giao trinh tri tue nhan tao dhqg ha noi

8/2/2019 Giao Trinh Tri Tue Nhan Tao DHQG Ha Noi

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Mc lc

Phn I : Gii quyt vn bng tm kim

1.1 Chng I - Cc chin lc tm kim m1.1 Biu din vn trong khng gian trng thi

1.2 Cc chin lc tm kim1.3 Cc chin lc tm kim m1.3.1 Tm kim theo b rng1.3.2 Tm kim theo su1.3.3 Cc trng thi lp1.3.4 Tm kim su lp1.4 Quy vn v cc vn con. Tm kim trn th v/hoc

1.4.1 Quy vn v cc vn con1.4.2 th v/hoc1.4.3 Tm kim trn th v/hoc

Chng II - Cc chin lc tm kim kinh nghim2.1 Hm nh gi v tm kim kinh nghim2.2 Tm kim tt nht - u tin2.3 Tm kim leo i2.4 Tm kim beam

1.2 Chng III - Cc chin lc tm kim ti u3.1 Tm ng i ngn nht3.1.1 Thut ton A*3.1.2 Thut ton tm kim Nhnh-v-Cn1.2.1 3.2 Tm i tng tt nht1.2.1.1 3.2.1 Tm kim leo i3.2.2 Tm kim gradient3.2.3 Tm kim m phng luyn kim1.2.2 3.3 Tm kim m phng s tin ha. Thut ton di truyn

1.3 Chng IV - Tm kim c i th4.1 Cy tr chi v tm kim trn cy tr chi4.2 Chin lc Minimax4.3 Phng php ct ct Alpha-Beta

Phn II: Tri thc v lp lun

inh Mnh Tng Trang 1
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inh Mnh Tng Trang 2

inh Mnh Tng

Gio trnh

Tr tu Nhn to

Khoa CNTT - i Hc Quc Gia H Ni


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Phn I

Gii quyt vn bng tm kim

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

Vn tm kim, mt cch tng qut, c th hiu l tm mt i tng tha mn mt s ihi no , trong mt tp hp rng ln cc i tng. Chng ta c th k ra rt nhiu vn mvic 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 nhiunc i c php thc hin, ta phi tm ra cc nc i dn ti tnh th kt cuc m ta l ngithng.

Chng minh nh l cng c th xem nh vn tm kim. Cho mt tp cc tin v cclut 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 giiquyt cc vn v c p dng rng ri trong cc lnh vc nghin cu khc ca Tr Tu NhnTo. 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 gihng 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.

inh Mnh Tng Trang 3


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Chng I

Cc chin lc tm kim m

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

Trong chng ny, chng ti s nghin cu cc chin lc tm kim m (blind search): tmkim theo b rng (breadth-first search) v tm kim theo su (depth-first search). Hiu qu cacc 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 xcnh khng gian tm kim. Khng gian tm kim bao gm tt c cc i tng m ta cn quan tmtm kim. N c th l khng gian lin tc, chng hn khng gian cc vct thc n chiu; n cngc 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 vicgii 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 cchs dng khi nim trng thi v ton t (php bin i trng thi). Chng hn, mt khch du lchc trong tay bn mng li giao thng ni cc thnh ph trong mt vng lnh th (hnh 1.1), dukhch 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 trngthi kt thc. Khi ang mt thnh ph, chng hn thnh ph D anh ta c th i theo cc conng ni ti cc thnh ph C, F v G. Cc con ng ni cc thnh ph s c biu din bicc 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 ltm 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 tont, 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 tsau:

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 dyton 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 cth xem nh mt nh x R: UU. 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 nhiuvn (chng hn cc loi c) c th c nhiu trng thi ch v ta khng th xc nh trc ccc trng thi ch. Ni chung trong phn ln cc vn hay, ta ch c th m t cc trng thich l cc trng thi tha mn mt s iu kin no .

inh Mnh Tng Trang 4


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

Chng ta c th biu din khng gian trng thi bng th nh hng, trong mi nhca 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 sl 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 svn .

V d 1: Bi ton 8 s. Chng ta c bng 3x3 v tm qun mang s hiu t 1 n 8 cxp vo tm , cn li mt trng, chng hn nh trong hnh 2 bn tri. Trong tr chi ny, bn cth chuyn dch cc qun cch trng ti trng . Vn ca bn l tm ra mt dy ccchuyn dch bin i cnh hung ban u (hnh 1.2 bn tri) thnh mt cnh hung xc nhno , 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 ktthc 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 sangphi). R rng l, cc ton t ny ch l cc ton t b phn; chng hn, t trng thi ban u (hnh1.2 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

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cc trng thi ca vn l hon ton khng n gin. Vic tm ra dng biu din tt cho cc trngthi ng vai tr ht sc quan trng trong qu trnh gii quyt mt vn . C th ni rng, nu tatm 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 ngnmt con sng, cng mt chic thuyn ch c mt hoc hai ngi. Hy tm cch a mi ngiqua sng sao cho khng li bn b sng k cp nhiu hn triu ph. ng nhin trong biton ny, cc ton t tng ng vi cc hnh ng ch 1 hoc 2 ngi qua sng. Nhng y tacn lu rng, khi hnh ng xy ra (lc thuyn ang bi qua sng) th bn b sng thuyn vadi ch, s k cp khng c nhiu hn s triu ph. Tip theo ta cn quyt nh ci g l trngthi ca vn . y ta khng cn phn bit cc nh triu ph v cc tn cp, m ch s lngca 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 bkia, k = 1 nu thuyn b t ngn v k = 0 nu thuyn b hu ngn. Nh vy, khng gian trngthi 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 giantrng 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 nom 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 trngthi k u, hoc v c sinh ra t trng thi u bi ton t R. Qu trnh p dng cc ton t sinhra cc trng thi k u c gi l pht trin trng thi u. Chng hn, trong bi ton ton s, phttrin 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 thvic tm li gii ca vn c quy v vic tm ng i t trng thi ban u ti mt trng thikt 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 thich no . C hai k thut tm kim m, l tm kim theo b rng v tm kim theo su.

inh Mnh Tng Trang 6


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T tng ca tm kim theo b rng l cc trng thi c pht trin theo th t m chngc 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 rnghoc 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 thigian. 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 davo 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 cctrng 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 gicc 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 tmkim l cy m cc nh c gn bi cc trng thi ca khng gian trng thi. Gc ca cy tmkim tng ng vi trng thi ban u. Nu mt nh ng vi trng thi u, th cc nh con ca nng vi cc trng thi v k u. Hnh 1.4a l th biu din mt khng gian trng thi vi trng thi

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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 xydng 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 lca cy tng ng vi cc trng thi cha c pht trin. Bc tip theo ph thuc vo chinlc 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 vtm kim theo su. Trong tm kim theo b rng, ti mi bc ta s chn trng thi pht trinl 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 trngthi ch, do ta cn lu li vt ca ng i. Ta c th s dng hm father lu li cha cami 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:

procedure Breadth_First_Search;

begin

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

2. loop do

2.1 ifL rngthen

{thng bo tm kim tht bi;stop};

2.2Loi trng thi u u danh sch L;

2.3 ifu 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)


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trng hp bi ton v nghim v khng gian trng thi hu hn, thut ton s dng v cho thngbo v nghim.

nh gi tm kim theo b rng

By gi ta nh gi thi gian v b nh m tm kim theo b rng i hi. Gi s rng, mitrng thi khi c pht trin s sinh ra b trng thi k. Ta s gi b l nhn t nhnh. Gi s rng,nghim ca bi ton l ng i c di d. Bi nhiu nghim c th c tm ra ti mt nh btk mc d ca cy tm kim, do s nh cn xem xt tm ra nghim l:

1 + b + b2 + ... + bd-1 + k

Trong k c th l 1, 2, ..., bd. Do s ln nht cc nh cn xem xt l:

1 + b + b2 + ... + bd

Nh vy, phc tp thi gian ca thut ton tm kim theo b rng l O(b d). phc tpkhng gian cng l O(bd), bi v ta cn lu vo danh sch L tt c cc nh ca cy tm kim mc d, s cc nh ny l bd.

thy r tm kim theo b rng i hi thi gian v khng gian ln ti mc no, ta xttrng hp nhn t nhnh b = 10 v su d thay i. Gi s pht hin v kim tra 1000 trngthi cn 1 giy, v lu gi 1 trng thi cn 100 bytes. Khi thi gian v khng gian m thut toni hi c cho trong bng sau:

su d Thi gian Khng gian

4 11 giy 1 megabyte

6 18 giy 111 megabytes

8 31 gi 11 gigabytes10 128 ngy 1 terabyte

12 35 nm 111 terabytes

14 3500 nm 11.111 terabytes

1.6.2 Tm kim theo su

Nh ta bit, t tng ca chin lc tm kim theo su l, ti mi bc trng thic chn pht trin l trng thi c sinh ra sau cng trong s cc trng thi ch pht trin.Do thut ton tm kim theo su l hon ton tng t nh thut ton tm kim theo b rng,

ch c mt iu khc l, ta x l danh sch L cc trng thi ch pht trin khng phi nh hng im nh ngn xp. C th l trong bc 2.4 ca thut ton tm kim theo b rng, ta cn sa li lt v vo u danh sch L.

Sau y chng ta s a ra cc nhn xt so snh hai chin lc tm kim m:

Thut ton tm kim theo b rng lun lun tm ra nghim nu bi ton c nghim. Songkhng phi vi bt k bi ton c nghim no thut ton tm kim theo su cng tm ra nghim!

Nu bi ton c nghim v khng gian trng thi hu hn, th thut ton tm kim theo su stm ra nghim. Tuy nhin, trong trng hp khng gian trng thi v hn, th c th n khng tmra nghim, l do l ta lun lun i xung theo su, nu ta i theo mt nhnh v hn m nghimkhng nm trn nhnh th thut ton s khng dng. Do ngi ta khuyn rng, khng nn p

dng tm kim theo d su cho cc bi ton c cy tm kim cha cc nhnh v hn.

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phc tp ca thut ton tm kim theo su.

Gi s rng, nghim ca bi ton l ng i c di d, cy tm kim c nhn t nhnh lb v c chiu cao l d. C th xy ra, nghim l nh ngoi cng bn phi trn mc d ca cy tmkim, do phc tp thi gian ca tm kim theo su trong trng hp xu nht l O(bd),tc l cng nh tm kim theo b rng. Tuy nhin, trn thc t i vi nhiu bi ton, tm kimtheo su thc s nhanh hn tm kim theo b rng. L do l tm kim theo b rng phi xem xtton b cy tm kim ti mc d-1, ri mi xem xt cc nh mc d. Cn trong tm kim theo su, c th ta ch cn xem xt mt b phn nh ca cy tm kim th tm ra nghim.

nh gi phc tp khng gian ca tm kim theo su ta c nhn xt rng, khi tapht trin mt nh u trn cy tm kim theo su, ta ch cn lu cc nh cha c pht trinm chng l cc nh con ca cc nh nm trn ng i t gc ti nh u. Nh vy i vi cytm kim c nhn t nhnh b v su ln nht l d, ta ch cn lu t hn db nh. Do phctp khng gian ca tm kim theo su lO(db), trong khi tm kim theo b rng i hikhng gian nh O(bd)!

1.6.3 Cc trng thi lpNh ta thy trong mc 1.2, cy tm kim c th cha nhiu nh ng vi cng mt trng

thi, cc trng thi ny c gi l trng thi lp. Chng hn, trong cy tm kim hnh 4b, cctrng thi C, E, F l cc trng thi lp. Trong th biu din khng gian trng thi, cc trng thilp ng vi cc nh c nhiu ng i dn ti n t trng thi ban u. Nu th c chu trnh thcy tm kim s cha cc nhnh vi mt s nh lp li v hn ln. Trong cc thut ton tm kims lng ph rt nhiu thi gian pht trin li cc trng thi m ta gp v pht trin. V vytrong qu trnh tm kim ta cn trnh pht sinh ra cc trng thi m ta pht trin. Chng ta cth p dng mt trong cc gii php sau y:

1. Khi pht trin nh u, khng sinh ra cc nh trng vi cha ca u.

2. Khi pht trin nh u, khng sinh ra cc nh trng vi mt nh no nm trn ng idn ti u.

3. Khng sinh ra cc nh m n c sinh ra, tc l ch sinh ra cc nh mi.

Hai gii php u d ci t v khng tn nhiu khng gian nh, tuy nhin cc gii php nykhng trnh c ht cc trng thi lp.

thc hin gii php th 3 ta cn lu cc trng thi pht trin vo tp Q, lu cc trngthi ch pht trin vo danh sch L. ng nhin, trng thi v ln u c sinh ra nu n khngc trong Q v L. Vic lu cc trng thi pht trin v kim tra xem mt trng thi c phi lnu c sinh ra khng i hi rt nhiu khng gian v thi gian. Chng ta c th ci t tp Q

bi bng bm (xem [ ]).1.6.4 Tm kim su lp

Nh chng ta nhn xt, nu cy tm kim cha nhnh v hn, khi s dng tm kim theo su, ta c th mc kt nhnh v khng tm ra nghim. khc phc hon cnh , ta tmkim theo su ch ti mc d no ; nu khng tm ra nghim, ta tng su ln d+1 v li tmkim theo su ti mc d+1. Qu trnh trn c lp li vi d ln lt l 1, 2, ... dn mt sumax no . Nh vy, thut ton tm kim su lp (iterative deepening search) s s dng th tctm kim su hn ch (depth_limited search) nh th tc con. l th tc tm kim theo su,nhng ch i ti su d no ri quay ln.

Trong th tc tm kim su hn ch, d l tham s su, hm depth ghi li su ca minh

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procedureDepth_Limited_Search(d);

begin

1. Khi to danh sch L ch cha trng thi ban u u 0;

depth(u0)

0;2. loop do

2.1 ifL rngthen

{thng bo tht bi; stop};

2.2Loi trng thi u u danh sch L;


{thng bo thnh cng; stop};

2.4 ifdepth(u)


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Tm li, tm kim su lp c phc tp thi gian l O(bd) (nh tm kim theo b rng), vc phc tp khng gian l O(biu din) (nh tm kim theo su). Ni chung, chng ta nn pdng tm kim su lp cho cc vn c khng gian trng thi ln v su ca nghim khng

bit trc.

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

1.7.1 Quy vn v cc vn con:

Trong mc 1.1, chng ta nghin cu vic biu din vn thng qua cc trng thi v ccton t. Khi vic tm nghim ca vn c quy v vic tm ng trong khng gian trngthi. Trong mc ny chng ta s nghin cu mt phng php lun khc gii quyt vn , datrn vic quy vn v cc vn con. Quy vn v cc vn con (cn gi l rt gn vn )l mt phng php c s dng rng ri nht gii quyt cc vn . Trong i sng hngngy, cng nh trong khoa hc k thut, mi khi gp mt vn cn gii quyt, ta vn thng cgng tm cch a n v cc vn n gin hn. Qu trnh rt gn vn s c tip tc choti khi ta dn ti cc vn con c th gii quyt c d dng. Sau y chng ta xt mt s vn.

Vn tnh tch phn bt nh

Gi s ta cn tnh mt tch phn bt nh, chng hn (xex+ x3) dx. Qu trnh chng ta vnthng lm tnh tch phn bt nh l nh sau. S dng cc quy tc tnh tch phn (quy tc tnhtch phn ca mt tng, quy tc tnh tch phn tng phn...), s dng cc php bin i bin s, cc

php bin i cc hm (chng hn, cc php bin i lng gic),... a tch phn cn tnh vtch phn ca cc hm s s cp m chng ta bit cch tnh. Chng hn, i vi tch phn (xex+ x3) dx, p dng quy tc tch phn ca tng ta a v hai tch phn xexdx v x3dx. pdng quy tc tch phn tng phn ta a tch phn xexdx v tch phn exdx. Qu trnh trn cth biu din bi th trong hnh 1.5.

Cc tch phn exdx v x3dx l cc tch phn c bn c trong bng tch phn. Kt hpcc kt qu ca cc tch phn c bn, ta nhn c kt qu ca tch phn cho.

Chng ta c th biu din vic quy mt vn v cc vn con c bi cc trng thi v cc

ton t. y, bi ton cn gii l trng thi ban u. Mi cch quy bi ton v cc bi ton conc biu din bi mt ton t, ton t AB, C biu din vic quy bi ton A v hai bi ton B vC. Chng hn, i vi bi ton tnh tch phn bt nh, ta c th xc nh cc ton t dng:

(f1 + f2) dx f1 dx, f2 dx v u dv v du



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Cc trng thi kt thc l cc bi ton s cp (cc bi ton bit cch gii). Chng hn,trong bi ton tnh tch phn, cc tch phn c bn l cc trng thi kt thc. Mt iu cn lu l,trong khng gian trng thi biu din vic quy vn v cc vn con, cc ton t c th l atr, n bin i mt trng thi thnh nhiu trng thi khc.

Vn tm ng i trn bn giao thngBi ton ny c pht trin nh bi ton tm ng i trong khng gian trng thi (xem

1.1), trong mi trng thi ng vi mt thnh ph, mi ton t ng vi mt con ng ni, nithnh ph ny vi thnh ph khc. By gi ta a ra mt cch biu din khc da trn vic quyvn v cc vn con. Gi s ta c bn giao thng trong mt vng lnh th (xem hnh 1.6).Gi s ta cn tm ng i t thnh ph A ti thnh ph B. C con sng chy qua hai thnh ph Ev G v c cu qua sng mi thnh ph . Mi ng i t A n B ch c th qua E hoc G.

Nh vy bi ton tm ng i t A n B c quy v:1) Bi ton tm ng i t A n B qua E (hoc)2) Bi ton tm ng i t A n b qua G.

Mi mt trong hai bi ton trn li c th phn nh nh sau1) Bi ton tm ng i t A n B qua E c quy v:

1.1 Tm ng i t A n E (v)1.2 Tm ng i t E n B.

2) Bi ton tm ng i t A n B qua G c quy v:2.1 Tm ng i t A n G (v)2.2 Tm ng i t G n B.

Qu trnh rt gn vn nh trn c th biu din di dng th ( th v/hoc) tronghnh 1.7. y mi bi ton tm ng i t mt thnh ph ti mt thnh ph khc ng vi mttrng thi. Cc trng thi kt thc l cc trng thi ng vi cc bi ton tm ng i, chng hn tA n C, hoc t D n E, bi v c ng ni A vi C, ni D vi E.



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1.7.2 th v/hoc

Khng gian trng thi m t vic quy vn v cc vn con c th biu din di dng th nh hng c bit c gi l th v/hoc. th ny c xy dng nh sau:

Mi bi ton ng vi mt nh ca th. Nu c mt ton t quy mt bi ton v mt biton khc, chng hn R : a b, th trong th s c cung gn nhn i t nh a ti nh b. i vimi ton t quy mt bi ton v mt s bi ton con, chng hn R : a b, c, d ta a vo mt nhmi a1, nh ny biu din tp cc bi ton con {b, c, d} v ton t R : a b, c, d c biu din

bi th hnh 1.8.

V d: Gi s chng ta c khng gian trng thi sau:

Trng thi ban u (bi ton cn gii) l a.

Tp cc ton t quy gm:

R1 : a d, e, f

R2 : a d, kR3 : a g, hR4 : d b, cR5 : fiR6 : fc, jR7 : ke, lR8 : kh

Tp cc trng thi kt thc (cc bi ton s cp) l T = {b, c, e, j, l}.



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Khng gian trng thi trn c th biu din bi th v/hoc trong hnh 1.9. Trong th, cc nh, chng hn a1, a2, a3 c gi l nh v, cc nh chng hn a, f, k c gi l nhhoc. L do l, nh a1 biu din tp cc bi ton {d, e, f} v a1 c gii quyt nu d v e v fc gii quyt. Cn ti nh a, ta c cc ton t R1, R2, R3 quy bi ton a v cc bi ton con khc

nhau, do a c gii quyt nu hoc a1 = {d, e, f}, hoc a2 = {d, k}, hoc a3 = {g, h} c giiquyt.

Ngi ta thng s dng th v/hoc dng rt gn. Chng hn, th v/hoc tronghnh 1.9 c th rt gn thnh th trong hnh 1.10. Trong th rt gn ny, ta s ni chng hnd, e, f l cc nh k nh a theo ton t R1, cn d, k l cc nh k a theo ton t R2.

Khi c cc ton t rt gn vn , th bng cch p dng lin tip cc ton t, ta c th

a bi ton cn gii v mt tp cc bi ton con. Chng hn, trong v d trn nu ta p dng ccton t R1, R4, R6, ta s quy bi ton a v tp cc bi ton con {b, c, e, f}, tt c cc bi ton conny u l s cp. T cc ton t R1, R4 v R6 ta xy dng c mt cy trong hnh 1.11a, cy nyc gi l cy nghim. Cy nghim c nh ngha nh sau:

Cy nghim l mt cy, trong :

Gc ca cy ng vi bi ton cn gii.

Tt c cc l ca cy l cc nh kt thc (nh ng vi cc bi ton s cp).

Nu u l nh trong ca cy, th cc nh con ca u l cc nh k u theo mt ton t no .

Cc nh ca th v/hoc s c gn nhn gii c hoc khng gii c.

Cc nhgii c c xc nh quy nh sau:



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Cc nh kt thc l cc nhgii c.

Nu u khng phi l nh kt thc, nhng c mt ton t R sao cho tt c cc nh k u theoR u gii c th ugii c.

Cc nh khng gii c c xc nh quy nh sau: Cc nh khng phi l nh kt thc v khng c nh k, l cc nh khng gii c.

Nu u khng phi l nh kt thc v vi mi ton t R p dng c ti u u c mt nhv k u theo R khng gii c, th u khng gii c.

Ta c nhn xt rng, nu bi ton agii c th s c mt cy nghim gc a, v ngc linu c mt cy nghim gc a th a gii c. Hin nhin l, mt bi ton gii c c th cnhiu cy nghim, mi cy nghim biu din mt cch gii bi ton . Chng hn trong v d nu, bi ton a c hai cy nghim trong hnh 1.11.

Th t gii cc bi ton con trong mt cy nghim l nh sau. Bi ton ng vi nh u ch

c gii sau khi tt c cc bi ton ng vi cc nh con ca u c gii. Chng hn, vi cynghim trong hnh 1.11a, th t gii cc bi ton c th l b, c, d, j, f, e, a. ta c th s dng th tcsp xp topo (xem [ ]) sp xp th t cc bi ton trong mt cy nghim. ng nhin ta cngc th gii quyt ng thi cc bi ton con cng mt mc trong cy nghim.

Vn ca chng ta by gi l, tm kim trn th v/hoc xc nh c nh ng vibi ton ban u l gii c hay khng gii c, v nu n gii c th xy dng mt cynghim cho n.

1.7.3 Tm kim trn th v/hoc

Ta s s dng k thut tm kim theo su trn th v/hoc nh du cc nh. Ccnh s c nh du gii c hoc khng gii c theo nh ngha quy v nh gii cv khng gii c. Xut pht t nh ng vi bi ton ban u, i xung theo su, nu gpnh u l nh kt thc th n c nh du gii c. Nu gp nh u khng phi l nh kt thcv t u khng i tip c, th u c nh du khng gii c. Khi i ti nh u, th t u ta lnlt i xung cc nh v k u theo mt ton t R no . Nu nh du c mt nh v khnggii c th khng cn i tip xung cc nh v cn li. Tip tc i xung cc nh k u theo mtton t khc. Nu tt c cc nh k u theo mt ton t no c nh du gii c th u sc nh du gii c v quay ln cha ca u. Cn nu t u i xung cc nh k n theo miton t u gp cc nh k c nh du khng gii c, th u c nh du khng gii cv quay ln cha ca u.

Ta s biu din th tc tm kim theo su v nh du cc nh trnh by trn bi hm

quy Solvable(u). Hm ny nhn gi tr true nu u gii c v nhn gi tr false nu u khnggii c. Trong hm Solvable(u), ta s s dng:

Bin Ok. Vi mi ton t R p dng c ti u, bin Ok nhn gi tr true nu tt c ccnh v k u theo R u gii c, v Ok nhn gi tr false nu c mt nh v k u theo R khnggii c.

Hm Operator(u) ghi li ton t p dng thnh cng ti u, tc l Operator(u) = R nu minh v k u theo R u gii c.

function Solvable(u);

begin

1. ifu l nh kt thc then



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{Solvable true; stop};

2. ifu khng l nh kt thc v khng c nh kthen

{Solvable(u) false; stop};

3. for mi ton t R p dng c ti u do{Ok true;

for mi v k u theo R do

ifSolvable(v) = false then {Okfalse; exit};

ifOkthen

{Solvable(u) true; Operator(u)R; stop}}

4. Solvable(u)false;

end;

Nhn xt Hon ton tng t nh thut ton tm kim theo su trong khng gian trng thi (mc

1.3.2), thut ton tm kim theo su trn th v/hoc s xc nh c bi ton ban u lgii c hay khng gii c, nu cy tm kim khng c nhnh v hn. Nu cy tm kim cnhnh v hn th cha chc thut ton dng, v c th n b xa ly khi i xung nhnh v hn.Trong trng hp ny ta nn s dng thut ton tm kim su lp (mc 1.3.3).

Nu bi ton ban u gii c, th bng cch s dng hm Operator ta s xy dng ccy nghim.



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Chng II

Cc chin lc tm kim kinh nghim

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

Trong chng I, chng ta nghin cu vic biu din vn trong khng gian trng thi vcc k thut tm kim m. Cc k thut tm kim m rt km hiu qu v trong nhiu trng hpkhng th p dng c. Trong chng ny, chng ta s nghin cu cc phng php tm kimkinh nghim (tm kim heuristic), l cc phng php s dng hm nh gi hng dn stm kim.

Hm nh gi v tm kim kinh nghim:

Trong nhiu vn , ta c th s dng kinh nghim, tri thc ca chng ta v vn nh

gi cc trng thi ca vn . Vi mi trng thi u, chng ta s xc nh mt gi tr s h(u), s nynh gi s gn ch ca trng thi u. Hm h(u) c gi l hm nh gi. Chng ta s s dnghm nh gi hng dn s tm kim. Trong qu trnh tm kim, ti mi bc ta s chn trngthi pht trin l trng thi c gi tr hm nh gi nh nht, trng thi ny c xem l trngthi c nhiu ha hn nht hng ti ch.

Cc k thut tm kim s dng hm nh gi hng dn s tm kim c gi chung lcc k thut tm kim kinh nghim (heuristic search). Cc giai on c bn gii quyt vn

bng tm kim kinh nghim nh sau:

1. Tm biu din thch hp m t cc trng thi v cc ton t ca vn .

2. Xy dng hm nh gi.

3. Thit k chin lc chn trng thi pht trin mi bc.

Hm nh gi

Trong tm kim kinh nghim, hm nh gi ng vai tr cc k quan trng. Chng ta c xydng c hm nh gi cho ta s nh gi ng cc trng thi th tm kim mi hiu qu. Nuhm nh gi khng chnh xc, n c th dn ta i chch hng v do tm kim km hiu qu.

Hm nh gi c xy dng ty thuc vo vn . Sau y l mt s v d v hm nhgi:

Trong bi ton tm kim ng i trn bn giao thng, ta c th ly di ca ngchim bay t mt thnh ph ti mt thnh ph ch lm gi tr ca hm nh gi.

Bi ton 8 s. Chng ta c th a ra hai cch xy dng hm nh gi.

Hm h1: Vi mi trng thi u th h1(u) l s qun khng nm ng v tr ca n trong trngthi ch. Chng hn trng thi ch bn phi hnh 2.1, v u l trng thi bn tri hnh 2.1, thh1(u) = 4, v cc qun khng ng v tr l 3, 8, 6 v 1.



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Hm h2: h2(u) l tng khong cch gia v tr ca cc qun trong trng thi u v v tr ca ntrong trng thi ch. y khong cch c hiu l s t nht cc dch chuyn theo hng hocct a mt qun ti v tr ca n trong trng thi ch. Chng hn vi trng thi u v trng thich nh trong hnh 2.1, ta c:

h2(u) = 2 + 3 + 1 + 3 = 9

V qun 3 cn t nht 2 dch chuyn, qun 8 cn t nht 3 dch chuyn, qun 6 cn t nht 1dch chuyn v qun 1 cn t nht 3 dch chuyn.

Hai chin lc tm kim kinh nghim quan trng nht l tm kim tt nht - u tin (best-first search) v tm kim leo i (hill-climbing search). C th xc nh cc chin lc ny nhsau:

Tm kim tt nht u tin = Tm kim theo b rng + Hm nh gi

Tm kim leo i = Tm kim theo su + Hm nh gi

Chng ta s ln lt nghin cu cc k thut tm kim ny trong cc mc sau.

Tm kim tt nht - u tin:

Tm kim tt nht - u tin (best-first search) l tm kim theo b rng c hng dn bihm nh gi. Nhng n khc vi tm kim theo b rng ch, trong tm kim theo b rng ta lnlt pht trin tt c cc nh mc hin ti sinh ra cc nh mc tip theo, cn trong tm

kim tt nht - u tin ta chn nh pht trin l nh tt nht c xc nh bi hm nh gi(tc l nh c gi tr hm nh gi l nh nht), nh ny c th mc hin ti hoc cc mctrn.



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V d: Ta li xt th khng gian trng thi trong hnh 2.2. Qu trnh tm kim leo i ctin hnh nh sau. u tin pht trin nh A sinh ra cc nh con C, D, E. Trong cc nh nychn D pht trin, v n sinh ra cc nh con B, G. Qu trnh tm kim kt thc. Cy tm kimleo i c cho trong hnh 2.4.

Trong th tc tm kim leo i c trnh by di y, ngoi danh sch L lu cc trng thich c pht trin, chng ta s dng danh sch L1 lu gi tm thi cc trng thi k trng thi

u, khi ta pht trin u. Danh sch L1 c sp xp theo th t tng dn ca hm nh gi, ri cchuyn vo danh sch L sao trng thi tt nht k u ng danh sch L.

procedureHill_Climbing_Search;begin1. Khi to danh sch L ch cha trng thi ban u ;2. loop do

2.1 ifL rngthen{thng bo tht bi; stop};

2.2 Loi trng thi u u danh sch L;2.3 ifu l trng thi kt thc then

{thng bo thnh cng; stop};2.3 for mi trng thi v k u do t v vo L 1;2.5 Sp xp L1 theo th t tng dn ca hm nh gi;2.6 Chuyn danh sch L1 vo u danh sch L;

end;

Tm kim beam

Tm kim beam (beam search) ging nh tm kim theo b rng, n pht trin cc nh

mt mc ri pht trin cc nh mc tip theo. Tuy nhin, trong tm kim theo b rng, ta phttrin tt c cc nh mt mc, cn trong tm kim beam, ta hn ch ch pht trin k nh tt nht(cc nh ny c xc nh bi hm nh gi). Do trong tm kim beam, bt k mc nocng ch c nhiu nht k nh c pht trin, trong khi tm kim theo b rng, s nh cn phttrin mc d l bd (b l nhn t nhnh).



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V d: Chng ta li xt th khng gian trng thi trong hnh 2.2. Chn k = 2. Khi cytm kim beam c cho nh hnh 2.5. Cc nh c gch di l cc nh c chn phttrin mi mc.



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Chng III

Cc chin lc tm kim ti u

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

Vn tm kim ti u, mt cch tng qut, c th pht biu nh sau. Mi i tng x trongkhng gian tm kim c gn vi mt s o gi tr ca i tng f(x), mc tiu ca ta l tmi tng c gi tr f(x) ln nht (hoc nh nht) trong khng gian tm kim. Hm f(x) c gi lhm mc tiu. Trong chng ny chng ta s nghin cu cc thut ton tm kim sau:

Cc k thut tm ng i ngn nht trong khng gian trng thi: Thut ton A*, thut tonnhnh_v_cn.

Cc k thut tm kim i tng tt nht: Tm kim leo i, tm kim gradient, tm kim

m phng luyn kim. Tm kim bt chc s tin ha: thut ton di truyn.

1.8 Tm ng i ngn nht.

Trong cc chng trc chng ta nghin cu vn tm kim ng i t trng thi banu ti trng thi kt thc trong khng gian trng thi. Trong mc ny, ta gi s rng, gi phi tr a trng thi a ti trng thi b (bi mt ton t no ) l mt s k(a,b) 0, ta s gi s ny l di cung (a,b) hoc gi tr ca cung (a,b) trong th khng gian trng thi. di ca cccung c xc nh ty thuc vo vn . Chng hn, trong bi ton tm ng i trong bn giao thng, gi ca cung (a,b) chnh l di ca ng ni thnh ph a vi thnh ph b. ding c xc nh l tng di ca cc cung trn ng i. Vn ca chng ta trong mc

ny, tm ng i ngn nht t trng thi ban u ti trng thi ch. Khng gian tm kim ybao gm tt c cc ng i t trng thi ban u ti trng thi kt thc, hm mc tiu c xcnh y l di ca ng i.

Chng ta c th gii quyt vn t ra bng cch tm tt c cc ng i c th c t trngthi ban u ti trng thi ch (chng hn, s sng cc k thut tm kim m), sau so snh di ca chng, ta s tm ra ng i ngn nht. Th tc tm kim ny thng c gi l th tc

bo tng Anh Quc (British Museum Procedure). Trong thc t, k thut ny khng th p dngc, v cy tm kim thng rt ln, vic tm ra tt c cc ng i c th c i hi rt nhiuthi gian. Do ch c mt cch tng hiu qu tm kim l s dng cc hm nh gi hngdn s tm kim. Cc phng php tm kim ng i ngn nht m chng ta s trnh by u l

cc phng php tm kim heuristic.Gi s u l mt trng thi t ti(c dng i t trng thi ban u u0 ti u). Ta xc nh

hai hm nh gi sau:

g(u) l nh gi di ng i ngn nht t u0 ti u (ng i t u0 ti trng thi u khngphi l trng thi ch c gi l ng i mt phn, phn bit vi ng i y , lng i t u0 ti trng thi ch).

h(u) l nh gi di ng i ngn nht t u ti trng thi ch.

Hm h(u) c gi l chp nhn c (hoc nh gi thp) nu vi mi trng thi u, h(u) di ng i ngn nht thc t t u ti trng thi ch. Chng hn trong bi ton tm ng

i ngn nht trn bn giao thng, ta c th xc nh h(u) l di ng chim bay t u ti ch.



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Ta c th s dng k thut tm kim leo i vi hm nh gi h(u). Tt nhin phng phpny ch cho php ta tm c ng i tng i tt, cha chc l ng i ti u.

Ta cng c th s dng k thut tm kim tt nht u tin vi hm nh gi g(u). Phngphp ny s tm ra ng i ngn nht, tuy nhin n c th km hiu qu.

tng hiu qu tm kim, ta s dng hm nh gi mi :

f(u) = g(u) + h(u)

Tc l, f(u) l nh gi di ng i ngn nht qua u t trng thi ban u ti trng thikt thc.

1.8.1 Thut ton A*

Thut ton A* l thut ton s dng k thut tm kim tt nht u tin vi hm nh gif(u).

thy c thut ton A* lm vic nh th no, ta xt th khng gian trng thi tronghnh 3.1. Trong , trng thi ban u l trng thi A, trng thi ch l B, cc s ghi cnh cccung l di ng i, cc s cnh cc nh l gi tr ca hm h.u tin, pht trin nh A sinhra cc nh con C, D, E v F. Tnh gi tr ca hm f ti cc nh ny ta c:

g(C) = 9, f(C) = 9 + 15 = 24, g(D) = 7, f(D) = 7 + 6 = 13,g(E) = 13, f(E) = 13 + 8 = 21, g(F) = 20, f(F) = 20 +7 = 27

Nh vy nh tt nht l D (v f(D) = 13 l nh nht). Pht trin D, ta nhn c cc nhcon H v E. Ta nh gi H v E (mi):

g(H) = g(D) + di cung (D, H) = 7 + 8 = 15, f(H) = 15 + 10 = 25.

ng i ti E qua D c di:g(E) = g(D) + di cung (D, E) = 7 + 4 = 11.

Vy nh E mi c nh gi l f(E) = g(E) + h(E) = 11 + 8 = 19. Trong s cc nh cho phttrin, th nh E vi nh gi f(E) = 19 l nh tt nht. Pht trin nh ny, ta nhn c cc nh



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con ca n l K v I. Chng ta tip tc qu trnh trn cho ti khi nh c chn pht trin lnh kt thc B, di ng i ngn nht ti B l g(B) = 19. Qu trnh tm kim trn c m t

bi cy tm kim trong hnh 3.2, trong cc s cnh cc nh l cc gi tr ca hm nh gi f(u).

procedureA*;

begin

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

2. loop do

2.1 ifL rngthen

{thng bo tht bi; stop};

2.2 Loi trng thi u u danh sch L;

2.3 ifu l trng thi ch then

{thng bo thnh cng; stop}

2.4 for mi trng thi v k u do

{g(v)g(u) + k(u,v);

f(v) g(v) + h(v);

t v vo danh sch L;}2.5 Sp xp L theo th t tng dn ca hm f sao cho

trng thi c gi tr ca hm f nh nht u danh sch;

end;

Chng ta a ra mt s nhn xt v thut ton A*.

Ngi ta chng minh c rng, nu hm nh gi h(u) l nh gi thp nht (trng hp

c bit, h(u) = 0 vi mi trng thi u) th thut ton A* l thut ton ti u, tc l nghim m ntm ra l nghim ti u. Ngoi ra, nu di ca cc cung khng nh hn mt s dng no th thut ton A* l thut ton y theo ngha rng, n lun dng v tm ra nghim.

Chng ta chng minh tnh ti u ca thut ton A*.

Gi s thut ton dng li nh kt thc G vi di ng i t trng thi ban u u0 tiG l g(G). V G l nh kt thc, ta c h(G) = 0 v f(G) = g(G) + h(G) = g(G). Gi s nghim tiu l ng i t u0 ti nh kt thc G1 vi di l. Gi s ng i ny thot ra khi cy tmkim ti nh l n (Xem hnh 3.3). C th xy ra hai kh nng: n trng vi G1 hoc khng. Nu n l

G1 th v G c chn pht trin trc G1, nn f(G) f(G1), do g(G) g(G1) = l. Nu n



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G1 th do h(u) l hm nh gi thp, nn f(n) = g(n) + h(n) l. Mt khc, cng do G c chn pht trin trc n, nn f(G) f(n), do , g(G) l. Nh vy, ta chng minh c rng dica ng i m thut ton tm ra g(G) khng di hn di l ca ng i ti u. Vy n l di ng i ti u.

Trong trng hp hm nh gi h(u) = 0 vi mi u, thut ton A* chnh l thut ton tmkim tt nht u tin vi hm nh gi g(u) m ta ni n.

Thut ton A* c chng t l thut ton hiu qu nht trong s cc thut ton y v ti u cho vn tm kim ng i ngn nht.

1.8.2 Thut ton tm kim nhnh-v-cn.

Thut ton nhnh_v_cn l thut ton s dng tm kim leo i vi hm nh gi f(u).

Trong thut ton ny, ti mi bc khi pht trin trng thi u, th ta s chn trng thi ttnht v (f(v) nh nht) trong s cc trng thi k u pht trin bc sau. i xung cho ti khigp trng thi v l ch, hoc gp trng thi v khng c nh k, hoc gp trng thi v m f(v) ln

hn di ng i ti u tm thi, tc l ng i y ngn nht trong s cc ng i y m ta tm ra. Trong cc trng hp ny, ta khng pht trin nh v na, hay ni cch khc, tact i cc nhnh cy xut pht t v, v quay ln cha ca v tip tc i xung trng thi tt nhttrong cc trng thi cn li cha c pht trin.

V d: Chng ta li xt khng gian trng thi trong hnh 3.1. Pht trin nh A, ta nhn ccc nh con C, D, E v F, f(C) = 24, f(D) = 13, f(E) = 21, f(F) = 27. Trong s ny D l tt nht,

pht trin D, sinh ra cc nh con H v E, f(H) = 25, f(E) = 19. i xung pht trin E, sinh ra ccnh con l K v I, f(K) = 17, f(I) = 18. i xung pht trin K sinh ra nh B vi f(B) = g(B) = 21.i xung B, v B l nh ch, vy ta tm c ng i ti u tm thi vi di 21. T B quayln K, ri t K quay ln cha n l E. T E i xung J, f(J) = 18 nh hn di ng i tm thi

(l 21). Pht trin I sinh ra cc con K v B, f(K) = 25, f(B) = g(B) = 19. i xung nh B, v nhB l ch ta tm c ng i y mi vi di l 19 nh hn di ng i ti u tmthi c (21). Vy di ng i ti u tm thi by gi l 19. By gi t B ta li quay ln ccnh cn li cha c pht trin. Song cc nh ny u c gi tr hm nh gi ln hn 19, do khng c nh no c pht trin na. Nh vy, ta tm c ng i ti u vi di 19. Cytm kim c biu din trong hnh 3.4.

Thut ton nhnh_v_cn s c biu din bi th tc Branch_and_Bound. Trong th tcny, bin cost c dng lu di ng i ngn nht. Gi tr ban u ca cost l s ln,

hoc di ca mt ng i y m ta bit.



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procedureBranch_and_Bound;

begin

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

Gn gi tr ban u cho cost;

2. loop do

2.1 ifL rngthen stop;

2.2 Loi trng thi u u danh sch L;


ifg(u) y then {y g(y); Quay li 2.1};

2.4 iff(u) > y then Quay li 2.1;

2.5 for mi trng thi v k u do

{g(v)g(u) + k(u,v);

f(v) g(v) + h(v);t v vo danh sch L1};

2.6 Sp xp L1 theo th t tng ca hm f;2.7 Chuyn L1 vou danh sch L sao cho trng thi

u L1 tr thnh u L;

end;

Ngi ta chng minh c rng, thut ton nhnh_v_cn cng l thut ton y v tiu nu hm nh gi h(u) l nh gi thp v c di cc cung khng nh hn mt s dng no .

1.9 Tm i tng tt nht

Trong mc ny chng ta s xt vn tm kim sau. Trn khng gian tm kim U c xcnh hm gi (hm mc tiu) cost, ng vi mi i tng x U vi mt gi tr s cost(x), s nyc gi l gi tr ca x. Chng ta cn tm mt i tng m ti hm gi tr ln nht, ta gi itng l i tng tt nht. Gi s khng gian tm kim c cu trc cho php ta xc nh ckhi nim ln cn ca mi i tng. Chng hn, U l khng gian trng thi th ln cn ca trngthi u gm tt c cc trng thi v k u; nu U l khng gian cc vect thc n-chiu th ln cn cavect x = (x1, x2, ... xn) gm tt c cc vect gn x theo khong cch clit thng thng.

Trong mc ny, ta s xt k thut tm kim leo i tm i tng tt nht. Sau ta s xtk thut tm kim gradient (gradient search). l k thut leo i p dng cho khng gian tmkim l khng gian cc vect thc n-chiu v hm gi l l hm kh vi lin tc. Cui cng ta s

nghin cu k thut tm kim m phng luyn kim( simulated annealing).1.9.1 Tm kim leo i

K thut tm kim leo i tm kim i tng tt nht hon ton ging nh k thut tmkim leo i tm trng thi kt thc xt trong mc 2.3. Ch khc l trong thut ton leo i mc 2.3, t mt trng thi ta "leo ln" trng thi k tt nht (c xc nh bi hm gi), tip tccho ti khi t ti trng thi ch; nu cha t ti trng thi ch m khng leo ln c na, thta tip tc "tt xung" trng thi trc n, ri li leo ln trng thi tt nht cn li. Cn y, tmt nh u ta ch leo ln nh tt nht v (c xc nh bi hm gi cost) trong ln cn u nu nhny "cao hn" nh u, tc l cost(v) > cost(u). Qu trnh tm kim s dng li ngay khi ta khngleo ln nh cao hn c na.



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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.

procedureHill_Climbing;begin

1. umt i tng ban u no ;

2. ifcost(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 hngi 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*) lln 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 thtc leo i xut pht t mt dy cc i tng ban u c chn ngu nhin v lu li nghim ttnht 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 hmgi. Nu mt cong ch c mt s t cc i a phng, th k thut leo i s tm ra rt nhanh cci ton cc. Song c nhng vn m mt cong ca hm gi ta nh lng nhm vy, khi sdng 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) cahm 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 sdng nh xc nh tiu chun dng. Bng cch ly cc bc nh theo hng ca vectgradient chng ta s tm c im cc i a phng, l im m ti f = 0, hoc tmc im rt gn vi cc i a phng.


=xn

,...,2x

,x1

ffff


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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 uton 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 xutpht t cc im khc nhau, ta thc hin mt s bc tt xung nhm thot ra khi cc im cci 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 lncn n. Cn by gi, trong tm kim m phng luyn kim, ta chn ngu nhin mt trng thi vtrong ln cn u. Nu trng thi v c chn tt hn u (cost(v) > cost(u)) th ta i ti v, cn nukhng ta ch i ti v vi mt xc sut no . Xc sut ny gim theo hm m ca xu catrng thi v. Xc sut ny cn ph thuc vo tham s nhit T. Nhit T cng cao th bc iti trng thi xu cng c kh nng c thc hin. Trong qu trnh tm kim, tham s nhit Tgim dn ti khng. Khi T gn khng, thut ton hot ng gn ging nh leo i, hu nh nkhng 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

t0;

u trng thi ban u no ;

Tnhit ban u;

repeat

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

ifcost(v) > cost(u) then u v

else u v vi xc sut e/T;

Tg(T, t);

tt + 1;

until T nh

end;

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 cngcho cc bi ton ti u c ln.

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

Thut 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 vnhn 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, cutrc gien ca n mang mt phn cu trc gien ca cha v m. Ngoi ra, trong qu trnh tin ha, cth xy ra hin tng t bin, cu trc gien ca c th con c th cha cc gien m c cha v mu khng c.



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Trong TTDT, mi c th c m ha bi mt cu trc d liu m t cu trc gien ca cth , ta s gi n l nhim sc th (chroniosome). Mi nhim sc th c to thnh t cc nv 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 onpht trin s c gi l mt th h. T th h ban u c to ra, TTDT bt chc chn lc tnhin v di truyn bin i cc th h. TTDT s dng cc ton t c bn sau y bin icc th h.

Ton t ti sinh (reproduction) (cn c gi l ton t chn lc (selection)). Cc c thtt c chn lc a vo th h sau. S la chn ny c thc hin da vo thch nghivi mi trng ca mi c th. Ta s gi hm ng mi c th vi thch nghi ca n l hmthch 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 TTDTl nh sau:

procedure Genetic_Algorithm;

begin

t0;

Khi to th h ban u P(t);

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

repeattt + 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. Khithut 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 thchnghi 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. Vimi c th xi, ta tnh thch nghi ca n f(xi). Tnh tng cc thch nghi ca tt c cc c thtrong qun th:


=

=n

1i

f(xi)F


31/58


Mi ln chn lc, ta thc hin hai bc sau:Sinh ra mt s thc ngu nhin q trong khong (0, F);xkl 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 chiathnh n phn, mi phn ng vi thch nghi ca mt c th (hnh 3.5). Mt mi tn ch vo bnhxe. 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 nngc chn. Cc c th c thch nghi cao c th c mt hay nhiu bn sao, cc c th c thchnghi 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 raxc sut lai ghp pc. xc sut ny cho ta hy vng c pc.n c th c lai ghp (n l c ca qunth).

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. tbin l thay i trng thi mt s gien no trong nhim sc th. Mi gien chu t bin vi xcsut 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.


= k

ixif

14)(


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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 lica 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]. sdng 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) = x 2.Qun th ban u gm 4 c th (c ca qun th l n = 4). Thc hin qu trnh chn lc, ta nhnc kt qu trong bng sau. Trong bng ny, ta thy c th 2 c thch nghi cao nht (576) nnn c chn 2 ln, c th 3 c thch nghi thp nht (64) khng c chn ln no. Mi c th1 v 4 c chn 1 ln.

Bng kt qu chn lc

S liuc th

Qun thban u

x thch nghif(x) = x2

S ln cchn

1 0 1 1 0 1 13 169 1

2 1 1 0 0 0 24 576 2

3 0 1 0 0 0 8 64 0

4 1 0 0 1 1 19 361 1

Thc hin qa trnh lai ghp vi xc sut lai ghp pc = 1, c 4 c th sau chn lc u clai ghp. Kt qu lai ghp c cho trong bng sau. Trong bng ny, chui th nht c lai ghpvi 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 chnlc imghp Qun th sau laighp x thch nghif(x) = x2

0 1 1 0 | 1 4 0 1 1 0 0 2 144

1 1 0 0 | 0 4 1 1 0 0 1 5 625

1 1 | 0 0 0 2 1 1 0 1 1 7 729

1 0 | 0 1 1 2 1 0 0 0 0 6 256

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



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l qun th sau lai ghp. Trong th h ban u, thch nghi cao nht l 576, thch nghi trungbnh 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 hmthc 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 vohm mc tiu.

Trn c s cu trc ca nhim sc th, thit k cc ton t di truyn (lai ghp, t bin) choph 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 l0,03. 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 thct, 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 nhcc ton t di truyn. Qu trnh sinh ra th h mi P(t) t th h c P(t - 1) cng c nhiu cchchn 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 susc hn v thut ton tin ho, bn c c th tm c [ ], [ ] v [ ] . [ ] v [ ] c xem l cc schhay nht vit v TTDT. [ ] cho ta ci nhn tng qut v s pht trin gn y ca TTDT.



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Chng IV

Tm kim c i th----------------------------

Nghin cu my tnh chi c xut hin rt sm. Khng lu sau khi my tnh lp trnhc ra i vo nm 1950, Claude Shannon vit chng trnh chi c u tin. cc nh nghincu Tr Tu Nhn To nghin cu vic chi c, v rng my tnh chi c l mt bng chngr 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 thca anh ta c gi l en. Mc tiu ca chng ta l nghin cu chin lc chn nc i choTrng (My tnh cm qun Trng).

Chng ta s xt cc tr chi hai ngi vi cc c im sau. Hai ngi chi thay phin nhaua ra cc nc i tun theo cc lut i no , cc lut ny l nh nhau cho c hai ngi. inhnh l c vua, trong c vua hai ngi chi c th p dng cc lut i con tt, con xe, ... a ranc 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 nhtrong chi bi, ngi chi khng th bit cc ngi chi khc cn nhng con bi g). Vn chic c th xem nh vn tm kim nc i, ti mi ln n lt mnh, ngi chi phi tm trongs rt nhiu nc i hp l (tun theo ng lut i), mt nc i tt nht sao cho qua mt dync i thc hin, anh ta ginh phn thng. Tuy nhin vn tm kim y s phc tp hnvn tm kim m chng ta xt trong cc chng trc, bi v y c i th, ngi chikhng bit c i th ca mnh s i nc no trong tng lai. Sau y chng ta s pht biuchnh xc hn vn tm kim ny.

Vn chi c c th xem nh vn tm kim trong khng gian trng thi. Mi trng thil 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 siu 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) hocha. Do , ta c th xc nh hm kt cuc l hm nhn gi tr 1 ti cc trng thi kt thc lthng (i vi Trng), -1 ti cc trng thi kt thc l thua (i vi Trng) v 0 ti cc trng thikt thc ha. Trong mt s tr chi khc, chng hn tr chi tnh im, hm kt cuc c th nhngi tr nguyn trong khong [-k, k] vi k l mt s nguyn dng no .



35/58


Nh vy vn ca Trng l, tm mt dy nc i sao cho xen k vi cc nc i ca ento 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 trngthi 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 nhng 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 nhnc t u do Trng (en) thc hin nc i hp l no . Do , trn cng mt mc ca cy ccnh 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 qunen, 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 khibn c. Ai a hai qun ca mnh ra khi bn c trc s thng, hoc to ra tnh th bt iphng 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 hinmt 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 mTrng (en) va chn, en (Trng) s phi chn i ti mt trong cc nh Trng (en) w l conca v. Qu trnh trn s dng li khi t ti mt nh l l ca cy.



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Gi s Trng cn tm nc i ti nh u. Nc i ti u cho Trng l nc i dn ti nhcon ca v l nh tt nht (cho Trng) trong s cc nh con ca u. Ta cn gi thit rng, n lti th chn nc i t v, en cng s chn nc i tt nht cho anh ta. Nh vy, chn nci 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 cngln cng tt cho Trng, nh c gi tr cng nh cng tt cho en. xc nh gi tr cc nh cacy tr chi gc u, ta i t mc thp nht ln gc u. Gi s v l nh trong ca cy v gi tr ccnh con ca n c xc nh. Khi nu v l nh Trng th gi tr ca n c xc nh lgi tr ln nht trong cc gi tr ca cc nh con. Cn nu v l nh en th gi tr ca n l gi trnh 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 ghicnh mi nh. nh i l Trng, nn gi tr ca n l max(3,-2) = 3, nh d l nh en, nn gi trca n l min(2, 3, 4) = 2.

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

functionMaxVal(u);

begin

ifu l nh kt thc thenMaxVal(u) f(u)

elseMaxVal(u) max{MinVal(v) | v l nh con ca u}

end;

functionMinVal(u);begin

ifu l nh kt thc thenMinVal(u) f(u)

elseMinVal(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 thtc chn nc i cho trng ti nh u. Trong th tc Minimax(u,v), v l bin lu li trng thi mTrng chn i ti t u.

procedure Minimax(u, v);

begin



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val-;

for mi w l nh con ca u do

ifval


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V d 2: By gi ta a ra mt cch nh gi cc trng thi trong tr chi Dodgem. Miqun Trng mt v tr trn bn c c cho mt gi tr tng ng trong bng bn tri hnh 4.4.Cn mi qun en mt v tr s c cho mt gi tr tng ng trong bng bn phi hnh 4.4:

Ngoi ra, nu qun Trng cn trc tip mt qun en, n c thm 40 im, nu cn gintip n c thm 30 im (Xem hnh 4.5). Tng t, nu qun en cn trc tip qun Trng n

c thm -40 im, cn cn gin tip n c thm -30 im.

p dng cc qui tc trn, ta tnh c gi tr ca trng thi bn tri hnh 4.6 l 75, gi trca trng thi bn phi hnh v l -5.

Trong cnh nh gi trn, ta xt n v tr ca cc qun v mi tng quan gia cc qun.Mt cch n gin hn ch khng gian tm kim l, khi cn xc nh nc i cho Trngti u, ta ch xem xt cy tr chi gc u ti cao h no . p dng th tc Minimax cho cy trchi gc u, cao h v s dng gi tr ca hm nh gi cho cc l ca cy , chng ta s tmc nc i tt cho Trng ti u.

1.13 Phng php ct ct alpha - beta

Trong chin lc tm kim Minimax, tm kim nc i tt cho Trng ti trng thi u, chod ta hn ch khng gian tm kim trong phm vi cy tr chi gc u vi cao h, th s nh cacy tr chi ny cng cn rt ln vi h 3. Chng hn, trong c vua, nhn t nhnh trong cy trchi trung bnh khong 35, thi gian i hi phi a ra nc i l 150 giy, vi thi gian ny trnmy tnh thng thng chng trnh ca bn ch c th xem xt cc nh trong su 3 hoc 4.Mt ngi chi c trnh trung bnh cng c th tnh trc c 5, 6 nc hoc hn na, v do chng trnh ca bn mi t trnh ngi mi tp chi!



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Khi nh gi nh u ti su h, mt thut ton Minimax i hi ta phi nh gi tt c ccnh ca cy gc u ti su h. Song ta c th gim bt s nh cn phi dnh gi m vn khngnh hng g n s nh gi nh u. Phng php ct ct alpha-beta cho php ta ct b cc nhnhkhng cn thit cho s nh gi nh u.

T tng ca k thut ct ct alpha-beta l nh sau: Nh li rng, chin lc tm kimMinimax l chin lc tm kim theo su. Gi s trong qu trnh tm kim ta i xung nh a lnh Trng, nh a c ngi anh em v c nh gi. Gi s cha ca nh a l b v b c ngianh em u d c nh gi, v gi s cha ca b l c (Xem hnh 4.7). Khi ta c gi tr nh c(nh Trng) t nht l gi tr ca u, gi tr ca nh b (nh en) nhiu nht l gi tr v. Do , nueval(u) > eval(v), ta khng cn i xung nh gi nh a na m vn khng nh hng g dnnh gi nh c. Hay ni cch khc ta c th ct b cy con gc a. Lp lun tng t cho trnghp 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 nhhin 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 camt nh en. Gi tr ca v s c cp nht trong qu trnh tm kim. v c s dngnh 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).

functionMaxVal(u, , );

begin

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

then MaxValeval(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

ifthen exit};

MaxVal;

end;

functionMinVal(u, , );

begin

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

thenMinValeval(u)



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else for mi nh v l con ca u do

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

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

ifthen exit};MinVal;

end;

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

procedureAlpha_beta(u,v);

begin

-;

;

for mi nh w l con ca u doif 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 cnhcc l l gi tr ca hm nh gi. p dng chin lc Minimax v k thut ct ct, ta xc nhc nc i tt nht cho Trng ti u, l nc i dn ti nh v c gi tr 10. Cnh mi nh tacng 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:



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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 trithc. 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.



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5.1. Biu din tri thc

Con ngi sng trong mi trng c th nhn thc c th gii nh cc gic quan (tai, mtv cc b phn khc), s dng cc tri thc tch lu c v nh kh nng lp lun, suy din, conngi c th a ra cc hnh ng hp l cho cng vic m con ngi ang lm. Mt mc tiuca Tr tu nhn to ng dng l thit k cc tc nhn thng minh (intelligent agent) cng c khnng nh con ngi. Chng ta c th hiu tc nhn thng minh l bt c ci g c th nhn thcc mi trng thng qua cc b cm nhn (sensors) v a ra hnh ng hp l p ng li mitrng thng qua b phn hnh ng (effectors). Cc robots, cc softbot (software robot), cc hchuyn gia,... l cc v d v tc nhn thng minh. Cc tc nhn thng minh cn phi c tri thc vth 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 nhnc cc thng tin a vo, tc nhn cn c kh nng suy din a ra cc cu tr li, cc hnhng hp l, ng n. Nhim v ny c thc hin bi b suy din. B suy din l thnh phnc bn khc ca cc h tri thc. Nh vy h tri thc bo tr mt CSTT v c trang b mt thtc suy din. Mi khi tip nhn c cc s kin t mi trng, th tc suy din thc hin qutrnh lin kt cc s kin vi cc tri thc trong CSTT rt ra cc cu tr li, hoc cc hnh nghp 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 dngc tri thc, c th x l tri thc, chng ta cn biu din tri thc di dng thun tin cho mytnh. 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 thxem nh s m ha ca mt s hiu bit ca chng ta v th gii hin thc. Ngn ng biu din trithc (cng nh mi ngn ng hnh thc khc) gm hai thnh phn c bn l c php v ngngha.

C php ca mt ngn ng bao gm cc k hiu v cc quy tc lin kt cc k hiu (cclut 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 liuthch hp c ci t trong mt vng nh no ca my tnh, l biu din bn trong. Bnthn 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 lmt cu vit ng c php. Ng ngha ca ngn ng ny cho php ta hiu rng, nu x, y, z, ngvi 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 nhnvi s nguyn z.

Ngoi hai thnh phn c php v ng ngha, ngn ng biu din tri thc cn c cungcp c ch suy din. Mt lut suy din (rule of inference) cho php ta suy ra mt cng thc t mttp 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 hocsuy din l mt qu trnh p dng cclut 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:



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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 cckt lun, th tc suy din i hi t thi gian tnh ton v t khng gian nh. Ngi ta cngmong 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-orderpredicate logic hoc first-order predicate calculus) - mt ngn ng biu din tri thc, bi vlogic v t cp mt c kh nng biu din tng i tt, v hn na n l c s cho nhiungn ng biu din tri thc khc, chng hn ton hon cnh (situation calculus) hoc logicthi gian khong cp mt (first-order interval tempral logic). Nhng trc ht chng ta snghin cu logic mnh (propositional logic hoc propositional calculus). N l ngn ngrt n gin, c kh nng biu din hn ch, song thun tin cho ta a vo nhiu khi nimquan 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.



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Cc cng thc l cc k hiu mnh s c gi l cc cu n hoc cu phn t. Cccng 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...Amtrong 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 thctrong th gii hin thc no . iu c thc hin bng cch kt hp mnh vi s kinno 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 vicc s kin trong th gii thc c gi l mt minh ha (interpretation ). Chng hn minh haca 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 SPi 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 gitr 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 lng v Q l ng th cu P ko theo Q l ng, cn khi P l ng Q l sai th cu P ko theoQ l sai. Nhng nu P sai v Q ng , hoc P sai Q sai th P ko theo Q l ng hay sai ? Nuchng ta xut pht t gi thit sai, th chng ta khng th khng nh g v kt lun. Khng c l



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do g ni rng, nu P sai v Q ng hoc P sai v Q sai th P ko theo Q l sai. Do trongtrng 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


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True True False True False

True True True True True

Hnh 5.2 Bng chn l cho cng thc (P=>Q) S

Cn lu rng, mt cng thc cha n bin, th s cc minh ha ca n l 2n , tc l bngchn l c 2n dng. Nh vy vic kim tra mt cng thc c tho c hay khng bng phng

php bng chn l, i hi thi gian m. Cook (1971) chng minh rng, vn kim tra mtcng thc trong logic mnh c tho c hay khng l vn NP-y .

Chng ta s ni rng (tho c, khng tho c) nu hi ca chng G1.......Gm l vngchc (tho c, khng tho c). Mt m hnh ca tp cng thc G l m hnh ca tp cng

thc G1.......Gm .

5.3 Dng chun tc

Trong mc ny chng ta s xt vic chun ha cc cng thc, a cc cng thc v dngthun li cho vic lp lun, suy din. Trc ht ta s xt cc php bin i tng ng. S dngcc php bin i ny, ta c th a mt cng thc bt k v cc dng chun tc.

5.3.1 S tng ng ca cc cng thc

Hai cng thc A v B c xem l tng ngnu chng c cng mt gi tr chn l

trong mi minh ha. ch A tng ng vi B ta vit A B bng phng php bng chn l,d dng chng minh c s tng ng ca cc cng thc sau y :

A=>B lA v B A< = > 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)



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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 cccng 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 mtcng 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 (lRS) ((lP v Q)vlR) ( (lP v Q) v S) (l P v Q vlR) (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 vlR) (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 cutuyn.

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 culP1 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



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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 trthnh dng lP1 v......v lPm hay tng ng l(P1^...^ Pm ). Cn ch rng, khng phi mi cngthc u c th biu din di dng hi ca cc cu Horn. Tuy nhin trong cc ng dng, c s trithc thng l mt tp no cc cu Horn (tc l mt tp no cc lut if-then).

5.4 Lut suy din

Mt 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 nicch 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 thcmi m n l h qu logic ca cc cng thc trong c s tri thc. iu c thc hin bngcc thc hin cc lut suy din (rule of inference). Lut suy din ging nh mt th tc m chngta 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 cngthc 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 , i ,, l cc cng thc :

1. Lut Modus Ponens

=>,

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

2. Lut Modus Tollens

=>,ll

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

3. Lut bc cu

=>,=>

=>



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T hai ko theo, m kt lun ca n l ca ko theo th nht trng vi gi thit ca ko theoth hai, ta suy ra ko theo mi m gi thit ca n l gi thit ca ko theo th nht, cn kt lunca n l kt lun ca ko theo th hai.

4. Lut loi b hi

1.......i........mi

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

5. Lut a vo hi

1,.......,i,........m1.......i....... m

T mt danh sch cc cng thc, ta suy ra hi ca chng.

6. Lut a vo tuyn

i1v.......vi.v.......vm

T mt cng thc, ta suy ra mt tuyn m mt trong cc hng t ca cc tuyn l cng thc.

7. Lut gii

v ,l v

v T 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 thitca 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 ul 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 v , l v ng th kt lun v cng ng.

Do lut gii l lut suy in tin cy.

v l v v

False False False False True False

False False True False True True

False True False True False False

False True True True True True



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True False False True True True

True False True True True True

True True False True False TrueTrue True True True True True

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 ciu ).

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 cngthc c suy ra cng thc mi bng mt dy p dng cc lut suy din. Cc cng thc choc gi l cc tin . Cc cng thc c suy ra c gi l cc nh l. Dy cc lut c pdng 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 lutsuy din ngi ta thit k ln cc th tc suy din (cn c gi l th tc chng minh) t cctri 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 dinno (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



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Phng php chng minh bc b (refutation proof hoc proof by contradiction) l mtphng php thng xuyn c s dng trong cc chng minh ton hc. T tng ca phngphp ny l nh sau : chng minh P ng, ta gi s P sai ( thm P vo cc gi thit ) v dnti 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 cngthc H l h qu logic ca G. iu tng ng vi chng minh cng thc G1^....^Gm -> Hl 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 cnu t Gta suy ra hai mnh i lp nhau. Vic chng minh cng thc H l h qu logic catp 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 cuc bit quan trng.

*0 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 Bn

trong Ai, Bj v C l cc literal.

*1 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 cv kt qu nhn c khi p dng lut gii cho hai cu c gi l gii thc ca chng. Giithc ca hai cu A v B c k hiu l res(A,B). Chng hn, hai cu tuyn gii c nu mt cucha 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.



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Gi s G l mt tp cc cu tuyn ( Bng cch chun ho ta c th a mt tp cc cngthc v mt tp cc cu tuyn ). Ta s k hiu R(G ) l tp cu bao gm cc cu thuc G v tt ccc 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 suyt 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 thac. 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 chay khng . Th tc ny c gi l th tc gii.

procedure Resolution ;

Input : tp G cc cu tuyn ;begin

1.Repeat

1.1 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 cuca G l hu hn. Do s cc cu tuyn thnh lp c t cc literal l hu hn. V vy chc 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



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h qu ca mt tp cng thc cho hay khng bng phng php chng minh bc b. V vylut 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;

Begin

1. 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 nhnc cu rng []. Vy P l h qu logic ca cc cu (1) --(6).



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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 rta dng v phong ph. Chng ta c th lit k ra rt nhiu v d v i tng, tnh cht, quan h.

*2 i tng : mt ci bn, mt ci nh, mt ci cy, mt con ngi, mt con s. ...*3 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. ..*4 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 )...*5 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 cci 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 thuctnh 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 trongmt 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 khnng biu din ca n ( n cho php ta biu din tri thc v th gii vi cc i tng, cc thuctnh ca i tng v cc quan h ca i tng), v hn na, n l c s cho nhiu ngn nglogic khc.



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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 tmt 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 ( n 1). 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 theonhau).

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 thc

Cc 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( t 1, t2, ..., tn) l hngthc. 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 hngthc c th.

1.21 Cc cng thc phn t

Chng ta s biu din cc tnh cht ca i tng, hoc cc quan h ca i tng bi cccng 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 (

giao trinh tri tue nhan tao dhqg ha noi

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