1 u dn xut ra

c t S-dn xut ( S*

anbncn).

Tht vy, nu n = 1 v s dng cc qui tc S aBC, aB ab, bB bb, bC bc,cC cc th s thu c abc.

Vi n > 1 th s dng n-1 ln S aSBC v 1 ln S aBC chng ta s can(BC)n. Tip theo chng ta c th s dng CB BC hoc aB ab sau thm

bB bb, bC bc, cC cc. S dng mt trong nhng qui tc trn chng ta snhn c cc xu c cc bin B, C nm c bn phi. Khi xy ra cc trnghp sau:

+ Nu B c dn ht v bn tri ca C th sau ch c th p dng bB bb chon khi loi ht bin B v sau s dng bC bc, cC cc cui cng c c anbncn.

+ Gi s bin C c chuyn thnh c trc khi tt c cc bin B c chuyn

thnh b, ngha l an(BC)n*

anb

ic, i < n v cha c bin B ln bin C. V trong

anb

icch c l cha cc bin cha kt thc v v c ng trc nn trong cc

qui tc ca G ch c cC cc l c th c p dng. Nu khng cn dn xuttip c na (khng c qui tc no c th p dng) v bt u bng B th khngth dn xut c thnh xu khng cn bin. Ngc li, bt u bng C th lpli vic p dng qui tc cC cc cho n khi gp phi t nht bin B, dnganbickB, i, k < n. Trng hp ny cng tng t nh trn, khng th s dng

c qui tc no loi b c bin B. Do vy, trng hp th 2 ny khng thdn xut ra c cu ca L(G).


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Kt hp c 2 trng hp chng ta c {anbncn | n 1} L(G) v cui cng l

{anbncn | n 1} = L(G)

V d 3.11 Xy dng vn phm G sinh ra {xx | x {a, b}*}.

Chng ta d on vn phm G = ({S, S1, S2, S3, A, B}, {a, b}, P, S), trong Pgm cc qui tc sau:

P1: S S1 S2 S3, P8: Ba aB,P2: S1S2 aS1A, P9: Bb bB,P3: S1S2 bS1B, P10: aS2 S2a,P4: AS3 S2aS3, P11: bS2 S2b,P5: BS3 S2bS3, P12: S1S2,P6: Aa aA, P13: S3.P7: Ab bA,

Ch : Trc khi kho st ngn ng sinh bi G chng ta lu mt s tnh chtsau:

1. Ch c P1l S-dn xut,

2. Khi s dng S1S2 aS1A, chng ta c th b sung a(k hiu kt thc) vobn tri ca S1v bin A vo bn phi. A c s dng ch ra rng chngta b sung avo bn phi ca S1. Qui tc AS3 S2aS3 c s dng a thm avo bn tri ca S2.

3. Tng t, s dng qui tc S1S2 bS1B b sung bvo bn tri ca S1v

bin B v bn phi. Qui tc BS3 S2bS3 c s dng a thm bvobn tri ca S2,

4. S2ng vai tr nhphn t trung tm,

5. a thm cc k hiu kt thc vo th ch s dng cc qui tc P 2P5.

6. Cc qui tc P6 - P9ch lm nhim v hon i v tr ca cc k hiu kt thcvi cha kt thc,

7. P10, P11thc hin chuyn S2sang bn tri,

8. P12, P13c s dng loi b hon ton cc bin S1, S2, S3.

t L = {xx | x {a, b}*}. Trc tin chng ta ch ra rng L L(G).

Chng ta c

S S1S2S3 aS1AS3 aS1S2aS3hoc (3.1)

S S1S2S3 bS1BS3 bS1S2bS3 (3.2)

Gis xx, x {a, b}*. Chng ta c th p dng (3.1) hoc (3.2) tu thuc vo khiu bt u ca x l a hay l b. Xt trng hp hai k t u ca x l ab (cc

trng hp khc cng tng t), chng ta c dn xut S*

aS1S2aS3 abS1BaS3

abS1aBS3 abS1aS2bS3 abS1S2abS3. Lp li cch thc hin nh trn vi


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mi k hiu trong x, chng ta nhn c x S1S2xS3., sau s dng P12, P13 suydn ra xx, ngha l L L(G).

chng minh chiu ngc li L(G) L, chng ta hy ch 3 bc u (3.1) v

(3.2) ca mi qu trnh dn xut trong G lun cho aS1S2aS3hoc bS1S2bS3.Chng ta tho lun v cc cch dn xut tip t aS1S2aS3(trng hp cn li tng t) cc mt cu trong L(G). Nhng qui tc c th p dng y l S1S2, S3,S1S2 aS1A, S1S2 bS1B. Xt cc trng hp c th xy ra.

Trng hp 1: S dng S1S2vo aS1S2aS3 chng ta c aaS3 = aaS3. Tiptheo ch c th s dng S3 c aa L.

Trng hp 2: S dng S3vo xu aS1S2aS3 chng ta c aS1S2a = aS1S2a.Nu s dng tip S1S2 th kt qu thu c l aa L. Ngc li c th sdng S

1S

2 aS

1A (hoc S

1S

2 bS

1B) th s nhn c aaS

1Aa (bbS

1Bb tng

ng). Trong G s khng c qui tc no c th p dng c dn xut tip mloi b c cc bin S1, A (B tng ng).

Trng hp 3: S dng S1S2 aS1A i vi aS1S2aS3 chng ta c aaS1AaS3.aaS1AaS3 aaS1aAS3 a

2S1aS2aS3 a

2S1S2a

2S3. n y d nhn ra qu trnh

dn xut lp li qu trnh trn c anS1S2anS3ri s dng S1S2, S3

suy ra ana

n L.

Trng hp 4: S dng S1S2 bS1B i vi aS1S2aS3 chng ta c abS1BaS3. Dnxut tng t nhtrn ta c xu abS1S2abS3v abab L.

Tip tc dn xut, thc hin mt trong bn trng hp nu trn, chng ta ssuy ra xu dng xS1S2xS3v do vy xx L.

T kt lun L(G) = L.

V d 3.12 Cho trc G c cc qui tc dn xut S aSa | bSb | aa | bb | . Hychng minh rng:

(i) Mi xu (cu) ca L(G) u c di chn,

(ii) S cc xu c di 2*n l 2n.

Mnh u (i) d khng nh v khi p dng bt k mt qui tc no ca G cngu thay mt k hiukhng kt thc (S) bng hai k hiu kt thc (a hoc b) (trS ) v c nhiu nht mt k hiu khng kt thc. Do vy, qua mi ln p dngmt qui tc th di ca xu c dn ra tng ln 2 tr qui tc S .

chng minh mnh (ii), chng ta xt xu w c di 2*n. Da vo cc qui tcca G, chng ta thy w c dng a1 a2 an an a2 a1, aic dng a hoc b. Vy stt c cc xu nhth s l 2n.

3.3 Phn loi cc ngn ng ca Chomsky

Trong nh ngha ca vn phm G =(V

N, ,

P, S); V

Nl tp cc bin, l tp

ch ci v S VN. tin li cho vic nghin cu v kho st cc ngn ng c


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Gi thit qui np: Gi s tnh cht trn ng vi w v |w| < k. Chng ta cn ch rarng chng cng ng vi w vi |w| = k.

Trc tin, nu S*

w th dy dn xut ny phi bt u bng qui tc S aB

hoc S bA. Trng hp s dng S aB th w = aw1, vi |w1| < k v B*

w1 .

Theo gi thit qui np th w1 c s cc ch b ln hn s cc ch a l 1. T suy raw c s cc ch a bng s cc ch b. Tng t i vi qui tc th hai.

Chng ta chng minh iu kin (ch nu): w c s cc ch a bng s cc ch

b v |w| = k th S*

w. Bit rng tron

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