phn i

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Mc lc

Phn I : Gii quyt vn bng tm kim

1.1 Chng I - Cc chin lc tm kim m

1.1 Biu din vn trong khng gian trng thi

1.2 Cc chin lc tm kim

1.3 Cc chin lc tm kim m

1.3.1 Tm kim theo b rng

1.3.2 Tm kim theo su

1.3.3 Cc trng thi lp

1.3.4 Tm kim su lp

1.4 Quy vn v cc vn con. Tm kim trn th v/hoc

1.4.1 Quy vn v cc vn con

1.4.2 th v/hoc

1.4.3 Tm kim trn th v/hoc

Chng II - Cc chin lc tm kim kinh nghim

2.1 Hm nh gi v tm kim kinh nghim

2.2 Tm kim tt nht - u tin

2.3 Tm kim leo i

2.4 Tm kim beam

1.2 Chng III - Cc chin lc tm kim ti u

3.1 Tm ng i ngn nht

3.1.1 Thut ton A*

3.1.2 Thut ton tm kim Nhnh-v-Cn

1.2.1 3.2 Tm i tng tt nht

1.2.1.1 3.2.1 Tm kim leo i

3.2.2 Tm kim gradient

3.2.3 Tm kim m phng luyn kim

1.2.2 3.3 Tm kim m phng s tin ha. Thut ton di truyn

1.3 Chng IV - Tm kim c i th

4.1 Cy tr chi v tm kim trn cy tr chi

4.2 Chin lc Minimax

4.3 Phng php ct ct Alpha-Beta

Phn II: Tri thc v lp lun

Phn IGii quyt vn bng tm kim

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Vn tm kim, mt cch tng qut, c th hiu l tm mt i tng tha mn mt s i hi no , trong mt tp hp rng ln cc i tng. Chng ta c th k ra rt nhiu vn m vic gii quyt n c quy v vn tm kim.

Cc tr chi, chng hn c vua, c car c th xem nh vn tm kim. Trong s rt nhiu nc i c php thc hin, ta phi tm ra cc nc i dn ti tnh th kt cuc m ta l ngi thng.

Chng minh nh l cng c th xem nh vn tm kim. Cho mt tp cc tin v cc lut suy din, trong trng hp ny mc tiu ca ta l tm ra mt chng minh (mt dy cc lut suy din c p dng) c a n cng thc m ta cn chng minh.

Trong cc lnh vc nghin cu ca Tr Tu Nhn To, chng ta thng xuyn phi i u vi vn tm kim. c bit trong lp k hoch v hc my, tm kim ng vai tr quan trng.

Trong phn ny chng ta s nghin cu cc k thut tm kim c bn c p dng gii quyt cc vn v c p dng rng ri trong cc lnh vc nghin cu khc ca Tr Tu Nhn To. Chng ta ln lt nghin cu cc k thut sau:

Cc k thut tm kim m, trong chng ta khng c hiu bit g v cc i tng hng dn tm kim m ch n thun l xem xt theo mt h thng no tt c cc i tng pht hin ra i tng cn tm.

Cc k thut tm kim kinh nghim (tm kim heuristic) trong chng ta da vo kinh nghim v s hiu bit ca chng ta v vn cn gii quyt xy dng nn hm nh gi hng dn s tm kim.

Cc k thut tm kim ti u.

Cc phng php tm kim c i th, tc l cc chin lc tm kim nc i trong cc tr chi hai ngi, chng hn c vua, c tng, c car.

Chng I

Cc chin lc tm kim m

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

Trong chng ny, chng ti s nghin cu cc chin lc tm kim m (blind search): tm kim theo b rng (breadth-first search) v tm kim theo su (depth-first search). Hiu qu ca cc phng php tm kim ny cng s c nh gi.

1.4 Biu din vn trong khng gian trng thi

Mt khi chng ta mun gii quyt mt vn no bng tm kim, u tin ta phi xc nh khng gian tm kim. Khng gian tm kim bao gm tt c cc i tng m ta cn quan tm tm kim. N c th l khng gian lin tc, chng hn khng gian cc vct thc n chiu; n cng c th l khng gian cc i tng ri rc.

Trong mc ny ta s xt vic biu din mt vn trong khng gian trng thi sao cho vic gii quyt vn c quy v vic tm kim trong khng gian trng thi.

Mt phm vi rng ln cc vn , c bit cc cu , cc tr chi, c th m t bng cch s dng khi nim trng thi v ton t (php bin i trng thi). Chng hn, mt khch du lch c trong tay bn mng li giao thng ni cc thnh ph trong mt vng lnh th (hnh 1.1), du khch ang thnh ph A v anh ta mun tm ng i ti thm thnh ph B. Trong bi ton ny, cc thnh ph c trong cc bn l cc trng thi, thnh ph A l trng thi ban u, B l trng thi kt thc. Khi ang mt thnh ph, chng hn thnh ph D anh ta c th i theo cc con ng ni ti cc thnh ph C, F v G. Cc con ng ni cc thnh ph s c biu din bi cc ton t. Mt ton t bin i mt trng thi thnh mt trng thi khc. Chng hn, trng thi D s c ba ton t dn trng thi D ti cc trng thi C, F v G. Vn ca du khch by gi s l tm mt dy ton t a trng thi ban u A ti trng thi kt thc B.

Mt v d khc, trong tr chi c vua, mi cch b tr cc qun trn bn c l mt trng thi. Trng thi ban u l s sp xp cc qun lc bt u cuc chi. Mi nc i hp l l mt ton t, n bin i mt cnh hung trn bn c thnh mt cnh hung khc.

Nh vy mun biu din mt vn trong khng gian trng thi, ta cn xc nh cc yu t sau:

Trng thi ban u.

Mt tp hp cc ton t. Trong mi ton t m t mt hnh ng hoc mt php bin i c th a mt trng thi ti mt trng thi khc.

Tp hp tt c cc trng thi c th t ti t trng thi ban u bng cch p dng mt dy ton t, lp thnh khng gian trng thi ca vn .

Ta s k hiu khng gian trng thi l U, trng thi ban u l u0 (u0 ( U). Mi ton t R c th xem nh mt nh x R: U(U. Ni chung R l mt nh x khng xc nh khp ni trn U.

Mt tp hp T cc trng thi kt thc (trng thi ch). T l tp con ca khng gian U. Trong vn ca du khch trn, ch c mt trng thi ch, l thnh ph B. Nhng trong nhiu vn (chng hn cc loi c) c th c nhiu trng thi ch v ta khng th xc nh trc c cc trng thi ch. Ni chung trong phn ln cc vn hay, ta ch c th m t cc trng thi ch l cc trng thi tha mn mt s iu kin no .

Khi chng ta biu din mt vn thng qua cc trng thi v cc ton t, th vic tm nghim ca bi ton c quy v vic tm ng i t trng thi ban u ti trng thi ch. (Mt ng i trong khng gian trng thi l mt dy ton t dn mt trng thi ti mt trng thi khc).

Chng ta c th biu din khng gian trng thi bng th nh hng, trong mi nh ca th tng ng vi mt trng thi. Nu c ton t R bin i trng thi u thnh trng thi v, th c cung gn nhn R i t nh u ti nh v. Khi mt ng i trong khng gian trng thi s l mt ng i trong th ny.

Sau y chng ta s xt mt s v d v cc khng gian trng thi c xy dng cho mt s vn .

V d 1: Bi ton 8 s. Chng ta c bng 3x3 v tm qun mang s hiu t 1 n 8 c xp vo tm , cn li mt trng, chng hn nh trong hnh 2 bn tri. Trong tr chi ny, bn c th chuyn dch cc qun cch trng ti trng . Vn ca bn l tm ra mt dy cc chuyn dch bin i cnh hung ban u (hnh 1.2 bn tri) thnh mt cnh hung xc nh no , chng hn cnh hung trong hnh 1.2 bn phi.

Trong bi ton ny, trng thi ban u l cnh hung bn tri hnh 1.2, cn trng thi kt thc bn phi hnh 1.2. Tng ng vi cc quy tc chuyn dch cc qun, ta c bn ton t: up (y qun ln trn), down (y qun xung di), left (y qun sang tri), right (y qun sang phi). R rng l, cc ton t ny ch l cc ton t b phn; chng hn, t trng thi ban u (hnh 1.2 bn tri), ta ch c th p dng cc ton t down, left, right.

Trong cc v d trn vic tm ra mt biu din thch hp m t cc trng thi ca vn l kh d dng v t nhin. Song trong nhiu vn vic tm hiu c biu din thch hp cho cc trng thi ca vn l hon ton khng n gin. Vic tm ra dng biu din tt cho cc trng thi ng vai tr ht sc quan trng trong qu trnh gii quyt mt vn . C th ni rng, nu ta tm c dng biu din tt cho cc trng thi ca vn , th vn hu nh c gii quyt.

V d 2: Vn triu ph v k cp. C ba nh triu ph v ba tn cp bn b t ngn mt con sng, cng mt chic thuyn ch c mt hoc hai ngi. Hy tm cch a mi ngi qua sng sao cho khng li bn b sng k cp nhiu hn triu ph. ng nhin trong bi ton ny, cc ton t tng ng vi cc hnh ng ch 1 hoc 2 ngi qua sng. Nhng y ta cn lu rng, khi hnh ng xy ra (lc thuyn ang bi qua sng) th bn b sng thuyn va di ch, s k cp khng c nhiu hn s triu ph. Tip theo ta cn quyt nh ci g l trng thi ca vn . y ta khng cn phn bit cc nh triu ph v cc tn cp, m ch s lng ca h bn b sng l quan trng. biu din cc trng thi, ta s dng b ba (a, b, k), trong a l s triu ph, b l s k cp bn b t ngn vo cc thi im m thuyn b ny hoc b kia, k = 1 nu thuyn b t ngn v k = 0 nu thuyn b hu ngn. Nh vy, khng gian trng thi cho bi ton triu ph v k cp c xc nh nh sau:

Trng thi ban u l (3, 3, 1).

Cc ton t. C nm ton t tng ng vi hnh ng thuyn ch qua sng 1 triu ph, hoc 1 k cp, hoc 2 triu ph, hoc 2 k cp, hoc 1 triu ph v 1 k cp.

Trng thi kt thc l (0, 0, 0).

1.5 Cc chin lc tm kim

Nh ta thy trong mc 1.1, gii quyt mt vn bng tm kim trong khng gian trng thi, u tin ta cn tm dng thch hp m t cc trng thi cu vn . Sau cn xc nh:

Trng thi ban u.

Tp cc ton t.

Tp T cc trng thi kt thc. (T c th khng c xc nh c th gm cc trng thi no m ch c ch nh bi mt s iu kin no ).

Gi s u l mt trng thi no v R l mt ton t bin i u thnh v. Ta s gi v l trng thi k u, hoc v c sinh ra t trng thi u bi ton t R. Qu trnh p dng cc ton t sinh ra cc trng thi k u c gi l pht trin trng thi u. Chng hn, trong bi ton ton s, pht trin trng thi ban u (hnh 2 bn tri), ta nhn c ba trng thi k (hnh 1.3).

Khi chng ta biu din mt vn cn gii quyt thng qua cc trng thi v cc ton t th vic tm li gii ca vn c quy v vic tm ng i t trng thi ban u ti mt trng thi kt thc no .

C th phn cc chin lc tm kim thnh hai loi:

Cc chin lc tm kim m. Trong cc chin lc tm kim ny, khng c mt s hng dn no cho s tm kim, m ta ch pht trin cc trng thi ban u cho ti khi gp mt trng thi ch no . C hai k thut tm kim m, l tm kim theo b rng v tm kim theo su.

T tng ca tm kim theo b rng l cc trng thi c pht trin theo th t m chng c sinh ra, tc l trng thi no c sinh ra trc s c pht trin trc.

Trong nhiu vn , d chng ta pht trin cc trng thi theo h thng no (theo b rng hoc theo su) th s lng cc trng thi c sinh ra trc khi ta gp trng thi ch thng l cc k ln. Do cc thut ton tm kim m km hiu qu, i hi rt nhiu khng gian v thi gian. Trong thc t, nhiu vn khng th gii quyt c bng tm kim m.

Tm kim kinh nghim (tm kim heuristic). Trong rt nhiu vn , chng ta c th da vo s hiu bit ca chng ta v vn , da vo kinh nghim, trc gic, nh gi cc trng thi. S dng s nh gi cc trng thi hng dn s tm kim: trong qu trnh pht trin cc trng thi, ta s chn trong s cc trng thi ch pht trin, trng thi c nh gi l tt nht pht trin. Do tc tm kim s nhanh hn. Cc phng php tm kim da vo s nh gi cc trng thi hng dn s tm kim gi chung l cc phng php tm kim kinh nghim.

Nh vy chin lc tm kim c xc nh bi chin lc chn trng thi pht trin mi bc. Trong tm kim m, ta chn trng thi pht trin theo th t m ng c sinh ra; cn trong tm kim kinh nghim ta chn trng thi da vo s nh gi cc trng thi.

Cy tm kim

Chng ta c th ngh n qu trnh tm kim nh qu trnh xy dng cy tm kim. Cy tm kim l cy m cc nh c gn bi cc trng thi ca khng gian trng thi. Gc ca cy tm kim tng ng vi trng thi ban u. Nu mt nh ng vi trng thi u, th cc nh con ca n ng vi cc trng thi v k u. Hnh 1.4a l th biu din mt khng gian trng thi vi trng thi ban u l A, hnh 1.4b l cy tm kim tng ng vi khng gian trng thi .

Mi chin lc tm kim trong khng gian trng thi tng ng vi mt phng php xy dng cy tm kim. Qu trnh xy dng cy bt u t cy ch c mt nh l trng thi ban u. Gi s ti mt bc no trong chin lc tm kim, ta xy dng c mt cy no , cc l ca cy tng ng vi cc trng thi cha c pht trin. Bc tip theo ph thuc vo chin lc tm kim m mt nh no trong cc l c chn pht trin. Khi pht trin nh , cy tm kim c m rng bng cch thm vo cc nh con ca nh . K thut tm kim theo b rng (theo su) tng ng vi phng php xy dng cy tm kim theo b rng (theo su).

1.6 Cc chin lc tm kim m

Trong mc ny chng ta s trnh by hai chin lc tm kim m: tm kim theo b rng v tm kim theo su. Trong tm kim theo b rng, ti mi bc ta s chn trng thi pht trin l trng thi c sinh ra trc cc trng thi ch pht trin khc. Cn trong tm kim theo su, trng thi c chn pht trin l trng thi c sinh ra sau cng trong s cc trng thi ch pht trin.

Chng ta s dng danh sch L lu cc trng thi c sinh ra v ch c pht trin. Mc tiu ca tm kim trong khng gian trng thi l tm ng i t trng thi ban u ti trng thi ch, do ta cn lu li vt ca ng i. Ta c th s dng hm father lu li cha ca mi nh trn ng i, father(v) = u nu cha ca nh v l u.

1.6.1 Tm kim theo b rng

Thut ton tm kim theo b rng c m t bi th tc sau:

procedureBreadth_First_Search;

begin

1. Khi to danh sch L ch cha trng thi ban u;

2. loop do

2.1 if L rng then

{thng bo tm kim tht bi; stop};

2.2 Loi trng thi u u danh sch L;

2.3 if u l trng thi kt thc then

{thng bo tm kim thnh cng; stop};

2.4 for mi trng thi v k u do {

t v vo cui danh sch L;

father(v) cost(u). Qu trnh tm kim s dng li ngay khi ta khng leo ln nh cao hn c na.

Trong th tc leo i di y, bin u lu nh hin thi, bin v lu nh tt nht (cost(v) nh nht) trong cc nh ln cn u. Khi thut ton dng, bin u s lu trong i tng tm c.

procedure Hill_Climbing;

begin

1. u ( mt i tng ban u no ;

2. if cost(v) > cost(u) then u ( v else stop;

end;

Ti u a phng v ti u ton cc

R rng l, khi thut ton leo i dng li ti i tng u*, th gi ca n cost(u*) ln hn gi ca tt c cc i tng nm trong ln cn ca tt c cc i tng trn ng i t i tng ban u ti trng thi u*. Do nghim u* m thut ton leo i tm c l ti u a phng. Cn nhn mnh rng khng c g m bo nghim l ti u ton cc theo ngha l cost(u*) l ln nht trn ton b khng gian tm kim.

nhn c nghim tt hn bng thut ton leo i, ta c th p dng lp li nhiu ln th tc leo i xut pht t mt dy cc i tng ban u c chn ngu nhin v lu li nghim tt nht qua mi ln lp. Nu s ln lp ln th ta c th tm c nghim ti u.

Kt qu ca thut ton leo i ph thuc rt nhiu vo hnh dng ca mt cong ca hm gi. Nu mt cong ch c mt s t cc i a phng, th k thut leo i s tm ra rt nhanh cc i ton cc. Song c nhng vn m mt cong ca hm gi ta nh lng nhm vy, khi s dng k thut leo i i hi rt nhiu thi gian.

1.9.2 Tm kim gradient

Tm kim gradient l k thut tm kim leo i tm gi tr ln nht (hoc nh nht) ca hm kh vi lin tc f(x) trong khng gian cc vect thc n-chiu. Nh ta bit, trong ln cn nh ca im x = (x1,...,xn), th hm f tng nhanh nht theo hng ca vect gradient:

Do t tng ca tm kim gradient l t mt im ta i ti im ln cn n theo hng ca vect gradient.

procedure Gradient_Search;

begin

x ( im xut pht no ;

repeat

x ( x + ((f(x);

until |(f| < (;

end;

Trong th tc trn, ( l hng s dng nh nht xc nh t l ca cc bc, cn ( l hng s dng nh xc nh tiu chun dng. Bng cch ly cc bc nh theo hng ca vect gradient chng ta s tm c im cc i a phng, l im m ti (f = 0, hoc tm c im rt gn vi cc i a phng.1.9.3 Tm kim m phng luyn kim:

Nh nhn mnh trn, tm kim leo i khng m bo cho ta tm c nghim ti u ton cc. cho nghim tm c gn vi ti u ton cc, ta p dng k thut leo i lp xut pht t cc im c la chn ngu nhin. By gi thay cho vic lun lun leo ln i xut pht t cc im khc nhau, ta thc hin mt s bc tt xung nhm thot ra khi cc im cc i a phng. chnh l t tng ca k thut tm kim m phng luyn kim.Trong tm kim leo i, khi mt trng thi u ta lun lun i ti trng thi tt nht trong ln cn n. Cn by gi, trong tm kim m phng luyn kim, ta chn ngu nhin mt trng thi v trong ln cn u. Nu trng thi v c chn tt hn u (cost(v) > cost(u)) th ta i ti v, cn nu khng ta ch i ti v vi mt xc sut no . Xc sut ny gim theo hm m ca xu ca trng thi v. Xc sut ny cn ph thuc vo tham s nhit T. Nhit T cng cao th bc i ti trng thi xu cng c kh nng c thc hin. Trong qu trnh tm kim, tham s nhit T gim dn ti khng. Khi T gn khng, thut ton hot ng gn ging nh leo i, hu nh n khng thc hin bc tt xung. C th ta xc nh xc sut i ti trng thi xu v t u l e(/T, y ( = cost(v) - cost(u).

Sau y l th tc m phng luyn kim.

procedure Simulated_Anneaning;

begin

t ( 0;

u ( trng thi ban u no ;

T ( nhit ban u;

repeat

v ( trng thi c chn nhu nhin trong ln cn u;

if cost(v) > cost(u) then u ( v

else u ( v vi xc sut e(/T;

T ( g(T, t);

t ( t + 1;

until T nhend;

Trong th tc trn, hm g(T, t) tha mn iu kin g(T, t) < T vi mi t, n xc nh tc gim ca nhit T. Ngi ta chng minh c rng, nu nhit T gim chm, th thut ton s tm c nghim ti u ton cc. Thut ton m phng luyn kim c p dng thnh cng cho cc bi ton ti u c ln.

1.10 Tm kim m phng s tin ha. Thut ton di truynThut ton di truyn (TTDT) l thut ton bt chc s chn lc t nhin v di truyn. Trong t nhin, cc c th khe, c kh nng thch nghi tt vi mi trng s c ti sinh v nhn bn cc th h sau. Mi c th c cu trc gien c trng cho phm cht ca c th . Trong qu trnh sinh sn, cc c th con c th tha hng cc phm cht ca c cha v m, cu trc gien ca n mang mt phn cu trc gien ca cha v m. Ngoi ra, trong qu trnh tin ha, c th xy ra hin tng t bin, cu trc gien ca c th con c th cha cc gien m c cha v m u khng c.

Trong TTDT, mi c th c m ha bi mt cu trc d liu m t cu trc gien ca c th , ta s gi n l nhim sc th (chroniosome). Mi nhim sc th c to thnh t cc n v c gi l gien. Chng hn, trong cc TTDT c in, cc nhim sc th l cc chui nh phn, tc l mi c th c biu din bi mt chui nh phn.

TTDT s lm vic trn cc qun th gm nhiu c th. Mt qun th ng vi mt giai on pht trin s c gi l mt th h. T th h ban u c to ra, TTDT bt chc chn lc t nhin v di truyn bin i cc th h. TTDT s dng cc ton t c bn sau y bin i cc th h.

Ton t ti sinh (reproduction) (cn c gi l ton t chn lc (selection)). Cc c th tt c chn lc a vo th h sau. S la chn ny c thc hin da vo thch nghi vi mi trng ca mi c th. Ta s gi hm ng mi c th vi thch nghi ca n l hm thch nghi (fitness function). Ton t lai ghp (crossover). Hai c th cha v m trao i cc gien to ra hai c th con. Ton t t bin (mutation). Mt c th thay i mt s gien to thnh c th mi.Tt c cc ton t trn khi thc hin u mang tnh ngu nhin. Cu trc c bn ca TTDT l nh sau:

procedure Genetic_Algorithm;

begin

t ( 0;

Khi to th h ban u P(t);

nh gi P(t) (theo hm thch nghi);

repeat

t ( t + 1;

Sinh ra th h mi P(t) t P(t-1) bi

Chn lc

Lai ghp

t bin;

nh gi P(t);

until iu kin kt thc c tha mn;

end;

Trong th tc trn, iu kin kt thc vng lp c th l mt s th h ln no , hoc thch nghi ca cc c th tt nht trong cc th h k tip nhau khc nhau khng ng k. Khi thut ton dng, c th tt nht trong th h cui cng c chn lm nghim cn tm.

By gi ta s xt chi tit hn ton t chn lc v cc ton t di truyn (lai ghp, t bin) trong cc TTDT c in.

1. Chn lc: Vic chn lc cc c th t mt qun th da trn thch nghi ca mi c th. Cc c th c thch nghi cao c nhiu kh nng c chn. Cn nhn mnh rng, hm thch nghi ch cn l mt hm thc dng, n c th khng tuyn tnh, khng lin tc, khng kh vi. Qu trnh chn lc c thc hin theo k thut quay bnh xe.

Gi s th h hin thi P(t) gm c n c th {x1,..,xn}. S n c gi l c ca qun th. Vi mi c th xi, ta tnh thch nghi ca n f(xi). Tnh tng cc thch nghi ca tt c cc c th trong qun th:

Mi ln chn lc, ta thc hin hai bc sau:

Sinh ra mt s thc ngu nhin q trong khong (0, F);

xk l c th c chn, nu k l s nh nht sao cho

Vic chn lc theo hai bc trn c th minh ha nh sau: Ta c mt bnh xe c chia thnh n phn, mi phn ng vi thch nghi ca mt c th (hnh 3.5). Mt mi tn ch vo bnh xe. Quay bnh xe, khi bnh xe dng, mi tn ch vo phn no, c th ng vi phn c chn.

R rng l vi cch chn ny, cc c th c th c thch nghi cng cao cng c kh nng c chn. Cc c th c thch nghi cao c th c mt hay nhiu bn sao, cc c th c thch nghi thp c th khng c mt th h sau (n b cht i).

2. Lai ghp: Trn c th c chn lc, ta tn hnh ton t lai ghp. u tin ta cn a ra xc sut lai ghp pc. xc sut ny cho ta hy vng c pc.n c th c lai ghp (n l c ca qun th).

Vi mi c th ta thc hin hai bc sau:

Sinh ra s thc ngu nhin r trong on [0, 1];

Nu r < pc th c th c chn lai ghp

T cc c th c chn lai ghp, ngi ta cp i chng mt cch ngu nhin. Trong trng hp cc nhim sc th l cc chui nh phn c di c nh m, ta c th thc hin lai ghp nh sau: Vi mi cp, sinh ra mt s nguyn ngu nhin p trn on [0, m -1], p l v tr im ghp. Cp gm hai nhim sc th

a = (a1 , ... , ap , ap+1 , ... , am)

a = (b1 , ... , bp , bp+1 , ... , bm)c thay bi hai con l:

a' = (a1 , ... , ap , bp+1 , ... , bm)

b' = (b1 , ... , bp , ap+1 , ... , am)3. t bin: Ta thc hin ton t t bin trn cc c th c c sau qu trnh lai ghp. t bin l thay i trng thi mt s gien no trong nhim sc th. Mi gien chu t bin vi xc sut pm. Xc sut t bin pm do ta xc nh v l xc sut thp. Sau y l ton t t bin trn cc nhim sc th chui nh phn.

Vi mi v tr i trong nhim sc th:

a = (a1 , ... , ai , ... , am)

Ta sinh ra mt s thc nghim ngu nhin pi trong [0,1]. Qua t bin a c bin thnh a nh sau:a' = (a'1 , ... , a'i , ... , a'm)

Trong :

a'i = ai nu pi ( pm1 - ai

nu pi < pm

Sau qu trnh chn lc, lai ghp, t bin, mt th h mi c sinh ra. Cng vic cn li ca thut ton di truyn by gi ch l lp li cc bc trn.

V d: Xt bi ton tm max ca hm f(x) = x2 vi x l s nguyn trn on [0,31]. s dng TTDT, ta m ho mi s nguyn x trong on [0,31] bi mt s nh phn di 5, chng hn, chui 11000 l m ca s nguyn 24. Hm thch nghi c xc nh l chnh hm f(x) = x2. Qun th ban u gm 4 c th (c ca qun th l n = 4). Thc hin qu trnh chn lc, ta nhn c kt qu trong bng sau. Trong bng ny, ta thy c th 2 c thch nghi cao nht (576) nn n c chn 2 ln, c th 3 c thch nghi thp nht (64) khng c chn ln no. Mi c th 1 v 4 c chn 1 ln.

Bng kt qu chn lc

S liu c thQun th ban ux thch nghi f(x) = x2S ln c chn

10 1 1 0 1131691

21 1 0 0 0245762

30 1 0 0 08640

41 0 0 1 1193611

Thc hin qa trnh lai ghp vi xc sut lai ghp pc = 1, c 4 c th sau chn lc u c lai ghp. Kt qu lai ghp c cho trong bng sau. Trong bng ny, chui th nht c lai ghp vi chui th hai vi im ghp l 4, hai chui cn li c lai ghp vi nhau vi im ghp l 2.

Bng kt qu lai ghp

Qun th sau chn lcim ghpQun th sau lai ghpx thch nghi f(x) = x2

0 1 1 0 | 140 1 1 0 02144

1 1 0 0 | 041 1 0 0 15625

1 1 | 0 0 021 1 0 1 17729

1 0 | 0 1 121 0 0 0 06256

thc hin qu trnh t bin, ta chn xc sut t bin pm= 0,001, tc l ta hy vng c 5.4.0,001 = 0,02 bit c t bin. Thc t s khng c bit no c t bin. Nh vy th h mi l qun th sau lai ghp. Trong th h ban u, thch nghi cao nht l 576, thch nghi trung bnh 292. Trong th h sau, thch nghi cao nht l 729, trung bnh l 438. Ch qua mt th h, cc c th tt ln rt nhiu.Thut ton di truyn khc vi cc thut ton ti u khc cc im sau:

TTDT ch s dng hm thch hng dn s tm kim, hm thch nghi ch cn l hm thc dng. Ngoi ra, n khng i hi khng gian tm kim phi c cu trc no c.

TTDT lm vic trn cc nhim sc th l m ca cc c th cn tm.

TTDT tm kim t mt qun th gm nhiu c th.

Cc ton t trong TTDT u mang tnh ngu nhin.

gii quyt mt vn bng TTDT, chng ta cn thc hin cc bc sau y:

Trc ht ta cn m ha cc i tng cn tm bi mt cu trc d liu no . Chng hn, trong cc TTDT c in, nh trong v d trn, ta s dng m nh phn.

Thit k hm thch nghi. Trong cc bi ton ti u, hm thch nghi c xc nh da vo hm mc tiu.

Trn c s cu trc ca nhim sc th, thit k cc ton t di truyn (lai ghp, t bin) cho ph hp vi cc vn cn gii quyt.

Xc nh c ca qun th v khi to qun th ban u.

Xc nh xc sut lai ghp pc v xc sut t bin. Xc sut t bin cn l xc sut thp. Ngi ta (Goldberg, 1989) khuyn rng nn chn xc sut lai ghp l 0,6 v xc sut t bin l 0,03. Tuy nhin cn qua th nghim tm ra cc xc sut thch hp cho vn cn gii quyt.

Ni chung thut ng TTDT l ch TTDT c in, khi m cu trc ca cc nhim sc th l cc chui nh phn vi cc ton t di truyn c m t trn. Song trong nhiu vn thc t, thun tin hn, ta c th biu din nhim sc th bi cc cu trc khc, chng hn vect thc, mng hai chiu, cy,... Tng ng vi cu trc ca nhim sc th, c th c nhiu cch xc nh cc ton t di truyn. Qu trnh sinh ra th h mi P(t) t th h c P(t - 1) cng c nhiu cch chn la. Ngi ta gi chung cc thut ton ny l thut ton tin ha (evolutionary algorithms) hoc chng trnh tin ha (evolution program).

Thut ton tin ha c p dng trong cc vn ti u v hc my. hiu bit su sc hn v thut ton tin ho, bn c c th tm c [ ], [ ] v [ ] . [ ] v [ ] c xem l cc sch hay nht vit v TTDT. [ ] cho ta ci nhn tng qut v s pht trin gn y ca TTDT.

Chng IV

Tm kim c i th

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

Nghin cu my tnh chi c xut hin rt sm. Khng lu sau khi my tnh lp trnh c ra i vo nm 1950, Claude Shannon vit chng trnh chi c u tin. cc nh nghin cu Tr Tu Nhn To nghin cu vic chi c, v rng my tnh chi c l mt bng chng r rng v kh nng my tnh c th lm c cc cng vic i hi tr thng minh ca con ngi. Trong chng ny chng ta s xt cc vn sau y:

Chi c c th xem nh vn tm kim trong khng gian trng thi.

Chin lc tm kim nc i Minimax.

Phng php ct ct (-(, mt k thut tng hiu qu ca tm kim Minimax.

1.11 Cy tr chi v tm kim trn cy tr chi.

Trong chng ny chng ta ch quan tm nghin cu cc tr chi c hai ngi tham gia, chng hn cc loi c (c vua, c tng, c ca r...). Mt ngi chi c gi l Trng, i th ca anh ta c gi l en. Mc tiu ca chng ta l nghin cu chin lc chn nc i cho Trng (My tnh cm qun Trng).

Chng ta s xt cc tr chi hai ngi vi cc c im sau. Hai ngi chi thay phin nhau a ra cc nc i tun theo cc lut i no , cc lut ny l nh nhau cho c hai ngi. in hnh l c vua, trong c vua hai ngi chi c th p dng cc lut i con tt, con xe, ... a ra nc i. Lut i con tt Trng xe Trng, ... cng nh lut i con tt en, xe en, ... Mt c im na l hai ngi chi u c bit thng tin y v cc tnh th trong tr chi (khng nh trong chi bi, ngi chi khng th bit cc ngi chi khc cn nhng con bi g). Vn chi c c th xem nh vn tm kim nc i, ti mi ln n lt mnh, ngi chi phi tm trong s rt nhiu nc i hp l (tun theo ng lut i), mt nc i tt nht sao cho qua mt dy nc i thc hin, anh ta ginh phn thng. Tuy nhin vn tm kim y s phc tp hn vn tm kim m chng ta xt trong cc chng trc, bi v y c i th, ngi chi khng bit c i th ca mnh s i nc no trong tng lai. Sau y chng ta s pht biu chnh xc hn vn tm kim ny.

Vn chi c c th xem nh vn tm kim trong khng gian trng thi. Mi trng thi l mt tnh th (s b tr cc qun ca hai bn trn bn c).

Trng thi ban u l s sp xp cc qun c ca hai bn lc bt u cuc chi.

Cc ton t l cc nc i hp l.

Cc trng thi kt thc l cc tnh th m cuc chi dng, thng c xc nh bi mt s iu kin dng no .

Mt hm kt cuc (payoff function) ng mi trng thi kt thc vi mt gi tr no . Chng hn nh c vua, mi trng thi kt thc ch c th l thng, hoc thua (i vi Trng) hoc ha. Do , ta c th xc nh hm kt cuc l hm nhn gi tr 1 ti cc trng thi kt thc l thng (i vi Trng), -1 ti cc trng thi kt thc l thua (i vi Trng) v 0 ti cc trng thi kt thc ha. Trong mt s tr chi khc, chng hn tr chi tnh im, hm kt cuc c th nhn gi tr nguyn trong khong [-k, k] vi k l mt s nguyn dng no .

Nh vy vn ca Trng l, tm mt dy nc i sao cho xen k vi cc nc i ca en to thnh mt ng i t trng thi ban u ti trng thi kt thc l thng cho Trng.

thun li cho vic nghin cu cc chin lc chn nc i, ta biu din khng gian trng thi trn di dng cy tr chi.

Cy tr chi

Cy tr chi c xy dng nh sau. Gc ca cy ng vi trng thi ban u. Ta s gi nh ng vi trng thi m Trng (en) a ra nc i l nh Trng (en). Nu mt nh l Trng (en) ng vi trng thi u, th cc nh con ca n l tt c cc nh biu din trng thi v, v nhn c t u do Trng (en) thc hin nc i hp l no . Do , trn cng mt mc ca cy cc nh u l Trng hc u l en, cc l ca cy ng vi cc trng thi kt thc.

V d: Xt tr chi Dodgen (c to ra bi Colin Vout). C hai qun Trng v hai qun en, ban u c xp vo bn c 3*3 (Hnh v). Qun en c th i ti trng bn phi, trn hoc di. Qun Trng c th i ti trng bn tri, bn phi, trn. Qun en nu ct ngoi cng bn phi c th i ra khi bn c, qun Trng nu hng trn cng c th i ra khi bn c. Ai a hai qun ca mnh ra khi bn c trc s thng, hoc to ra tnh th bt i phng khng i c cng s thng.

Gi s en i trc, ta c cy tr chi c biu din nh trong hnh 4.2.

1.12 Chin lc Minimax

Qu trnh chi c l qu trnh Trng v en thay phin nhau a ra quyt nh, thc hin mt trong s cc nc i hp l. Trn cy tr chi, qu trnh s to ra ng i t gc ti l. Gi s ti mt thi im no , ng i dn ti nh u. Nu u l nh Trng (en) th Trng (en) cn chn i ti mt trong cc nh en (Trng) v l con ca u. Ti nh en (Trng) v m Trng (en) va chn, en (Trng) s phi chn i ti mt trong cc nh Trng (en) w l con ca v. Qu trnh trn s dng li khi t ti mt nh l l ca cy.

Gi s Trng cn tm nc i ti nh u. Nc i ti u cho Trng l nc i dn ti nh con ca v l nh tt nht (cho Trng) trong s cc nh con ca u. Ta cn gi thit rng, n lt i th chn nc i t v, en cng s chn nc i tt nht cho anh ta. Nh vy, chn nc i ti u cho Trng ti nh u, ta cn phi xc nh gi tr cc nh ca cy tr chi gc u. Gi tr ca cc nh l (ng vi cc trng thi kt thc) l gi tr ca hm kt cuc. nh c gi tr cng ln cng tt cho Trng, nh c gi tr cng nh cng tt cho en. xc nh gi tr cc nh ca cy tr chi gc u, ta i t mc thp nht ln gc u. Gi s v l nh trong ca cy v gi tr cc nh con ca n c xc nh. Khi nu v l nh Trng th gi tr ca n c xc nh l gi tr ln nht trong cc gi tr ca cc nh con. Cn nu v l nh en th gi tr ca n l gi tr nh nht trong cc gi tr ca cc nh con.

V d: Xt cy tr chi trong hnh 4.3, gc a l nh Trng. Gi tr ca cc nh l s ghi cnh mi nh. nh i l Trng, nn gi tr ca n l max(3,-2) = 3, nh d l nh en, nn gi tr ca n l min(2, 3, 4) = 2.

Vic gn gi tr cho cc nh c thc hin bi cc hm qui MaxVal v MinVal. Hm MaxVal xc nh gi tr cho cc nh Trng, hm MinVal xc nh gi tr cho cc nh en.

function MaxVal(u);

begin

if u l nh kt thc then MaxVal(u) ( f(u)else MaxVal(u) ( max{MinVal(v) | v l nh con ca u}end;

function MinVal(u);

begin

if u l nh kt thc then MinVal(u) ( f(u)else MinVal(u) ( min{MaxVal(v) | v l nh con ca u}end;

Trong cc hm quy trn, f(u) l gi tr ca hm kt cuc ti nh kt thc u. Sau y l th tc chn nc i cho trng ti nh u. Trong th tc Minimax(u,v), v l bin lu li trng thi m Trng chn i ti t u.

procedure Minimax(u, v);

begin

val ( -(;

for mi w l nh con ca u do

if val eval(v), ta khng cn i xung nh gi nh a na m vn khng nh hng g dn nh gi nh c. Hay ni cch khc ta c th ct b cy con gc a. Lp lun tng t cho trng hp a l nh en, trong trng hp ny nu eval(u) < eval(v) ta cng c th ct b cy con gc a.

ci t k thut ct ct alpha-beta, i vi cc nh nm trn ng i t gc ti nh hin thi, ta s dng tham s ( ghi li gi tr ln nht trong cc gi tr ca cc nh con nh gi ca mt nh Trng, cn tham s ( ghi li gi tr nh nht trong cc nh con nh gi ca mt nh en. Gi tr ca ( v ( s c cp nht trong qu trnh tm kim. ( v ( c s dng nh cc bin a phng trong cc hm MaxVal(u, (, () (hm xc nh gi tr ca nh Trng u) v Minval(u, (, () (hm xc nh gi tr ca nh en u).

function MaxVal(u, (, ();

begin

if u l l ca cy hn ch hoc u l nh kt thc

thenMaxVal ( eval(u)else for mi nh v l con ca u do

{( ( max[(, MinVal(v, (, ()];

// Ct b cc cy con t cc nh v cn li

if ( ( ( then exit};

MaxVal ( (;

end;

function MinVal(u, (, ();

begin

if u l l ca cy hn ch hoc u l nh kt thc

then MinVal ( eval(u)else for mi nh v l con ca u do

{( ( min[(, MaxVal(v, (, ()];

// Ct b cc cy con t cc nh v cn li

if ( ( ( then exit};

MinVal ( (;

end;

Thut ton tm nc i cho Trng s dng k thut ct ct alpha-beta, c ci t bi th tc Alpha_beta(u,v), trong v l tham bin ghi li nh m Trng cn i ti t u.

procedure Alpha_beta(u,v);

begin

( ( -(;

( ( (;

for mi nh w l con ca u do

if ( ( MinVal(w, (, () then

{( ( MinVal(w, (, ();

v ( w;}

end;

V d. Xt cy tr chi gc u (nh Trng) gii hn bi cao h = 3 (hnh 4.8). S ghi cnh cc l l gi tr ca hm nh gi. p dng chin lc Minimax v k thut ct ct, ta xc nh c nc i tt nht cho Trng ti u, l nc i dn ti nh v c gi tr 10. Cnh mi nh ta cng cho gi tr ca cp tham s ((, (). Khi gi cc hm MaxVal v MinVal xc nh gi tr ca nh . Cc nhnh b ct b c ch ra trong hnh:

Phn II:

Tri thc v lp lun

Chng V:

Logic mnh Trong chng ny chng ta s trnh by cc c trng ca ngn ng biu din tri thc. Chng ta s nghin cu logic mnh , mt ngn ng biu din tri thc rt n gin, c kh nng biu din hp, nhng thun li cho ta lm quen vi nhiu khi nim quan trng trong logic, c bit trong logic v t cp mt s c nghin cu trong cc chng sau.

5.1. Biu din tri thc

Con ngi sng trong mi trng c th nhn thc c th gii nh cc gic quan (tai, mt v cc b phn khc), s dng cc tri thc tch lu c v nh kh nng lp lun, suy din, con ngi c th a ra cc hnh ng hp l cho cng vic m con ngi ang lm. Mt mc tiu ca Tr tu nhn to ng dng l thit k cc tc nhn thng minh (intelligent agent) cng c kh nng nh con ngi. Chng ta c th hiu tc nhn thng minh l bt c ci g c th nhn thc c mi trng thng qua cc b cm nhn (sensors) v a ra hnh ng hp l p ng li mi trng thng qua b phn hnh ng (effectors). Cc robots, cc softbot (software robot), cc h chuyn gia,... l cc v d v tc nhn thng minh. Cc tc nhn thng minh cn phi c tri thc v th gii hin thc mi c th a ra cc quyt nh ng n.

Thnh phn trung tm ca cc tc nhn da trn tri thc (knowledge-based agent), cn c gi l h da trn tri thc (knowledge-based system) hoc n gin l h tri thc, l c s tri thc. C s tri thc (CSTT) l mt tp hp cc tri thc c biu din di dng no . Mi khi nhn c cc thng tin a vo, tc nhn cn c kh nng suy din a ra cc cu tr li, cc hnh ng hp l, ng n. Nhim v ny c thc hin bi b suy din. B suy din l thnh phn c bn khc ca cc h tri thc. Nh vy h tri thc bo tr mt CSTT v c trang b mt th tc suy din. Mi khi tip nhn c cc s kin t mi trng, th tc suy din thc hin qu trnh lin kt cc s kin vi cc tri thc trong CSTT rt ra cc cu tr li, hoc cc hnh ng hp l m tc nhn cn thc hin. ng nhin l, khi ta thit k mt tc nhn gii quyt mt vn no th CSTT s cha cc tri thc v min i tng c th . my tnh c th s dng c tri thc, c th x l tri thc, chng ta cn biu din tri thc di dng thun tin cho my tnh. l mc tiu ca biu din tri thc.

Tri thc c m t di dng cc cu trong ngn ng biu din tri thc. Mi cu c th xem nh s m ha ca mt s hiu bit ca chng ta v th gii hin thc. Ngn ng biu din tri thc (cng nh mi ngn ng hnh thc khc) gm hai thnh phn c bn l c php v ng ngha.

C php ca mt ngn ng bao gm cc k hiu v cc quy tc lin kt cc k hiu (cc lut c php) to thnh cc cu (cng thc) trong ngn ng. Cc cu y l biu din ngoi, cn phn bit vi biu din bn trong my tnh. Cc cu s c chuyn thnh cc cu trc d liu thch hp c ci t trong mt vng nh no ca my tnh, l biu din bn trong. Bn thn cc cu cha cha ng mt ni dung no c, cha mang mt ngha no c.

Ng ngha ca ngn ng cho php ta xc nh ngha ca cc cu trong mt min no ca th gii hin thc. Chng hn, trong ngn ng cc biu thc s hc, dy k hiu (x+y)*z l mt cu vit ng c php. Ng ngha ca ngn ng ny cho php ta hiu rng, nu x, y, z, ng vi cc s nguyn, k hiu + ng vi php ton cng, cn * ng vi php chia, th biu thc (x+y)*z biu din qu trnh tnh ton: ly s nguyn x cng vi s nguyn y, kt qu c nhn vi s nguyn z.

Ngoi hai thnh phn c php v ng ngha, ngn ng biu din tri thc cn c cung cp c ch suy din. Mt lut suy din (rule of inference) cho php ta suy ra mt cng thc t mt tp no cc cng thc. Chng hn, trong logic mnh , lut modus ponens t hai cng thc A v AB suy ra cng thc B. Chng ta s hiu lp lun hoc suy din l mt qu trnh p dng cc lut suy din t cc tri thc trong c s tri thc v cc s kin ta nhn c cc tri thc mi. Nh vy chng ta xc nh:1.14 Ngn ng biu din tri thc = C php + Ng ngha + C ch suy din.

1.15 Mt ngn ng biu din tri thc tt cn phi c kh nng biu din rng, tc l c th m t c mi iu m chng ta mun ni. N cn phi hiu qu theo ngha l, i ti cc kt lun, th tc suy din i hi t thi gian tnh ton v t khng gian nh. Ngi ta cng mong mun ngn ng biu din tri thc gn vi ngn ng t nhin.

1.16 Trong sch ny, chng ta s tp trung nghin cu logic v t cp mt (first-order predicate logic hoc first-order predicate calculus) - mt ngn ng biu din tri thc, bi v logic v t cp mt c kh nng biu din tng i tt, v hn na n l c s cho nhiu ngn ng biu din tri thc khc, chng hn ton hon cnh (situation calculus) hoc logic thi gian khong cp mt (first-order interval tempral logic). Nhng trc ht chng ta s nghin cu logic mnh (propositional logic hoc propositional calculus). N l ngn ng rt n gin, c kh nng biu din hn ch, song thun tin cho ta a vo nhiu khi nim quan trng trong logic.

5.2. C php v ng ngha ca logic mnh .

5.2.1 C php:

C php ca logic mnh rt n gin, n cho php xy dng nn cc cng thc. C php ca logic mnh bao gm tp cc k hiu v tp cc lut xy dng cng thc.

1. Cc k hiu

Hai hng logic True v False.

Cc k hiu mnh (cn c gi l cc bin mnh ): P, Q,...

Cc kt ni logic , , , , .

Cc du m ngoc (v ng ngoc).

2. Cc quy tc xy dng cc cng thc

Cc bin mnh l cng thc.

Nu A v B l cng thc th:

(AB) (c A hi B hoc A v B)(AB) (c A tuyn B hoc A hoc B) (A) (c ph nh A)(AB) (c A ko theo B hoc nu A th B)(AB) (c A v B ko theo nhau)l cc cng thc.

Sau ny cho ngn gn, ta s b i cc cp du ngoc khng cn thit. Chng hn, thay cho ((AB)C) ta s vit l (AB)C.

Cc cng thc l cc k hiu mnh s c gi l cc cu n hoc cu phn t. Cc cng thc khng phi l cu n s c gi l cu phc hp. Nu P l k hiu mnh th P v TP c gi l literal, P l literal dng, cn TP l literal m. Cu phc hp c dng A1...Am trong Ai l cc literal s c gi l cu tuyn (clause).5.2.2 Ng ngha:

Ng ngha ca logic mnh cho php ta xc nh thit lp ngha ca cc cng thc trong th gii hin thc no . iu c thc hin bng cch kt hp mnh vi s kin no trong th gii hin thc. Chng hn, k hiu mnh P c th ng vi s kin Paris l th nc Php hoc bt k mt s kin no khc. Bt k mt s kt hp cc k hiu mnh vi cc s kin trong th gii thc c gi l mt minh ha (interpretation ). Chng hn minh ha ca k hiu mnh P c th l mt s kin (mnh ) Paris l th nc Php . Mt s kin ch c th ng hoc sai. Chng hn, s kin Paris l th nc Php l ng, cn s kin S Pi l s hu t l sai.

Mt cch chnh xc hn, cho ta hiu mt minh ha l mt cch gn cho mi k hiu mnh mt gi tr chn l True hoc False. Trong mt minh ha, nu k hiu mnh P c gn gi tr chn l True/False (P Q (P ko theo Q ), P l gi thit, cn Q l kt lun. Trc quan cho php ta xem rng, khi P l ng v Q l ng th cu P ko theo Q l ng, cn khi P l ng Q l sai th cu P ko theo Q l sai. Nhng nu P sai v Q ng , hoc P sai Q sai th P ko theo Q l ng hay sai ? Nu chng ta xut pht t gi thit sai, th chng ta khng th khng nh g v kt lun. Khng c l do g ni rng, nu P sai v Q ng hoc P sai v Q sai th P ko theo Q l sai. Do trong trng hp P sai th P ko theo Q l ng d Q l ng hay Q l sai.Bng chn l cho php ta xc nh ngu nhin cc cu phc hp. Chng hn ng ngha ca cc cu PQ trong minh ha {P B (A=>B) (B=>A)

l(lA) A

1.17 Lut De Morgan

l(A v B) lA lB

l(A B) lA v lB

1.18 Lut giao hon

A v B B v A

A B B A

1.19 Lut kt hp

(A v B) v C Av( B v C)

(A B) C A ( B C)

1.20 Lut phn phi

A (B v C) (A B ) v (A C)

A v (B C) (A v B ) (A v C)

5.3.2 Dng chun tc :

Cc cng thc tng ng c th xem nh cc biu din khc nhau ca cng mt s kin. d dng vit cc chng trnh my tnh thao tc trn cc cng thc, chng ta s chun ha cc cng thc, a chng v dng biu din chun c gi l dng chun hi. Mt cng thc dng chun hi, c dng A1 v ... .v Am trong cc Ai l literal . Chng ta c th bin i mt cng thc bt k v cng thc dng chun hi bng cch p dng cc th tc sau.

B cc du ko theo (=>) bng cch thay (A=>B) bi (lAvB).

Chuyn cc du ph nh (l) vo st cc kt hiu mnh bng cch p dng lut De Morgan v thay l(lA) bi A .

p dng lut phn phi, thay cc cng thc c dng Av(BC) bi (A v B) ( A v B ) .

V d : Ta chun ha cng thc ( P => Q) v l(R v lS) :

(P => Q) v l(R v lS) (lP v Q) v (lR S) ((lP v Q)vlR) ( (lP v Q) v S) (l P v Q v lR) (lP v Q v S). Nh vy cng thc (P=> Q) v l(R v lS) c a v dng chun hi (lP v Q v lR) (lP v Q v S).

Khi biu din tri thc bi cc cng thc trong logic mnh , c s tri thc l mt tp no cc cng thc. Bng cch chun ho cc cng thc, c s tri thc l mt tp no cc cu tuyn.

Cc cu Horn:

trn ta ch ra, mi cng thc u c th a v dng chun hi, tc l cc hi ca cc tuyn, mi cu tuyn c dng

lP1 v........v lPm v Q1 v.....v Qm

trong Pi , Qi l cc k hiu mnh (literal dng) cu ny tng ng vi cu

lP1 v........v lPm => v Q1 v.....v Qm ???? p1^ .... ^ pm => Q

Dng cu ny c gi l cu Kowalski (do nh logic Kowalski a ra nm 1971).

Khi n 0, n=1, cu Horn c dng :

P1 ..... Pm => Q

Trong Pi, Q l cc literal dng. Cc Pi c gi l cc iu kin (hoc gi thit), cn Q c gi l kt lun (hoc h qu ). Cc cu Horn dng ny cn c gi l cc lut if ... then v c biu din nh sau :

If P1 and ....and Pm then Q .

Khi m=0, n=1 cu Horn tr thnh cu n Q, hay s kin Q. Nu m>0, n=0 cu Horn tr thnh dng lP1 v......v lPm hay tng ng l(P1^...^ Pm ). Cn ch rng, khng phi mi cng thc u c th biu din di dng hi ca cc cu Horn. Tuy nhin trong cc ng dng, c s tri thc thng l mt tp no cc cu Horn (tc l mt tp no cc lut if-then).

5.4 Lut suy dinMt cng thc H c xem l h qa logic (logical consequence) ca mt tp cng thc G ={G1,.....,Gm} nu trong bt k minh ha no m {G1,.....,Gm} ng th H cng ng, hay ni cch khc bt k mt m hnh no ca G cng l m hnh ca H.

Khi c mt c s tri thc, ta mun s dng cc tri thc trong c s ny suy ra tri thc mi m n l h qu logic ca cc cng thc trong c s tri thc. iu c thc hin bng cc thc hin cc lut suy din (rule of inference). Lut suy din ging nh mt th tc m chng ta s dng sinh ra mt cng thc mi t cc cng thc c. Mt lut suy din gm hai phn : mt tp cc iu kin v mt kt lun. Chng ta s biu din cc lut suy din di dng phn s , trong t s l danh sch cc iu kin, cn mu s l kt lun ca lut, tc l mu s l cng thc mi c suy ra t cc cng thc t s.

Sau y l mt s lut suy din quan trng trong logic mnh . Trong cc lut ny a, ai , b, g l cc cng thc :

1. Lut Modus Ponens

a=>b,ab

T mt ko theo v gi thit ca ko theo, ta suy ra kt lun ca n.

2. Lut Modus Tollens

a=>b,lblaT mt ko theo v ph nh kt lun ca n, ta suy ra ph nh gi thit ca ko theo.

3. Lut bc cu

a=>b,b=>ga=>gT hai ko theo, m kt lun ca n l ca ko theo th nht trng vi gi thit ca ko theo th hai, ta suy ra ko theo mi m gi thit ca n l gi thit ca ko theo th nht, cn kt lun ca n l kt lun ca ko theo th hai.

4. Lut loi b hi

a1.......

symbol 97 \f "Symbol" \s 14ai........

symbol 97 \f "Symbol" \s 14am

ai

T mt hi ta a ra mt nhn t bt k ca hi .

5. Lut a vo hi

a1,.......,ai,........ama1.......

symbol 97 \f "Symbol" \s 14ai.......

symbol 97 \f "Symbol" \s 14amT mt danh sch cc cng thc, ta suy ra hi ca chng.

6. Lut a vo tuyn

aia1v.......vai.v.......vamT mt cng thc, ta suy ra mt tuyn m mt trong cc hng t ca cc tuyn l cng thc .

7. Lut gii

a v b,lb v g

a v gT hai tuyn, mt tuyn cha mt hng t i lp vi mt hng t trong tuyn kia, ta suy ra tuyn ca cc hng t cn li trong c hai tuyn.

Mt lut suy din c xem l tin cy (secured) nu bt k mt m hnh no ca gi thit ca lut cng l m hnh kt lun ca lut. Chng ta ch quan tm n cc lut suy din tin cy.

Bng phng php bng chn l, ta c th kim chng c cc lut suy din nu trn u l tin cy. Bng chn l ca lut gii c cho trong hnh 5.3. T bng ny ta thy rng , trong bt k mt minh ha no m c hai gi thit a v b , lb v g ng th kt lun a v g cng ng. Do lut gii l lut suy in tin cy.

abga v blb v ga v g

FalseFalseFalseFalseTrueFalse

FalseFalseTrueFalseTrueTrue

FalseTrueFalseTrueFalseFalse

FalseTrueTrueTrueTrueTrue

TrueFalseFalseTrueTrueTrue

TrueFalseTrueTrueTrueTrue

TrueTrueFalseTrueFalseTrue

TrueTrueTrueTrueTrueTrue

Hnh 5.3 Bng chn l chng minh tnh tin cy ca lut gii.

Ta c nhn xt rng, lut gii l mt lut suy din tng qut, n bao gm lut Modus Ponens, lut Modus Tollens, lut bc cu nh cc trng hp ring. (Bn c d dng chng minh c iu ).

Tin nh l chng minh.

Gi s chng ta c mt tp no cc cng thc. Cc lut suy din cho php ta t cc cng thc c suy ra cng thc mi bng mt dy p dng cc lut suy din. Cc cng thc cho c gi l cc tin . Cc cng thc c suy ra c gi l cc nh l. Dy cc lut c p dng dn ti nh l c gi l mt chng minh ca nh l. Nu cc lut suy din l tin cy, th cc nh l l h qu logic ca cc tin .

V d: Gi s ta c cc cng thc sau :

Q S => G v H (1)

P => Q

(2)

R => S

(3)

P

(4)

R

(5)

T cng thc (2) v (4), ta suy ra Q (Lut Modus Ponens) . Li p dng lut Modus Ponens, t (3) v (5) ta suy ra S . T Q, S ta suy ra QS (lut a vo hi ). T (1) v QS ta suy ra G v H. Cng thc G v H c chng minh.

Trong cc h tri thc, chng hn cc h chuyn gia, h lp trnh logic,..., s dng cc lut suy din ngi ta thit k ln cc th tc suy din (cn c gi l th tc chng minh) t cc tri thc trong c s tri thc ta suy ra cc tri thc mi p ng nhu cu ca ngi s dng.

Mt h hnh thc (formal system) bao gm mt tp cc tin v mt tp cc lut suy din no (trong ngn ng biu din tri thc no ).

Mt tp lut suy din c xem l y , nu mi h qu logic ca mt tp cc tin u chng minh c bng cch ch s dng cc lut ca tp .

Phng php chng minh bc b

Phng php chng minh bc b (refutation proof hoc proof by contradiction) l mt phng php thng xuyn c s dng trong cc chng minh ton hc. T tng ca phng php ny l nh sau : chng minh P ng, ta gi s P sai ( thm ( P vo cc gi thit ) v dn ti mt mu thun. Sau y ta s trnh by c s ny.Gi s chng ta c mt tp hp cc cng thc G ={G1,.....,Gm} ta cn chng minh cng thc H l h qu logic ca G . iu tng ng vi chng minh cng thc G1^....^Gm -> H l vng chc. Thay cho chng minh G1^..... ^Gm =>H l vng chc, ta chng minh G1^....^Gm ^( H l khng tha mn c. Tc l ta chng minh tp G= ( G1,.......,Gm,( H ) l khng tha c nu t Gta suy ra hai mnh i lp nhau. Vic chng minh cng thc H l h qu logic ca tp cc tiu G bng cch chng minh tnh khng tha c ca tp cc tiu c thm vo ph nh ca cng thc cn chng minh, c gi l chng minh bc b.

5.5 Lut gii, chng minh bc b bng lut gii

thun tin cho vic s dng lut gii, chng ta s c th ho lut gii trn cc dng cu c bit quan trng.

* Lut gii trn cc cu tuyn

A1 v. . ............. vAm v C

( C v B1 v.. ............. v Bn

A1 v.. ......... v Am v B1 v.... v Bntrong Ai, Bj v C l cc literal.

* Lut gii trn cc cu Horn:

Gi s Pi, Rj, Q v S l cc literal. Khi ta c cc lut sau :

P1 ^. ..............^Pm ^ S => Q,

R1 ^. .............^ Rn => S

P1 ^........^Pm ^ R1 ^...... ^ Rn =>Q

Mt trng hp ring hay c s dng ca lut trn l :

P1 ^...............^ Pm ^ S => Q,

S

P1 ^................^Pm => Q

Khi ta c th p dng lut gii cho hai cu, th hai cu ny c gi l hai cu gii c v kt qu nhn c khi p dng lut gii cho hai cu c gi l gii thc ca chng. Gii thc ca hai cu A v B c k hiu l res(A,B). Chng hn, hai cu tuyn gii c nu mt cu cha mt literal i lp vi mt literal trong cu kia. Gii thc ca hai literal i lp nhau (P v ( P) l cu rng, chng ta s k hiu cu rng l [] , cu rng khng tho c.

Gi s G l mt tp cc cu tuyn ( Bng cch chun ho ta c th a mt tp cc cng thc v mt tp cc cu tuyn ). Ta s k hiu R(G ) l tp cu bao gm cc cu thuc G v tt c cc cu nhn c t G bng mt dy p dng lut gii.

Lut gii l lut y chng minh mt tp cu l khng tha c. iu ny c suy t nh l sau :

nh l gii:

Mt tp cu tuyn l khng tha c nu v ch nu cu rng [] ( R(G ).

nh l gii c ngha rng, nu t cc cu thuc G , bng cch p dng lut gii ta dn ti cu rng th G l khng tha c, cn nu khng th sinh ra cu rng bng lut gii th G tha c. Lu rng, vic dn ti cu rng c ngha l ta dn ti hai literal i lp nhau P v ( P ( tc l dn ti mu thun ).

T nh l gii, ta a ra th tc sau y xc nh mt tp cu tuyn G l tha c hay khng . Th tc ny c gi l th tc gii.

procedure Resolution ;

Input : tp G cc cu tuyn ;

begin1.Repeat1.1 Chn hai cu A v B thuc G ;

1.2 if A v B gii c then tnh Res ( A,B ) ;

1.3 if Res (A,B ) l cu mi then thm Res ( A,B ) vo G ;

until

nhn c [] hoc khng c cu mi xut hin ;

2. if nhn c cu rng then thng bo G khng tho ce lse thng bo G tho c ;

end;Chng ta c nhn xt rng, nu G l tp hu hn cc cu th cc literal c mt trong cc cu ca G l hu hn. Do s cc cu tuyn thnh lp c t cc literal l hu hn. V vy ch c mt s hu hn cu c sinh ra bng lut gii. Th tc gii s dng li sau mt s hu hn bc.

Ch s dng lut gii ta khng th suy ra mi cng thc l h qu logic ca mt tp cng thc cho. Tuy nhin, s dng lut gii ta c th chng minh c mt cng thc bt k c l h qu ca mt tp cng thc cho hay khng bng phng php chng minh bc b. V vy lut gii c xem l lut y cho bc b.

Sau y l th tc chng minh bc b bng lut gii

Procedure Refutation_Proof ;

input : Tp G cc cng thc ;

Cng thc cn chng minh H;

Begin1. Thm (H vo G ;

2. Chuyn cc cng thc trong G v dng chun hi ;

3. T cc dng chun hi bc hai, thnh lp tp cc cu tuyn g ;4. p dng th tc gii cho tp cu G ;5. if G khng tho c then thng bo H l h qu logic

else thng bo H khng l h qu logic ca G ;

end;V d: Gi gi G l tp hp cc cu tuyn sau

( A v ( B v P (1)

( C v ( D v P (2)

( E v C (3)

A (4)

E (5)

D (6)

Gi s ta cn chng minh P. Thm vo G cu sau:

( P (7)

p dng lut gii cho cu (2) v (7) ta c cu:

( C v ( D (8)

T cu (6) v (8) ta nhn c cu:

( C (9)

T cu (3) v (9) ta nhn c cu:

( E (10)

Ti y xut hin mu thun, v cu (5) v (10) i lp nhau. T cu (5) v (10) ta nhn c cu rng []. Vy P l h qu logic ca cc cu (1) --(6).

CHNG VI :

LOGIC V T CP MT

Logic mnh cho php ta biu din cc s kin, mi k hiu trong logic mnh c minh ha nh l mt s kin trong th gii hin thc, s dng cc kt ni logic ta c th to ra cc cu phc hp biu din cc s kin mang ngha phc tp hn. Nh vy kh nng biu din ca logic mnh ch gii hn trong phm vi th gii cc s kin.

Th gii hin thc bao gm cc i tng, mi i tng c nhng tnh cht ring phn bit n vi cc i tng khc. Cc i tng li c quan h vi nhau. Cc mi quan h rt a dng v phong ph. Chng ta c th lit k ra rt nhiu v d v i tng, tnh cht, quan h.

* i tng : mt ci bn, mt ci nh, mt ci cy, mt con ngi, mt con s. ...

* Tnh cht : Ci bn c th c tnh cht : c bn chn, lm bng g, khng c ngn ko. Con s c th c tnh cht l s nguyn, s hu t, l s chnh phng. ..

* Quan h : cha con, anh em, b bn (gia con ngi ); ln hn nh hn, bng nhau (gia cc con s ) ; bn trong, bn ngoi nm trn nm di (gia cc vt )...

* Hm : Mt trng hp ring ca quan h l quan h hm. Chng hn, v mi ngi c mt m, do ta c quan h hm ng mi ngi vi m ca n.

Logic v t cp mt l m rng ca logic mnh . N cho php ta m t th gii vi cc i tng, cc thuc tnh ca i tng v cc mi quan h gia cc i tng. N s dng cc bin ( bin i tng ) ch mt i tng trong mt min i tng no . m t cc thuc tnh ca i tng, cc quan h gia cc i tng, trong logic v t, ngi ta da vo cc v t ( predicate). Ngoi cc kt ni logic nh trong logic mnh , logic v t cp mt cn s dng cc lng t. Chng hn, lng t " (vi mi) cho php ta to ra cc cu ni ti mi i tng trong mt min i tng no .

Chng ny dnh cho nghin cu logic v t cp mt vi t cch l mt ngn ng biu din tri thc. Logic v t cp mt ng vai tr cc k quan trng trong biu din tri thc, v kh nng biu din ca n ( n cho php ta biu din tri thc v th gii vi cc i tng, cc thuc tnh ca i tng v cc quan h ca i tng), v hn na, n l c s cho nhiu ngn ng logic khc.

6.1 C php v ng ngha ca logic v t cp mt.

6.1.1 C php.

Cc k hiu.

Logic v t cp mt s dng cc loi k hiu sau y.

Cc k hiu hng: a, b, c, An, Ba, John,...

Cc k hiu bin: x, y, z, u, v, w,...

Cc k hiu v t: P, Q, R, S, Like, Havecolor, Prime,...

Mi v t l v t ca n bin ( n

0). Chng hn Like l v t ca hai bin, Prime l v t mt bin. Cc k hiu v t khng bin l cc k hiu mnh .

Cc k hiu hm: f, g, cos, sin, mother, husband, distance,...

Mi hm l hm ca n bin ( n1). Chng hn, cos, sin l hm mt bin, distance l hm ca ba bin.

Cc k hiu kt ni logic: ( hi), (tuyn), ( ( ph nh), (ko theo), (ko theo nhau).

Cc k hiu lng t: " ( vi mi), $ ( tn ti).

Cc k hiu ngn cch: du phy, du m ngoc v du ng ngoc.

Cc hng thcCc hng thc ( term) l cc biu thc m t cc i tng. Cc hng thc c xc nh quy nh sau.

Cc k hiu hng v cc k hiu bin l hng thc.

Nu t1, t2, t3, ..., tn l n hng thc v f l mt k hiu hm n bin th f( t1, t2, ..., tn) l hng thc. Mt hng thc khng cha bin c gi l mt hng thc c th ( ground term).

Chng hn, An l k hiu hng, mother l k hiu hm mt bin, th mother (An) l mt hng thc c th.

1.21 Cc cng thc phn tChng ta s biu din cc tnh cht ca i tng, hoc cc quan h ca i tng bi cc cng thc phn t ( cu n).

Cc cng thc phn t ( cu n) c xc nh quy nh sau.

Cc k hiu v t khng bin ( cc k hiu mnh ) l cu n.

Nu t1, t2,...,tn l n hng thc v p l v t ca n bin th p( t1,t2,...,tn) l cu n.

Chng hn, Hoa l mt k hiu hng, Love l mt v t ca hai bin, husband l hm ca mt bin, th Love ( Hoa, husband( Hoa)) l mt cu n.

1.21.1 Cc cng thc

T cng thc phn t, s dng cc kt ni logic v cc lng t, ta xy dng nn cc cng thc (cc cu).

Cc cng thc c xc nh quy nh sau:

Cc cng thc phn t l cng thc.

Nu G v H l cc cng thc, th cc biu thc (G H), (G H), (( G), (GH), (GH) l cng thc.

Nu G l mt cng thc v x l bin th cc biu thc ( " x G), ($ x G) l cng thc.

Cc cng thc khng phi l cng thc phn t s c gi l cc cu phc hp. Cc cng thc khng cha bin s c gi l cng thc c th. Khi vit cc cng thc ta s b i cc du ngoc khng cn thit, chng hn cc du ngoc ngoi cng.

Lng t ph dng (") cho php m t tnh cht ca c mt lp cc i tng, ch khng phi ca mt i tng, m khng cn phi lit k ra tt c cc i tng trong lp. Chng hn s dng v t Elephant(x) (i tng x l con voi ) v v t Color(x, Gray) (i tng x c mu xm) th cu tt c cc con voi u c mu xm c th biu din bi cng thc "x (Elephant(x) Color(x, Gray)).

Lng t tn ti ($) cho php ta to ra cc cu ni n mt i tng no trong mt lp i tng m n c mt tnh cht hoc tho mn mt quan h no . Chng hn bng cch s dng cc cu n Student(x) (x l sinh vin) v Inside(x, P301), (x trong phng 301), ta c th biu din cu C mt sinh vin phng 301 bi biu thc $x (Student(x) Inside(x,P301).

Mt cng thc l cng thc phn t hoc ph nh ca cng thc phn t c gi l literal. Chng hn, Play(x, Football), ( Like( Lan, Rose) l cc literal. Mt cng thc l tuyn ca cc literal s c gi l cu tuyn. Chng hn, Male(x) ( Like(x, Foodball) l cu tuyn.

Trong cng thc ( "x G), hoc $x G trong G l mt cng thc no , th mi xut hin ca bin x trong cng thc G c gi l xut hin buc. Mt cng thc m tt c cc bin u l xut hin buc th c gi l cng thc ng.

V d: Cng thc "xP( x, f(a, x)) ( (y Q(y) l cng thc ng, cn cng thc "x P( x, f(y, x)) khng phi l cng thc ng, v s xut hin ca bin y trong cng thc ny khng chu rng buc bi mt lng t no c (S xut hin ca y gi l s xut hin t do).

Sau ny chng ta ch quan tm ti cc cng thc ng.

6.1.2 Ng ngha.Cng nh trong logic mnh , ni n ng ngha l chng ta ni n ngha ca cc cng thc trong mt th gii hin thc no m chng ta s gi l mt minh ha.

xc nh mt minh ho, trc ht ta cn xc nh mt min i tng ( n bao gm tt c cc i tng trong th gii hin thc m ta quan tm).

Trong mt minh ho, cc k hiu hng s c gn vi cc i tng c th trong min i tng cc k hiu hm s c gn vi mt hm c th no . Khi , mi hng thc c th s ch nh mt i tng c th trong min i tng. Chng hn, nu An l mt k hiu hng, Father l mt k hiu hm, nu trong minh ho An ng vi mt ngi c th no , cn Father(x) gn vi hm; ng vi mi x l cha ca n, th hng thc Father(An) s ch ngi cha ca An .

Ng ngha ca cc cu n .

Trong mt minh ho, cc k hiu v t s c gn vi mt thuc tnh, hoc mt quan h c th no . Khi mi cng thc phn t (khng cha bin) s ch nh mt s kin c th. ng nhin s kin ny c th l ng (True) hoc sai (False). Chng hn, nu trong minh ho, k hiu hng Lan ng vi mt c gi c th no , cn Student(x) ng vi thuc tnh x l sinh vin th cu Student (Lan) c gi tr chn l l True hoc False tu thuc trong thc t Lan c phi l sinh vin hay khng.

Ng ngha ca cc cu phc hp.

Khi xc nh c ng ngha ca cc cu n, ta c th thc hin c ng ngha ca cc cu phc hp (c to thnh t cc cu n bng cch lin kt cc cu n bi cc kt ni logic) nh trong logic mnh .

V d: Cu Student(Lan) ( Student(An) nhn gi tr True nu c hai cu Student(Lan) v Student(An) u c gi tr True, tc l c Lan v An u l sinh vin.

Cu Like(Lan, Rose) Like(An, Tulip) l ng nu cu Like(Lan, Rose) l ng hoc cu Like(An, Tulip) l ng.

Ng ngha ca cc cu cha cc lng t.

Ng ngha ca cc cu "x G, trong G l mt cng thc no , c xc nh nh l ng ngha ca cng thc l hi ca tt c cc cng thc nhn c t cng thc G bng cch thay x bi mt i tng trong min i tng. Chng hn, nu min i tng gm ba ngi {Lan, An, Hoa} th ng ngha ca cu "x Student(x) c xc nh l ng ngha ca cu Student(Lan) Student(An) Student(Hoa). Cu ny ng khi v ch khi c ba cu thnh phn u ng, tc l c Lan, An, Hoa u l sinh vin.

Nh vy, cng thc "x G l ng nu v ch nu mi cng thc nhn c t G bng cch thay x bi mt i tng trong min i tng u ng, tc l G ng cho tt c cc i tng x trong min i tng.

Ng ngha ca cng thc $x G c xc nh nh l ng ngha ca cng thc l tuyn ca tt c cc cng thc nhn c t G bng cch thay x bi mt i tng trong min i tng. Chng hn, nu ng ngha ca cu Younger(x,20) l x tr hn 30 tui v min i tng gm ba ngi {Lan, An, Hoa} th ng ngha ca cu $x Yourger(x,20) l ng ngha ca cu Yourger(Lan,20) Yourger(An,20) Yourger(Hoa,20). Cu ny nhn gi tr True nu v ch nu t nht mt trong ba ngi Lan, An, Hoa tr hn 20.

Nh vy cng thc $x G l ng nu v ch nu mt trong cc cng thc nhn c t G bng cch thay x bng mt i tng trong min i tng l ng.

Bng cc phng php trnh by trn, ta c th xc nh c gi tr chn l ( True, False ) ca mt cng thc bt k trong mt minh ho. (Lu rng, ta ch quan tm ti cc cng thc ng ).

Sau khi xc nh khi nim minh ho v gi tr chn l ca mt cng thc trong mt minh ho, c th a ra cc khi nim cng thc vng chc ( tho c, khng tho c ), m hnh ca cng thc ging nh trong logic mnh .

Cc cng thc tng ng

Cng nh trong logic mnh , ta ni hai cng thc G v H tng ng ( vit l G H ) nu chng cng ng hoc cng sai trong mt minh ho. Ngoi cc tng ng bit trong logic mnh , trong logic v t cp mt cn c cc tng ng khc lin quan ti cc lng t. Gi s G l mt cng thc, cch vit G(x) ni rng cng thc G c cha cc xut hin ca bin x. Khi cng thc G(y) l cng thc nhn c t G(x) bng cch thay tt c cc xut hin ca x bi y. Ta ni G(y) l cng thc nhn c t G(x) bngcch t tn li ( bin x c i tn li l y ).

Chng ta c cc tng ng sau y:

1. "x G(x) "y G(y)

$x G(x) $y G(y)

t tn li bin i sau lng t ph dng ( tn ti ), ta nhn c cng thc tng ng .

2. ( ("x G(x)) $x ( ( G(x))

( ( $x G(x)) "x ( ( G(x))

3. "x (G(x) H(x)) "x G(x) "x H(x)

$x (G(x) H(x)) $x G(x) $x H(x)

v d : "x Love(x, Husband(x)) "y Love(y, Husband(y)).

Tr tu Nhn to

Gio trnh

inh Mnh Tng

Khoa CNTT - i Hc Quc Gia H Ni

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