Gio trnh Kin trc my tnh v H iu hnh 1 I HC NNG TRNG I HC BCH
KHOAKHOA CNG NGH THNG TIN NGN NG HNH THC & TMTGio trnh Kin trc
my tnh v H iu hnh 2 Mc tiu gio trnh 1. Cung cp nhng kin thc c bn v
ngn ng, vn phm v tmt. 2. Cungcpccphngphpphn tch t vng, phn tch c
php. 3. C s cho vic tm hiu cc ngn ng lp trnh. 4. Rn luyn k nng lp
trnh cho sinh vin TRNG I HC BCH KHOA NNG Gii thiu Gio trnh Kin trc
my tnh v H iu hnh 3 Ni dung gio trnh CHNG 1. M U CHNG 2. TMT HU HN
CHNG 3. BIU THC V VN PHM CHNH QUI CHNG 4. VN PHM V NGN NG PHI NG
CNH CHNG 5. TMT Y XUNG CHNG 6. MY TURING TRNG I HC BCH KHOA NNG Gii
thiu Gio trnh Kin trc my tnh v H iu hnh 4 CHNG 1. M U TRNG I HC BCH
KHOA NNG Mt s vn v ngn ng Khi nim vn phm Khi nim tmt Gio trnh Kin
trc my tnh v H iu hnh 5 CHNG 1. M U TRNG I HC BCH KHOA NNG 1. Mt s
vn v ngn ng 1.1. Xu- B ch (bng ch) l tp hp hu hn cc k hiu V d:{0,1}
b ch gm 2 k hiu 0 v 1 {a,b,c,,z} b ch gm cc k hiu a z Tp cc ch ci
ting vit Gio trnh Kin trc my tnh v H iu hnh 6 CHNG 1. M U TRNG I HC
BCH KHOA NNG 1. Mt s vn v ngn ng 1.1. Xu - Xu trn b ch V l 1 dy cc
k hiu ca V V d:0110 l xu trn b ch {0,1} a, ab, giathanh l xu trn b
ch {a,b,,z} - di xu l s cc k hiu trong xu K hiu: di xu x l |x| V
d:|01110|=5 Gio trnh Kin trc my tnh v H iu hnh 7 CHNG 1. M U TRNG I
HC BCH KHOA NNG 1. Mt s vn v ngn ng 1.1. Xu - Xu rng l xu c di bng
0 K hiu:c, |c|=0 - Tp tt c cc xu trn V l V*, {c}_V* V+ =V *-{c} V*:
tp v hn m c V d:V={a,b}V*={c,a,b,aa,bb,ab,ba,} Gio trnh Kin trc my
tnh v H iu hnh 8 CHNG 1. M U TRNG I HC BCH KHOA NNG 1. Mt s vn v
ngn ng 1.1. Xu - Cc php ton trn xu Ghp tip: cho 2 xu x,y. Ghp tip
ca x, y l x.y hay xy l 1 xu vit x trc, ri n y sau ch khng c du cch.
V d:x=01 y=0110 xy=010110 Gio trnh Kin trc my tnh v H iu hnh 9 CHNG
1. M U TRNG I HC BCH KHOA NNG 1. Mt s vn v ngn ng 1.1. Xu o ngc xu
x (xr): xu c vit theo th t ngc li ca xu x V d: x=0101 xr =1010 Ch :
cr= c, 1r =1 - Xu x m x=xr th x l xu hnh thp (xu i xng) V d: x=0110
xr=0110, x: xu hnh thp Gio trnh Kin trc my tnh v H iu hnh 10 CHNG
1. M U TRNG I HC BCH KHOA NNG 1. Mt s vn v ngn ng 1.1. Xu Ly tha xu
x (xn): xu nhn c bng cch ghp tip chnh xu x n ln. xn = x.x.x...x (n
ln x) x0 = c vi mi xu x V d: x=01 x3 =010101 10= c Gio trnh Kin trc
my tnh v H iu hnh 11 CHNG 1. M U TRNG I HC BCH KHOA NNG 1. Mt s vn
v ngn ng 1.2. Ngn ng - Ngn ng L trn b ch V l tp hp cc xu trn V,L_V*
- Cc php ton trn ngn ng V ngn ng l tp hp nn c cc php ton tp hp:
(giao), (hp), -(hiu), b Ghp tip 2 ngn ng Gio trnh Kin trc my tnh v
H iu hnh 12 CHNG 1. M U TRNG I HC BCH KHOA NNG 1. Mt s vn v ngn ng
1.2. Ngn ng - Cc php ton trn ngn ng: Cho L1, L2 trn b ch V Php
giao: L1L2 = {x | xeL1 v xeL2} Php hp: L1L2 = {x | xeL1 hoc xeL2}
Php hiu: L1- L2 = {x | xeL1 v xeL2} Php b: L=V*- L Gio trnh Kin trc
my tnh v H iu hnh 13 CHNG 1. M U TRNG I HC BCH KHOA NNG 1. Mt s vn
v ngn ng 1.2. Ngn ng Ghp tip 2 ngn ng Cho 2 ngn ng L1, L2. Ta gi
ghp tip L1.L2 (L1L2) ca L1 v L2 l mt tp hp L1L2={xy/(xeL1) v
(yeL2)} x.x=x2; x.x.x=x3; x0=c; xi=xi-1x L0={c}; Li=Li-1.L -
L*=L0L1L2; L+=L1L2 Gio trnh Kin trc my tnh v H iu hnh 14 CHNG 1. M
U TRNG I HC BCH KHOA NNG 1. Mt s vn v ngn ng 1.3. Biu din ngn ng
Ngn ng n gin - Phng php lit k: ngn ng c s xu l hu hn v c th xc nh
c. V d: ngn ng l cc s t nhin nh hn 20 v ln hn 12 L={13, 14, 15, 16,
17, 18, 19} Gio trnh Kin trc my tnh v H iu hnh 15 CHNG 1. M U TRNG
I HC BCH KHOA NNG 1. Mt s vn v ngn ng 1.3. Biu din ngn ng Ngn ng n
gin - Phng php s dng tn t P(x): ngn ng m cc xu c cng cc c im. V d:
ngn ng l cc s thc nh hn 5. L={x/ (xe R) v (x0S1==>01
S==>0S1==>00S11==>0011
S==>0S1==>00S11==>000S111==>000111 Vy L(G)={0n1n | vi
n>=0} (1) (2) (1)(2) (1) (1) (2) (1) (1) (1)(2) Gio trnh Kin trc
my tnh v H iu hnh 25 CHNG 1. M U TRNG I HC BCH KHOA NNG 2. Vn
phm2.3. Ngn ng c sinh ra t vn phm - V d 2: Cho G: SaAAaA | b
L(G)={anb | n>0} Gio trnh Kin trc my tnh v H iu hnh 26 CHNG 1. M
U TRNG I HC BCH KHOA NNG 2. Vn phm2.4. Phn cp vn phm ca Chomsky -
Nu cc sn xut u c dng Aa |aB vi A,B eA;a eE: vn phm chnh quy (VP loi
3) - Nu cc sn xut c dng Ao vi AeA; oe(EA)*: vn phm phi ng cnh (VP
loi 2) - Nu cc sn xut c dng o| vi o, |e(EA)*: vn phm cm ng cnh (VP
loi 1) - Nu khng c hn ch g trn sn xut: vn phm t do (VP loi 0) Gio
trnh Kin trc my tnh v H iu hnh 27 CHNG 1. M U TRNG I HC BCH KHOA
NNG 2. Vn phm2.4. Phn cp vn phm ca Chomsky Lu : - Vn phm loi 3 l
trng hp c bit ca vn phm loi 2. - Vn phm loi 2 l trng hp c bit ca vn
phm loi 1. - Vn phm loi 1 l trng hp c bit ca vn phm loi 0. Gio trnh
Kin trc my tnh v H iu hnh 28 CHNG 1. M U TRNG I HC BCH KHOA NNG 3.
Khi nim tmt- B gm: tp cc trng thi v cc iu khin dch chuyn t trng thi
ny sang trng thi khc khi nhn d liu vo. - tmt biu din hot ng ca bng
in - tmt on nhn t kha int Tt Bt n cng tc i in intint Gio trnh Kin
trc my tnh v H iu hnh 29 CHNG 2. TMT HU HN TRNG I HC BCH KHOA NNG
tmt hu hn n nh(DFA) tmt hu hn khng n nh(NFA) S tng ng ca DFA v NFA
ng dng Gio trnh Kin trc my tnh v H iu hnh 30 CHNG 2. TMT HU HN TRNG
I HC BCH KHOA NNG 1. tmt hu hn n nh(Deterministic finite automata
DFA) 1.1. M t - tmt hu hn l mt ci my on nhn xu gm: Mt bng vo c chia
thnh nhiu , mi cha mt k hiu ca xu vo Mt u c, mi thi im tr vo mt trn
bng Gio trnh Kin trc my tnh v H iu hnh 31 CHNG 2. TMT HU HN TRNG I
HC BCH KHOA NNG 01011 1. tmt hu hn n nh 1.1. M t Mt b iu khin Q gm
cc trng thi, ti mi thi im n c mt trng thi c xc nh qua hm chuyn trng
thi Bng vo q B iu khin u c Gio trnh Kin trc my tnh v H iu hnh 32
CHNG 2. TMT HU HN TRNG I HC BCH KHOA NNG 1. tmt hu hn n nh 1.1. M t
- Ti mt thi im, trng thi q b iu khin v k hiu m u c ang c s xc nh
trng thi tip theo b iu khin. - Mi ln c xong mt , u c chuyn sang phi
mt . - Trng thi u tin b iu khin: trng thi bt u ca tmt Gio trnh Kin
trc my tnh v H iu hnh 33 CHNG 2. TMT HU HN TRNG I HC BCH KHOA NNG
1. tmt hu hn n nh 1.2. nh ngha: M(, Q, , q0, F) : b ch vo Q: tp hu
hn cc trng thi q0 e Q: trng thi u F _ Q: tp cc trng thi kt thc : hm
chuyn trng thi c dng (q,a)=pVi q,p e Q, a e Vi mi (q,a)=p ch c mt
trng thi p duy nht Gio trnh Kin trc my tnh v H iu hnh 34 CHNG 2.
TMT HU HN TRNG I HC BCH KHOA NNG 1. tmt hu hn n nh 1.2. nh ngha:- V
d: DFA M(, Q, , q0, F) : {0,1} Q: {0,1} q0: 0 F: {1} tp cc trng thi
kt thc : (0,0)=0, (0,1)=1, (1,1)=1 v (1,0)=0 Gio trnh Kin trc my
tnh v H iu hnh 35 CHNG 2. TMT HU HN TRNG I HC BCH KHOA NNG 1.
Otomat hu hn n nh 1.3. Biu din cc hm chuyn trng thi Dng bng: s dng
ma trn c: - Chs hng: trng thi - Ch s ct: k hiu vo - Gi tr ti hng q,
ct a l trng thi p, sao cho (q,a)=p Gio trnh Kin trc my tnh v H iu
hnh 36 CHNG 2. TMT HU HN TRNG I HC BCH KHOA NNG 1. Otomat hu hn n
nh 1.3. Biu din cc hm chuyn trng thi Dng bng:V d: c hm chuyn ca mt
Otomat nh sau:(1,a)=2, (2,b)=2, (2,c)=2 abc 12 222 Gio trnh Kin trc
my tnh v H iu hnh 37 CHNG 2. TMT HU HN TRNG I HC BCH KHOA NNG 1.
Otomat hu hn n nh 1.3. Biu din cc hm chuyn trng thi Biu dch chuyn:-
Mi trng thi qeQ c t trong cc vng trn. - Trng thi bt u q0 c thm du
> u. - Trng thi kt thc qeF c t trong vng trn kp. - Cc cung ni t
trng thi q sang trng thi p c mang cc nhn ae, c ngha (q,a)=pGio trnh
Kin trc my tnh v H iu hnh 38 CHNG 2. TMT HU HN TRNG I HC BCH KHOA
NNG 1. tmat hu hn n nh 1.3. Biu din cc hm chuyn trng thi Biu dch
chuyn:V d: c hm chuyn ca mt Otomat nh sau:(1,a)=2, (2,b)=2, (2,c)=2
1 2 a b c Gio trnh Kin trc my tnh v H iu hnh 39 CHNG 2. TMT HU HN
TRNG I HC BCH KHOA NNG 1. tmt hu hn n nh 1.4. Hm chuyn trng thi m
rng - trng thi q1c xong xu x=a1a2a3...ai-1=yai-1 chuyn sang trng
thi qi - C o(q1,a1)=q2; o(q2,a2)=q3;..., o(qi-1,ai-1)=qi : ta c th
biu din nh sau: o*(q1,x)=qi
- o*(q1,x)= o(o*(q1,y),ai-1)=qi: hm chuyn trng thi m rng. Gio
trnh Kin trc my tnh v H iu hnh 40 CHNG 2. TMT HU HN TRNG I HC BCH
KHOA NNG 1. tmt hu hn n nh 1.4. Hm chuyn trng thi m rng Cho :
(0,0)=0, (0,1)=1, (1,1)=1 v (1,0)=0 Xc nh o*(0,01101)=? o(0,0) = 0
o*(0,01) = o(o(0,0),1) = 1 o*(0,011) = o(o*(0,01),1) = 1 o*(0,0110)
= o(o*(0,011),0) = 0 o*(0,01101) = o(o*(0,0110),1) = 1 Gio trnh Kin
trc my tnh v H iu hnh 41 CHNG 2. TMT HU HN TRNG I HC BCH KHOA NNG
1. tmt hu hn n nh 1.5. Hot ng on nhn xu - T trng thi bt u, c tng k
hiu vo t tri sang phi. Mi ln c th xc nh li trng thi qua hm chuyn
trng thi. - Nu c xong xu vo m ri vo trng thi kt thc th xu vo c on
nhn ngc li xu vo khng c on nhn - Nu khng th c xong xu vo th xu vo
khng c on nhn. Gio trnh Kin trc my tnh v H iu hnh 42 CHNG 2. TMT HU
HN TRNG I HC BCH KHOA NNG 1. tmt hu hn n nh 1.5. Hot ng on nhn xu -
Xu x c on nhn bi tmat DFA o*(q0,x)=peF- Cho DFA
M({a,b},{0,1,2,3,4},o,0,{4}), :(0,a)=1, (1,b)=2, (2,a)=3,
(2,b)=4(3,a)=3, (3,b)=4, (4,b)=4xu x=abaabbb c c on nhn k0? Gio
trnh Kin trc my tnh v H iu hnh 43 CHNG 2. TMT HU HN TRNG I HC BCH
KHOA NNG 1. tmt hu hn n nh 1.5. Hot ng on nhn xu o(0,a) = 1
o*(0,ab) = o(o(0,a),b) = 2 o*(0,aba) = o(o*(0,ab),a) = 3 o*(0,abaa)
= o(o*(0,aba),a) = 3 o*(0,abaab) = o(o*(0,abaa),b) = 4 o*(0,abaabb)
= o(o*(0,abaab),b) = 4 o*(0,abaabbb) = o(o*(0,abaabb),b) = 4eF, xu
x c on nhn ab 12 0a 3b a 4b b Gio trnh Kin trc my tnh v H iu hnh 44
CHNG 2. TMT HU HN TRNG I HC BCH KHOA NNG 1. tmt hu hn n nh 1.5. Hot
ng on nhn xu Ta c th vit li nh sau: 0abaabbb 1baabbb
2aabbb3abbb3bbb4bb4b 4eF, xu x c on nhn ab 12 0a 3b a 4b b Gio trnh
Kin trc my tnh v H iu hnh 45 CHNG 2. TMT HU HN TRNG I HC BCH KHOA
NNG 1. tmt hu hn n nh 1.6. Ngn ng c on nhn bi DFA - Cho DFA M(, Q,
, q0, F), ngn ng L c on nhn bi M c xc nh nh sau: L(M)={xeE*|
o*(q0,x)=peF} - V d: v tmat on nhn ngn ng L gm cc xu l s nh phn c
di >=2 0120|10|1 0|1 Gio trnh Kin trc my tnh v H iu hnh 46 CHNG
2. TMT HU HN TRNG I HC BCH KHOA NNG 1. tmt hu hn n nh Trng thi khng
chp nhn c - Trng thi khng c mi tn i vo ch c m tn i ra (tr trng thi
bt u). - Trng thi ch c mi tn i vo khng c mi tn i ra (tr trng thi kt
thc q qGio trnh Kin trc my tnh v H iu hnh 47 CHNG 2. TMT HU HN TRNG
I HC BCH KHOA NNG 1. tmt hu hn n nh Bi tp: (1) Cho M, xu x: kim tra
xu x c c on nhn hay khng? x: aabbca, abbbca, abbaca, aaaca
01234abca c ab Gio trnh Kin trc my tnh v H iu hnh 48 CHNG 2. TMT HU
HN TRNG I HC BCH KHOA NNG 1. tmt hu hn n nh Bi tp: (2) Xy dng DFA
on nhn ngn ng L(M) sau: - L(M)={abcnbmvi n>=0, m>0} -
L(M)={0(10)n vi n>0} - L(M)={0n1m vi n>=0, m>0} -
L(M)={0n1m vi n>0, m>0} Gio trnh Kin trc my tnh v H iu hnh 49
CHNG 2. TMT HU HN TRNG I HC BCH KHOA NNG 1. tmt hu hn n nh Bi tp:
(3) Xy dng tmat on nhn L(M) l: - Mt s nh phn chn c di xu ln hn 2. -
Mt s nguyn c du, khng du. - Mt s nguyn khng du ca NNLT C - Mt s thc
tnh c du, khng du. Gio trnh Kin trc my tnh v H iu hnh 50 CHNG 2.
TMT HU HN TRNG I HC BCH KHOA NNG 2. tmat hu hn khng n nh
(Nondeterministic finite automata NFA) 2.1.nh ngha:M(, Q, , q0, F)
: b ch vo Q: tp hu hn cc trng thi q0 e Q: trng thi u F _ Q: tp cc
trng thi kt thc : hm chuyn trng thi c dng (q,a)=P Vi qeQ, ae(c),
P_Q Gio trnh Kin trc my tnh v H iu hnh 51 CHNG 2. TMT HU HN TRNG I
HC BCH KHOA NNG 2. tmat hu hn khng n nh V d: tmt on nhn cc xu c di
chn trn b ch {0,1} 0|10|1 12 0c Gio trnh Kin trc my tnh v H iu hnh
52 CHNG 2. TMT HU HN TRNG I HC BCH KHOA NNG 2. tmat hu hn khng n nh
S khc nhau gia DFA v NFA - NFA: (q,a)=P, vi qeQ, ae(c), P_Q DFA:
(q,a)=p, vi q,p e Q, a e - NFA: c th c dch chuyn khi c c DFA: khng
th dch chuyn khi c c Gio trnh Kin trc my tnh v H iu hnh 53 CHNG 2.
TMT HU HN TRNG I HC BCH KHOA NNG 2. tmat hu hn khng n nh 2.2.Hm dch
chuyn m rng - o(q0,a1)={q1, q2} v o(q1,a2)={q3,q4} o(q2,a2)={q5,q6}
- o*(q0,a1a2)= o(q1,a2)o(q2,a2)={q3,q4,q5,q6} - C o*(q,x)={p1,
p2,...,pk} tho*(q,xa)= vi xeE*, aeE kiia p1) , (= oGio trnh Kin trc
my tnh v H iu hnh 54 CHNG 2. TMT HU HN TRNG I HC BCH KHOA NNG 2.
tmat hu hn khng n nh 2.2.Hm dch chuyn m rng - Cho NFA,
o*(0,01001)=? o(0,0)={1} o*(0,01)= o(1,1)={1,2} o*(0,010)=
o(1,0)o(2,0) ={1} o*(0,0100)= o(1,0) ={1} o*(0,01001)= o(1,1)
={1,2}01 12 0 0|1 Gio trnh Kin trc my tnh v H iu hnh 55 CHNG 2. TMT
HU HN TRNG I HC BCH KHOA NNG 2. tmat hu hn khng n nh2.3.on nhn xu x
bi NFA - Xu x c on nhn bi tmat o*(q0,x)F=u - Theo v d
trc:o*(0,01001)= {1,2} {1,2}F ={2} = u: xu 01001 c on nhn Gio trnh
Kin trc my tnh v H iu hnh 56 CHNG 2. TMT HU HN TRNG I HC BCH KHOA
NNG 2. tmat hu hn khng n nh2.4.Ngn ng c on nhn bi NFA - Cho NFA M(,
Q, , q0, F), ngn ng L c on nhn bi M c xc nh nh sau: L(M)={xeE*|
o*(q0,x) F=u} - Theo v d trc cc xu bt u bng 0 v kt thc bng 1 c on
nhn: 00111, 0100101, 0111, 010101,... Gio trnh Kin trc my tnh v H
iu hnh 57 CHNG 2. TMT HU HN TRNG I HC BCH KHOA NNG 3. Xy dng DFA t
NFA - Cho NFA M(N, QN, N, q0, FN) th DFA M(N, QD, D, q0, FD) - Xc
nh QD, D, FD To tt c cc tp con T t tp QN Xc nh oD(T,a)=oN(qi,a) vi
qi eT, aeE Mi tp T tng ng vi 1 trng thi ca DFA Loi b cc trng thi
khng chp nhn c Trng thi tng ng vi tp T c cha trng thi kt thc s l
trng thi kt thc ca DFA Gio trnh Kin trc my tnh v H iu hnh 58 CHNG
2. TMT HU HN TRNG I HC BCH KHOA NNG 3. Xy dng DFA t NFA - V d: 01
12 00|1 01 {0}{0,1}{0} {1}{2} *{2} {0,1}{0,1}{0,2} *{0,2}{0,1}{0}
*{1,2}{2} *{1,2,0}{0,1}{0,2} 01 q0 q3 q0 q1 q2 *q2 q3 q3 q4 * q4 q3
q0 * q5 q2 * q6 q3q4 Gio trnh Kin trc my tnh v H iu hnh 59 CHNG 2.
TMT HU HN TRNG I HC BCH KHOA NNG 3. Xy dng DFA t NFA - V d: 01 q0
q3 q0 q1 q2 *q2 q3 q3 q4 *q4 q3 q0 *q5 q2 *q6 q3q4 0 q301 q4q0 10 1
0 1 1 q1 q2 q5 q6 1 Gio trnh Kin trc my tnh v H iu hnh 60 CHNG 2.
TMT HU HN TRNG I HC BCH KHOA NNG 3. Xy dng DFA t NFA - Cc trng thi
q1, q2, q5, q6: khng chp nhn c - q4 tng ng {0,2}: l trng thi kt thc
01 q0 q3 q0 q3 q3 q4 *q4 q3 q0 q301 q4q0 10 1 0 Gio trnh Kin trc my
tnh v H iu hnh 61 CHNG 2. TMT HU HN TRNG I HC BCH KHOA NNG 4. ng
dng - on nhn t kha, s, ..., t t - DFA on nhn kha float - NFA on nhn
s nguyn c du 5 0f 1l 2o 3a 4t +| - |c 120|..|9 00|..|9 Gio trnh Kin
trc my tnh v H iu hnh 62 CHNG 2. TMT HU HN TRNG I HC BCH KHOA NNG
Bi tp (1) V NFA on nhn- cc s nh phn c di l bi s ca 4. - cc t 110,
101. - Cc t 0(101)+1 (2) Chuyn NFA thnh DFA - Cc NFA cu (1) Gio
trnh Kin trc my tnh v H iu hnh 63 CHNG 2. TMT HU HN TRNG I HC BCH
KHOA NNG Bi tp - NFA hnh v sau: - NFA c hm chuyn sau: 3 120 0|1
0|10 00|1 oa b cd 0{1,2}{1}{2} 1{0}{2}{0,1} *2 Gio trnh Kin trc my
tnh v H iu hnh 64 CHNG 2. TMT HU HN TRNG I HC BCH KHOA NNG Bi tp -
NFA c hm chuyn sau: o01 0{1,3}{1} *1{2}{1,2} 2{3}{0} *3{0} Gio trnh
Kin trc my tnh v H iu hnh 65 CHNG 3. BIU THC & VN PHM CHNH QUY
TRNG I HC BCH KHOA NNG Biu thc chnh quy Ngn ng chnh quy Vn phm chnh
quiGio trnh Kin trc my tnh v H iu hnh 66 CHNG 3. BIU THC & VN
PHM CHNH QUY TRNG I HC BCH KHOA NNG 1. Biu Thc chnh quy 1.1.nh
ngha: Cho b ch E, biu thc chnh quy (BTCQ) trn E c nh ngha qui nh
sau: - u l BTCQ - c l BTCQ - aeE, a l BTCQ - Vi r, s l BTCQ th
(r+s), rs, r*, s* l BTCQ C th vit (r+s) hay (rs) Gio trnh Kin trc
my tnh v H iu hnh 67 CHNG 3. BIU THC & VN PHM CHNH QUY TRNG I
HC BCH KHOA NNG 1. Biu Thc chnh quy 1.1.nh ngha: V d: Cho E={0,1}
ta c: - u, c, 0, 1 l BTCQ - (0+1), 01, 0*, 1*, (0+1)01, (0+1)0*,
(01)*, ((0+1)01)* l BTCQ Gio trnh Kin trc my tnh v H iu hnh 68 CHNG
3. BIU THC & VN PHM CHNH QUY TRNG I HC BCH KHOA NNG 1. Biu Thc
chnh quy 1.1.nh ngha: V d: - (a+b+c) l BTCQ ch nh 3 xu a, b, c trn
b ch {a, b, c} - (a+b)* l BTCQ ch nh mi xu trn {a, b} - aa*bb* hay
a+b+ l BTCQ ch nh cc xu bt u bng mt s k hiu a, tip theo l mt s k
hiu b trong k hiu a v b xut hin t nht 1 ln Gio trnh Kin trc my tnh
v H iu hnh 69 CHNG 3. BIU THC & VN PHM CHNH QUY TRNG I HC BCH
KHOA NNG 1. Biu Thc chnh quy 1.1.nh ngha: V d: - (b+c)(a+ab)* l
BTCQ ch nh cc xu trn {a, b} khng cha bb - (0+1)*1 l BTCQ ch nh cc
xu l s nh phn l Gio trnh Kin trc my tnh v H iu hnh 70 CHNG 3. BIU
THC & VN PHM CHNH QUY TRNG I HC BCH KHOA NNG 1. Biu Thc chnh
quy 1.2.Th t u tin ca cc php ton - Php bao ng (*) - Php ghp tip (.)
- Php hp (+) V d: 10*+0 :(1(0)*)+0 Gio trnh Kin trc my tnh v H iu
hnh 71 CHNG 3. BIU THC & VN PHM CHNH QUY TRNG I HC BCH KHOA NNG
1. Biu Thc chnh quy 1.3.Tnh cht i s- Tnh giao hon:r + s = s + r -
Tnh kt hp:(r + s) + t = r + (s + t) r(st)=(rs)tGio trnh Kin trc my
tnh v H iu hnh 72 CHNG 3. BIU THC & VN PHM CHNH QUY TRNG I HC
BCH KHOA NNG 1. Biu Thc chnh quy 1.3.Tnh cht i s- Tnh cht phn b r
(s + t) = rs + rt (r + s)t = rt + st - Mt s tnh cht khc (r*)* =
r*cr = rc = rc*= c r + r = rr* = r+ + c r+ = rr* = r*r (r*s*)* =(r
+ s)* Gio trnh Kin trc my tnh v H iu hnh 73 CHNG 3. BIU THC &
VN PHM CHNH QUY TRNG I HC BCH KHOA NNG 1. Biu Thc chnh quy 1.4.Ngn
ng c ch nh bi BTCQNgn ng L(r) c ch nh bi BTCQ r bt k l c xy dng da
trn quy tc - L(c ) = {c}- L(a) = {a} - L(r+s)=L(r)L(s) - L(rs) =
L(r)L(s) - L(r*) = (L(r))* Gio trnh Kin trc my tnh v H iu hnh 74
CHNG 3. BIU THC & VN PHM CHNH QUY TRNG I HC BCH KHOA NNG 1. Biu
Thc chnh quy 1.4.Ngn ng c ch nh bi BTCQ V d: xy dng BTCQ ch nh ngn
ng L c t nht mt k hiu a v 1 k hiu b trn {a, b} - L(r) =
L(r1)L(r2)=L(r1+r2) - L(r1): cc xu r1 c dng w1aw2bw3 - L(r2): cc xu
r2 c dng w1bw2aw3 - BTCQ ch nh L: (a+b)*a(a+b)*b(a+b)* +
(a+b)*b(a+b)*a(a+b)* Gio trnh Kin trc my tnh v H iu hnh 75 CHNG 3.
BIU THC & VN PHM CHNH QUY TRNG I HC BCH KHOA NNG 1. Biu Thc
chnh quy Bi tp (1) Xy dng BTCQ ch nh cc ngn ng sau: - Tp hp cc xu
trn {a,b} c di chia ht cho 3 - Tp hp cc xu trn {a, b, c} ch cha 1 k
hiu a, cn li l cc k hiu b v c - Tp hp cc s nh phn c tn cng l 11 -
Tp hp cc s nguyn k0 du chia ht cho 5 Gio trnh Kin trc my tnh v H iu
hnh 76 CHNG 3. BIU THC & VN PHM CHNH QUY TRNG I HC BCH KHOA NNG
1. Biu Thc chnh quy Bi tp (2) M t ngn ng c ch nh bi BTCQ sau: -
(a+b+c+d)*(a+d) - 1(0+1)(0+1)(0+1) - ((0+1)(0+1))+
- a(a+b)*a - (ab)* + (ba)* Gio trnh Kin trc my tnh v H iu hnh 77
CHNG 3. BIU THC & VN PHM CHNH QUY TRNG I HC BCH KHOA NNG 2. Ngn
ng chnh quy 2.1.Khi nim - Ngn ng chnh quy l ngn ng c biu din bi biu
thc chnh qui. - Ngn ng chnh qui c on nhn bi tmt hu hn. - Ngn ng c
sn sinh t vn phm chnh quy l ngn ng chnh qui Gio trnh Kin trc my tnh
v H iu hnh 78 CHNG 3. BIU THC & VN PHM CHNH QUY TRNG I HC BCH
KHOA NNG 2. Ngn ng chnh quy 2.2.tmat NFA on nhn ngn ng c biu din bi
BTCQ - BTCQ l a - BTCQ l c a c Gio trnh Kin trc my tnh v H iu hnh
79 CHNG 3. BIU THC & VN PHM CHNH QUY TRNG I HC BCH KHOA NNG 2.
Ngn ng chnh quy 2.2.tmat NFA on nhn ngn ng c biu din bi BTCQ - BTCQ
l rs c s
rGio trnh Kin trc my tnh v H iu hnh 80 CHNG 3. BIU THC & VN
PHM CHNH QUY TRNG I HC BCH KHOA NNG 2. Ngn ng chnh quy 2.2.tmat NFA
on nhn ngn ng c biu din bi BTCQ - BTCQ l (r+s) c r s c c c Gio trnh
Kin trc my tnh v H iu hnh 81 CHNG 3. BIU THC & VN PHM CHNH QUY
TRNG I HC BCH KHOA NNG 2. Ngn ng chnh quy 2.2.tmat NFA on nhn ngn
ng c biu din bi BTCQ - BTCQ l r*
r
c c c c Gio trnh Kin trc my tnh v H iu hnh 82 CHNG 3. BIU THC
& VN PHM CHNH QUY TRNG I HC BCH KHOA NNG 2. Ngn ng chnh quy
2.2.tmat NFA on nhn ngn ng c biu din bi BTCQ - V d: xy dng NFA on
nhn (0+1)(01)* 0 1 c
c c c Gio trnh Kin trc my tnh v H iu hnh 83 CHNG 3. BIU THC
& VN PHM CHNH QUY TRNG I HC BCH KHOA NNG 2. Ngn ng chnh quy
2.2.tmat on nhn ngn ng c biu din bi BTCQ - V d: (0+1)(01)*
c c c 0 c 1 Gio trnh Kin trc my tnh v H iu hnh 84 CHNG 3. BIU
THC & VN PHM CHNH QUY TRNG I HC BCH KHOA NNG 2. Ngn ng chnh quy
0 1 c
c c c
c c c 0 c 1 00|1 c 1 20 c 31 Gio trnh Kin trc my tnh v H iu hnh
85 CHNG 3. BIU THC & VN PHM CHNH QUY TRNG I HC BCH KHOA NNG 2.
Ngn ng chnh quy 2.3.Tnh cht ng ca ngn ng chnh quy Cho L1(r) v L2(s)
l ngn ng chnh quy th: - L1(r)L2(s): ngn ng chnh quy- L1(r) L2(s):
ngn ng chnh quy - L1(r).L2(s): ngn ng chnh quy - (L1(r))*: ngn ng
chnh quy Gio trnh Kin trc my tnh v H iu hnh 86 CHNG 3. BIU THC
& VN PHM CHNH QUY TRNG I HC BCH KHOA NNG 2. Ngn ng chnh quy Bi
tp (1) V NFA on nhn ngn ng c biu din bi BTCQ sau: - (01)*(0+10*) -
(10+(0+01)1*0) - (0+1)*(11+00)(0+1)* Gio trnh Kin trc my tnh v H iu
hnh 87 CHNG 3. BIU THC & VN PHM CHNH QUY TRNG I HC BCH KHOA NNG
2. Ngn ng chnh quy Bi tp (2) V DFA on nhn ngn ng c biu din bi BTCQ
sau: - 1*01+ - 10*11* - (a+b)c*ab* Gio trnh Kin trc my tnh v H iu
hnh 88 CHNG 3. BIU THC & VN PHM CHNH QUY TRNG I HC BCH KHOA NNG
3. Vn phm chnh quy 3.1.nh ngha vn phm tuyn tnh tri - Mt vn phm G=(,
, s, p) c gi l vn phm tuyn tnh tri nu tt c cc sn xut c dng ABw hay
AB vi A,Be; we* - V d: SS a | S b | a Gio trnh Kin trc my tnh v H
iu hnh 89 CHNG 3. BIU THC & VN PHM CHNH QUY TRNG I HC BCH KHOA
NNG 3. Vn phm chnh quy 3.2.nh ngha vn phm tuyn tnh phi - Mt vn phm
G=(, , s, p) c gi l vn phm tuyn tnh phi nu tt c cc sn xut c dng AwB
hay AB vi A,Be; we* - V d: Sa A Aa A | b A | c Gio trnh Kin trc my
tnh v H iu hnh 90 CHNG 3. BIU THC & VN PHM CHNH QUY TRNG I HC
BCH KHOA NNG 3. Vn phm chnh quy 3.3.nh ngha vn phm chnh quy - Vn
phm tuyn tnh tri, vn phm tuyn tnh phi l vn phm chnh quy. Gio trnh
Kin trc my tnh v H iu hnh 91 CHNG 3. BIU THC & VN PHM CHNH QUY
TRNG I HC BCH KHOA NNG 3. Vn phm chnh quy 3.4.Xy dng NFA t vn phm
tuyn tnh phi Cho vn phm chnh qui G=(G , , s, p) xy dng NFA M=(Q, ,
q0, o, F) on nhn ngn ng c sinh ra t G - = G - Vi Ae th AeQ - q0 = S
Gio trnh Kin trc my tnh v H iu hnh 92 CHNG 3. BIU THC & VN PHM
CHNH QUY TRNG I HC BCH KHOA NNG 3. Vn phm chnh quy 3.4.Xy dng NFA t
vn phm tuyn tnh phi - Nu Aa1a2...ai th o*(A,a1a2...ai)=qi : thm qi
vo Q v F - Nu Aa1a2...ai th t vo o cc dch chuyn trng thi v thm vo Q
cc trng thi mi sao cho o*(A,a1a2...ai)=qia1a2 A ... qi
ai Gio trnh Kin trc my tnh v H iu hnh 93 CHNG 3. BIU THC &
VN PHM CHNH QUY TRNG I HC BCH KHOA NNG 3. Vn phm chnh quy 3.4.Xy
dng NFA t vn phm tuyn tnh phi - Nu Aa1a2...aiB th t vo o cc dch
chuyn trng thi v thm vo Q cc trng thi mi sao cho o*(A,a1a2...ai)=B
- Nu Ac th thm A vo F a1a2 A ... Bai Gio trnh Kin trc my tnh v H iu
hnh 94 CHNG 3. BIU THC & VN PHM CHNH QUY TRNG I HC BCH KHOA NNG
3. Vn phm chnh quy 3.4.Xy dng NFA t vn phm tuyn tnh phi - V d: Sa A
Aa A | b A | ca a |b S A Gio trnh Kin trc my tnh v H iu hnh 95 CHNG
3. BIU THC & VN PHM CHNH QUY TRNG I HC BCH KHOA NNG 3. Vn phm
chnh quy 3.4.Xy dng vn phm chnh quy t NFA Cho NFA M=(Q, , q0, o, F)
xy dng vn phm chnh qui G=(G , , s, p) - G = - =Q - S=q0
- qap nu o(q,a)=p- qc nu q e F Gio trnh Kin trc my tnh v H iu
hnh 96 CHNG 3. BIU THC & VN PHM CHNH QUY TRNG I HC BCH KHOA NNG
3. Vn phm chnh quy 3.4. Xy dng vn phm chnh quy t NFAV d:- G = {a,b}
- ={S, A} - S=S- SaA v o(S,a)=A ; AaA v o(A,a)=A - AbA v o(A,b)=A;
Ac v A e F a a |b S A Gio trnh Kin trc my tnh v H iu hnh 97 CHNG 3.
BIU THC & VN PHM CHNH QUY TRNG I HC BCH KHOA NNG 3. Vn phm chnh
quy Bi tp (1) V otomat NFA t G sau: - S0S| 1S | 1 - S + A |- A A0
A| 1A| ..| 9A |0|..|9 - SabA AbB| B Bc Gio trnh Kin trc my tnh v H
iu hnh 98 CHNG 3. BIU THC & VN PHM CHNH QUY TRNG I HC BCH KHOA
NNG 3. Vn phm chnh quy Bi tp (2) Xy dng G t NFA sau: D BC0 0|1 0|10
A0|1 Gio trnh Kin trc my tnh v H iu hnh 99 CHNG 3. BIU THC & VN
PHM CHNH QUY TRNG I HC BCH KHOA NNG 4. Cc tiu ha DFA 4.1. Trng thi
tng ng - Vi weE*c o*(p,w)=ql
v o* (q,w)=qt m qt, ql cng kt thchoc cng khng kt thc - {q,p} v
{p,k} cc cp trng thi tng ng th cp {q,k} cng tng ng (t/c bt cu) -
Hai trng thi khng tng ng c gi l hai trng thi phn bit. {p, q} l cp
trng thi tng ng. Gio trnh Kin trc my tnh v H iu hnh 100 CHNG 3. BIU
THC & VN PHM CHNH QUY TRNG I HC BCH KHOA NNG 4. Cc tiu ha DFA
4.1. Trng thi tng ng - Xy dng bng nh du cp trng thi phn bit (1) Nu
p l trng thi khng kt thc v q l trng thi kt thc th {p,q} l trng thi
phn bit. (2) Vi aeE c o(p,a)=ql v o(q,a)=qt m {qt, ql} l cp trng
thi phn bit th {p, q} cp trng thi phn bit. Gio trnh Kin trc my tnh
v H iu hnh 101 CHNG 3. BIU THC & VN PHM CHNH QUY TRNG I HC BCH
KHOA NNG 4. Cc tiu ha DFA 4.1. Trng thi tng ng V d:xy dng bng nh du
cp trng thi phn bit cho otomat sau ACBDEFG1 0 0 1 0,1 0 1 0,1 0,1 0
1 Gio trnh Kin trc my tnh v H iu hnh 102 CHNG 3. BIU THC & VN
PHM CHNH QUY TRNG I HC BCH KHOA NNG 4. Cc tiu ha DFA 4.1. Trng thi
tng ng - Ta c B, D, E, G: trng thi kt thc A,C,F: trng thi khng kt
thc p dng (qt1) nn cc cp sau l phn bit: {A,B}, {A,D}, {A,E}, {A,G},
{C,B}, {C,D}, {C,E}, {C,G},{F,B}, {F,D}, {F,E}, {F,G} Gio trnh Kin
trc my tnh v H iu hnh 103 CHNG 3. BIU THC & VN PHM CHNH QUY
TRNG I HC BCH KHOA NNG 4. Cc tiu ha DFA 4.1. Trng thi tng ng - nh
du x vo cc l cp trng thi phn bit BX CX DXX EXX FXXX GXXX ABCDEF Gio
trnh Kin trc my tnh v H iu hnh 104 CHNG 3. BIU THC & VN PHM
CHNH QUY TRNG I HC BCH KHOA NNG 4. Cc tiu ha DFA 4.1. Trng thi tng
ng - p dng (qt2) i vi tng cp trng thi phn bit {A,B}, {A,D}, {A,E},
{A,G}: v A l trng thi bt u nn khng c qt2. {C,B}: c to ra t cng trng
thi A nn k0 c (qt2) {C,D}: o(A,1)=C v o(B,1)=D nn {A,B} phn bit (
nh du ri) Gio trnh Kin trc my tnh v H iu hnh 105 CHNG 3. BIU THC
& VN PHM CHNH QUY TRNG I HC BCH KHOA NNG 4. Cc tiu ha DFA 4.1.
Trng thi tng ng {C,E}: k0 c a tha (qt2) {C,G}: o(A,1)=C v o(E,1)=G
nn {A,E} phn bit ( nh du ri) {F,B}: k0 c a tha (qt2) {F,D}:
o(C,1)=F v o(B,1)=D nn {C,B} phn bit ( nh du ri) {F,E}: k0 c a tha
(qt2) Gio trnh Kin trc my tnh v H iu hnh 106 CHNG 3. BIU THC &
VN PHM CHNH QUY TRNG I HC BCH KHOA NNG 4. Cc tiu ha DFA 4.1. Trng
thi tng ng {F,G}:o(E,1)=G v o(C,1)=F nn {C,E} phn bit ( nh du
ri)o(G,1)=G v o(C,1)=F nn {G,C} phn bit ( nh du ri) o(F,1)=F v
o(G,1)=G nn {F,G} phn bit ( nh du ri) Ta k0 tm thm c cp trng thi
phn bit no Gio trnh Kin trc my tnh v H iu hnh 107 CHNG 3. BIU THC
& VN PHM CHNH QUY TRNG I HC BCH KHOA NNG 4. Cc tiu ha DFA 4.1.
Trng thi tng ng - Ta c bng nh du trng thi phn bit sau BX CX DXX EXX
FXXX GXXX ABCDEF Gio trnh Kin trc my tnh v H iu hnh 108 CHNG 3. BIU
THC & VN PHM CHNH QUY TRNG I HC BCH KHOA NNG 4. Cc tiu ha DFA
4.1. Trng thi tng ng - Ta c cc cp trng thi tng ng sau: {A,C},
{A,F}, {B,D}, {B,E}, {B,G}, {C,F}, {D,E}, {D,G}, {E,G} Gio trnh Kin
trc my tnh v H iu hnh 109 CHNG 3. BIU THC & VN PHM CHNH QUY
TRNG I HC BCH KHOA NNG 4. Cc tiu ha DFA 4.2. Hai otomat tng ng -
Hai otomat cng on nhn mt ngn ng th tng ng. - Hai DFA tng ng th cp
trng thi u tng ng. - Cc tiu ha DFA: tm DFA tng ng c s trng thi t
nht. Gio trnh Kin trc my tnh v H iu hnh 110 CHNG 3. BIU THC &
VN PHM CHNH QUY TRNG I HC BCH KHOA NNG 4. Cc tiu ha DFA 4.3. Thut
ton - Loi b cc trng thi khng chp nhn c - Xc nh tt c cc cp trng thi
tng ng - Chia cc trng thi thnh cc nhm, sao cho: cc trng thi trong
cng mt nhm tng ng nhau Khng c cp trng thi no 2 nhm khc nhau l tng
ng.Gio trnh Kin trc my tnh v H iu hnh 111 CHNG 3. BIU THC & VN
PHM CHNH QUY TRNG I HC BCH KHOA NNG 4. Cc tiu ha DFA 4.3. Thut ton
- omin(S,a)=T khi qeS th o(q,a)=peT - Trng thi u Mmin l trng thi c
cha trng thi u ca M - Trng thi kt thc Mmin l nhng trng thi c cha
trng thi kt thc ca M Gio trnh Kin trc my tnh v H iu hnh 112 CHNG 3.
BIU THC & VN PHM CHNH QUY TRNG I HC BCH KHOA NNG 4. Cc tiu ha
DFA 4.3. Thut ton V d: Cc tiu ha DFA v d trc - Ta c cc cp trng thi
tng ng sau: {A,C}, {A,F}, {B,D}, {B,E}, {B,G}, {C,F}, {D,E}, {D,G},
{E,G} - To thnh cc nhm trng thi tng ng: {A,C,F}, {B,D,E,G} ACF 1 0
BDEG 0|1 Gio trnh Kin trc my tnh v H iu hnh 113 CHNG 3. BIU THC
& VN PHM CHNH QUY TRNG I HC BCH KHOA NNG 4. Cc tiu ha DFA Bi
tp:Cc tiu cc DFA sau: (1) o01 ABF BGC *CAC DCG EHF FCG GGE HGC Gio
trnh Kin trc my tnh v H iu hnh 114 CHNG 3. BIU THC & VN PHM
CHNH QUY TRNG I HC BCH KHOA NNG 4. Cc tiu ha DFA Bi tp:Cc tiu cc
DFA sau: (2) o01 ABE BDC *CCC DEC *EDC Gio trnh Kin trc my tnh v H
iu hnh 115 CHNG 4. VN PHM & NGN NG PHI NG CNH TRNG I HC BCH
KHOA NNG Vn phm phi ng cnh S nhp nhng trong vn phm phi ng cnh Ngn
ng phi ng cnh Dng chun ca vn phm phi ng cnhGio trnh Kin trc my tnh
v H iu hnh 116 CHNG 4. VN PHM & NGN NG PHI NG CNH TRNG I HC BCH
KHOA NNG 1. Vn phm phi ng cnh nh ngha: G=(, , s, p) trong : : tp hu
hn cc k hiu kt thc. : tp hu hn cc k hiu cha kt thc. s: k hiu bt u;
se p: tp hu hn cc sn xut c dngAovi Ae v oe()* Gio trnh Kin trc my
tnh v H iu hnh 117 CHNG 4. VN PHM & NGN NG PHI NG CNH TRNG I HC
BCH KHOA NNG 1. Vn phm phi ng cnh V d:SS(S)S | cGio trnh Kin trc my
tnh v H iu hnh 118 CHNG 4. VN PHM & NGN NG PHI NG CNH TRNG I HC
BCH KHOA NNG 1. Vn phm phi ng cnh Suy dn tri, suy dn phi - Nu lun
lun thay th k hiu cha kt thc bn tri nht gi l suy dn tri.- Tng t ta
c suy dn phi - V d: vit suy dn tri, suy dn phi to ra xu a+a*a ca vn
phm sau:EE+E |E*E| (E) |a Gio trnh Kin trc my tnh v H iu hnh 119
CHNG 4. VN PHM & NGN NG PHI NG CNH TRNG I HC BCH KHOA NNG 1. Vn
phm phi ng cnh Cy suy dn: cy tho mn cc iu kin:- Mi nt c 1 nhn: k
hiu kt thc hoc cha kt thc - Nhn ca nt gc: k hiu bt u - Nhn ca nt l:
k hiu kt thc - Nu mt nt c nhn A c cc nt con ca n t tri sang phi c
nhn x1, x2, x3, xn th Ax1x2x3xn e p Gio trnh Kin trc my tnh v H iu
hnh 120 CHNG 4. VN PHM & NGN NG PHI NG CNH TRNG I HC BCH KHOA
NNG 1. Vn phm phi ng cnh Cy suy dn - Suy dn tri to ra cy suy dn
tri.- Suy dn phi to ra cy suy dn phi - V d: v cy suy dn tri, cy suy
dn phi to ra xu a+a*a ca vn phm sau:EE+E |E*E| (E) |a Gio trnh Kin
trc my tnh v H iu hnh 121 CHNG 4. VN PHM & NGN NG PHI NG CNH
TRNG I HC BCH KHOA NNG 2. S nhp nhng trong vn phm phi ng cnh Khi
nim vn phm nhp nhng- Cho vn phm phi ng cnh G. Nu -xeL(G)c sinh ra t
2 cy suy dn khc nhau th G c gi l vn phm nhp nhng - V d:EE+E | E*E |
(E) | a Gio trnh Kin trc my tnh v H iu hnh 122 CHNG 4. VN PHM &
NGN NG PHI NG CNH TRNG I HC BCH KHOA NNG 2. S nhp nhng trong vn phm
phi ng cnh Kh s nhp nhng - a vo vn phm lut u tin - t tha s chung
Gio trnh Kin trc my tnh v H iu hnh 123 CHNG 4. VN PHM & NGN NG
PHI NG CNH TRNG I HC BCH KHOA NNG 2. S nhp nhng trong vn phm phi ng
cnh Kh s nhp nhng - V d: SaS | aSbS | c Gio trnh Kin trc my tnh v H
iu hnh 124 CHNG 4. VN PHM & NGN NG PHI NG CNH TRNG I HC BCH
KHOA NNG 3. Ngn ng phi ng cnh - Tp hp cc xu c sinh ra t vn phm phi
ng cnh l ngn ng phi ng cnh Gio trnh Kin trc my tnh v H iu hnh 125
CHNG 4. VN PHM & NGN NG PHI NG CNH TRNG I HC BCH KHOA NNG 4.
Dng chun ca vn phm phi ng cnh Khi nimVn phm phi ng cnh k0 cha: K
hiu tha Sn xut c Sn xut n v c gi l vn phm phi ng cnh dng chunGio
trnh Kin trc my tnh v H iu hnh 126 CHNG 4. VN PHM & NGN NG PHI
NG CNH TRNG I HC BCH KHOA NNG 4. Dng chun ca vn phm phi ng cnh Xc
nh v kh k hiu tha - K hiu tha c 2 loi: K hiu v sinh K hiu k0 t n c
- K hiu AeA c gi l k hiu v sinh khi AoeE* - K hiu AeA c gi l k hiu
k0 t n c khi S+oA| Gio trnh Kin trc my tnh v H iu hnh 127 CHNG 4.
VN PHM & NGN NG PHI NG CNH TRNG I HC BCH KHOA NNG 4. Dng chun
ca vn phm phi ng cnh Xc nh v kh k hiu tha - V d: xc nh k hiu tha ca
VP sau: (1) S0S | 1S | A| c A0A (2) S0S | 1S | c A0A | 1 Gio trnh
Kin trc my tnh v H iu hnh 128 CHNG 4. VN PHM & NGN NG PHI NG
CNH TRNG I HC BCH KHOA NNG 4. Dng chun ca vn phm phi ng cnh Xc nh v
kh k hiu tha - Xc nh k hiu k0 v sinh: c k0 v sinh aeE k0 v sinh Vi
Ao m mi k hiu thuc o k0 v sinh th A k0 v sinh. - K hiu k0 phi l k
hiu k0 v sinh th l k hiu v sinh Gio trnh Kin trc my tnh v H iu hnh
129 CHNG 4. VN PHM & NGN NG PHI NG CNH TRNG I HC BCH KHOA NNG
4. Dng chun ca vn phm phi ng cnh Xc nh v kh k hiu tha - Xc nh k hiu
t n c: S l k hiu t n c Vi Ao m A l k hiu t n c th mi k hiu thuc o l
k hiu t n c. - K hiu k0 phi l k hiu t n th l k hiu k0 t n c Gio
trnh Kin trc my tnh v H iu hnh 130 CHNG 4. VN PHM & NGN NG PHI
NG CNH TRNG I HC BCH KHOA NNG 4. Dng chun ca vn phm phi ng cnh Xc
nh v kh k hiu tha - Kh k hiu tha: Loi b tt c cc k hiu v sinh v cc
sn xut cha n. Loi b tt c cc k hiu k0 t n c v cc sn xut cha n. Gio
trnh Kin trc my tnh v H iu hnh 131 CHNG 4. VN PHM & NGN NG PHI
NG CNH TRNG I HC BCH KHOA NNG 4. Dng chun ca vn phm phi ng cnh Xc
nh v kh k hiu tha - V d: Kh k hiu tha ca cc vn phm: (1) S0S | 1S |
A| c A0A (2) S0S | 1S | c A0A | 1 Gio trnh Kin trc my tnh v H iu
hnh 132 CHNG 4. VN PHM & NGN NG PHI NG CNH TRNG I HC BCH KHOA
NNG 4. Dng chun ca vn phm phi ng cnh Kh sn xut c - A l bin trit tiu
khi A+c - Xc nh bin trit tiu: A c th A: bin trit tiu AC1C2...Cn nu
C1, C2,...,Cn l bin trit tiu th A l bin trit tiu Gio trnh Kin trc
my tnh v H iu hnh 133 CHNG 4. VN PHM & NGN NG PHI NG CNH TRNG I
HC BCH KHOA NNG 4. Dng chun ca vn phm phi ng cnh Kh sn xut c - Cho
G(E, A, S, p), kh sn xut c ta cG(E, A, S, p) vi p xc nh nh sau: Vi
mi AB1B2...Bn ep ta thm cc Ax1x2...xn vo p trong mi xi thay th Bi
tha mn: Nu Bi l bin k0 trit tiu c th xi=Bi Nu Bi l bin trit tiu th
xi=c v xi=Bi Khng cho tt c cc xi=c Gio trnh Kin trc my tnh v H iu
hnh 134 CHNG 4. VN PHM & NGN NG PHI NG CNH TRNG I HC BCH KHOA
NNG 4. Dng chun ca vn phm phi ng cnh Kh sn xut c - V d: kh sn xut c
ca vn phm : SAB AaAB | c BbAB | c Gio trnh Kin trc my tnh v H iu
hnh 135 CHNG 4. VN PHM & NGN NG PHI NG CNH TRNG I HC BCH KHOA
NNG 4. Dng chun ca vn phm phi ng cnh Kh sn xut c - Ta c S, A, B: l
bin trit tiu - Xt SAB cA l bin trit tiu nn thay A bng c v A B l bin
trit tiu nn thay B bng c v B ta c: SAB | B | A Gio trnh Kin trc my
tnh v H iu hnh 136 CHNG 4. VN PHM & NGN NG PHI NG CNH TRNG I HC
BCH KHOA NNG 4. Dng chun ca vn phm phi ng cnh Kh sn xut c - Xt AaAB
ta c: AaAB | aB | aA | a - Xt BbAB ta c: BbAB| bB | bA | b Vy ta c
vn phm sau: SAB | B | A AaAB | aB | aA | a BbAB| bB | bA | b Gio
trnh Kin trc my tnh v H iu hnh 137 CHNG 4. VN PHM & NGN NG PHI
NG CNH TRNG I HC BCH KHOA NNG 4. Dng chun ca vn phm phi ng cnh Kh
sn xut c Bi tp: kh sn xut c ca cc vn phm sau: (1) S0S |1S | c (2)
SS(S)S | c (3) Sa S b | b S a | c Gio trnh Kin trc my tnh v H iu
hnh 138 CHNG 4. VN PHM & NGN NG PHI NG CNH TRNG I HC BCH KHOA
NNG 4. Dng chun ca vn phm phi ng cnh Sn xut n v - Sn xut c dng AB c
gi l sn xut n v; vi A,BeA - Cp (A,B) cp bin ca sn xut n v Gio trnh
Kin trc my tnh v H iu hnh 139 CHNG 4. VN PHM & NGN NG PHI NG
CNH TRNG I HC BCH KHOA NNG 4. Dng chun ca vn phm phi ng cnh Kh sn
xut n v Cho G(E, A, S, p), kh sn xut n v ta cG(E, A, S, p) vi p xc
nh nh sau: - Thm cc sn xut khng n v vo p - Xc nh cc cp bin (A,B) m
A+B (ch s dng cc sn xut n v) - Vi mi cp bin (A,B) xc nh trn, thm vo
p cc sn xut Ao; vi Bo l sn xut khng n v trong G Gio trnh Kin trc my
tnh v H iu hnh 140 CHNG 4. VN PHM & NGN NG PHI NG CNH TRNG I HC
BCH KHOA NNG 4. Dng chun ca vn phm phi ng cnh Kh sn xut n v - V d:
Kh sn xut n v cho vn phm sau: SaA |A | bB AB | a BA | ab | bb Gio
trnh Kin trc my tnh v H iu hnh 141 CHNG 4. VN PHM & NGN NG PHI
NG CNH TRNG I HC BCH KHOA NNG 4. Dng chun ca vn phm phi ng cnh Kh
sn xut n v - Ta c cc sn xut khng n v: SaA| bB Aa Bab |bb - Ta c cc
cp bin (S,A), (S,B), (A,B), (B,A) tha mn S+A; S+B; A+B; B+A cp
(S,A), c Aa nn thm: Sa Gio trnh Kin trc my tnh v H iu hnh 142 CHNG
4. VN PHM & NGN NG PHI NG CNH TRNG I HC BCH KHOA NNG 4. Dng
chun ca vn phm phi ng cnh Kh sn xut n v cp (S,B), c Bab v Bbb nn
thm: Sab |bb cp (A,B), c Bab v Bbb nn thm: Aab |bb cp (B,A), c Aa
nn thm: Ba Gio trnh Kin trc my tnh v H iu hnh 143 CHNG 4. VN PHM
& NGN NG PHI NG CNH TRNG I HC BCH KHOA NNG 4. Dng chun ca vn
phm phi ng cnh Kh sn xut n v Vy ta c vn phm sau: SaA | a | ab |bb
|bB Aa | ab | bb Bab |bb | a Gio trnh Kin trc my tnh v H iu hnh 144
CHNG 4. VN PHM & NGN NG PHI NG CNH TRNG I HC BCH KHOA NNG 4.
Dng chun ca vn phm phi ng cnh Kh sn xut n v Bi tp: kh sn xut n v
cho vn phm sau: (1) S0A | 1 A | A AS | 0 | 1 (2) EE+T | T TT*F | F
F(E) | a Gio trnh Kin trc my tnh v H iu hnh 145 CHNG 5. TMT Y XUNG
TRNG I HC BCH KHOA NNG tmat y xung (PDA) Ngn ng c on nhn PDA S tng
ng gia PDA v vn phm phi ng cnh tmat y xung n nhGio trnh Kin trc my
tnh v H iu hnh 146 CHNG 5. TMT Y XUNG TRNG I HC BCH KHOA NNG 01011
1. tmat y xung (Push Down Automat - PDA) 1.1. M tBng vo q B iu khin
u c Ngn xp Gio trnh Kin trc my tnh v H iu hnh 147 CHNG 5. TMT Y
XUNG TRNG I HC BCH KHOA NNG 1. tmat y xung 1.1. M tTi mt thi im b
iu khin: Trng thi q c mt k hiu trn bng vo (c: k0 c) Nhn k hiu nh
ngn xpxc nh trng thi tip theo v quyt nh hnh ng lin quan n ngn xp
Gio trnh Kin trc my tnh v H iu hnh 148 CHNG 5. TMT Y XUNG TRNG I HC
BCH KHOA NNG 1. tmat y xung 1.1. M tC 2 cch tmt y xung on nhn xu
vo: c xong xu vo v tmat trng thi kt thc c xong xu vo v ngn xp rng
Gio trnh Kin trc my tnh v H iu hnh 149 CHNG 5. TMT Y XUNG TRNG I HC
BCH KHOA NNG 1. tmat y xung 1.2. nh ngha: M(, Q, I, z, , q0, F) : b
ch vo Q: tp hu hn cc trng thi q0 e Q: trng thi u F _ Q: tp cc trng
thi kt thc I : tp k hiu trn ngn xp z e I : k hiu u tin nh ngn xp :
hm chuyn trng thi dng (q,a,x)={(p,o )} Vi q,p e Q, a e {c}, xe I, o
e I* Gio trnh Kin trc my tnh v H iu hnh 150 CHNG 5. TMT Y XUNG TRNG
I HC BCH KHOA NNG 1. tmat y xung 1.2. nh ngha: M(, Q, I, z, , q0,
F) : hm chuyn trng thi dng (q,a,x)={(p,o )} Vi q,p e Q, a e {c}, xe
I, o e I* X: k hiu nh ngn xp o: chui k hiu thay th x nh ngn xp o=
c: k hiu x trn nh ngn xp c ly ra o=x: ngn xp khng thay i. o=yz: ly
x ra, y z vo, y y vo. Gio trnh Kin trc my tnh v H iu hnh 151 CHNG
5. TMT Y XUNG TRNG I HC BCH KHOA NNG 1. tmat y xung 1.2. nh ngha V
d: Otomt y xung on nhn cc xu anbn vi n>=0: M(, Q, I, z, , q0, F)
: {a,b} Q: {q0, q1, q2, q3} q0: trng thi u {q3}: tp cc trng thi kt
thc {1,z} : tp k hiu trn ngn xp z : k hiu u tin nh ngn xp Gio trnh
Kin trc my tnh v H iu hnh 152 CHNG 5. TMT Y XUNG TRNG I HC BCH KHOA
NNG 1. tmat y xung 1.2. nh ngha (q0,a,z)={(q1,1z )} (q0,c,z)={(q3,c
)} (q1,a,1)={(q1,11)} (q1,b,1)={(q2,c)} (q2,b,1)={(q2,c )}
(q2,c,z)={(q3,c )} Gio trnh Kin trc my tnh v H iu hnh 153 CHNG 5.
TMT Y XUNG TRNG I HC BCH KHOA NNG 1. tmat y xung 1.3. Biu din PDA
bng biu dch chuyn- Cc trng thi c t trong cc vng trn - Trng thi u c
du > gn pha trc - Trng thi kt thc t trong vng trn kp -
(q,a,x)=(p,o ): t trng thi q sang trng thi p c nhn a, x|o - K hiu u
nh ngn xp qui c l z Gio trnh Kin trc my tnh v H iu hnh 154 CHNG 5.
TMT Y XUNG TRNG I HC BCH KHOA NNG 1. tmat y xung 1.3. Biu din PDA
bng biu dch chuyn- V d: v tmat PDA on nhn xu anbn vi n>=0 a,z|1z
a,1|11 b,1|cc,z|c b,1|c c,z|c q0 q1 q2 q3 Gio trnh Kin trc my tnh v
H iu hnh 155 CHNG 5. TMT Y XUNG TRNG I HC BCH KHOA NNG 1. tmat y
xung 1.4. Cu hnh ca PDACu hnh ca PDA l b (q,w, o): - q: trng thi -
w: phn xu vo s c - o: ni dung ca ngn xp (nh-y: tri-phi) Gio trnh
Kin trc my tnh v H iu hnh 156 CHNG 5. TMT Y XUNG TRNG I HC BCH KHOA
NNG 1. tmat y xung 1.4. Cu hnh ca PDA V d: Cu hnh ca PDA v d trc on
nhn xu: aaabbb - (q0,aaabbb,z) |- (q1,aabbb,1z) |- (q1,abbb,11z) |-
(q1,bbb,111z) |- (q2,bb,11z) |- (q2,b,1z) |- (q2,c,z) |- (q3,c,c)-
C ngha (q0,aaabbb,z) |-* (q3,c,c) : xu vo c on nhn Gio trnh Kin trc
my tnh v H iu hnh 157 CHNG 5. TMT Y XUNG TRNG I HC BCH KHOA NNG 2.
Ngn ng c on nhn bi PDA 2.1. on nhn bi trng thi kt thc- Ngn ng L c
PDA M on nhn bi trng thi kt thc l: L(M)={w | (q0,w,z ) |-* (qf
,c,o)} vi qf eF, oeI* Gio trnh Kin trc my tnh v H iu hnh 158 CHNG
5. TMT Y XUNG TRNG I HC BCH KHOA NNG 2. Ngn ng c on nhn bi PDA 2.2.
on nhn bi ngn xp rng- Ngn ng L c PDA M on nhn bi ngn xp rng :
L(M)={w | (q0,w,z ) |-* (qk ,c,c)} vi qk eQ Gio trnh Kin trc my tnh
v H iu hnh 159 CHNG 5. TMT Y XUNG TRNG I HC BCH KHOA NNG 2. Ngn ng
c on nhn bi PDA Bi tp: v PDA on nhn xu x l:(1)anbmcm vi n,m>=0
(2) anbmcn vi n,m>0 (3)Cc cp du () tng thch (4)S nh phn c s ch s
0 bng s ch s 1 (5)biu thc s hc dng hu t c 1 s hng a (6) S nh phn c
s ch s 0 gp i s ch s 1 v xu khc cGio trnh Kin trc my tnh v H iu hnh
160 CHNG 5. TMT Y XUNG TRNG I HC BCH KHOA NNG 2. Ngn ng c on nhn bi
PDA 2.3. Chuyn PDA t on nhn bi ngn xp rng sang on nhn bi trng thi
kt thc Cho PDA M(, Q, I, z, , q0) on nhn bi ngn xp rng thnh M(, Q,
I, z, , q0, F) on nhn bi trng thi kt thc c: - Q=Q{q0,qf} - F={qf} -
I=I {z} Gio trnh Kin trc my tnh v H iu hnh 161 CHNG 5. TMT Y XUNG
TRNG I HC BCH KHOA NNG 2. Ngn ng c on nhn bi PDA 2.3. Chuyn PDA t
on nhn bi ngn xp rng sang on nhn bi trng thi kt thc - c xc nh: M q0
q0 c,z|zz qf c,z| c c,z| c c,z| c Gio trnh Kin trc my tnh v H iu
hnh 162 CHNG 5. TMT Y XUNG TRNG I HC BCH KHOA NNG 2. Ngn ng c on
nhn bi PDA 2.3. Chuyn PDA t on nhn bi ngn xp rng sang on nhn bi
trng thi kt thc V d:q1 q0
a,z|1z a,1|11b,1|c c,z|c Gio trnh Kin trc my tnh v H iu hnh 163
CHNG 5. TMT Y XUNG TRNG I HC BCH KHOA NNG 2. Ngn ng c on nhn bi PDA
2.3. Chuyn PDA t on nhn bi ngn xp rng sang on nhn bi trng thi kt
thc V d:q1 q0
a,z|1z a,1|11b,1|c c,z|c c,z|zz qf q0 q0 q1
a,z|1za,1|11b,1|c c,z|c c,z|c Gio trnh Kin trc my tnh v H iu hnh
164 CHNG 5. TMT Y XUNG TRNG I HC BCH KHOA NNG 2. Ngn ng c on nhn bi
PDA 2.3. Chuyn PDA t on nhn bi ngn xp rng sang on nhn bi trng thi
kt thc V d:q1 q0
a,z|1za,1|11b,1|c c,z|c Gio trnh Kin trc my tnh v H iu hnh 165
CHNG 5. TMT Y XUNG TRNG I HC BCH KHOA NNG 2. Ngn ng c on nhn bi PDA
2.3. Chuyn PDA t on nhn bi ngn xp rng sang on nhn bi trng thi kt
thc V d: (ab)n vi n>0 q1 q0
q1
a,z|zzb,z|c a,z|zz q3 c,z|c Gio trnh Kin trc my tnh v H iu hnh
166 CHNG 5. TMT Y XUNG TRNG I HC BCH KHOA NNG 2. Ngn ng c on nhn bi
PDA 2.4. Chuyn PDA t on nhn bi trng thi cui sang on nhn bi ngn xp
rngCho PDA M(, Q, I, z, , q0, F) on nhn bi ngn xp rng thnh M(,Q,I,
z, ,q0) on nhn bi trng thi kt thc c: - Q=Q{q0,qk} - I=I {z} Gio
trnh Kin trc my tnh v H iu hnh 167 CHNG 5. TMT Y XUNG TRNG I HC BCH
KHOA NNG 2. Ngn ng c on nhn bi PDA 2.4. Chuyn PDA t on nhn bi trng
thi cui sang on nhn bi ngn xp rng- c xc nh: M q0 c,z|zz qk c,oeI| c
q0 c,oeI| c c,oeI| c Gio trnh Kin trc my tnh v H iu hnh 168 CHNG 5.
TMT Y XUNG TRNG I HC BCH KHOA NNG 2. Ngn ng c on nhn bi PDA 2.4.
Chuyn PDA t on nhn bi trng thi cui sang on nhn bi ngn xp rngV d:
anbm vi n,m>=0 q1 q0 a,z|1z a,1|11 b,z|1z b,1|11 b,1|11 c,z|z
Gio trnh Kin trc my tnh v H iu hnh 169 CHNG 5. TMT Y XUNG TRNG I HC
BCH KHOA NNG 2. Ngn ng c on nhn bi PDA 2.4. Chuyn PDA t on nhn bi
trng thi cui sang on nhn bi ngn xp rngV d: anbn vi n>=0 b,1|11
a,z|1z a,1|11 b,z|1z b,1|11 qf q1 q0 c,z|c c,1|c c,z|z c,z|c c,1|c
Gio trnh Kin trc my tnh v H iu hnh 170 CHNG 5. TMT Y XUNG TRNG I HC
BCH KHOA NNG 3. tmat y xung n nh - tmt y xung tha mn: (1) o(q,a,X)
ch cha mt gi tr duy nht (2) o(q,a,X) khng rng th o(q,c,X) phi rng
Gio trnh Kin trc my tnh v H iu hnh 171 CHNG 5. TMT Y XUNG TRNG I HC
BCH KHOA NNG 4. S tng ng gia PDA v vn phm phi ng cnh 4.1. Chuyn vn
phm phi ng cnh thnh PDA - B ch (bng ch) l tp hp hu hn cc k hiu V
d:{0,1} b ch gm 2 k hiu 0 v 1 {a,b,c,,z} b ch gm cc k hiu a z Tp cc
ch ci ting vit Gio trnh Kin trc my tnh v H iu hnh 172 CHNG 5. TMT Y
XUNG TRNG I HC BCH KHOA NNG 4. S tng ng gia PDA v vn phm phi ng cnh
4.2. Chuyn PDA thnh vn phm phi ng cnh - B ch (bng ch) l tp hp hu hn
cc k hiu V d:{0,1} b ch gm 2 k hiu 0 v 1 {a,b,c,,z} b ch gm cc k
hiu a z Tp cc ch ci ting vit Gio trnh Kin trc my tnh v H iu hnh 173
CHNG 6. MY TURING TRNG I HC BCH KHOA NNG 1. Mt s vn v ngn ng 1.1.
Xu- B ch (bng ch) l tp hp hu hn cc k hiu V d:{0,1} b ch gm 2 k hiu
0 v 1 {a,b,c,,z} b ch gm cc k hiu a z Tp cc ch ci ting vit Gio trnh
Kin trc my tnh v H iu hnh 174 CHNG 6. MY TURING TRNG I HC BCH KHOA
NNG 1. Mt s vn v ngn ng 1.1. Xu- B ch (bng ch) l tp hp hu hn cc k
hiu V d:{0,1} b ch gm 2 k hiu 0 v 1 {a,b,c,,z} b ch gm cc k hiu a z
Tp cc ch ci ting vit
