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8/12/2019 ti liu thut ton tr tu nhn to

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Cc thut ton v

gii thut

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CHNG 1 : THUT TON THUT GII

I. KHI NIM THUT TON THUT GII

II. THUT GII HEURISTIC

III. CC PHNG PHP TM KIM HEURISTIC

III.1. Cu trc chung ca bi ton tm kim

III.2. Tm kim chiu su v tm kim chiu rng

III.3. Tm kim leo i

III.4. Tm kim u tin ti u (best-first search)

III.5. Thut gii AT

III.6. Thut gii AKT

III.7. Thut gii A*

III.8. V dminh ha hot ng ca thut gii A*

III.9. Bn lun vA*

III.10. ng dng A* gii bi ton Ta-canh

III.11. Cc chin lc tm kim lai

I. TNG QUAN THUT TON THUT GII

Trong qu trnh nghin cu gii quyt cc vn bi ton, ngi ta a ra nhngnhn xt nhsau:

C nhiu bi ton cho n nay vn cha tm ra mt cch gii theo kiu thut ton

v cng khng bit l c tn ti thut ton hay khng.

C nhiu bi ton c thut ton gii nhng khng chp nhn c v thigian gii theo thut ton qu ln hoc cc iu kin cho thut ton kh png.

C nhng bi ton c gii theo nhng cch gii vi phm thut ton nhng vnchp nhn c.

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Tnhng nhn nh trn, ngi ta thy rng cn phi c nhng i mi cho khi nimthut ton. Ngi ta mrng hai tiu chun ca thut ton: tnh xc nh v tnh ngn. Vic mrng tnh xc nh i vi thut ton c thhin qua cc gii thut quy v ngu nhin. Tnh ng ca thut ton by gikhng cn bt buc i vi mt scch gii bi ton, nht l cc cch gii gn ng. Trong thc tin c nhiu trng hp

ngi ta chp nhn cc cch gii thng cho kt qutt (nhng khng phi lc no cngtt) nhng t phc tp v hiu qu. Chng hn nu gii mt bi ton bng thut ton tiu i hi my tnh thc hin nhiu nm th chng ta c thsn lng chp nhn mt giiphp gn ti u m chcn my tnh chy trong vi ngy hoc vi gi.

Cc cch gii chp nhn c nhng khng hon ton p ng y cc tiu chun cathut ton thng c gi l cc thut gii. Khi nim mrng ny ca thut ton mca cho chng ta trong vic tm kim phng php gii quyt cc bi ton ct ra.

Mt trong nhng thut gii thng c cp n v sdng trong khoa hc tr tu

nhn to l cc cch gii theo kiu Heuristic

II. THUT GII HEURISTIC

Thut gii Heuristic l mt smrng khi nim thut ton. N thhin cch gii biton vi cc c tnh sau:

Thng tm c li gii tt (nhng khng chc l li gii tt nht)

Gii bi ton theo thut gii Heuristic thng ddng v nhanh chnga ra kt quhn so vi gii thut ti u, v vy chi ph thp hn.

Thut gii Heuristic thng thhin kh tnhin, gn gi vi cch suynghv hnh ng ca con ngi.

C nhiu phng php xy dng mt thut gii Heuristic, trong ngi ta thngda vo mt snguyn l cbn nhsau:

Nguyn l vt cn thng minh:Trong mt bi ton tm kim no , khikhng gian tm kim ln, ta thng tm cch gii hn li khng gian tm kimhoc thc hin mt kiu d tm c bit da vo c th ca bi ton nhanhchng tm ra mc tiu.

Nguyn l tham lam (Greedy):Ly tiu chun ti u (trn phm vi ton cc)ca bi ton lm tiu chun chn la hnh ng cho phm vi cc bca tngbc (hay tng giai on) trong qu trnh tm kim li gii.

Nguyn l tht:Thc hin hnh ng da trn mt cu trc ththp lca khng gian kho st nhm nhanh chng t c mt li gii tt.

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Hm Heuristic:Trong vic xy dng cc thut gii Heuristic, ngi ta thngdng cc hm Heuristic. l cc hm nh gi th, gi trca hm phthucvo trng thi hin ti ca bi ton ti mi bc gii. Nhgi trny, ta c thchn c cch hnh ng tng i hp l trong tng bc ca thut gii.

Bi ton hnh trnh ngn nht ng dng nguyn l Greedy

Bi ton: Hy tm mt hnh trnh cho mt ngi giao hng i qua n im khc nhau, miim i qua mt ln v trvim xut pht sao cho tng chiu di on ng cn i lngn nht. Gisrng c con ng ni trc tip tgia hai im bt k.

Tt nhin ta c thgii bi ton ny bng cch lit k tt ccon ng c thi, tnhchiu di ca mi con ng ri tm con ng c chiu di ngn nht. Tuy nhin,cch gii ny li c phc tp 0(n!) (mt hnh trnh l mt hon vca n im, do ,tng shnh trnh l slng hon vca mt tp n phn tl n!). Do , khi si ltng th scon ng phi xt stng ln rt nhanh.

Mt cch gii n gin hn nhiu v thng cho kt qutng i tt l dng mt thutgii Heuristic ng dng nguyn l Greedy. Ttng ca thut gii nhsau:

Tim khi u, ta lit k tt cqung ng tim xut pht cho n n il ri chn i theo con ng ngn nht.

Khi i n mt i l, chn i n i l ktip cng theo nguyn tc trn.Ngha l lit k tt ccon ng ti l ta ang ng n nhng i l cha in. Chn con ng ngn nht. Lp li qu trnh ny cho n lc khng cn il no i.

Bn c thquan st hnh sau thy c qu trnh chn la. Theo nguyn l Greedy, taly tiu chun hnh trnh ngn nht ca bi ton lm tiu chun cho chn la cc b. Tahy vng rng, khi i trn n on ng ngn nht th cui cng ta sc mt hnh trnhngn nht.iu ny khng phi lc no cng ng. Vi iu kin trong hnh tip theo ththut gii cho chng ta mt hnh trnh c chiu di l 14 trong khi hnh trnh ti u l 13.Kt quca thut gii Heuristic trong trng hp ny chlch 1 n vso vi kt qutiu. Trong khi , phc tp ca thut gii Heuristic ny chl 0(n2).

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Hnh : Gii bi ton sdng nguyn l Greedy

Tt nhin, thut gii theo kiu Heuristic i lc li a ra kt qukhng tt, thm ch rttnhtrng hp hnh sau.

Bi ton phn vic ng dng ca nguyn l tht

Mt cng ty nhn c hp ng gia cng m chi tit my J1, J2, Jm. Cng ty c n mygia cng ln lt l P1, P2, Pn. Mi chi tit u c thc gia cng trn bt kmyno. Mt khi gia cng mt chi tit trn mt my, cng vistip tc cho n lc honthnh, khng thbct ngang. gia cng mt vic J1trn mt my bt kta cn dngmt thi gian tng ng l t1. Nhim vca cng ty l phi lm sao gia cng xong tonbn chi tit trong thi gian sm nht.

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Chng ta xt bi ton trong trng hp c 3 my P1, P2, P3v 6 cng vic vi thi gian lt1=2, t2=5, t3=8, t4=1, t5=5, t6=1. ta c mt phng n phn cng (L) nhhnh sau:

Theo hnh ny, ti thi im t=0, ta tin hnh gia cng chi tit J2trn my P1, J5trn P2v J1ti P3. Ti thi im t=2, cng vic J1c hon thnh, trn my P3ta gia cng tipchi tit J4. Trong lc , hai my P1v P2 vn ang thc hin cng vic u tin mnh Sphn vic theo hnh trn c gi l lc GANTT. Theo lc ny, ta thythi gian hon thnh ton b6 cng vic l 12. Nhn xt mt cch cm tnh ta thyrng phng n (L) va thc hin l mt phng n khng tt. Cc my P1v P2c qunhiu thi gian rnh.

Thut ton tm phng n ti u L0cho bi ton ny theo kiu vt cn c phc tp cO(mn) (vi m l smy v n l scng vic). By gita xt n mt thut gii Heuristicrt n gin (phc tp O(n)) gii bi ton ny.

Sp xp cc cng vic theo thtgim dn vthi gian gia cng.

Ln lt sp xp cc vic theo tht vo my cn dnhiu thi giannht.

Vi ttng nhvy, ta sc mt phng n L* nhsau:

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R rng phng n L* va thc hin cng chnh l phng n ti u ca trng hp nyv thi gian hon thnh l 8, ng bng thi gian ca cng vic J3. Ta hy vng rng mtgii Heuristic n gin nhvy sl mt thut gii ti u. Nhng tic thay, ta ddnga ra c mt trng hp m thut gii Heuristic khng a ra c kt quti u.

Nu gi T* l thi gian gia cng xong n chi tit my do thut gii Heuristic a ra vT0l thi gian ti u th ngi ta chng minh c rng

, M l smy

Vi kt quny, ta c thxc lp c sai sm chng ta phi gnh chu nu dng

Heuristic thay v tm mt li gii ti u. Chng hn vi smy l 2 (M=2) ta c ,v chnh l sai scc i m trng hp trn gnh chu. Theo cng thc ny, smy cng ln th sai scng ln.

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ny sang trng thi Tkc biu din di dng cc con snm trn cung ni gia hai nt tng trng chohai trng thi.

a scc bi ton thuc dng m chng ta ang m tu c thc biu din didng th. Trong , mt trng thi l mt nh ca th. Tp hp S bao gm tt ccc trng thi chnh l tp hp bao gm tt cnh ca th. Vic bin i ttrng thiTi-1sang trng thi Ti l vic i tnh i din cho Ti-1sang nh i din cho Titheocung ni gia hai nh ny.

III.2. Tm kim chiu su v tm kim chiu rng

bn c c thhnh dung mt cch cthbn cht ca thut gii Heuristic, chng tanht thit phi nm vng hai chin lc tm kim cbn l tm kim theo chiu su(Depth First Search) v tm kim theo chiu rng (Breath First Search). Sdchng tadng tchin lc m khng phi lphng phpl bi v trong thc t, ngi ta hunhchng bao givn dng mt trong hai kim tm kim ny mt cch trc tip mkhng phi sa i g.

III.2.1. Tm kim chiu su (Depth-First Search)

Trong tm kim theo chiu su, ti trng thi (nh) hin hnh, ta chn mt trng thi ktip (trong tp cc trng thi c thbin i thnh ttrng thi hin ti) lm trng thihin hnh cho n lc trng thi hin hnh l trng thi ch. Trong trng hp ti trngthi hin hnh, ta khng thbin i thnh trng thi ktip th ta squay lui (back-tracking) li trng thi trc trng thi hin hnh (trng thi bin i thnh trng thi hinhnh) chn ng khc. Nu trng thi trc ny m cng khng thbin i cna th ta quay lui li trng thi trc na v cth. Nu quay lui n trng thi khiu m vn tht bi th kt lun l khng c li gii. Hnh nh sau minh ha hot ngca tm kim theo chiu su.

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Hnh : Hnh nh ca tm kim chiu su. N chlu "mrng" trng thi c chn m khng "mrng" cc trng thi khc (nt mu trng trong hnh v).

III.2.2. Tm kim chiu rng (Breath-First Search)

Ngc li vi tm kim theo kiu chiu su, tm kim chiu rng mang hnh nh ca vtdu loang. Ttrng thi ban u, ta xy dng tp hp S bao gm cc trng thi ktip(m ttrng thi ban u c thbin i thnh). Sau , ng vi mi trng thi Tk trongtp S, ta xy dng tp Sk bao gm cc trng thi ktip ca Tkri ln lt bsung cc

Sk vo S. Qu trnh ny clp li cho n lc S c cha trng thi kt thc hoc S khngthay i sau khi bsung tt cSk.

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Hnh : Hnh nh ca tm kim chiu rng. Ti mt bc, mi trng thi u c mrng, khng bst trng thi no.

Chiu su Chiu rng

Tnh hiu qu Hiu qukhi li gii nm su trongcy tm kim v c mt phng nchn hng i chnh xc. Hiu quca chin lc phthuc vo

phng n chn hng i. Phngn cng km hiu quth hiu quca chin lc cng gim. Thunli khi mun tm chmt li gii.

Hiu qukhi li gii nmgn gc ca cy tm kim.Hiu quca chin lc

phthuc vo su cali gii. Li gii cng xagc th hiu quca chinlc cng gim. Thun likhi mun tm nhiu ligii.

Lng bnhsdng

lu trcc trng thi

Chlu li cc trng thi cha xt

n.

Phi lu ton bcc trng

thi.

Trng hp xu nht Vt cn ton b Vt cn ton b.

Trng hp tt nht Phng n chn hng i tuyt ichnh xc. Li gii c xc nhmt cch trc tip.

Vt cn ton b.

Tm kim chiu su v tm kim chiu rng u l cc phng php tm kim c hthngv chc chn tm ra li gii. Tuy nhin, do bn cht l vt cn nn vi nhng bi ton ckhng gian ln th ta khng thdng hai chin lc ny c. Hn na, hai chin lcny u c tnh cht "m qung" v chng khng ch n nhng thng tin (tri thc) trng thi hin thi v thng tin vch cn t ti cng mi quan hgia chng. Cc trithc ny v cng quan trng v rt c ngha thit kcc thut gii hiu quhn mta sp sa bn n.

III.3. Tm kim leo i

III.3.1. Leo i n gin

Tm kim leo i theo ng ngha, ni chung, thc cht chl mt trng hp c bitca tm kim theo chiu su nhng khng thquay lui. Trong tm kim leo i, vic lachn trng thi tip theo c quyt nh da trn mt hm Heuristic.

Hm Heuristic l g ?

Thut ng"hm Heuristic" mun ni ln iu g? Chng c g gh gm. Bn quen vin ri! n gin chl mt c lng vkhnng dn n li gii tnh ttrng thi (khong cch gia trng thi hin ti v trng thi ch). Ta squy c gi hm ny lhtrong sut gio trnh ny. i lc ta cng cp n chi ph ti u thc stmttrng thi dn n li gii. Thng thng, gi trny l khng thtnh ton c (v tnh

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c ng ngha l bit con ng n li gii !) m ta chdng n nhmt cssuy lun vmt l thuyt m thi ! Hm h, ta quy c rng, lun trra kt qul mt skhng m. bn c thc snm c ngha ca hai hm ny, hy quan st hnh sautrong minh ha chi ph ti u thc sv chi ph c lng.

Hnh Chi ph c lng h = 6 v chi ph ti u thc sh = 4+5 = 9 (i theo ng 1-3-7)

Bn ang trong mt thnh phxa lm khng c bn trong tay v ta mun i vokhu trung tm? Mt cch suy nghn gin, chng ta snhm vo hng nhng ta caoc ca khu trung tm!

Ttng

1)Nu trng thi bt u cng l trng thi ch th thot v bo l tm c li gii.Ngc li, t trng thi hin hnh (Ti) l trng thi khi u (T0)

2) Lp li cho n khi t n trng thi kt thc hoc cho n khi khng tn ti mttrng thi tip theo hp l(Tk) ca trng thi hin hnh :

a. t Tk l mt trng thi tip theo hp lca trng thi hin hnh Ti.

b. nh gi trng thi Tk mi :

b.1.Nu l trng thi kt thc th trvtrny v thot.

b.2.Nu khng phi l trng thi kt thc nhng tt hn trng thi

hin hnh th cp nht n thnh trng thi hin hnh.b.3.Nu n khng tt hn trng thi hin hnh th tip tc vnglp.

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M gi

Ti:= T0; Stop :=FALSE;

WHILEStop=FALSE DO BEGIN

IFTi TG THENBEGIN

; Stop:=TRUE;

END;

ELSE BEGIN

Better:=FALSE;

WHILE (Better=FALSE) AND (STOP=FALSE) DO BEGINIF THEN BEGIN

; Stop:=TRUE;END;

ELSE BEGIN

Tk := ;

IF THEN BEGIN

Ti :=Tk; Better:=TRUE;

END;

END;

END; {WHILE}

END; {ELSE}

END;{WHILE}

Mnh "h(Tk) tt hn h(Ti)" ngha l g? y l mt khi nim chung chung. Khi cit thut gii, ta phi cung cp mt nh ngha tng minh vtt hn.Trong mt strng hp, tt hn l nhhn : h(Tk) < h(Ti); mt strng hp khc tt hn l lnhn h(Tk) > h(Ti)...Chng hn, i vi bi ton tm ng i ngn nht gia hai im.Nu dng hm h l hm cho ra khong cch theo ng chim bay gia vtr hin ti(trng thi hin ti) v ch n (trng thi ch) th tt hn ngha l nhhn.

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Vn cn lm r ktip l thno l ?Mt trng thi ktip hp ll trng thi cha c xt n. Gish ca trng thi hin ti Ti c gi trlh(Ti) = 1.23 v tTi ta c thbin i sang mt trong 3 trng thi ktip ln lt l Tk1,Tk2, Tk3vi gi trcc hm h tng ng l h(Tk1) = 1.67, h(Tk2) = 2.52, h(Tk3) = 1.04.u tin, Tk sc gn bng Tk1, nhng v h(Tk) = h(Tk1) > h(Ti) nn Tk khng

c chn. Ktip l Tk sc gn bng Tk2v cng khng c chn. Cui cng thTk3c chn. Nhng gish(Tk3) = 1.3 th cTk3cng khng c chn v mnh sc gi trTRUE. Gii thch ny c vhin nhinnhng c lcn thit trnh nhm ln cho bn c.

thy r hot ng ca thut gii leo i. Ta hy xt mt bi ton minh ha sau. Cho 4khi lp phngging nhau A, B, C, D. Trong cc mt (M1), (M2), (M3), (M4),(M5), (M6) c thc t bng 1 trong 6 mu (1), (2), (3), (4), (5), (6). Ban u cc khilp phng c xp vo mt hng. Mi mt bc, ta chc xoay mt khi lpphng quanh mt trc (X,Y,Z) 900theo chiu bt k(ngha l ngc chiu hay thunchiu kim ng hcng c). Hy xc nh sbc quay t nht sao cho tt ccc mt

ca khi lp phng trn 4 mt ca hng l c cng mu nhhnh v.

Hnh : Bi ton 4 khi lp phng

gii quyt vn , trc ht ta cn nh ngha mt hm Gdng nh gi mt tnhtrng cthc phi l li gii hay khng? Bn c c thddng a ra mt ci t cahm G nhsau :

IF(Gtri + Gphi + Gtrn + Gdi + Gtrc + Gsau) = 16 THEN

G:=TRUE

ELSE

G:=FALSE;

Trong , Gphilslng cc mtc cng mu ca mt bn phi ca hng. Tng tcho Gtri, Gtrn, Ggia, Gtrc, Gsau. Tuy nhin, do cc khi lp phng A,B,C,D lhon ton tng tnhau nn tng quan gia cc mt ca mi khi l ging nhau. Do ,

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nu c 2 mt khng i nhau trn hng ng mu th 4 mt cn li ca hng cng ngmu. T ta chcn hm G c nh ngha nhsau l :

IFGphi + Gdi = 8 THEN

G:=TRUE

ELSE

G:=FALSE;

Hm h(c lng khnng dn n li gii ca mt trng thi) sc nh ngha nhsau :

h= Gtri+ Gphi+ Gtrn+ Gdi

Bi ton ny n gin thut gii leo i c thhot ng tt. Tuy nhin, khng philc no ta cng may mn nhth!

n y, c thchng ta sny sinh mt tng. Nu chn trng thi tt hn lmtrng thi hin ti th ti sao khng chn trng thi tt nht ? Nhvy, c lta snhanhchng dn n li gii hn! Ta sbn lun vvn : "liu ci tin ny c thc sgipchng ta dn n li gii nhanh hn hay khng?" ngay sau khi trnh by xong thut giileo i dc ng.

III.3.2. Leo i dc ng

Vcbn, leo i dc ng cng ging nhleo i, chkhc im l leo i dc ngsduyt tt ccc hng i c thv chn i theo trng thi tt nht trong scc trngthi ktip c thc (trong khi leo i chchn i theo trng thi ktip u tin tthntrng thi hin hnh m n tm thy).

Ttng

1)Nu trng thi bt u cng l trng thi ch th thot v bo l tm c li gii. Ngc li, ttrng thi hin hnh (Ti) l trng thi khi u (T0)

2) Lp li cho n khi t n trng thi kt thc hoc cho n khi (Ti) khng tn ti mt trng thi ktip(Tk) no tt hn trng thi hin ti (Ti)

a) t S bng tp tt ctrng thi ktip c thc ca Tiv tt hn Ti.

b) Xc nh Tkmax l trng thi tt nht trong tp S

t Ti = Tkmax

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M gi

Ti:= T0;

Stop :=FALSE;

WHILEStop=FALSE DO BEGIN

IFTi TG THENBEGIN

;

STOP :=TRUE;

END;

ELSE BEGIN

Best:=h(Ti);

Tmax:= Ti;

WHILE DOBEGIN

Tk := ;

IF THEN BEGIN

Best :=h(Tk);

Tmax:= Tk;

END;

END;

IF (Best>Ti) THEN

Ti:= Tmax;

ELSE BEGIN

;

STOP:=TRUE;

END;

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END; {ELSE IF}

END;{WHILE STOP}

III.3.3. nh gi

So vi leo i n gin, leo i dc ng c u im l lun lun chn hng c trinvng nht i. Liu iu ny c m bo leo i dc ng lun tt hn leo i n ginkhng? Cu trli l khng. Leo i dc ng chtt hn leo i n gin trong mt strng hp m thi. chn ra c hng i tt nht, leo i dc ng phi duyt quatt ccc hng i c thc ti trng thi hin hnh. Trong khi , leo i n gin chchn i theo trng thi u tin tt hn (so vi trng thi hin hnh) m n tm ra c.Do , thi gian cn thit leo i dc ng chn c mt hng i sln hn so vileo i n gin. Tuy vy, do lc no cng chn hng i tt nht nn leo i dc ngthng stm n li gii sau mt sbc t hn so vi leo i n gin. Ni mt cchngn gn, leo i dc ng stn nhiu thi gian hn cho mt bc nhng li i t bc

hn; cn leo i n gin tn t thi gian hn cho mt bc i nhng li phi i nhiubc hn. y chnh l yu tc v mt gia hai thut gii nn ta phi cn nhc klng khi la chn thut gii.

Chai phng php leo ni n gin v leo ni dc ng u c khnng tht bi trongvic tm li gii ca bi ton mc d li gii thc shin hu. Chai gii thut u cthkt thc khi t c mt trng thi m khng cn trng thi no tt hn na c thpht sinh nhng trng thi ny khng phi l trng thi ch. iu ny sxy ra nuchng trnh t n mt im cc i a phng, mt on n iu ngang.

im cc i a phng (a local maximum) : l mt trng thi tt hn tt cln cn ca

n nhng khng tt hn mt strng thi khc xa hn. Ngha l ti mt im cc ia phng, mi trng thi trong mt ln cnca trng thi hin ti uxu hntrngthi hin ti. Tuy c dng vca li gii nhng cc cc i a phng khng phi l ligii thc s. Trong trng hp ny, chng c gi l nhng ngn i thp.

on n iu ngang (a plateau): l mt vng bng phng ca khng gian tm kim,trong , ton bcc trng thi ln cn u c cng gi tr.

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Hnh : Cc tnh hung kh khn cho tm kim leo o.

i ph vi cc cc im ny, ngi ta a ra mt sgii php. Ta stm hiu 2trong scc gii php ny. Nhng gii ny, khng thc sgii quyt trn vn vn mchl mt phng n cu nguy tm thi m thi.

Phng n u tin l kt hp leo i v quay lui. Ta squay lui li cc trng thi trc v thi theo hng khc. Thao tc ny hp l nu ti cc trng thi trc c mthng i tt m ta bqua trc . y l mt cch kh hay i ph vi cc imcc i a phng. Tuy nhin, do c im ca leo i l "bc sau cao hn bc trc"nn phng n ny stht bi khi ta xut pht tmt im qu cao hoc xut pht tmtnh i m n c li gii cn phi i qua mt "thung lng" tht su nhtronghnh sau.

Hnh : Mt trng hp tht bi ca leo o kt hp quay lui.

Cch thhai l thc hin mt bc nhy vt theo hng no thn mt vng mica khng gian tm kim. Nm na l "bc" lin tc nhiu "bc" (chng hn 5,7,10, )m tm thi "qun" i vic kim tra "bc sau cao hn bc trc". Tip cn c vhiuqukhi ta gp phi mt on n iu ngang. Tuy nhin, nhy vt cng c ngha l ta bqua chi tin n li gii thc s. Trong trng hp chng ta ang ng kh gnli gii, vic nhy vt sa chng ta sang mt vtr hon ton xa l, m t, c thsdn chng ta n mt rc ri kiu khc. Hn na, sbc nhy l bao nhiu v nhy theohng no l mt vn phthuc rt nhiu vo c im khng gian tm kim ca biton.

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Hnh Mt trng hp kh khn cho phng n "nhy vt".

Leo ni l mt phng php cc bbi v n quyt nh slm g tip theo da vo mt

nh gi vtrng thi hin ti v cc trng thi ktip c thc (tt hntrng thi hinti, trng thi tt nht tt hn trng thi hin ti) thay v phi xem xt mt cch ton dintrn tt ccc trng thi i qua. Thun li ca leo ni l t gp sbng nthp hnso vi cc phng php ton cc. Nhng n cng ging nhcc phng php cc bkhc chl khng chc chn tm ra li gii trong trng hp xu nht.

Mt ln na, ta khng nh li vai tr quyt nh ca hm Heuristic trong qu trnh tmkim li gii. Vi cng mt thut gii (nhleo i chng hn), nu ta c mt hmHeuristic tt hn th kt qusc tm thy nhanh hn. Ta hy xt bi ton vcc khic trnh by hnh sau. Ta c hai thao tc bin i l:

+ Ly mt khi nh mt ct bt kv t n ln mt chtrng to thnh mtct mi. Lu l chc thto ra ti a 2 ct mi.

+ Ly mt khi nh mt ct v t n ln nh mt ct khc

Hy xc nh sthao tc t nht bin i ct cho thnh ct kt qu.

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TTNT

Hnh : Trng thi khi u v trng thi kt thc

Gisban u ta dng mt hm Heuristic n gin nhsau :

H1: Cng 1 im cho mi khi vtr ng so vi trng thi ch. Tr1 imcho mi khi t vtr sai so vi trng thi ch.

Dng hm ny, trng thi kt thc sc gi trl 8 v c8 khi u c t vtrng. Trng thi khi u c gi trl 4 (v n c 1 im cng cho cc khi C, D, E, F, G,H v 1 im trcho cc khi A v B). Chc thc mt di chuyn ttrng thi khi u, l dch chuyn khi A xung to thnh mt ct mi (T1).

iu sinh ra mt trng thi vi sim l 6(v vtr ca khi A by gisinh ra 1im cng hn l mt im tr). Thtc leo ni schp nhn sdch chuyn . T

trng thi mi T1, c ba di chuyn c ththc hin dn n ba trng thi Ta, Tb, Tccminh ha trong hnh di. Nhng trng thi ny c sim l : h(Ta)= 4; h(Tb) = 4 vh(Tc) = 4

T1 TA TB TC

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TTNT

Hnh Cc trng thi c tht c tT1

Thtc leo ni stm dng bi v tt ccc trng thi ny c sim thp hn trng thihin hnh. Qu trnh tm kim chdng li mt trng thi cc i a phng m khngphi l cc i ton cc.

Chng ta c thli cho chnh gii thut leo i v tht bi do khng tm nhntng qut tm ra li gii. Nhng chng ta cng c thli cho hm Heuristic v cgng sa i n. Gista thay hm ban u bng hm Heuristic sau y :

H2: i vi mi khi phtrng (khi phtrl khi nm bn di khi hinti), cng 1 im, ngc li tr1 im.

Dng hm ny, trng thi kt thc c sim l 28 v B nm ng vtr v khng c khi

phtrno, C ng vtr c 1 im cng vi 1 im do khi phtrB nm ng vtrnn C c 2 im, D c 3 im, ....Trng thi khi u c sim l28. Vic dichuyn A xung to thnh mt ct mi lm sinh ra mt trng thi vi sim l h(T1) =21v A khng cn 7 khi sai pha di n na. Ba trng thi c thpht sinh tip theoby gic cc im sl : h(Ta)=28; h(Tb)=16 v h(Tc) = 15. Lc ny thtc leoni dc ng schn di chuyn n trng thi Tc, c mt khi ng. Qua hm H2ny ta rt ra mt nguyn tc : tt hnkhng chc ngha l c nhiu u imhn m cnphi t khuyt imhn. Hn na, khuyt im khng c ngha chl ssai bit ngay timt vtr m cn l skhc bit trong tng quan gia cc vtr. R rng l ng vmtkt qu, cng mt thtc leo i nhng hm H1btht bi (do chbit nh gi u im)cn hm H2mi ny li hot ng mt cch hon ho (do bit nh gi cu im v

khuyt im).

ng tic, khng phi lc no chng ta cng thit kc mt hm Heuristic hon honhth. V vic nh gi u im kh, vic nh gi khuyt im cng kh v tinh thn. Chng hn, xt li vn mun i vo khu trung tm ca mt thnh phxa l. hm Heuristic hiu qu, ta cn phi a cc thng tin vcc ng mt chiu v cc ngct, m trong trng hp mt thnh phhon ton xa lth ta kh hoc khng thbitc nhng thng tin ny.

n y, chng ta hiu r bn cht ca hai thut gii tip cn theo chin lc tm kimchiu su. Hiu quca chai thut gii leo i n gin v leo i dc ng phthuc

vo :

+ Cht lng ca hm Heuristic.

+ c im ca khng gian trng thi.

+ Trng thi khi u.

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TTNT

Khi u, chc mt nt (trng thi) A nn n sc mrng to ra 3 nt mi B,C vD. Cc con sdi nt l gi trcho bit tt ca nt. Con scng nh, nt cng tt.Do D l nt c khnng nht nn n sc mrng tip sau nt A v sinh ra 2 nt ktip l E v F. n y, ta li thy nt B c vc khnng nht (trong cc nt B,C,E,F)nn ta schn mrng nt B v to ra 2 nt G v H. Nhng li mt ln na, hai nt G, H

ny c nh gi t khnng hn E, v thsch li trvE. E c mrng v ccnt c sinh ra tE l I v J. bc ktip, J sc mrng v n c khnng nht.Qu trnh ny tip tc cho n khi tm thy mt li gii.

Lu rng tm kim ny rt ging vi tm kim leo i dc ng, vi 2 ngoi l. Trongleo ni, mt trng thi c chn v tt ccc trng thi khc bloi b, khng bao gichng c xem xt li. Cch xl dt khot ny l mt c trng ca leo i. TrongBFS, ti mt bc, cng c mt di chuyn c chn nhng nhng ci khc vn cgili, ta c thtrli xt sau khi trng thi hin ti trnn km khnng hnnhng trng thi c lu tr. Hn na, ta chn trng thi tt nht m khng quan tmn n c tt hn hay khng cc trng thi trc . iu ny tng phn vi leo i v

leo i sdng nu khng c trng thi tip theo no tt hn trng thi hin hnh.ci t cc thut gii theo kiu tm kim BFS, ngi ta thng cn dng 2 tp hp sau:

OPEN: tp cha cc trng thi c sinh ra nhng cha c xt n (v ta chnmt trng thi khc). Thc ra, OPENl mt loi hng i u tin (priority queue) mtrong , phn tc u tin cao nht l phn ttt nht. Ngi ta thng ci t hngi u tin bng Heap. Cc bn c ththam kho thm trong cc ti liu vCu trc dliu vloi dliu ny.

CLOSE : tp cha cc trng thi c xt n. Chng ta cn lu trnhng trng thiny trong bnhphng trng hp khi mt trng thi mi c to ra li trng vimt trng thi m ta xt n trc . Trong trng hp khng gian tm kim c dngcy th khng cn dng tp ny.

Thut gii BEST-FIRST SEARCH

1. t OPENcha trng thi khi u.

2. Cho n khi tm c trng thi ch hoc khng cn nt no trong OPEN, thc hin :

2.a. Chn trng thi tt nht (Tmax) trong OPEN (v xa Tmaxkhi OPEN)

2.b.Nu Tmax l trng thi kt thc th thot.

2.c.Ngc li, to ra cc trng thi ktip Tk c thc ttrng thi Tmax. i vi mitrng thi ktip Tk thc hin :

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TTNT

Tnh f(Tk); Thm Tk vo OPEN

BFS kh n gin. Tuy vy, trn thc t, cng nhtm kim chiu su v chiu rng,him khi ta dng BFS mt cch trc tip. Thng thng, ngi ta thng dng cc phinbn ca BFS l AT, AKT v A*

Thng tin vqu khv tng lai

Thng thng, trong cc phng n tm kim theo kiu BFS, tt fca mt trng thic tnh da theo 2 hai gi trm ta gi l l gv h. hchng ta bit, l mt clng vchi ph ttrng thi hin hnh cho n trng thi ch (thng tin tng lai). Cngl "chiu di qung ng" i ttrng thi ban u cho n trng thi hin ti(thng tin qu kh). Lu rng gl chi ph thc s(khng phi chi ph c lng). dhiu, bn hy quan st hnh sau :

Hnh 6.14 Phn bit khi nim g v h

Kt hp g v h thnh f(f = g + h) sthhin mt c lng v"tng chi ph" cho conng ttrng thi bt u n trng thi kt thc dc theo con ng i qua trng thihin hnh. thun tin cho thut gii, ta quy c l gv hu khng m v cng nhngha l cng tt.

III.5. Thut gii AT

Thut gii AT

l mt phng php tm kim theo kiu BFS vi tt ca nt l gi trhm g tng chiu di con ng i ttrng thi bt u n trng thi hin ti.

Thut gii AT



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TTNT

2.a. Chn trng thi (Tmax) c gi trg nhnhttrong OPEN (v xa Tmax khiOPEN)


2.c.Ngc li, to ra cc trng thi ktip Tk c thc ttrng thi Tmax. i vi mi

trng thi ktip Tk thc hin :

g(Tk) = g(Tmax) + cost(Tmax, Tk);

Thm Tk vo OPEN.

* V chsdng hm g (m khng dng hm c lng h) fsnh gi tt ca mt trng thi nn tacng c thxem AT chl mt thut ton.

III.6. Thut gii AKT

(Algorithm for Knowlegeable Tree Search)

Thut gii AKTmrng AT bng cch sdng thm thng tin c lng h. tt camt trng thi f l tng ca hai hm g v h.

Thut gii AKT



2.a. Chn trng thi (Tmax) c gi trf nhnhttrong OPEN (v xa Tmax khiOPEN)


2.c.Ngc li, to ra cc trng thi ktip Tk c thc ttrng thi Tmax. i vi mitrng thi ktip Tk thc hin :

g(Tk) = g(Tmax) + cost(Tmax, Tk);

Tnh h(Tk)

f(Tk) = g(Tk) + h(Tk);

Thm Tk vo OPEN.

III.7. Thut gii A*

A* l mt phin bn c bit ca AKT p dng cho trng hp th. Thut gii A* csdng thm tp hp CLOSElu trnhng trng hp c xt n. A*mrngAKTbng cch bsung cch gii quyt trng hp khi "m" mt nt m nt ny csn trong OPEN hoc CLOSE. Khi xt n mt trng thi Ti bn cnh vic lu tr3 gi

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TTNT

trcbn g,h, f phn nh tt ca trng thi , A*cn lu trthm hai thng ssau :

1. Trng thi cha ca trng thi Ti (k hiu l Cha(Ti): cho bit trng thi dn n trngthi Ti. Trong trng hp c nhiu trng thi dn n Ti

th chn Cha(Ti) sao cho chi ph

i ttrng thi khi u n Ti l thp nht, ngha l :

g(Ti) = g(Tcha) + cost(Tcha, Ti) l thp nht.

2.Danh sch cc trng thi ktip ca Ti: danh sch ny lu trcc trng thi ktipTk ca Ti sao cho chi ph n Tk thng qua Ti ttrng thi ban u l thp nht. Thccht th danh sch ny c thc tnh ra tthuc tnh Cha ca cc trng thi c lutr. Tuy nhin, vic tnh ton ny c thmt nhiu thi gian (khi tp OPEN, CLOSEc mrng) nn ngi ta thng lu trra mt danh sch ring. Trong thut ton sauy, chng ta skhng cp n vic lu trdanh sch ny. Sau khi hiu r thut ton,bn c c thddng iu chnh li thut ton lu trthm thuc tnh ny.

1.t OPEN chcha T0. t g(T0) = 0, h(T0) = 0 v f(T0) = 0.t CLOSE l tp hp rng.

2. Lp li cc bc sau cho n khi gp iu kin dng.

2.a.Nu OPEN rng : bi ton v nghim, thot.

2.b.Ngc li, chn Tmax trong OPEN sao cho f(Tmax) l nhnht

2.b.1.Ly Tmax ra khi OPENv a Tmax vo CLOSE.

2.b.2.Nu Tmaxchnh l TGth thot v thng bo li gii l Tmax.

2.b.3.Nu Tmax khng phi l TG. To ra danh sch tt ccc trng thi ktip ca Tmax. Gi mt trng thi ny l Tk. Vi mi Tk, lm cc bc sau :

2.b.3.1.Tnh g(Tk) = g(Tmax) + cost(Tmax, Tk).

2.b.3.2.Nu tn ti Tk trong OPEN trng vi Tk.

Nu g(Tk) < g(Tk) th

t g(Tk) = g(Tk)

Tnh li f(Tk)

t Cha(Tk) = Tmax

2.b.3.3.Nu tn ti Tk trong CLOSE trng vi Tk.

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TTNT

Nu g(Tk) < g(Tk) th

t g(Tk) = g(Tk)

Tnh li f(Tk)

t Cha(Tk) = Tmax

Lan truynsthay i gi trg, f cho tt ccctrng thi ktip ca Ti (tt ccc cp) clu trtrong CLOSE v OPEN.

2.b.3.4.Nu Tkcha xut hin trong cOPENln CLOSEth :

Thm Tk vo OPEN

Tnh : f' (Tk) = g(Tk)+h(Tk).

C mt sim cn gii thch trong thut gii ny. u tin l vic sau khi tm thytrng thi ch TG, lm sao xy dng li c "con ng" tT0n TG. Rt ngin, bn chcn ln ngc theo thuc tnh Cha ca cc trng thi c lu trtrongCLOSE cho n khi t n T0. chnh l "con ng" ti u i tTG n T0(hay nicch khc l tT0n TG).

im thhai l thao tc cp nht li g(Tk) , f(Tk) v Cha(Tk) trong bc 2.b.3.2 v2.b.3.3. Cc thao tc ny thhin ttng : "lun chn con ng ti u nht". Nhchng ta bit, gi trg(Tk) nhm lu trchi ph ti u thc stnh tT0n Tk. Do, nu chng ta pht hin thy mt "con ng" khc tt hn thng qua Tk (c chi phnhhn) con ng hin ti c lu trth ta phi chn "con ng" mi tt hn ny.

Trng hp 2.b.3.3 phc tp hn. V tTk nm trong tp CLOSE nn tTk ta lutrcc trng thi con ktip xut pht tTk. Nhng g(Tk) thay i dn n gi trgca cc trng thi con ny cng phi thay i theo. V n lt cc trng thi con ny lic thc cc cc trng thi con tip theo ca chng v cthcho n khi mi nhnh ktthc vi mt trng thi trong OPEN (ngha l khng c trng thi con no na). thchin qu trnh cp nht ny, ta hy thc hin qu trnh duyt theo chiu su vi im khiu l Tk. Duyt n u, ta cp nht li gca cc trng thi n ( dng cng thcg(T) = g(Cha(T)) +cost(Cha(T), T) )v v thgi trfca cc trng thi ny cng thayi theo.

Mt ln na, xin nhc li rng, bn c thcho rng tp OPEN lu trcc trng thi "s

c xem xt n sau" cn tp CLOSE lu trcc trng thi " c xt n ri".

C thbn scm thy kh lng tng trc mt thut gii di nhth. Vn c lstrnn sng sa hn khi bn quan st cc bc gii bi ton tm ng i ngn nht trn thbng thut gii A* sau y.

III.8. V dminh ha hot ng ca thut gii A*

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TTNT

Chng ta sminh ha hot ng ca thut gii A* trong vic tm kim ng i ngnnht tthnh phArad n thnh phBucharestca Romania. Bn cc thnh phcaRomania c cho trong thsau. Trong mi nh ca thca l mt thnh ph,gia hai nh c cung ni ngha l c ng i gia hai thnh phtng ng. Trng sca cung chnh l chiu di (tnh bng km) ca ng i ni hai thnh phtng ng,

chiu di theo ng chim bay mt thnh phn Bucharest c cho trong bng kmtheo.

Hnh : Bng ca Romania vi khong cch ng tnh theo km

Bng : Khong cch ng chim bay tmt thnh phn Bucharest.

Chng ta schn hm hchnh l khong cch ng chim bay cho trong bng trn vhm chi ph cost(Ti, Ti+1)chnh l chiu di con ng ni tthnh phTi v Ti+1.

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TTNT

Sau y l tng bc hot ng ca thut ton A* trong vic tm ng i ngn nht tArad n Bucharest.

Ban u :

OPEN ={(Arad,g=0,h=0,f=0)}

CLOSE ={}

Do trong OPEN chcha mt thnh phduy nht nn thnh phny sl thnh phttnht. Ngha l Tmax =Arad.Ta ly Arad ra khi OPEN v a vo CLOSE.

OPEN ={}

CLOSE ={(Arad,g=0,h=0,f=0)}

TArad c thi n c 3 thnh phl Sibiu, Timisoara v Zerind. Ta ln lt tnhgi trf, g v h ca 3 thnh phny. Do c3 nt mi to ra ny cha c nt cha nn banu nt cha ca chng u l Arad.

h(Sibiu) =253

g(Sibiu) =g(Arad)+cost(Arad,Sibiu)

=0+140=140

f(Sibiu) =g(Sibiu)+h(Sibiu)

=140+253 =393

Cha(Sibiu) =Arad

h(Timisoara) =329

g(Timisoara) =g(Arad)+cost(Arad, Timisoara)

=0+118=118

f(Timisoara) =g(Timisoara)+ h(Timisoara)

=118+329 =447

Cha(Timisoara) =Arad

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h(Zerind) =374

g(Zerind) =g(Arad)+cost(Arad, Zerind)

=0+75=75

f(Zerind) =g(Zerind)+h(Zerind)

=75+374 =449

Cha(Zerind) =Arad

Do c3 nt Sibiu, Timisoara, Zerind u khng c trong cOPEN v CLOSE nn ta bsung 3 nt ny vo OPEN.

OPEN ={(Sibiu,g=140,h=253,f=393,Cha=Arad)

(Timisoara,g=118,h=329,f=447,Cha=Arad)

(Zerind,g=75,h=374,f=449,Cha=Arad)}

CLOSE ={(Arad,g=0,h=0,f=0)}

Hnh : Bc 1, nt c ng ngoc vung (nh[Arad]) l nt trong tp CLOSE, ngcli l trong tp OPEN.

Trong tp OPEN, nt Sibiu l nt c gi trf nhnht nn ta schn Tmax =Sibiu. Taly Sibiu ra khi OPEN v a vo CLOSE.

OPEN ={(Timisoara,g=118,h=329,f=447,Cha=Arad)

(Zerind,g=75,h=374,f=449,Cha=Arad)}

CLOSE ={(Arad,g=0,h=0,f=0)

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TTNT

(Sibiu,g=140,h=253,f=393,Cha=Arad)}

TSibiu c thi n c 4 thnh phl : Arad, Fagaras, Oradea, Rimnicu. Ta ln lttnh cc gi trg, h, f cho cc nt ny.

h(Arad) =366

g(Arad) =g(Sibiu)+cost(Sibiu,Arad)

=140+140=280

f(Arad) =g(Arad)+h(Arad)

=280+366 =646

h(Fagaras) =178

g(Fagaras) =g(Sibiu)+cost(Sibiu, Fagaras) =140+99=239

f(Fagaras) =g(Fagaras)+ h(Fagaras)

=239+178=417

h(Oradea) =380

g(Oradea) =g(Sibiu)+cost(Sibiu, Oradea)

=140+151 =291

f(Oradea) =g(Oradea)+ h(Oradea)

=291+380 =671

h(R.Vilcea) =193

g(R.Vilcea) =g(Sibiu)+cost(Sibiu, R.Vilcea)

=140+80 =220

f(R.Vilcea) =g(R.Vilcea)+ h(R.Vilcea)

=220+193 =413

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TTNT

Nt Arad c trong CLOSE. Tuy nhin, do g(Arad) mi c to ra (c gi tr280) lnhn g(Arad) lu trong CLOSE (c gi tr0) nn ta skhng cp nht li gi trg v f caArad lu trong CLOSE. 3 nt cn li : Fagaras, Oradea, Rimnicu u khng c trong cOPEN v CLOSE nn ta sa 3 nt ny vo OPEN, t cha ca chng l Sibiu. Nhvy, n bc ny OPEN cha tng cng 5 thnh ph.


(Zerind,g=75,h=374,f=449,Cha=Arad)

(Fagaras,g=239,h=178,f=417,Cha=Sibiu)

(Oradea,g=291,h=380,f=617,Cha=Sibiu)

(R.Vilcea,g=220,h=193,f=413,Cha=Sibiu)}


(Sibiu,g=140,h=253,f=393,Cha=Arad)}

Trong tp OPEN, nt R.Vilcea l nt c gi trf nhnht. Ta chn Tmax =R.Vilcea.Chuyn R.Vilcea tOPEN sang CLOSE. TR.Vilcea c thi n c 3 thnh phlCraiova, Pitesti v Sibiu. Ta ln lt tnh gi trf, g v h ca 3 thnh phny.

h(Sibiu)=

253g(Sibiu) =g(R.Vilcea)+ cost(R.Vilcea,Sibiu)

=220+80=300

f(Sibiu) =g(Sibiu)+h(Sibiu)

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TTNT

=300+253 =553

h(Craiova) =160

g(Craiova) =g(R.Vilcea)+ cost(R.Vilcea, Craiova)

=220+146=366

f(Craiova) =g(Fagaras)+h(Fagaras)

=366+160=526

h(Pitesti) =98

g(Pitesti) =g(R.Vilcea)+ cost(R.Vilcea, Pitesti)

=220+97 =317

f(Pitesti) =g(Oradea)+h(Oradea)

=317+98 =415

Sibiu c trong tp CLOSE. Tuy nhin, do g(Sibiu) mi (c gi trl 553) ln hng(Sibiu) (c gi trl 393) nn ta skhng cp nht li cc gi trca Sibiu c lutrong CLOSE. Cn li 2 thnh phl Pitesti v Craiova u khng c trong cOPEN v

CLOSE nn ta sa n vo OPEN v t cha ca chng l R.Vilcea.

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TTNT


(Zerind,g=75,h=374,f=449,Cha=Arad) (Fagaras,g=239,h=178,f=417,Cha=Sibiu)

(Oradea,g=291,h=380,f=617,Cha=Sibiu) (Craiova,g=366,h=160,f=526,Cha=R.Vilcea)

(Pitesti,g=317,h=98,f=415,Cha=R.Vilcea) }


(Sibiu,g=140,h=253,f=393,Cha=Arad)

(R.Vilcea,g=220,h=193,f=413,Cha=Sibiu) }

n y, trong tp OPEN, nt tt nht l Pitesti, tPitesti ta c thi n cR.Vilcea, Bucharest v Craiova. Ly Pitesti ra khi OPEN v t n vo CLOSE.Thc hin tip theo tng tnhtrn, ta skhng cp nht gi trf, g caR.Vilcea v Craiova lu trong CLOSE. Sau khi tnh ton f, g ca Bucharest, ta sa Bucharest vo tp OPEN, t Cha(Bucharest) =Pitesti.

h(Bucharest) =0

g(Bucharest) =g(Pitesti)+cost(Pitesti, Bucharest)

=317+100=418

f(Bucharest) =g(Fagaras)+h(Fagaras)

=417+0=417

bc ktip, ta schn c Tmax =Bucharest. V nhvy thut ton kt thc (thcra th ti bc ny, c hai ng cvin l Bucharest v Fagaras v u cng c f=417 ,nhng v Bucharest l ch nn ta su tin chn hn).

xy dng li con ng i tArad n Bucharest ta ln theo gi trCha c lu trkm vi f, g v h cho n lc n Arad.

Cha(Bucharest) =Pitesti

Cha(R.Vilcea) =Sibiu

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TTNT

Cha(Sibiu) =Arad

Vy con ng i ngn nht tArad n Bucharest l Arad, Sibiu, R.Vilcea, Pitesti,Bucharest.

Trong v dminh ha ny, hm hc cht lng kh tt v cu trc thkh n ginnn ta gn nhi thng n ch m t phi kho st cc con ng khc. y l mttrng hp n gin, trong trng hp ny, thut gii c dng dp ca tm kim chiusu.

n y, minh ha mt trng hp phc tp hn ca thut gii. Ta thsa i li cutrc thv quan st hot ng ca thut gii. Gista c thm mt thnh phtm gil TPv con ng gia Sibiuv TPc chiu di 100, con ng gia TP v Pitesticchiu di 60. V khong cch ng chim bay tTP n Bucharest l 174. Nhvy rrng, con ng ti u n Bucharest khng cn l Arad, Sibiu, R.Vilcea, Pitesti,Bucharest na m l Arad, Sibiu, TP, Pitesti, Bucharest.

Trong trng hp ny, chng ta vn tin hnh bc 1 nhtrn. Sau khi thc hin hinbc 2 (mrng Sibiu), chng ta c cy tm kim nhhnh sau. Lu l c thm nhnhTP.

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TTNT

R.Vilcea vn c gi trf thp nht. Nn ta mrng R.Vilcea nhtrng hp u tin.

Bc ktip ca trng hp n gin l mrng Pitesti c c kt qu. Tuy nhin,trong trng hp ny, TP c gi trf thp hn. Do , ta chn mrng TP. TTP ta chc 2 hng i, mt quay li Sibiu v mt n Pitesti. nhanh chng, ta skhng tnhton gi trca Sibiu v bit chc n sln hn gi trc lu trtrong CLOSE (v ingc li).

h(Pitesti) =98

g(Pitesti) =g(TP)+cost(TP, Pitesti)

=240+75=315

f(Pitesti) =g(TP)+h(Pitesti) =315+98=413

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TTNT

Pistestti xut hin trong tp OPEN v g(Pitesti) mi (c gi trl 315) thp hng(Pitesti) c(c gi tr317) nn ta phi cp nht li gi trca f,g, Cha ca Pitesti lutrong OPEN. Sau khi cp nht xong, tp OPEN v CLOSE snhsau :


(Zerind,g=75,h=374,f=449,Cha=Arad)

(Fagaras,g=239,h=178,f=417,Cha=Sibiu)

(Oradea,g=291,h=380,f=617,Cha=Sibiu)

(Craiova,g=366,h=160,f=526,Cha=R.Vilcea)

(Pitesti,g=315,h=98,f=413,Cha=TP)}


(Sibiu,g=140,h=253,f=393,Cha=Arad)

(R.Vilcea,g=220,h=193,f=413,Cha=Sibiu)

}

n y ta thy rng, ban u thut gii chn ng i n Pitesti qua R.Vilcea. Tuy

nhin, sau , thut gii pht hin ra con ng n Pitesti qua TP l tt hn nn n ssdng con ng ny. y chnh l trng hp 2.b.iii.2 trong thut gii.

Bc sau, chng ta schn mrng Pitesti nhbnh thng. Khi ln ngc theo thuctnh Cha, ta sc con ng ti u l Arad, Sibiu, TP, Pitesti, Bucharest.

III.9. Bn lun vA*

n y, c lbn hiu c thut gii ny. Ta c mt vi nhn xt kh th vvA*.u tin l vai tr ca g trong vic gip chng ta la chn ng i. N cho chng ta khnng la chn trng thi no mrng tip theo, khng chda trn vic trng thi

tt nhthno (thhin bi gi trh) m cn trn cscon ng ttrng thi khiu n trng thi hin ti tt ra sao. iu ny srt hu ch nu ta khng chquantm vic tm ra li gii hay khng m cn quan tm n hiu quca con ng dn nli gii. Chng hn nhtrong bi ton tm ng i ngn nht gia hai im. Bn cnhvic tm ra ng i gia hai im, ta cn phi tm ra mt con ng ngn nht. Tuynhin, nu ta chquan tm n vic tm c li gii(m khng quan tm n hiu quca con ng n li gii), chng ta c tht g=0 mi trng thi. iu ny sgip talun chn i theo trng thi c vgn nht vi trng thi kt thc (v lc ny fchph

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39/104


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TTNT

Hnh : h nh gi cao h

n y chng ta kt thc vic bn lun vthut gii A*, mt thut gii linh ng,tng qut, trong hm cha ctm kim chiu su, tm kim chiu rng v nhngnguyn l Heuristic khc. Chnh v thm ngi ta thng ni, A* chnh l thut giitiu biu cho Heuristic.

A* rt linh ng nhng vn gp mt khuyt im cbn ging nhchin lc tmkim chiu rng l tn kh nhiu bnhlu li nhng trng thi i qua nuchng ta mun n chc chn tm thy li gii ti u. Vi nhng khng gian tm kim lnnhth y khng phi l mt im ng quan tm. Tuy nhin, vi nhng khng gian tmkim khng l(chng hn tm ng i trn mt ma trn kch thc c106x 106) thkhng gian lu trl cmt vn hc ba. Cc nh nghin cu a ra kh nhiu cchng tip cn lai gii quyt vn ny. Chng ta stm hiu mt sphng nnhng quan trng nht, ta cn phi nm r vtr ca A* so vi nhng thut gii khc.

III.10. ng dng A* gii bi ton Ta-canh

Bi ton Ta-canh tng l mt tr chi kh phbin, i lc ngi ta cn gi y l biton 9-puzzle. Tr chi bao gm mt hnh vung kch thc 3x3 . C 8 c s, mi c mt st1 n 8. Mt cn trng. Mi ln di chuyn chc di chuyn mt nmcnh trng vpha trng. Vn l tmt trng thi ban u bt k, lm sao ac vtrng thi cui l trng thi m cc c sp ln lt t1 n 8 theo thtttri sang phi, ttrn xung di, cui dng l trng.

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TTNT

Cho n nay, ngoi tr2 gii php vt cn v tm kim Heuristic, ngi ta vn cha tmc mt thut ton chnh xc, ti u gii bi ton ny. Tuy nhin, cch gii theo thutgii A*li kh n gin v thng tm c li gii (nhng khng phi lc no cng tmc li gii). Nhn xt rng: Ti mi thi im ta chc ti a 4 c thdi chuyn. Vnl ti thi im , ta schn la di chuyn no? Chng hn hnh trn, ta nn di

chuyn (1), (2), (6), hay (7) ? Bi ton ny hon ton c cu trc thch hp c thgiibng A* (tng strng thi c thc ca bn cl n2! vi n l kch thc bn cv mitrng thi l mt hon vca tp n2con s).

Ti mt trng thi ang xt Tk, t d(i,j)l s cn di chuyn a con s (i,j) vng vtr ca n trng thi ch.

Hm c lng h ti trng thi Tk bt kbng tng ca cc d(i,j) sao cho vtr (i,j)khng phi l trng.

Nhvy i vi trng thi hnh ban u, hm f(Tk) sc gi trl

Fk=2+1+3+1+0+1+2+2=12

III.11. Cc chin lc tm kim lai

Chng ta bit qua 4 kiu tm kim : leo o (L), tm theo chiu su (MC), tm theochiu rng (BR) v tm kim BFS. Bn kiu tm kim ny c thc xem nh4 thicc ca khng gian lin tc bao gm cc chin lc tm kim khc nhau. gii thchiu ny r hn, stin hn cho chng ta nu nhn mt chin lc tm kim li gii dihai chiu sau :

Chiu khnng quay lui (R): l khnng cho php quay li xem xt nhngtrng thi xt n trc nu gp mt trng thi khng thi tip.

Chiu phm vi ca snh gi (S): scc trng thi xt n trong mi quytnh.

Hnh : Tng quan gia cc chin lc leo o, quay lui v tt nht

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TTNT

Theo hng R, chng ta thy leo o nm mt thi cc (n khng cho php quay linhng trng thi cha c xt n), trong khi tm kim quay lui v BFS mt thicc khc (cho php quay li tt ccc hng i cha xt n). Theo hng S chng tathy leo o v ln ngc nm mt thi cc (chtp trung vo mt phm vi hp trn tpcc trng thi mi to ra ttrng thi hin ti) v BFS nm mt thi cc khc (trong khi

BF xem xt ton btp cc con ng c, bao gm cnhng con ng mi c tora cng nhtt cnhng con ng khng c xt ti trc y trc mi mt quytnh).

Nhng thi cc ny c trc quan ha bng hnh trn. Vng in m biu din mt mtphng lin tc cc chin lc tm kim m n kt hp mt sc im ca mt trong bathi cc (leo o, chiu su, BFS) c c mt ha hp cc c tnh tnh ton cachng.

Nu chng ta khng bnhcn thit p dng thut ton BFS thun ty. Ta c thkt hp BFS vi tm theo chiu su gim bt yu cu bnh. Dnhin, ci gi m ta

phi trl slng cc trng thi c thxt n ti mt bc snhi. Mt loi kt hpnhthc chra trong hnh di. Trong hnh ny, thut gii BFS c p dng tinh ca thtm kim (biu din bng vng t tm) v tm kim theo chiu su cp dng ti y (biu din bi tam gic t nht). u tin ta p dng BFS vo trng thiban u T0mt cch bnh thng. BFS sthi hnh cho n mt lc no , slngtrng thi c lu trchim dng mt khng gian bnhvt qu mt mc cho phpno . n lc ny, ta sp dng tm kim chiu su xut pht ttrng thi tt nhtTmax trong OPEN cho ti khi ton bkhng gian con pha "di" trng thi cduyt ht. Nu khng tm thy kt qu, trng thi Tmax ny c ghi nhn l khng dnn kt quv ta li chn ra trng thi tt thhai trong OPEN v li p dng tm kimchiu su cho cho phn khng gian pha "di" trng thi ny....

Hnh : Chin lc lai BFS-MC trong , BFS p dng ti nh v MC ti y.

Mt cch kt hp khc l dng tm kim chiu su ti nh khng gian tm kim v BFSc dng ti y. Chng ta p dng tm kim chiu su cho ti khi gp mt trng thiTk m su (strng thi trung gian) ca n vt qu mt ngng d0no . Ti imny, thay v ln ngc trli, ta p dng kiu tm kim BFS cho phn khng gian pha"di" bt u tTk cho ti khi n trvmt gii php hoc khng tm thy. Nu n

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TTNT

khng tm thy kt qu, chng ta ln ngc trli v li dng BFS khi t su d0.Tham sd0sc chn sao cho bnhdng cho tm kim BFS trn khng gian "di"mc d0skhng vt qu mt hng scho trc. R rng ta ta khng dg xc nhc d0(v ni chung, ta kh nh gi c khng gian bi ton rng n mc no). Tuynhin, kiu kt hp ny li c mt thun li. Phn y khng gian tm kim thng cha

nhiu thng tin "bch" hn l phn nh. (Chng hn, tm ng i n khu trung tmca thnh ph, khi cng n gn khu trung tm y th bn cng ddng tin ntrung tm hn v c nhiu "du hiu" ca trung tm xut hin xung quanh bn!). Ngha l,cng tin vpha y ca khng gian tm kim, c lng h thng cng trnn chnhxc hn v do , cng ddn ta n kt quhn.

Hnh : Chin lc lai BFS-MC trong , MC p dng ti nh v BFS ti y.

Cn mt kiu kt hp phc tp hn na. Trong , BFS c thc hin cc bv chiu

su c thc hin ton cc. Ta bt u tm kim theo BFS cho ti khi mt slng bnhxc nh M0c dng ht. Ti im ny, chng ta xem tt cnhng trng thi trongOPEN nhnhng trng thi con trc tip ca trng thi ban u v chuyn giao chngcho tm kim chiu su. Tm kim chiu su schn trng thi tt nht trong nhng trngthi con ny v "bnh trng" n dng BFS, ngha l n chuyn trng thi chn chotm kim BFS cc bcho n khi mt lng bnhM0li c dng ht v trng thicon mi trong OPEN li tip tc c xem nhnt con ca nt "bnh trng"...Nu vic"bnh trng" bng BFS tht bi th ta quay lui li v chn nt con tt thhai ca tpOPEN trc , ri li tip tc bnh trng bng BFS...

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TTNT

Hnh : Chin lc lai BFS-MC trong , BFS c p dng cc bv chiu su c pdng ton cc.

C mt cch phi hp ni ting khc c gi l tm kim theo giai on c thc hinnhsau. Thay v lu trtrong bnhton bcy tm kim c sinh ra bi BFS, ta ch

gili cy con c trin vng nht. Khi mt lng bnhM0c dng ht, ta snhdu mt tp con cc trng thi trong OPEN(nhng trng thi c gi trhm f thp nht)gili; nhng ng i tt nht qua nhng trng thi ny cng sc ghi nhv ttcphn cn li ca cy bloi b. Qu trnh tm kim sau stip tc theo BFS cho tikhi mt lng bnhM0li c dng ht v cth. Chin lc ny c thc xemnhl mt slai ghp gia BF v leo o. Trong , leo o thun ty loi btt cnhng chgili phng n tt nht cn tm kim theo giai on loi btt cnhng chgili tpcc phng n tt nht.

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TTNT

A. TNG QUAN TR TUNHN TO

I. MU

Chto c nhng cmy thng minh nhcon ngi (thm ch thng minh hn con

ngi) l mt c mchy bng ca loi ngi thng ngn nm nay. Hn bn c cnnhn nh khoa hc Alan Turing cng nhng ng gp to ln ca ng trong lnh vc trtunhn to. Nng lc my tnh ngy cng mnh ml mt iu kin ht sc thun licho tr tunhn to. iu ny cho php nhng chng trnh my tnh p dng cc thutgii tr tunhn to c khnng phn ng nhanh v hiu quhn trc. Skin my tnhDeep Blue nh bi kin tng cvua thgii Casparov l mt minh chng hng hncho mt bc tin di trong cng cuc nghin cu vtr tunhn to. Tuyc thnh bic Casparov nhng Deep Blue l mt cmy chbit nh c! N thm ch khng cc tr thng minh sng ca mt a b bit ln ba nhnhn din c nhng ngithn, khnng quan st nhn bit thgii, tnh cm thng, ght, ... Ngnh tr tunhnto c nhng bc tin ng k, nhng mt tr tunhn to thc svn chc trong

nhng bphim khoa hc gitng ca Hollywood. Vy th ti sao chng ta vn nghincu vtr tunhn to? iu ny cng tng tnhc mchto vng ca cc nh gikim thut thi Trung C, tuy cha thnh cng nhng chnh qu trnh nghin cu lmsng tnhiu vn .

Mc d mc tiu ti thng ca ngnh TTNT l xy dng mt chic my c nng lc tduy tng tnhcon ngi nhng khnng hin ti ca tt ccc sn phm TTNT vncn rt khim tn so vi mc tiu ra. Tuy vy, ngnh khoa hc mi mny vnang tin bmi ngy v ang tra ngy cng hu dng trong mt scng vic i hitr thng minh ca con ngi. Hnh nh sau sgip bn hnh dung c tnh hnh cangnh tr tunhn to.

Trc khi bc vo tm hiu vtr tunhn to, chng ta hy nhc li mt nh nghac nhiu nh khoa hc chp nhn.

Mc tiu ca ngnh khoa hc tr tunhn to ?

To ra nhng chic my tnh c khnng nhn thc, suy lun v phn ng.

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TTNT

Nhn thc c hiu l khnng quan st, hc hi, hiu bit cng nhnhng kinhnghim vthgii xung quanh. Qu trnh nhn thc gip con ngi c tri thc. Suy lunl khnng vn dng nhng tri thc sn c phn ng vi nhng tnh hung hay nhngvn - bi ton gp phi trong cuc sng. Nhn thc v suy lun t a ra nhngphn ng thch hp l ba hnh vi c thni l c trng cho tr tuca con ngi. (D

nhin cn mt yu tna l tnh cm. Nhng chng ta skhng cp n y!). Do, cng khng c g ngc nhin khi mun to ra mt chic my tnh thng minh, ta cnphi trang bcho n nhng khnng ny. Cba khnng ny u cn n mt yu tcbn l tri thc.

Di gc nhn ca tp sch ny, xy dng tr tunhn to l tm cch biu din tri thc,tm cch vn dng tri thcgii quyt vn v tm cch bsung tri thcbngcch "pht hin" tri thc tcc thng tin sn c (my hc).

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TTNT

II. THNG TIN, DLIU V TRI THC

Tri thc l mt khi nim rt tru tng. Do , chng ta skhng cgng a ra mtnh ngha hnh thc chnh xc y. Thay vo , chng ta hy cng nhau cm nhnkhi nim "tri thc" bng cch so snh n vi hai khi nim khc l thng tin v dliu.

Nh bc hc ni ting Karan Sing tng ni rng "Chng ta ang ngp chm trong binthng tin nhng li ang kht tri thc". Cu ni ny lm ni bt skhc bit vlngln vcht gia hai khi nim thng tin v tri thc.

Trong ngcnh ca ngnh khoa hc my tnh, ngi ta quan nim rng dliu l cccon s, chci, hnh nh, m thanh... m my tnh c thtip nhn v xl. Bn thn dliu thng khng c ngha i vi con ngi. Cn thng tin l tt cnhng g m conngi c thcm nhn c mt cch trc tip thng qua cc gic quan ca mnh (khugic, vgic, thnh gic, xc gic, thgic v gic quan th6) hoc gin tip thng quacc phng tin kthut nhtivi, radio, cassette,... Thng tin i vi con ngi lun c

mt ngha nht nh no . Vi phng tin my tnh (m cthl cc thit bu ra),con ngi stip thu c mt phn dliu c ngha i vi mnh. Nu so vlng,dliu thng nhiu hn thng tin.

Cng c thquan nim thng tin l quan hgia cc dliu. Cc dliu c sp xptheo mt ththoc c tp hp li theo mt quan hno scha ng thng tin.Nu nhng quan hny c chra mt cch r rng th l cc tri thc. Chng hn :

Trong ton hc :

Bn thn tng con sring lnh1, 1, 3, 5, 2, 7, 11, ... l cc dliu. Tuy nhin, khi t

chng li vi nhau theo trt tnhdi y th gia chng bt u c mt mi lin h

Dliu : 1, 1, 2, 3, 5, 8, 13, 21, 34, ....

Mi lin hny c thc biu din bng cng thc sau : Un = Un-1+ Un-2.

Cng thc nu trn chnh l tri thc.

Trong vt l :

Bn sau y cho chng ta bit so vin tr(R), in th(U) v cng dng in(I) trong mt mch in.

I U R

5 10 2

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TTNT

2.5 20 8

4 12 3

7.3 14.6 2

Bn thn nhng con strong cc ct ca bn trn khng c my ngha nu ta tch richng ta. Nhng khi t knhau, chng cho thy c mt slin hno . V milin hny c thc din tbng cng thc n gin sau :

Cng thc ny l tri thc.

Trong cuc sng hng ngy :

Hng ngy, ngi nng dn vn quan st thy cc hin tng nng, ma, rm v chunchun bay. Rt nhiu ln quan st, h c nhn xt nhsau :

Chun chun bay thp th ma, bay cao th nng, bay va th rm.

Li nhn xt trn l tri thc.

C quan im trn cho rng chnhng mi lin htng minh(c thchng minh c)gia cc dliu mi c xem l tri thc. Cn nhng mi quan hkhng tng minh thkhng c cng nhn. y, ta cng c thquan nim rng, mi mi lin hgia ccdliu u c thc xem l tri thc, bi v, nhng mi lin hny thc stn ti.im khc bit l chng ta cha pht hin ra n m thi. R rng rng "d sao th trit cng vn xoay quanh mt tri" d tri thc ny c c Galil pht hin ra haykhng!

Nhvy, so vi dliu th tri thc c slng t hn rt nhiu. Thut ngt y khngchn gin l mt du nhhn bnh thng m lskt tinh hoc c ng li. Bn hyhnh dung dliu nhl nhng im trn mt phng cn tri thc chnh lphng trnhca ng cong ni tt cnhng im ny li. Chcn mtphng trnh ng cong ta

c thbiu din c v sim!. Cng vy, chng ta cn c nhng kinh nghim, nhnxt thng ng sliu thng k, nu khng, chng ta sngp chmtrong bin thng tinnhnh bc hc Karan Sing cnh bo!.

Ngi ta thng phn loi tri thc ra lm cc dng nhsau :

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TTNT

Tri thc skin : l cc khng nh vmt skin, khi nim no (trong mtphm vi xc nh). Cc nh lut vt l, ton hc, ... thng c xp vo loi ny.(Chng hn : mt tri mc ng ng, tam gic u c 3 gc 600, ...)

Tri thc thtc : thng dng din tphng php, cc bc cn tin hnh, trnh

thay ngn gn l cch gii quyt mt vn . Thut ton, thut gii l mt dng ca trithc thtc.

Tri thc m t: cho bit mt i tng, skin, vn , khi nim, ... c thy, cmnhn, cu to nhthno (mt ci bn thng c 4 chn, con ngi c 2 tay, 2 mt,...)

Tri thc Heuristic : l mt dng tri thc cm tnh. Cc tri thc thuc loi ny thngc dng c lng, phng on, v thng c hnh thnh thng qua kinh nghim.

Trn thc t, rt him c mt tr tum khng cn n tri thc (liu c thc mt ikin tng cvua m khng bit nh choc khng bit cc thcquan trng khng?).

Tuy tri thc khng quyt nh sthng minh (ngi bit nhiu nh l ton hn chachc gii ton gii hn!) nhng n l mt yu tcbn cu thnh tr thng minh.Chnh v vy, mun xy dng mt tr thng minh nhn to, ta cn phi c yu tcbnny. Ty t ra vn u tin l Cc phng php a tri thc vo my tnh cgi l biu din tri thc.

III. THUT TON MT PHNG PHP BIU DIN TRI THC?

Trc khi trli cu hi trn, bn hy thnghxem, liu mt chng trnh gii phngtrnh bc 2 c thc xem l mt chng trnh c tri thc hay khng? ... C ch! Vyth tri thc nm u? Tri thc vgii phng trnh bc hai thc cht c m ha

di dng cc cu lnh if..then..elsetrong chng trnh. Mt cch tng qut, c thkhngnh l tt ccc chng trnh my tnh t nhiu u c tri thc. chnh l tri thcca lp trnh vin c chuyn thnh cc cu lnh ca chng trnh. Bn sthc mc"nhvy ti sao a tri thc vo my tnh li l mt vn ? (v ttrc ti gichngta , ang v stip tc lm nhthm?)". ng nhththt, nhng vn nm ch, cc tri thc trong nhng chng trnh truyn thng l nhng tri thc "cng", ngha ln khng thc thm vo hay iu chnh mt khi chng trnh c bin dch.Mun iu chnh th chng ta phi tin hnh sa li m ngun ca chng trnh (ri sau bin dch li). M thao tc sa chng trnh th chc nhng lp trnh vin mi c thlm c. iu ny slm gim khnng ng dng chng trnh (v a sngi dngbnh thng u khng bit lp trnh).

Bn thnghxem, vi mt chng trnh htrra quyt nh (nhu tcphiu, u tbt ng sn chng hn), liu ngi dng c cm thy thoi mi khng khi mun a vochng trnh nhng kin thc ca mnh th anh ta phi chn mt trong hai cch l (1) tsa li m chng trnh!?(2) tm tc gica chng trnh nhngi ny sa li!?.Chai thao tc trn u khng thchp nhn c i vi bt kngi dng bnhthng no. Hcn c mt cch no chnh hc tha tri thc vo my tnh mtcch ddng, thun tin ging nhhang i thoi vi mt con ngi.

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TTNT

ht nc trong bnh 5

ht nc cn li tbnh 7 sang bnh 5

Mc y bnh 7

Trt ht qua bnh 5 cho n khi bnh 5 y.

Phn cn li chnh l snc cn ong.

Tuy nhin, vi nhng sliu khc, bn phi "my m" li tu tm ra quy trnh nc. Cth, mi mt trng hp sc mt cch nc hon ton khc nhau. Nhvy, nu c mt ai yu cu bn a ra mt cch lm tng qut th chnh bn cng slng tng (dnhin, ngoi trtrng hp bn bit trc cch gii theo tri thc mchng ta sp sa tm hiu y!).

n y, bn hy bnh tm kim li cch thc bn tm kim li gii cho mt trng hpcth. V cha tm ra mt quy tc cthno, bn sthc hin mt lot cc thao tc "cmtnh" nhong y mt bnh, trt mt bnh ny sang bnh kia, ht nc trong mt bnhra... va lm va nhm tnh xem cch lm ny c thi n kt quhay khng. Sau nhiuln th nghim, rt c thbn srt ra c mt skinh nghim nh"khi bnh 7 ync m bnh 5 cha y th hy n sang bnh 5 cho n khi bnh 5 y"... Vy th tisao bn li khng th"truyn" nhng kinh nghim ny cho my tnh v cho my tnh"my m" tm cc thao tc cho chng ta? iu ny hon ton c li, v my tnh c khnng "my m" hn hn chng ta! Nu nhng "kinh nghim" m chng ta cung cp chomy tnh khng gip chng ta tm c li gii, chng ta sthay thn bng nhng kinhnghim khc v li tip tc my tnh tm kim li gii!

Chng ta hy pht biu li bi ton mt cch hnh thc hn.

Khng lm mt tnh tng qut, ta lun c thgisrng VX


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xt cc lut ktip, nu ht lut, quay trli lut u tin. Qu trnh tip din cho n khit c iu kin kt thc ca bi ton.

Ba lut ny c m tnhsau :

(L1) Nu bnh X y th ht nc trong bnh X i.

(L2) Nu bnh Y rng th y nc vo bnh Y.

(L3) Nu bnh X khng y v bnh Y khng rng th hy trt nc t? bnh Y sang bnh X(cho n khi bnh X y hoc bnh Y ht nc).

Trn thc t, lc u gii trng hp tng qut ca bi ton ny, ngi ta dng n hn 15 lut (kinh nghim) khc nhau. Tuy nhin, sau ny, ngi ta rt gn li chcn 3 lut nhtrn.

Bn c thddng chuyn i cch gii ny thnh chng trnh nhsau :

. . .

x : = 0; y : = 0;

WHI LE ( ( x z) AND ( yz) ) DO BEGI N

I F ( x = Vx) THEN x : = 0;

I F ( y = 0) THEN ( y: = Vy) ;

I F ( y > 0) THEN BEGI N

k: = mi n( Vx - x, y) ;

x : = x + k;

y : = y - k;

END;

END;

. . .

Th"chy" chng trnh trn vi sliu cthl :

Vx = 3, Vy = 4 v z = 2

Ban u : x = 0, y = 0

Lut (L2) -> x = 0, y = 4

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Lut (L3) -> x = 3, y = 1

Lut (L1) -> x = 0, y = 1

Lut (L3) -> x = 1, y = 0

Lut (L2) -> x = 1, y = 4

Lut (L3) -> x = 3, y = 2

3 lut m chng ta ci t trong chng trnh trn c gi l cstri thc. Cncch thc tm kim li gii bng cch duyt tun ttng lut v p dng n c gi lng csuy din. Chng ta snh ngha chnh xc hai thut ngny cui mc.

Ngi ta chng minh c rng, bi ton ong nc chc li gii khi snc cn ong l mt bi sca c schung ln nht ca thtch hai bnh.

z = n USCLN(VX, VY) (vi n nguyn dng)

Cch gii quyt vn theo kiu ny khc so vi cch gii bng thut ton thng thngl chng ta khng a ra mt trnh tgii quyt vn cthm cha ra cc quy tcchung chung (di dng cc lut), my tnh sda vo (p dng cc lut) txydngmt quy trnh gii quyt vn . iu ny cng ging nhvic chng ta gii tonbng cch a ra cc nh l, quy tc lin quan n bi ton m khng cn phi chracch gii cth.

Vy th im th vnm im no? Bn sc thcm thy rng chng ta vn angdng tri thc "cng" ! (v cc tri thc vn l cc cu lnh IF c ci sn trong chngtrnh). Thc ra th chng trnh ca chng ta "mm" hn mt t ri y. Nu khngtin, cc bn hy quan st phin bn ktip ca chng trnh ny.

FUNCTI ON DK( L I NTEGER) : BOOLEAN;

BEGI N

CASE L OF

1 : DK : = ( x = Vx) ;

2 : DK : = ( y = 0) ;

3 : DK : = ( y>0) ;

END;

END;

PROCEDURE Thi Hanh( L I NTEGER) : BOOLEAN;

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BEGI N

CASE L OF

1 : x : = 0;

2: y : = Vy;

3 : BEGI N

k : = mi n( Vx- x, y) ;

x : = x+k;

y : = y- k;

END;

END;

END;

CONST SO_LUAT = 3;

BEGI N

WHI LE ( xz) AND ( yz) DO BEGI N

FOR i : =1 TO SO_LUAT DO

I F DK( L) THEN Thi Hanh(L) ;

END;

END.

on chng trnh chnh cng thi hnh bng cch ln lt xt qua 3 lnh IF nhchngtrnh u tin. Tuy nhin, y, biu thc iu kin c thay thbng hm DK v cchnh ng ng vi iu kin c thay thbng thtc ThiHanh. Tnh cht "mm"hn ca chng trnh ny thhin ch, nu mun bsung "tri thc", ta chphi iuchnh li cc hm DK v ThiHanh m khng cn phi sa li chng trnh chnh.

By gihy gisrng ta c hm v thtc c bit sau :

FUNCTION GiaTriBool(DK : String) : BOOLEAN;

PROCEDURE ThucHien(ThaoTac : String) ;

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hm GiaTriBool nhn vo mt chuiiu kin, n sphn tch chui, tnh ton ri trragi trBOOLEAN ca biu thc ny.

V d: GiaTriBoolean(60 ; 9;

CacLuat[ 1] . ThaoTac : = x: =0 ;

CacLuat[ 2] . ThaoTac: = y: =Vy ;

CacLuat [ 3] . ThaoTac: = k: =mi n( Vx- x, y) , x: =x+k, y: =y- k ;

END;

BEGI N

WHI LE ( xz) AND ( yz) DO BEGI N

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FOR i : =1 TO SO_LUAT DO

I F Gi aTr i Bool ean( CacLuat [ i ] . DK)

THEN ThucHi en(CacLuat [ i ] . ThaoTac) ;

END;

END.

Chng ta tm cho rng trong qu trnh chng trnh thi hnh, ta c thddng thay isphn tmng CacLuat (cc ngn nglp trnh sau ny nhVisual C++, Delphi ucho php iu ny). Vi chng trnh ny, khi mun sa i "tri thc", bn chcn thayi gi trmng Luat l xong.

Tuy nhin, ngi dng vn gp kh khn khi mun bsung hoc hiu chnh tri thc. Hcn phi nhp cc chui i loi nhx=0 hoc k:=min(Vx-x,y) ...Cc chui ny, tuy

c ngha i vi chng trnh nhng vn cn kh xa li vi ngi dng bnhthng. Chng ta cn gim bt "khong cch" ny li bng cch a ra nhng chui iukin hoc thao tc c ngha trc tipi vi ngi dng. Chng trnh sc chuyni li cc iu kin v thao tc ny sang dng ph hp vi chng trnh.

lm c iu trn. Chng ta cn phi lit k c cc trng thi v thao tc cbnca bi ton ny. Sau y l mt strng thi v thao tc cbn.

Trng thi cbn :

Bnh X y, Bnh X rng, Bnh X khng rng, Bnh X c n lt nc.

Thao tc

ht nc trong bnh, y nc trong bnh, nc tbnh A sang bnh B cho nkhi B y hoc A rng.

Lu rng ta khng thc thao tc "n lt nc tA sang B" v bi ton ginh rng ccbnh u khng c vch chia, hn na nu ta bit cch n lt nc tA sang B th li gii biton trthnh qu n gin.

"Mc y X"

"z lt nc tX sang Y"

V y l mt bi ton n gin nn bn c thdnhn thy rng, cc trng thi cbn v thao tcchng c g khc so vi cc iu kin m chng ta a ra.

Ktip, ta svit cc on chng trnh cho php ngi dng nhp vo cc lut (dngnu ... th ...) c hnh thnh tcc trng thi v iu kin cbn ny, ng thi tin

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hnh chuyn sang dng my tnh c thxl c nhv dtrn. Chng ta skhngbn n vic ci t cc on chng trnh giao tip vi ngi dng y.

Nhvy, so vi chng trnh truyn thng (c cu to thai "cht liu" cbn l dliu v thut ton), chng trnh tr tunhn to c cu to thai thnh phn l cs

tri thc (knowledge base) v ng csuy din (inference engine).

Cstri thc: l tp hp cc tri thc lin quan n vn m chng trnh quan tmgii quyt.

ng csuy din: l phng php vn dng tri thc trong cstri thc gii quytvn .

Nu xt theo quan nim biu din tri thc m ta va bn lun trn th cstri thc chl mt dng dliu c bit v ng csuy din cng chl mt dng ca thut ton cbit m thi. Tuy vy, c thni rng, cstri thc v ng csuy din l mt bc tinha mi ca dliu v thut ton ca chng trnh! Bn c thhnh dung ng csuy

dinging nhmt loi ng ctng qut, c chun hac thdng vn hnhnhiu loi xe mykhc nhau v cstri thc chnh l loi nhin liu c bit vnhnh loi ng cny !

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Cstri thc cng gp phi nhng vn tng tnhnhng csdliu khc nhstrng lp, tha, mu thun. Khi xy dng cstri thc, ta cng phi ch n nhngyu tny. Nhvy, bn cnh vn biu din tri thc, ta cn phi ra cc phngphp loi bnhng tri thc trng lp, tha hoc mu thun. Nhng thao tc ny sc thc hin trong qu trnh ghi nhn tri thc vo hthng. Chng ta scp n

nhng phng php ny trong phn tm hiu vcc lut dn.Hnh nh trn tm tt cho chng ta thy cu trc chung nht ca mt chng trnh tr tunhn to.

B. CC PHNG PHP BIU DIN TRI THC TRN MY TNH

V. LOGIC MNH

y c ll kiu biu din tri thc n gin nht v gn gi nht i vi chng ta. Mnhl mt khng nh, mt pht biu m gi trca n chc thhoc l ng hoc l sai.

V d:

pht biu "1+1=2" c gi trng.

pht biu "Mi loi c c thsng trn b" c gi trsai.

Gi trca mnh khng chphthuc vo bn thn mnh . C nhng mnh m gi trca n lun ng hoc sai bt chp thi gian nhng cng c nhng mnh m gi trca n li phthuc vo thi gian, khng gian v nhiu yu tkhc quan khc.Chng hn nhmnh : "Con ngi khng thnhy cao hn 5m vi chn trn" l ng

khi tri t , cn nhng hnh tinh c lc hp dn yu th c thsai.

Ta k hiu mnh bng nhng chci la tinh nha, b, c, ...

C 3 php ni cbn to ra nhng mnh mi tnhng mnh csl php hi(), giao() v phnh ()

Bn c chn hn tng sdng logic mnh trong chng trnh rt nhiu ln (nhtrongcu trc lnh IF ... THEN ... ELSE) biu din cc tri thc "cng" trong my tnh !

Bn cnh cc thao tc tnh ra gi trcc mnh phc tgi trnhng mnh con,

chng ta c c mt cchsuy din nhsau :

Modus Ponens: Nu mnh A l ng v mnh AB l ng th gi trca B sl ng.

Modus Tollens: Nu mnh AB l ng v mnh B l sai th gi trca A slsai.

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Cc php ton v suy lun trn mnh c cp nhiu n trong cc ti liu vton nn chng ta skhng i vo chi tit y.

VI. LOGIC VT

Biu din tri thc bng mnh gp phi mt trngi cbn l ta khng thcan thipvo cu trc ca mt mnh . Hay ni mt cch khc l mnh khng c cu trc .iu ny lm hn chrt nhiu thao tc suy lun . Do , ngi ta a vo khi nimvtv lng t(- vi mi, - tn ti) tng cng tnh cu trc ca mt mnh .

Trong logic vt, mt mnh c cu to bi hai thnh phn l cc i tng tri thcv mi lin hgia chng (gi l vt). Cc mnh sc biu din di dng :

Vt(, , , )

Nhvy biu din vca cc tri cy, cc mnh sc vit li thnh :

Cam c vNgt V(Cam, Ngt)

Cam c mu Xanh Mu (Cam, Xanh)

...

Kiu biu din ny c hnh thc tng tnhhm trong cc ngn nglp trnh,cc i tng tri thc chnh l cc tham sca hm, gi trmnh chnh l ktquca hm (thuc kiu BOOLEAN).

Vi vt, ta c thbiu din cc tri thc di dng cc mnh tng qut, l nhngmnh m gi trca n c xc nh thng qua cc i tng tri thc cu to nn n.

Chng hn tri thc : "A l bca B nu B l anh hoc em ca mt ngi con ca A"cthc biu din di dng vtnhsau :

B(A, B) = Tn ti Z sao cho : B(A, Z) v (Anh(Z, B) hoc Anh(B,Z))

Trong trng hp ny, mnh B(A,B) l mt mnh tng qut

Nhvy nu ta c cc mnh csl :

a)B("An", "Bnh") c gi trng (Anh l bca Bnh)

b) Anh("T", "Bnh") c gi trng (T l anh ca Bnh)

th mnh c) B("An", "T") sc gi trl ng. (An l bca T).

R rng l nu chsdng logic mnh thng thng th ta skhng thtm c mtmi lin hno gia c v a,b bng cc php ni mnh , , . T, ta cng khng

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thtnh ra c gi trca mnh c. Sdnhvy v ta khng ththhin tng minhtri thc "(A l bca B) nu c Z sao cho (A l bca Z) v (Z anh hoc em C)"didng cc mnh thng thng. Chnh c trng ca vt cho php chng ta thhinc cc tri thc dng tng qut nhtrn.

Thm mt sv dna cc bn thy r hn khnng ca vt:

Cu cch ngn "Khng c vt g l ln nht v khng c vt g l b nht!" c thcbiu din di dng vtnhsau :

LnHn(x,y) = x>y

NhHn(x,y) = x


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GT1, GT2, ..., GTn KL1, KL2, ..., KLm

Trong cc GTi v KLi l cc mnh c xy dng tcc bin mnh v 3 phpni cbn : , ,

B2 : Chuyn vcc GTi v KLi c dng phnh.

V d:

p q, (r s), g, p r s, p

p q, p r, p (r s), g, s

B3 :Nu GTi c php th thay thphp bng du ","

Nu KLi c php th thay thphp bng du ","

V d:

p q, r (p s) q, s

p, q, r, p s q, s

B4 :Nu GTi c php th tch thnh hai dng con.

Nu KLi c php th tch thnh hai dng con.

V d:

p, p q q

p, p q p, q q

B5 : Mt dng c chng minh nu tn ti chung mt mnh chai pha.

V d:

p, q

qc chng minhp, p q pp, qB6 :

a) Nu mt dng khng cn php ni hoc chai vv 2 vkhng c chung mtbin mnh th dng khng c chng minh.

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b) Mt vn c chng minh nu tt cdng dn xut tdng chun ban u uc chng minh.

VII.2 Thut gii Robinson

Thut gii ny hot ng da trn phng php chng minh phn chng.

Phng php chng minh phn chng

Chng minh php suy lun (a b) l ng (vi a l githit, b l kt lun).

Phn chng : gisb sai suy ra b l ng.

Bi ton c chng minh nu a ng v b ng sinh ra mt mu thun.

B1: Pht biu li githit v kt lun ca vn di dng chun nhsau :

GT1, GT2, ...,GTn KL1, KL2, .., KLm

Trong : GTi v KLj c xy dng tcc bin mnh v cc php ton : , ,

B2 :Nu GTi c php th thay bng du ","

Nu KLi c php th thay bng du ","

B3: Bin i dng chun B1 vthnh danh sch mnh nhsau :

{ GT1, GT2, ..., GTn , KL1, KL2, ..., KLm }

B4: Nu trong danh sch mnh bc 2 c 2 mnh i ngu nhau th bi tonc chng minh. Ngc li th chuyn sang B4. (a v a gi l hai mnh i ngunhau)

B5: Xy dng mt mnh mi bng cch tuyn mt cp mnh trong danh schmnh bc 2. Nu mnh mi c cc bin mnh i ngu nhau th cc bin c loi b.

V d: p qr s q

Hai mnh q, q l i ngu nn sc loi b

p r s

B6: Thay thhai mnh va tuyn trong danh sch mnh bng mnh mi.

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V d:

{ p q, r s q , w r, s q }

{ p r s , w r, s q }

B7: Nu khng xy dng c thm mt mnh mi no v trong danh sch mnh khng c 2 mnh no i ngu nhau th vn khng c chng minh.

V d: Chng minh rng

p q, q r, r s, u s p, u

B3: {p q, q r, r s, u s, p, u}B4 : C tt c6 mnh nhng cha c mnh no i ngu nhau.

B5 : tuyn mt cp mnh (chn hai mnh c bin i ngu). Chn hai mnh u :

p qqr p rDanh sch mnh thnh :

{p r, r s, u s, p, u }

Vn cha c mnh i ngu.

Tuyn hai cp mnh u tin

p r rs p sDanh sch mnh thnh {p s, u s, p, u }

Vn cha c hai mnh i ngu

Tuyn hai cp mnh u tin

p su sp u

Danh sch mnh thnh : {p u, p, u }

Vn cha c hai mnh i ngu

Tuyn hai cp mnh :

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p u u p

Danh sch mnh trthnh : {p, p }

C hai mnh i ngu nn biu thc ban u c chng minh.

VIII. BIU DIN TRI THC SDNG LUT DN XUT (LUTSINH)

VIII.1. Khi nim

Phng php biu din tri thc bng lut sinh c pht minh bi Newell v Simon tronglc hai ng ang cgng xy dng mt hgii bi ton tng qut. y l mt kiu biudin tri thc c cu trc. tng cbn l tri thc c thc cu trc bng mt cpiu kin hnh ng: "NU iu kinxy ra TH hnh ng sc thi hnh".Chng hn : NU n giao thng l TH bn khng c i thng, NU my tnh

mm khng khi ng c TH kim tra ngun in,

Ngy nay, cc lut sinh trnn phbin v c p dng rng ri trong nhiu hthng tr tunhn to khc nhau. Lut sinh c thl mt cng cm tgii quyt ccvn thc tthay cho cc kiu phn tch vn truyn thng. Trong trng hp ny,cc lut c dng nhl nhng chdn (tuy c thkhng hon chnh) nhng rt hu chtrgip cho cc quyt nh trong qu trnh tm kim, t lm gim khng gian tmkim. Mt v dkhc l lut sinh c thc dng bt chc hnh vi ca nhngchuyn gia. Theo cch ny, lut sinh khng chn thun l mt kiu biu din tri thctrong my tnh m l mt kiu biu din cc hnh vi ca con ngi.

Mt cch tng qut lut sinh c dng nhsau :

P1P2... Pn QTy vo cc vn ang quan tm m lut sinh c nhng ngngha hay cu to khcnhau :

Trong logic vt: P1, P2, ..., Pn, Q l nhng biu thc logic.

Trong ngn nglp trnh, mi mt lut sinh l mt cu lnh.

IF (P1AND P2AND .. AND Pn) THEN Q.

Trong l thuyt hiu ngn ngtnhin, mi lut sinh l mt php dch :

ONE mt.

TWO hai.

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JANUARY thng mt

biu din mt tp lut sinh, ngi ta thng phi chr hai thnh phn chnh sau :

(1) Tp cc skin F(Facts)

F = { f1, f2, ... fn}

(2) Tp cc quy tc R (Rules) p dng trn cc skin dng nhsau :

f1^ f2^ ... ^ fi q

Trong , cc fi , q u thuc F

V d: Cho 1 cstri thc c xc nh nhsau :

Cc skin : A, B, C, D, E, F, G, H, K

Tp cc quy tc hay lut sinh (rule)

R1 : A E

R2 : B D

R3 : H A

R4 : E G C

R5 : E K B

R6 : D E K C

R7 : G K F A

VIII.2. Cchsuy lun trn cc lut sinh

Suy din tin : l qu trnh suy lun xut pht tmt sskin ban u, xc nh ccskin c thc "sinh" ra tskin ny.

Skin ban u : H, K

R3 : H A {A, H. K }

R1 : A E { A, E, H, H }

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R5 : E K B { A, B, E, H, K }

R2 : B D { A, B, D, E, H, K }

R6 : D E K C { A, B, C, D, E, H, K }

Suy din li : l qu trnh suy lun ngc xut pht tmt sskin ban u, ta tmkim cc skin "sinh" ra skin ny. Mt v dthng gp trong thc tl xutpht tcc tnh trng ca my tnh, chn on xem my tnh bhng hc u.

V d:

Tp cc skin :

cng l "hng" hay "hot ng bnh thng" Hng mn hnh. Lng cp mn hnh. Tnh trng n cng l "tt" hoc "sng" C m thanh c cng. Tnh trng n mn hnh "xanh" hoc "chp " Khng sdng c my tnh. in vo my tnh "c" hay "khng"

Tp cc lut :

R1. Nu ( (cng "hng") hoc (cp mn hnh "lng")) th khng sdng c mytnh.

R2. Nu (in vo my l "c") v ( (m thanh c cng l "khng") hoc tnh trngn cng l "tt")) th (cng "hng").

R3. Nu (in vo my l "c") v (tnh trng n mn hnh l "chp ") th (cp mnhnh "lng").

xc nh c cc nguyn nhn gy ra skin "khng sdng c my tnh", taphi xy dng mt cu trc thgi l thAND/OR nhsau :

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Nhvy l xc nh c nguyn nhn gy ra hng hc l do cng hng hay cpmn hnh lng, hthng phi ln lt i vo cc nhnh kim tra cc iu kin nhin vo my "c", m thanh cng "khng"Ti mt bc, nu gi trcn xc nhkhng thc suy ra tbt kmt lut no, hthng syu cu ngi dng trc tipnhp vo. Chng hn nhbit my tnh c in khng, hthng shin ra mn hnhcu hi "Bn kim tra xem c in vo my tnh khng (kim tra n ngun)? (C/K)".thc hin c cchsuy lun li, ngi ta thng sdng ngn xp (ghi nhn linhng nhnh cha kim tra).

VIII.3. Vn ti u lut

Tp cc lut trong mt cstri thc rt c khnng tha, trng lp hoc mu thun. Dnhin l hthng c thli cho ngi dng vvic a vo hthng nhng tri thcnhvy. Tuy vic ti u mt cstri thc vmt tng qut l mt thao tc kh (v giacc tri thc thng c quan hkhng tng minh), nhng trong gii hn cstri thcdi dng lut, ta vn c mt sthut ton n gin loi bcc vn ny.

VIII.3.1. Rt gn bn phi

Lut sau hin nhin ng :

AB A (1)

Do lut

A BA CL hon ton tng ng vi

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A BCQuy tc rt gn : C thloi bnhng skin bn vphi nu nhng skin xuthin bn vtri. Nu sau khi rt gn m vphi trthnh rng th lut l lut hinnhin. Ta c thloi bcc lut hin nhin ra khi tri thc.

VIII.3.2. Rt gn bn tri

Xt cc lut :

(L1) A, B C (L2) A X (L3) X C

R rng l lut A, B C c thc thay thbng lut A C m khng lm nhhng n cc kt lun trong mi trng hp. Ta ni rng skin B trong lut (1) l d

tha v c thc loi bkhi lut dn trn.

VIII.3.3. Phn r v kt hp lut

Lut A B C

Tng ng vi hai lut

A C

B C

Vi quy tc ny, ta c thloi bhon ton cc lut c php ni HOC. Cc lut c phpni ny thng lm cho thao tc xl trnn phc tp.

VIII.3.4. Lut tha

Mt lut dn A B c gi l tha nu c thsuy ra lut ny tnhng lut cn li.

V d: trong tp cc lut gm {A B, B C, A C} th lut th3 l lut tha v nc thc suy ra t2 lut cn li.

VIII.3.5. Thut ton ti u tp lut dn

Thut ton ny sti u ha tp lut cho bng cch loi i cc lut c php niHOC, cc lut hin nhin hoc cc lut tha.

Thut ton bao gm cc bc chnh

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B1: Rt gn vphi

Vi mi lut r trong R

Vi mi skin A VPhi(r)

Nu A VTri(r) th Loi A ra khi vphi ca R.

Nu VPhi(r) rng th loi br ra khi hlut dn : R = R {r}

B2 : Phn r cc lut

Vi mi lut r : X1 X2 Xn Y trong R

Vi mi i t1 n n R := R + { Xi Y }

R := R {r}

B3 : Loi blut tha

Vi mi lut r thuc R

Nu VPhi(r) Baong(VTri(r), R-{r}) th R := R {r}

B4 : Rt gn vtri

Vi mi lut dn r : X : A1 A2, , An Y thuc R

Vi mi skin Ai thuc r

Gi lut r1 : X Ai Y

S = ( R {r} ) {r1}

Nu Baong( X Ai , S) Baong(X, R) th loi skin A ra khi X

VIII.4. u im v nhc im ca biu din tri thc bng lut

u im

Biu din tri thc bng lut c bit hu hiu trong nhng tnh hung hthng cn a ra

nhng hnh ng da vo nhng skin c thquan st c. N c nhng u imchnh yu sau y :

Cc lut rt dhiu nn c thddng dng trao i vi ngi dng(v n l mt trong nhng dng tnhin ca ngn ng).

C thddng xy dng c cchsuy lun v gii thch tcc lut.

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Vic hiu chnh v bo tr hthng l tng i ddng.

C thci tin ddng tch hp cc lut m.

Cc lut thng t phthuc vo nhau.

Nhc im

Cc tri thc phc tp i lc i hi qu nhiu (hng ngn) lut sinh.iu ny slm ny sinh nhiu vn lin quan n tc ln qun trhthng.

Thng k cho thy, ngi xy dng hthng tr tunhn to thch sdng lut sinh hn tt cphng php khc (dhiu, dci t) nn hthng tm mi cch biu din tri thc bng lut sinh cho d c phngphp khc thch hp hn! y l nhc im mang tnh chquan ca con

ngi.Cstri thc lut sinh ln slm gii hn khnng tm kim ca

chng trnh iu khin. Nhiu hthng gp kh khn trong vic nh gicc hda trn lut sinh cng nhgp kh khn khi suy lun trn lutsinh.

X. BIU DIN TRI THC SDNG MNG NGNGHA

X.1. Khi nim

Mng ngngha l mt phng php biu din tri thc u tin v cng l phng phpdhiu nht i vi chng ta. Phng php ny sbiu din tri thc di dng mt th, trong nh l cc i tng (khi nim) cn cc cung cho bit mi quan hgiacc i tng (khi nim) ny.

Chng hn : gia cc khi nim chch che, chim, ht, cnh, tc mt smi quan hnhsau :

Chch che l mt loi chim.

Chim bit ht

Chim c cnh

Chim sng trong t

Cc mi quan hny sc biu din trc quan bng mt thnhsau :

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Do mng ngngha l mt loi thcho nn n tha hng c tt cnhng mtmnh ca cng cny. Ngha l ta c thdng nhng thut ton ca thtrn mngngngha nhthut ton tm lin thng, tm ng i ngn nht, thc hin cc cchsuy lun. im c bit ca mng ngngha so vi ththng thng chnh l vicgn mt ngha (c, lm, l, bit, ...) cho cc cung. Trong thtiu chun, vic c mtcung ni gia hai nh chcho bit c slin hgia hai nh v tt ccc cung trongthu biu din cho cng mt loi lin h. Trong mng ngngha, cung ni gia hai

nh cn cho bit gia hai khi nim tng ng c slin hnhthno. Vic gn ngngha vo cc cung ca th gip gim bt c slng thcn phi dng biu din cc mi lin hgia cc khi nim. Chng hn nhtrong v dtrn, nu sdng ththng thng, ta phi dng n 4 loi thcho 4 mi lin h: mt thbiu din mi lin h"l", mt thcho mi lin h"lm", mt cho "bit" v mt cho"c".

Mt im kh th vca mng ngngha l tnh ktha. Bi v ngay ttrong khi nim,mng ngngha hm sphn cp (nhcc mi lin h"l") nn c nhiu nh trongmng mc nhin sc nhng thuc tnh ca nhng nh khc. Chng hn theo mng ngngha trn, ta c thddng trli "c" cho cu hi : "Chch che c lm tkhng?".

Ta c thkhng nh c iu ny v nh "chch che" c lin kt "l" vi nh "chim"v nh "chim" li lin kt "bit" vi nh "lm t" nn suy ra nh "chch che" cng clin kt loi "bit" vi nh "lm t". (Nu , bn snhn ra c kiu "suy lun" mta va thc hin bt ngun tthut ton "loang" hay "tm lin thng" trn th!). Chnhc tnh ktha ca mng ngngha cho php ta c ththc hin c rt nhiu phpsuy din tnhng thng tin sn c trn mng.

Tuy mng ngngha l mt kiu biu din trc quan i vi con ngi nhng khi avo my tnh, cc i tng v mi lin hgia chng thng c biu din di dngnhng pht biu ng t(nhvt). Hn na, cc thao tc tm kim trn mng ngnghathng kh khn (c bit i vi nhng mng c kch thc ln). Do , m hnh mng

ngngha c dng chyu phn tch vn . Sau , n sc chuyn i sangdng lut hoc frame thi hnh hoc mng ngngha sc dng kt hp vi mt sphng php biu din khc.

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X.2. u im v nhc im ca mng ngngha

u im

Mng ngngha rt linh ng, ta c thddng thm vo mng cc nh

hoc cung mi bsung cc tri thc cn thit.

Mng ngngha c tnh trc quan cao nn rt dhiu.

Mng ngngha cho php cc nh c ththa kcc tnh cht tccnh khc thng qua cc cung loi "l", t, c thto ra cc lin kt"ngm" gia nhng nh khng c lin kt trc tip vi nhau.

Mng ngngha hot ng kh tnhin theo cch thc con ngi ghinhn thng tin.

Nhc im

Cho n nay, vn cha c mt chun no quy nh cc gii hn cho ccnh v cung ca mng. Ngha l bn c thgn ghp bt kkhi nimno cho nh hoc cung!

Tnh tha k(vn l mt u im) trn mng sc thdn n nguy cmu thun trong tri thc. Chng hn, nu bsung thm nt "G" vomng nhhnh sau th ta c thkt lun rng "G" bit "bay"!. Sdc

iu ny l v c skhng r rng trong ngngha gn cho mt nt camng. Bn c c thphn i quan im v cho rng, vic sinh ra muthun l do ta thit kmng dchkhng phi do khuyt im camng!. Tuy nhin, xin lu rng, tnh tha ksinh ra rt nhiu mi lin"ngm" nn khnng ny sinh ra mt mi lin hkhng hp ll rt ln!

Hu nhkhng thbin din cc tri thc dng thtc bng mng ngngha v cc khinim vthi gian v trnh tkhng c thhin tng minh trn mng ngngha.

X.3. Mt v dtiu biu

D l mt phng php tng i cv c nhng yu im nhng mng ngnghavnc nhng ng dng v cng c o. Hai loi ng dng tiu biu ca mng ngngha lng dng xl ngn ngtnhin v ng dng gii bi ton tng.

V d1: Trong ng dng xl ngn ngtnhin, mng ngngha c thgip my tnhphn tch c cu trc ca cu t c thphn no "hiu" c ngha ca cu.Chng hn, cu "Chu ang c mt cun sch dy v ci khoi tr" c thc biudin bng mt mng ngngha nhsau :

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V d2 : Gii bi ton tam gic tng qut

Chng ta skhng i su vo v d1 v y l mt vn qu phc tp c thtrnhby trong cun sch ny. Trong v dny, chng ta skho st mt vn n gin hnnhng cng khng km phn c o. Khi mi hc lp trnh, bn thng c gio vincho nhng bi tp nhp mn i loi nh"Cho 3 cnh ca tam gic, tnh chiu di ccng cao", "Cho gc a, b v cnh AC. Tnh chiu di trung tuyn", ... Vi mi bi tpny, vic bn cn lm l ly giy bt ra tm cch tnh, sau khi xc nh cc bc tnhton, bn chuyn n thnh chng trnh. Nu c 10 bi, bn phi lm li vic tnh ton rilp trnh 10 ln. Nu c 100 bi, bn phi lm 100 ln. V tin bun cho bn l slngbi ton thuc loi ny l rt nhiu! Bi v mt tam gic c tt c22yu tkhc nhau!.Khng lmi ln gp mt bi ton mi, bn u phi lp trnh li? Liu c mt chngtrnh tng qut c thtng gii c tt c(vi ngn!) nhng bi ton tam gic thucloi ny khng? Cu trli l C ! V ngc nhin hn na, chng trnh ny li kh ngin. Bi ton ny sc gii bng mng ngngha.

C 22 yu tlin quan n cnh v gc ca tam gic. xc nh mt tam gic hay xy dng mt 1 tam gic ta cn c 3 yu ttrong phi c yu tcnh. Nhvy ckhong C322 -1 (khong vi ngn) cch xy dng hay xc nh mt tam gic. Theothng k, c khong 200 cng thc lin quan n cnh v gc 1 tam gic.

gii bi ton ny bng cng cmng ngngha, ta phi sdng khong 200 nh cha cng thc v khong 22 nh cha cc yu tca tam gic. Mng ngngha chobi ton ny c cu trc nhsau :

nh ca thbao gm hai loi :

nh cha cng thc (k hiu bng hnh chnht) nh cha yu tca tam gic (k hiu bng hnh trn)

Cung : chni tnh hnh trn n nh hnh chnht cho bit yu ttam gic xut hintrong cng thc no (khng c trng hp cung ni gia hai nh hnh trn hoc cungni gia hai nh hnh chnht).

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* Lu : trong mt cng thc lin hgia n yu tca tam gic, ta ginh rng nu bit gitrca n-1 yu tth stnh c gi trca yu tcn li. Chng hn nhtrong cng thc tng 3gc ca tam gic bng 1800th khi bit c hai gc, ta stnh c gc cn li.

Cchsuy din thc hin theo thut ton "loang" n gin sau :

B1 : Kch hot nhng nh hnh trn cho ban u (nhngyu t c gi tr)

B2 : Lp li bc sau cho n khi kch hot c tt cnhngnh ng vi nhng yu tcn tnh hoc khng thkch hotc bt knh no na.

Nu mt nh hnh chnht c cung ni vi nnh hnh trnm n-1nh hnh trn c kch hot th kch hot nhhnh trn cn li (v tnh gi trnh cn li ny thng quacng thc nh hnh chnht).

Gista c mng ngngha gii bi ton tam gic nhhnh sau

V d: "Cho hai gc , v chiu di cnh a ca tam gic. Tnh chiu di ng caohC". Vi mng ngngha cho trong hnh trn. Cc bc thi hnh ca thut ton nhsau :

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Bt u : nh , , aca thc kch hot.Cng thc (1) c kch hot (v, , a c kch hot). Tcng thc (1)tnh c cnh b. nh b c kch hot.

Cng thc (4) c kch hot (v , ). Tcng thc (4) tnh c gc Cng thc (2) c kch hot (v 3 nh , , bc kch hot). Tcngthc (2) tnh c cnh c. nh cc kch hot.

Cng thc (3) c kch hot (v 3 nh a, b, c c kch hot) . Tcngthc (3) tnh c din tch S. nh S c kch hot.

Cng thc (5) c kch hot (v 2 nh S, c c kch hot). Tcngthc (5) tnh c hC. nh hC c kch hot.

Gi trhC c tnh. Thut ton kt thc.

Vmt chng trnh, ta c thci t mng ngngha gii bi ton tam gic bng mtmng hai chiu A trong :

Ct : ng vi cng thc. Mi ct ng vi mt cng thc tam gic khcnhau (nh hnh chnht).

Dng: ng vi yu ttam gic. Mi dng ng vi mt yu ttam gickhc nhau (nh hnh trn).

Phn tA[i, j] = -1ngha l trong cng thc ng vi ctj c yu ttamgic ng vi ct i. Ngc li A[i,j] = 0.

thc hin thao tc "kch hot" mt nh hnh trn, ta t gi trca ton dng ng viyu ttam gic bng 1.

kim tra xem mt cng thc c n-1 yu thay cha (ngha l kim tra iu kin"nh hnh chnht c cung ni vi n nh hnh trn m n-1 nh hnh trn c kchhot"), ta chvic ly hiugia tngs c gi trbng 1 v tng s c gi tr-1trnct ng vi cng thc cn kim tra. Nu kt qubng n, th cng thc c n-1 yut.

Trli mng ngngha cho. Qu trnh thi hnh kch hot c din ra nhsau :

Mng biu din mng ngngha ban u

(1) (2) (3) (4) (5)

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-1 0 0 -1 0 -1 -1 0 -1 0 0 -1 0 -1 0a -1 0 -1 0 0

b -1 -1 -1 0 0

c 0 -1 -1 0 -1

S 0 0 -1 0 -1

hC 0 0 0 0 -1

Khi u : nh , , aca thc kch hot.(1) (2) (3) (4) (5)

1 0 0 1 0 1 1 0 1 0 0 -1 0 -1 0a 1 0 1 1 0

b -1 -1 -1 0 0

c 0 -1 -1 0 -1

S 0 0 -1 0 -1

hC 0 0 0 0 -1

Trn ct (1), hiu (1+1+1 (-1)) = 4 nn dng b sc kch hot.

(1) (2) (3) (4) (5)

1 0 0 1 0 1 1 0 1 0

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0 -1 0 -1 0a 1 0 1 1 0

b 1 1 1 0 0

c 0 -1 -1 0 -1

S 0 0 -1 0 -1

hC 0 0 0 0 -1

Trn ct (4), hiu (1+1+1 (-1)) = 4 nn dng sc kch hot.

(1) (2) (3) (4) (5)

1 0 0 1 0 1 1 0 1 0 0 1 0 1 0a 1 0 1 1 0

b 1 1 1 0 0

c 0 -1 -1 0 -1S 0 0 -1 0 -1

hC 0 0 0 0 -1

Trn ct (2), hiu (1+1+1 (1)) = 4 nn dng cc kch hot.

(1) (2) (3) (4) (5)

1 0 0 1 0 1 1 0 1 0 0 1 0 1 0A 1 0 1 1 0

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B 1 1 1 0 0

C 0 1 1 0 1

S 0 0 -1 0 -1

hC 0 0 0 0 -1

Trn ct (3), hiu (1+1+1 (-1)) = 4 nn dng Sc kch hot.

(1) (2) (3) (4) (5)

1 0 0 1 0 1 1 0 1 0 0 1 0 1 0a 1 0 1 1 0

b 1 1 1 0 0

c 0 1 1 0 1

S 0 0 1 0 1

hC 0 0 0 0 -1

Trn ct (5), hiu (1+1 (1)) = 3 nn dng hC c kch hot.

Khnng ca hthng ny khng chdng li vic tnh ra gi trcc yu tcn thit,vi mt cht sa i, chng trnh ny cn c tha ra cch gii hnh thc ca bi tonv thm ch cn c thchn c cch gii hnh thc ti u (ti u hiu theo ngha lcch gii sdng nhng cng thc n gin nht). Sdc thni nhvy v cch suylun ca ta trong bi ton ny l tm kim theo chiu rng. Do , khi t n kt qu, tac thc rt nhiu cch khc nhau. c thchn c gii php ti u, bn cn phinh ngha c "phc tp" ca mt cng thc. Mt trong nhng tiu chun thng

c dng l slng php nhn, chia, cng, tr, rt cn, tnh sin, cos, ... c p dngtrong cng thc. Cc php tnh sin, cos v rt cn c phc tp cao nht, kn l nhnchia v cui cng l cng tr. Cui cng bn c thci tin li phng php suy lunbng cch vn dng thut ton Avi c lng h=0c thchn ra c "ng i"ti u. Ta chn c lng h=0 v hai l do sau (1) khng gian bi ton nhnn ta khngcn phi gii hn rng tm kim (2) xy dng mt c lng nhvy l tng i khkhn, c bit l lm sao hthng khng nh gi qu cao h.

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Range(min gi tr) : (tng tnhkiu bin), cho bit gi trslot c thnhn nhngloi gi trg (nhsnguyn, sthc, chci, ...)

If added: m tmt hnh ng sc thi hnh khi mt gi trtrong slot c thmvo (hoc c hiu chnh). Thtc thng c vit di dng mt script.

If needed : c sdng khi slot khng c gi trno. Facet m tmt hm tnh ragi trca slot.

Frame : XE HI

Thuc lp:phng tin vn chuyn.

Tn nh sn xut : Audi

Quc gia ca nh sn xut : c

Model:5000 Turbo

Loi xe : Sedan

Trng lng : 3300lb

Slng ca : 4 (default)

Hp s: 3 stng

Slng bnh: 4 (default)

My (tham chiu n frame My)

Kiu : In-line, overhead cam

Sxy-lanh: 5

Khnng tng tc

0-60 : 10.4 giy

dm : 17.1 giy, 85 mph.

Frame MYXy-lanh : 3.19inchTlnn : 3.4incheXng :TurboCharger

lc : 140 hpM

XI.3. Tnh ktha

Trong thc t, mt hthng tr tunhn to thng sdng nhiu frame c lin ktvi nhau theo mt cch no . Mt trong nhng im th vca frame l tnh phn cp.c tnh ny

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