bai giang he co so tri thuc

8/8/2019 Bai Giang He Co So Tri Thuc

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Cc H c s tri th cKBS: Knowledge Based Systems

Nguy n nh ThunKhoa Cng ngh Thng tin

i h c Nha TrangEmail: [email protected]

Nha Trang 4-2007

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H c s tri th cChng 1:T ng quan v H c s tri th c Chng 2:Bi u di n v suy lu n tri th c Chng 3:H MYCIN Chng 4:H h c Chng 5:H th ng m cho cc bi n lin t c


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Ti li u tham kh o

[1] Rich Elaine.Artificial Intelligence.AddisonWesley 1983

[2] Robert I. Levine.Knowledge based systems. Wissenschafs Verlag, 1991

[3] Trung Tu n. H chuyn gia.NXB Giod c 1999

[4] Hong Kim. Gio trnh Cc h c s tri

th c. HQG TP H Ch Minh. 2002

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Chng 1: T ng quan v H c s tri th c

1.1 Khi ni m v H C s tri th c H c s tri th c l chng trnh my tnh c thi t k m hnh ho kh nng gi i

quy t v n c a chuyn gia con ng i.H CSTT l h th ng d a trn tri th c, chophp m hnh ho cc tri thc c a chuyngia, dng tri th c ny gi i quy t v nph c t p thu c cng l nh v c.Hai y u t quan tr ng trong H CSTT l: trith c chuyn gia v l p lu n, tng ng v ih th ng c 2 kh i chnh l C s tri th c v

ng c suy di n.


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1.1 Khi ni m v H CSTT (Ti p)

H Chuyn gia l m t lo i c s tri th c c thi t k cho m t l nh v c ng d ng c

th .V d : H Chuyn gia v ch n on b nhtrong Y khoa, H Chuyn gia ch n onh ng hc c a ng dy i n tho i,H Chuyn gia lm vic nh m t chuyn giath c th v cung c p cc ki n d a trnkinh nghim c a chuyn gia con ng i c avoH Chuyn gia.

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1.1 Khi ni m v H CSTT (Ti p)C s tri th c: Ch a cc tri th c chuyn suv l nh v c nh chuyn gia. C s tri th cbao g m: cc s ki n, cc lu t, cc khini m v cc quan h .

ng c suy di n: b x l tri th c theo mhnh ho theo cch l p lu n c a chuyn gia.

ng c ho t ng trn thng tin v v nang xt, so snh v i tri th c lu trong c s

tri th c r i r t r ak t lu n.K s tri th c (Knowledge Engineer): ng ithi t k , xy d ng v th nghi m H Chuyngia.


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4. B ti p nh n tri th c (Knowledge editor): Lmnhi m v thu nh n tri th c t chuyn gia con ng i(human expert), t k s tri th c v User thng qua ccyu c u v lu tr vo c s tri th c5. C s tri th c: Lu tr , bi u di n cc tri th c m h

m nh n, lm c s cho cc ho t ng c a h . C s tri th c bao g m cc s ki n (facts) v cc l t (rules).6. Vng nh lm vi c (working memory): M t ph nc a H Chuyn gia ch a cc s ki n c a v n ang

xt.

1.2 C u trc c a H Chuyn gia(ti p)

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1.3 H h tr ra quy t nhDSS (Decision Support System)

Ch c nng: H tr ra quy t nhHo t ng theo cch tng tc v i ng i s d ng

Cc tnh ch t c a DSS:H ng n cc quy t nh c a ng i qu n lUy n chuy n v i hon c nh

Tr l i cu h i trong tnh hungDo ng i s d ng kh i ng v ki m sot


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1.4 H h c

Trong nhi u tinh hu ng, s khng c s n tri th cnh:

K s tri th c c n thu nh n tri th c t chuyn gia l nhv c.

C n bi t cc lu t m t l nh v c c th . Bi ton khng c bi u di n t ng minh theo lu t, s

ki n hay cc quan h .C hai ti p c n cho h th ng h c:

H c t k hi u: bao g m vi c hnh th c ha, s a ch acc lu t t ng minh, s ki n v cc quan h .

H c t d li u s : c p d ng cho nh ng h th ng c m hnh d i d ng s lin quan n cc k thu t

nh m t i u cc tham s . H c theo d ng s bao g mm ng Neural nhn t o, thu t gi i di truy n, bi ton t i utruy n th ng. Cc k thu t h c theo s khng t o raCSTT t ng minh.

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1.5 H i u khi n m

M ha: Chuy n i gi tr r u vo thnhcc vector m Xc nh cc lu t h p thnh v thu t tonxc nh gi tr mGi i m : Phng php i m tr ng tm


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1.6 ng d ng c a H C s tri th c

1. Di n gi i (Interpretation): M ttnh hu ngcc d li u thu th p c

2. D bo (Prediction): a ra cc tri th c vd bo m t tnh hu ng: d bo gi c ,

3. Thi t k (Design): L a ch n c u hnh phh p, v d : s p x p cng vi c.

4. Ch n on (Diagnosis): Da vo cc d

li u quan st c, xc nh cc l i h nghc.

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1.6 ng d ng c a H C s tri th c(ti p)

5. V ch k ho ch (Planing): to l p ccphng n hnh ng.

6. D n d t (Monotoring): So snh d li u vcc k t qu ho t ng.

7. G r i (Debugging): M t cc phngphp kh c ph c c a h th ng.

8. Gi ng d y (Instruction): Sa ch a cc l ic a ng i h c trong qu trnh hc t p.9. i u khi n (Control): dn d t dng i u

t ng th c a h th ng.


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Chng 2: Bi u di n v suy lu n tri th c

2.1. M utri th c , l nh v c v bi u di n tri th c .

2.2. Cc lo i tri th c: c chia thnh 5 lo i1. Tri th c th t c: m t cch th c gi i quy t m t v n . Lo i

tri th c ny a ra gi i php th c hi n m t cng vi c no. Cc d ng tri th c th t c tiu bi u th ng l cc lu t,

chi n l c, lch trnh v th t c.2. Tri th c khai bo: cho bi t m t v n c th y nh th

no. Lo i tri th c ny bao g m cc pht bi u n gi n, d id ng cc kh ng nh logic ng ho c sai. Tri th c khai bo

cng c th l m t danh sch cc kh ng nh nh m m ty hn v i t ng hay m t khi ni m no .

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2.2. Cc lo i tri th c (ti p)3. Siu tri th c: m t tri th c v tri th c . Lo i tri th c ny

gip l a ch n tri th c thch h p nh t trong s cc tri th c khigi i quy t m t v n . Cc chuyn gia s d ng tri th c ny

i u ch nh hi u qu gi i quy t v n b ng cch h ngcc l p lu n v mi n tri th cckh nng hn c .

4. Tri th c heuristic:m t cc "m o " d n d t ti ntrnh l p lu n. Tri th c heuristic ltri th c khng b m mhon ton 100% chnh xc v k t qu gi i quy t v n . Ccchuyn gia th ng dng cc tri th c khoa h c nh s ki n,lu t, sau chuy n chng thnh cc tri th c heuristic

thu n ti n hn trong vi c gi i quy t m t s bi ton.5. Tri th c c c u trc:m t tri th c theo c u trc. Lo itri th c ny m t m hnh t ng quan h th ng theo quani m c a chuyn gia, bao g m khi ni m, khi nim con, v

cc i t ng; di n t ch c nng v m i lin h gi a cc trith c d a theo c u trc xc nh.


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V d : Hy phn lo i cc tri th c sau

1. Nha Trang l thnh ph p.2. B n Lan thch c sch.3. Modus Ponens.4. Modus Tollens.5. Thu t ton tm kim BFS, DFS6. Thu t gi i Greedy7. M t s cch chi u t ng trong vi c chi c t ng.8. H th ng cc khi ni m trong hnh h c.9. Cch t p vi t ch p.

10. Tm t t quy n sch v H chuyn gia.11. Ch n lo i c phi u mua c phi u.

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2.3. CC K THU T BI U DI N TRI TH C

2.3.1 B ba i t ng-Thu c tnh-Gi tr2.3.2 Cc lu t d n2.3.3 M ng ng ngh a2.3.4 Frames2.3.5 Logic


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2.3.1 B ba i t ng-Thu c tnh-Gi tr

M t s ki n c th c dng xc nh n gi tr c a m tthu c tnh xc nh c a m t vi i t ng. V d , m nh "qubng mu " xc nh n " " l gi tr thu c tnh "mu" ca it ng "qu bng". Kiu s ki n ny c g i l b ba it ng-Thu c tnh-Gi tr(O-A-V Object-Attribute-Value).

Hnh 2.1.Bi u di n tri th c theo b ba O-A-V

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2.3.1 B ba i t ng-Thu c tnh-Gi tr (ti p)

Trong cc s ki n O-A-V, mt i t ng c th c nhi u thu ctnh v i cc ki u gi tr khc nhau. Hn n a m t thu c tnhcng c th c m t hay nhi u gi tr. Chng c g i l cc ski n n tr (single-valued) hoc a tr (multi-valued). i u nycho php cc h tri th c linh ng trong vi c bi u di n cc trith c c n thi t.Cc s ki n khng ph i lc no cng b o m l ng hay saiv i ch c ch n hon ton. V th , khi xem xt cc s ki n,ng i ta cn s d ng thm m t khi ni m l tin c y .Phng php truy n th ng qu n l thng tin khng chcch n l s d ng nhn t ch c ch n CF (certainly factor). Khini m ny b t u t h th ng MYCIN (khong nm 1975),dng tr l i cho cc thng tin suy lun. Khi , trong s ki nO-A-V s c thm m t gi tr xc nh tin c y c a n l CF.


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2.3.2 Cc lu t d n

Lu t l c u trc tri th c dng lin k t thng tin bi t v i cc thng tin khc gip a ra cc suy lu n,k t lu n t nh ng thng tin bi t.Trong h th ng d a trn cc lu t, ng i ta thu th pcc tri th c l nh v c trong m t t p v lu chngtrong c s tri th c c a h th ng. H th ng dngcc lu t ny cng v i cc thng tin trong b nh gi i bi ton. Vic x l cc lu t trong h th ng d atrn cc lu t c qu n l b ng m t module g i lb suy di n .

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2.3.2 Cc lu t d n(ti p)

Cc d ng lu t c b n: 7 d ng1. Quan h :

IF Bnh i n h ngTHEN Xe s khng kh i ng c

2. L i khuyn:IF Xe khng kh i ng c

THEN i b3. H ng d nIF Xe khng kh i ng c AND H th ng nhin li u t tTHEN Kimt rah th ng i n


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2.3.2 Cc lu t d n(ti p)

4. Chi n l cIF Xe khng kh i ng cTHEN u tin hy ki m tra h th ng nhin li u, sau ki m t r a h th ng i n

5. Di n gi iIFXen AND ti ng ginTHEN ng c ho t ng bnh th ng

6. Ch n onIF S t cao AND hay ho AND HngTHEN Vim hng

7. Thi t kIFLn AND Da sngTHEN Nn ch n Xe Spacy AND Ch n mu sng

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2.3.2 Cc lu t d n(ti p) M r ng cho cc lu t

Trong m t s p d ng c n th c hi n cng m t php ton trn m t t p haycc i t ng gi ng nhau. Lc c n cc lu t cb i n .

V d : IF X l nhn vin AND Tui c a X > 65THEN X c thngh hu

Khi m nh pht bi u v s ki n, hay b n thn s ki n c th khng ch cch n, ng i ta dng h s ch c ch n CF. Lu t thi t l p quan h khngchnh xc gi a cc s ki n gi thi t v k t lu n c g i l lu t khng ch c ch n .

V d : IF L m pht CAO THEN Hu nh ch c ch n li su t s CAOLu t ny c vi t l i v i gi tr CF c th nh sau:

IF L m pht cao THEN Li sut cao, CF = 0.8D ng lu t ti p theo lsiu lu t: M t lu t v i ch c nng m t cch th c dng cc lu t khc. Siu

lu t s a ra chi n l c s d ng cc lu t theo l nh v c chuynd ng, thay v a ra thng tin m i.

V d : IF Xe khng khi ng AND H th ng i n lm vi c bnh th ngTHEN C th s d ng cc lu t lin quan n h th ng i n


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2.3.3 M ng ng ngh a

M ng ng ngh a l m t phng php bi u di n trith c dng th trong nt bi u di n i t ngv cung bi u di n quan h gi a cc i t ng.

Hnh 2.3."S l Chim" th hi n trn m ng ng ngh a

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2.3.3 M ng ng ngh a(ti p)

Hnh 2.4.Pht tri n m ng ng ngh a


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2.3.4 Frame

Hnh 2.6.C u trc frame

Hnh 2.7.Nhi u m c c a frame m t quan h ph c t p hn

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2.3.5 Logic1. Logic m nhIF Xe khng khi ng c (A)AND Khong cch t nh n ch lm l xa (B)THEN S tr gi lm (C)

Lu t trn c th bi u di n l i nh sau:AB C

2. Logic v tLogic vt , cng gi ng nh logic m nh , dng cck hi u th hi n tri th c. Nh ng k hi unyg mh ng s , v t , bi n v hm.


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2.4 SUY DIN D LI U

1. Modus ponens1. E12. E1 E23. E2N u c tin khc, c d ng E2 E3 th E3 c a vo

danh sch.

2. Modus tollens1. E22. E1 E23. E1

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2.4.2 Cc ho t ng c a H th ng Suy di n ti n


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V d v Suy di n ti n

Lu t 1. IFB nh nhn rt h ng ANDNghi vim nhimTHENTin r ng b nh nhn vim h ng, i ch a h ng.

Lu t 2 . IFNhi t b nh nhn qa 37THENB nh nhn b s t

Lu t 3 . IFB nh nhn m trn 1 tu n ANDB nh nhn s tTHENNghi b nh nhn vim nhim.

Thng tin t b nh nhn l: B nh nhn c nhi t 39 B nh nhn m hai tu n B nh nhn h ng rt

Khi h th ng th y gi thi t c a lu t kh p v i thng tin trong bnh , cu k t lu n c a lu t c b sung vo b nh .

Minh h a V d suy di n li

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C ch suy di nSuy di n v i logic m nh :1. Thu t ton suy di n ti n

Input: - T p lu t Rule= {r1, r2, ..., rm}- GT, KL

Output: Thng bo thnh cng nu GTKLNg c l i, thng bo khng thnh cng

Method:TD=GT;T=Loc(Rule, TD);While (KLTD) AND (T) Do

{r = Get(T);

TD=TD{q}; // r:leftqRule = Rule \ {r};T=Loc(Rule, TD);

}If KLTD THEN Return Trueelse Return False

V d : Rule ={r1:a c, r2:b d, r3:a e, r4:ad e, r5:b c f, r6:e fg}H i a b g?


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Thu t ton suy di n liIf KLGT THEN Return TrueElse {T ch=; V t =; First=1; Quaylui= False;}For Each qKL DO T ch=T ch{(q,0)};Repeat

first ++;((f,i)=Get(T ch);If (fGT) THEN{

j = Tmlu t(f,i,Rule); // rj: Leftj f

If (Tm c rj) THEN{ Vet = Vet{(f,j)};For Each t (Leftj \GT) DO T ch = T ch{((t,0)};

else{ Quaylui=True;

While (fKL) AND Quaylui DO{Repeat { (g,k)=Get(Vt);

T ch = T ch \ Leftk;}Until fLeftk;l=Tmlu t(g,k,Rule);

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Thu t ton suy di n liIf (Tm c rl) THEN

{ T ch = T ch \ Leftk ;For Each t (Leftl \GT) DO

T ch = T ch{((t,0)};V t = V t {(g,l)};Quaylui = False;

} //end if3 else f=g;} //end while

} //enf if2 }//end if1

Until (T ch =) OR ((fKL) and (First>2));If (fKL) then Return False else Return TRue;

V d : Rule ={r1:a c, r2:b d, r3:a e, r4:ad e, r5:b c f, r6:efg}H i a b g?V d 2: Rule ={r1:abc, r2:ahd, r3:bce, r4:adm,r5:abp, r6:pem}H i i) a b m? ii) a m?


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2.4.3u i m*Suy di n ti n

u i m chnh c a suy di n ti n l lm vic t t khi bi ton v b nch t i thu th p thng tin r i th y i u c n suy di n. Suy di n ti n cho ra kh i l ng l n cc thng tin t m t s thng tinban u. N sinh ra nhiu thng tin m i. Suy di n ti n l ti p c n l t ng i v i lo i bi ton c n gi i quy tcc nhi m v nh l p k ho ch, i u hnh i u khi n v di n d ch.

* Suy di n liM t trong ccu i m chnh c a suy di n li l ph h p v i bi tona ra gi thuy t r i xem hi u q a gi thi t c ng khng.

Suy di n li t p trung vo ch cho. N t o ra m t lo t cu h i ch

lin quan n v n ang xt, n hon c nh thu n ti n i v ing i dng.Khi suy din li mu n suy di n ci g t cc thng tin bi t, nch tm trn m t ph n c a c s tri th c thch ng i v i bi tonang xt.

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2.4.4 Nh c i m

* Suy din ti n M t nh c i m chnh c a h th ng suy di n ti n l khng c mnh n c r ng ch m t vi thng tin l quan trng. H th ng h icc cu h i c th h i m khng bi t r ng ch m t t cu i nk t lu n c. H th ng c th h i c cu khng lin quan. C th cc cu tr l i cng quan tr ng, nhng lm ng i dng lng tng khi phi trl i cc cu khng dnh n ch .

* Suy din liNh c i m c b n c a suy di n ny l n th ng ti p theodng suy di n, thay v ng ra ph i ng m sang nhnhkhc. Tuy nhin c th dng nhn t tin c y v cc lu t meta kh c ph c.


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Chng 3: H MYCIN

3.1 Gi i thi uMYCIN l mt h l p lu n trong y h c c hon t t v o nm 1970 t i

i h c Standford, Hoa K. y l m t h chuyn gia d a trn lu tv s ki n. MYCIN s d ng c ch l p lu t g n ng x l cclu t suy di n d a trn o ch c ch n. Ti p theo sau MYCIN, hEMYCIN ra i. EMYCIN l mt h chuyn gia t ng qut c t ol p b ng cch lo i ph n c s tri th c ra kh i h MYCIN. EMYCINcung c p m t c ch l p lu n v tu theo bi ton c th s bsung tri th c ring c a bi ton t o thnh h chuyn gia.

Cc c i m chnh:- S d ng k thu t suy di n li.- C k h nng phn tch tri thc v i u khi n.

- C tch h p Meta-Rule.-C th dng khi thiu thng tin ho c thng tin khng chc ch n.- D s d ng, giao ti p b ng ti ng Anh.- Cung c p cc ch c gi i thch: HOW, WHY.

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M t phin lm vi c c th v i MYCIN:

?Thng tin v b nh nhnMYCIN: Hv tn?User: Jack Durkin.MYCIN: Tui?User: 24.MYCIN: Gii tnh?User: Nam.MYCIN: Chng t c?User: CAUCASIAN.

? Xc nh b n ch t ly nhi mMYCIN: C bt k php c y vi khun cho Jack Durkin lin quan n b nhhi n t i?User: C.

MYCIN: Mu v t l y t u?User: CSF.MYCIN: Ngy gith c hi n c y:User: 5-6-77 8:09 AM..MYCIN: Gn y Jack Durkin c cc triuch ng nh: chong vn, hn m khng?User: C...


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CC THNH PHN C A H MYCIN

1. Chng trnh t v n: Cung c p cho cc Bc s cc l i khuyn ch n phng php i u tr thchh p b ng cch xc nh r cch th c i u tr b i ccd li u l y ra t cc phng th nghim lm sngthng qua cc cu tr l i c a bc s cho cu h i c amy tnh.2. Kh nng gi i thch c tc ng qua l i: Cho phpchng trnh t v n gi i thch cc kin th c c a nv cc phng php i u tr v ch ng minh cc chthch v cc phng php i u tr c bi t.3. Thu n p tri th c: cho php cc chuyn gia conng i trong l nh v c i u tr cc cn b nh truy nnhi m d y cho MYCIN cc lut quy t nh theophng php i u tr m h tm th y trong th c tlm sng.

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PH M VI S D NG C A H MYCIN

1. Ch n on nguyn nhn gy b nh: i v i cc bc s i u tr ,khi xt nghim cho b nh nhn c k t qu ch n on ch cch n m t 24-48 gi . Nhi u tr ng h p ph i i u tr c ngay khicha c k t lu n hon ch nh. MYCIN gip chn on nguynnhn gy b nh nhanh hn: khi g i chng trnh MYCIN, ccbc s tr l i cc cu h i v ti u s b nh nhn, b nh n, cck t qu xt nghi m, cc tri u ch ng, t MYCIN a rach n on b nh.

2. T o ra phng php i u tr: Sau khi nh n c cc cu tr l ic a bc s v tnh tr ng b nh nhn thng qua i tho i. Trongtr ng h p cu tr l i khng bi t ho c bi t khng ch c ch n,th MYCIN ssuy lu n t cc thng tin khng hon ch nh.3. D on di n bi n c a b nh: B ng cc cu h i HOW, WHY,MYCIN s gi i thch cc nguyn nhn v l do cho cc bc s .Sau khi vi c ch n on b nh v k n hon t t, bc s c ththeo di ton b qu trnh ch n on b nh c a MYCIN v qua theo di di n bi n c a b nh


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NGUYN NHN THNH CNG CA MYCIN

1. S c n thi t c a vi c t v n dng khng sinh c acc bc s : vo th i i m ny vi c l m d ng khngsinh em l i khng t phn ng ph .

2. C s tri th c c a MYCIN c thu n p t ccchuyn gia xu t s c nh t trong l nh v c.

3. MYCIN khng bao gi i n ngay k t lu n lunc thm cc thng tin ct y u qua m i b c.

4. MYCIN c hnh thnh t m t chng trnh tr tunhn t o c p d ng th c t (DENDRAL) v c th c hi n t i trung tm y t n i ti ng v i cc

tri th c m i nh t v b nh h c v d c h c.

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3.2 L THUYT V S CH C CH N

MB (Measure of Belief in): o s tin c yMD (Measure of Disbelief in): o s khng tin cyCF (Certainly Factor): Hs ch c ch nG i:MB(H/E) l o s tin c y c a gi thuy t H khi

c ch ng c E.MD(H/E) l o s khng tin cy c a gi thuy t Hkhi c ch ng c E.

Khi : 0


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3.2 L THUYT V S CH C CH N(ti p)

1. Lu t n gi n: If(e) then (c)CF(e) l o ch c ch n c a ch ng c .CF(r) l o ch c ch n c a lu t suy di n.Khi : CF(c) l o ch c ch n c a k t lu n s c tnh

b ng cng th c:CF(c) =CF(e) *CF(r)

2. Lu t ph c t p:If(e1 AND e2) then (c)

CF (e1 AND e2) = MIN(CF(e1), CF(e2))if (e1 OR e2) then (c)

CF (e1 OR e2) = MAX(CF(e1), CF(e2))

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3.2 L THUYT V S CH C CH N(ti p)

3. V i lu t: if ((e1 AND e2) OR e3) then (c)CF ((e1 AND e2) OR e3) = MAX(MIN(CF(e1), CF(e2)), CF(e3))4. CF(NOT e) = - CF(e)5. K t h p nhi u lu t c cng k t lu n:- Lu t 1: If(e1) then (c) v i CF(r1): o ch c ch n c a lu t 1- Lu t 2: If(e2) then (c) v i CF(r2): o ch c ch n c a lu t 2

V i CF(t1), CF(t2) l CF ca k t lu n c a lu t 1 v 2, khiCF(t1) v Cf(t2) u dng th:

Ct ng = CF(t1) + CF(t2) CF(t1) * CF(t2)Khi CF(t1) v Cf(t2) u m th:

Ct ng = CF(t1) + CF(t2) + CF(t1) * CF(t2)N u CF(t1) khc d u v i CF(t2) th:

Ct ng = (CF(t1) + CF(t2)) / (1 MIN(ABS(CF(t1)), ABS(CF(t2))))


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V d v l p lu n trong H MYCIN

V d : C 7 lu t sau y:r1: If(e1) Then (c1)CF(r1) = 0,8r2: If (e2) Then (c2)CF(r2) = 0,9r3: If (e3) Then (c2)CF(r3) = 0,7r4: If (e4) Then (c3)CF(r4) = 0,6r5: If (NOTe5) Then (c3)CF(r5) = 0,5r6: If (c2 ANDc3) Then (c4)CF(r6) = 0,9r7: If (c1 OR c4) Then (c5)CF(r7) = 0,8

B ng lu t ny t o thnh m ng suy di n hnh 3.1 v i c5 l gi thuy t c n h ng n.

46

Hnh 3.1. M ng suy di n


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L p lu n trn m ng suy di n

Gi s cc ch ng c e1, e2, e3, e4, e5 c o ch c ch n nh sau:

CF(e1) = 0,9CF(e2) = 0,9CF(e3) = -0,3CF(e4) = 0,4CF(e5) = -0,3

48

L p lu n trn m ng suy di n (ti p)Chng ta s l p lu n t cc CF c a ch ng c d nln gi thuy t c5 nh sau:D a vo lu t r1 tnh c CF(c1):CF(c1) =CF(e1) *CF(r1) = 0,8*0,9 = 0,72D a vo lu t r2, r3 tnh c CF(c2)V i lu t r2: CF(c2) =CF(e2) *CF(r2) = 0,9 * 0,9 =0,81V i lu t r3: CF(c2) =CF(e3) *CF(r3) = -0,3 * 0,7 = -0,21DoCF(c2) c a r2 tri d u v i CF(c2) c a r3, nn:CF(c2)t ng = (0,81 + (-0,21)) / (1-MIN(0,81, 0,21)) =0,74


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L p lu n trn m ng suy di n (ti p)D a vo lu t r4, r5 ta tnh c CF(c3)V i lu t r4:CF(c3) =CF(e4) *CF(r4) = 0,4 * 0,6 = 0, 24V i lu t r5:CF(c3) =CF(NOTe5)*CF(r5) = -CF(e5)*CF(r5) = 0,3*0,5 = 0,15DoCF(c3) c a r4 v CF(c3) c a r5 cng dng nnCF(c3)t ng = 0,24 + 0,15 0, 24 * 0, 15 = 0,354D a vo lu t r6 ta tnh c CF(c4):CF(c4) =MIN(CF(c2),CF(c3)) *CF(r6) =MIN(0,74, 0,354) * 0,9= 0,354 * 0,9 = 0,3186D a vo lu t r7 ta tnh c CF(c5)CF(c5) =MAX(CF(c1),CF(c4)) *CF(r7) =MAX(0,72, 0,3186) *0,8 = 0,576Nh th ch c ch n c a gi thuy t c5 l 0,576.

50

Chng 4 H h c4.1 M U

Cc chng tr c th o lu n v bi u di n v suy lu n trith c. Trong tr ng h p ny gi nh c s n tri th c vc th bi u di n t ng minh tri th c.Tuy v y trong nhi u tinh hu ng, s khng c s n tri th cnh: K s tri th c c n thu nh n tri th c t chuyn gia l nh v c. C n bi t cc lu t m t l nh v c c th . Bi ton khng c bi u di n t ng minh theo lu t, s ki n

hay cc quan h .

C hai ti p c n cho h th ng h c: H c t k hi u: bao g m vi c hnh th c ha, s a ch a cclu t t ng minh, s ki n v cc quan h .

H c t d li u s : c p d ng cho nh ng h th ng cm hnh d i d ng s lin quan n cc k thu t nh m t i ucc tham s . H c theo d ng s bao g m m ng Neural nhnt o, thu t gi i di truy n, bi ton t i u truy n th ng. Cc k thu t h c theo s khng t o ra CSTT t ng minh.


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4.2 CC HNH THC H C

1. H c v t: H ti p nh n cc kh ng nh c a cc quy tnh ng. Khi h t o ra m t quy t nh khng ng, h

s a ra cc lu t hay quan h ng m h s d ng.Hnh th c h c v t nh m cho php chuyn gia cung c ptri th c theo ki u tng tc.

2. H c b ng cch ch d n: Thay v a ra m t lu t c th c n p d ng vo tnh hu ng cho tr c, h th ng s c cung c p b ng cc ch d n t ng qut. V d : "gas

h u nh b thot ra t van thay v thot ra t ng d n".H th ng ph i t mnh ra cch bi n i t tr ut ng n cc lu t kh d ng.

3. H c b ng qui n p: H th ng c cung c p m t t pcc v d v k t lu n c rt ra t t ng v d . H lint c l c cc lu t v quan h nh m x l t ng v d m i.

52

4.2 CC HNH THC H C (Ti p)

4. H c b ng tng t : H th ng c cung c p p ng ngcho cc tc v tng t nhng khng gi ng nhau. H th ngc n lm thch ng p ng tr c nh m t o ra m t lu tm i c kh nng p d ng cho tnh hu ng m i.

5. H c d a trn gi i thch: H th ng phn tch t p cc l i gi i v d (v k t qu ) nh m n nh kh nng ng ho c sai v t ora cc gi i thch dng h ng d n cch gi i bi ton trongtng lai.

6. H c d a trn tnh hu ng: B t k tnh hu ng no c h th ng l p lu n u c lu tr cng v i k t qu cho d

ng hay sai. Khi g p tnh h ng m i, h th ng s lm thchnghi hnh vi lu tr v i tnh hu ng m i.

7. Khm ph hay h c khng gim st: Thay v c m c tiu t ngminh, h khm ph lin t c tm ki m cc m u v quan h trong d li u nh p. Cc v d v h c khng gim st bao g mgom c m d li u, h c nh n d ng cc c tnh c b n nh c nh t cc i m nh.


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V d v CC HNH THC H C

V d :- H MYCIN- M ng Neural nhn t o- Thu t ton h c Quinland- Bi ton nh n d ng- My chi c car, c t ng

54

4.3 THU T GI I Quinlan

- L thu t ton h c theo quy n p dng lu t, am c tiu.

- Do Quinlan a ra nm 1979.- t ng: Ch n thu c tnh quan tr ng nh t

t o cy quy t nh.- Thu c tnh quan tr ng nh t l thu c tnh

phn lo i B ng quan st thnh cc b ng consao cho t m i b ng con ny d phn tch tm quy lu t chung.


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4.3.1 THU T GI I A. Quinlan

ASingleGermanLarge3

BMarriedGermanSmall8

BMarriedItalianLarge7BSingleItalianLarge6

BMarriedGermanLarge5

BSingleItalianSmall4

ASingleFrenchLarge2

ASingleGermanSmall1

ConclusionFamilyNationalitySizeSTT

56

V i m i thu c tnh c a b ng quan st:

Xt vector V: c s chi u b ng s phn lo i V(Size=Small)= (ASmall, BSmall) ASmall=S quan st A c Size l Small / Tng s quan st c Size=Small BSmall= S quan st B c Size l Small / Tng s quan st c Size=Small

V(Size=Small)= (1/3 , 2/3)V(Size=Large)= (2/5 , 3/5)

V i thu c tnh NationalityV(Nat = German)= (2/4 , 2/4)V(Nat = French)= (1 , 0)V(Nat = Italian)= (0 , 1)

Thu c tnh Family:V(Family=Single)= (2/5 , 3/5)V(Family = Married)= (0, 1)


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V i m i thu c tnh c a b ng quan st:

Ch cn xt GermanThu c tnh Size:

V(Size=Small) = (1/2 , 1/2)V(Size=Large) = (1/2 , 1/2)

Thu c tnh Family:V(Family=Single) = (1, 0)V(Family=Married) = (0,1)

ConclusionFamilySizeSTT

BMarriedSmall4

BMarriedLarge3

ASingleLarge2

ASingleSmall1

Nationality

Italian French German

Single Married

58

V i m i thu c tnh c a b ng quan st(ti p)

Nationality

Italian French German

Single Married

Rule 1: If (Nationality IS Italian) then (Conclusion IS B)

Rule 2: If (Nationality IS French) then (Conclusion IS A)Rule 3: If (Nationality IS German) AND (Family IS Single)

then (Conclusion IS A)

Rule 4: If (Nationality IS German) AND (Family IS Married)then (Conclusion IS B)


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Thu t gi i: H c theo b t nh

UpSoftwareNoMidle7

UpSoftwareNoNew6

UpHardwareNoNew5

DownHardwareNoOld4

UpHardwareNoMidle3

DownSoftwareYesMidle2

DownSoftwareNoOld1

ProfitTypeCompetitionAgeStt

60

H c theo b t nh(ti p)b t nh c a X:

Tnh Entropy cho mi thu c tnh v ch n thu c tnhc Entropy nh nh t.

=

=k

1i2log-)( ii p p X E

=+=

==

==

=

=

=

=

8752.0811.0*4.0918.0*6.0) / (

811.043

log43

-41

log41

-) / (

918.062

log62

-64

log64

-) / (

),(log),(-) / (

22

22

k

1i2

nCompetitioC E

C E

C E

ac pac p AC E

YesnCompetitio

NonCompetitio

iiii


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H c theo b t nh(ti p)

Tng t :E(C/Age) = 0.4E(C/Type) = 1Age cho nhi u thngtin nh t

ProfitTypeCompetitionSTT

DownHardwareYes4

UpSoftwareNo3

UpHardwareNo2

DownSoftwareYes1

Age

Old Milde New

Down Competition UpNo Yes

Up Down

62

H c theo b t nh(ti p)Age

Old Milde New

Down Competition UpNo Yes

Up Down

Rule 1: If (Age IS Old) then (Profit IS Down)Rule 2: If (Age IS New) then (Profit IS Up)Rule 3: If (Age IS Midle) And (Competition IS No)

then (Profit IS Up)Rule 4: If (Age IS Midle) And (Competition IS Yes)

then (Profit IS Down)


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4.4 THU T GI I ILA(Inductive Learning Algorithm)

Xc nh d li u1. T p m u c li t k trong m t b ng, v i m i dng

tng ng m t m u, v m i c t th hi n m t thu ctnh trong m u.

2. T p m u c m m u, m i m u g m k thu c tnh vtrong c m t thu c tnh quy t nh. T ng s n ccgi tr c a thu c tnh quy t nh chnh l s l p c at p m u.

3. T p lu t R c gi tr kh i t o l

4. T t c cc dng trong b ng ban u cha cnh d u (ki m tra).

64

4.4 THU T GI I ILA(ti p) B c 1: Chia b ng m m u ban u thnh n b ng con.M i b ng con ng v i m t gi tr c a thu c tnh phn l pc a t p m u.

(* th c hi nccb c 2 n 8 cho m i b ng con*)B c 2 : Kh i t o bi n m k t h p thu c tnh j, j=1.B c 3 : V i m i b ng con ang kh o st, phn chiadanh sch cc thu c tnh theo cc t h p phn bi t, m it h p ng v i j thu c tnh phn bi t.B c 4 : V i m i t h p cc thu c tnh, tnh s l ng ccgi tr thu c tnh xu t hi n theo cng t h p thu c tnhtrong cc dng cha c nh d u c a b ng con angxt (m ng th i khng xu t hi n v i t h p thu c tnhny trn cc b ng cn l i). G i t h p u tin (trongb ng con) c s l n xu t hi n nhi u nh t l t h p l n nh t .


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4.4 THU T GI I ILA(ti p)

B c 5 : N u t h p l n nh t b ng , tng j ln 1 vquay l i b c 3.B c 6 : nh d u cc dng tho t h p l n nh t c ab ng con ang x l theo l p.B c 7 : Thm lu t m i vo t p lu t R, v i v tri l t pcc gi tr c a thu c tnh ng v i t h p l n nh t (k th p cc thu c tnh b ng ton t AND) v v ph i l gitr thu c tnh quy t nh tng ng.B c 8 : N u t t c cc dng u c nh d uphn l p, ti p t c th c hi n t b c 2 cho cc b ng concn l i. Ng c l i (n u cha nh d u h t cc dng) thquay l i b c 4. N u t t c cc b ng con c xt thk t thc, k t qu thu c l t p lu t c n tm.

66

Minh h a thu t gi i ILA

yesspheregreenlarge7

nopillarredlarge6

yespillargreenlarge5nowedgeredlarge4

yessphereredsmall3nowedgeredsmall2

yesbrick bluemedium1

DecisionShapeColorSizeM u s

B ng 4.1 T p m u h c cho bi ton phn l p i t ng


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Minh h a thu t gi i ILA(ti p): B c 1

B ng 4.2. V i n=2, Chia thnh hai bng con theo thu c tnh Decisionnopillarredlarge6 3

nowedgeredlarge4 2nowedgeredsmall2 1

DecisionShapeColorSizeM u s c m i

B ng con 2

yesspheregreenlarge7 4

yespillargreenlarge5 3

yessphereredsmall3 2

yesbrick bluemedium1 1

DecisionShapeColorSizeM u s c , m i

B ng con 1

68

Minh h a thu t gi i ILA(ti p)

p d ng b c 2 c a thu t gi i vo b ng con th nh t trong b ng trn. V i j =1, danh sch cc t h pthu c tnh g m c {Size}, {Color}, v {Shape}. V i t h p {Size}, gi trthu c tnh "medium" xuthi n trong b ng con th nh t nhng khng c trongb ng con th hai, do gi tr t h p l n nh t l"medium". Bi v cc gi tr thu c tnh "small" v"large" xu t hi n trong c hai b ng con, nn khng c xt trong b c ny. V i t h p {Size}, gi tr

thu c tnh "medium" ch b ng 1


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Xt tip cho t h p {Color} th gi trt h p l n nh tl b ng 2, ng v i thu c tnh "green", cn thuc tnh"blue" l bng 1. Tng t , v i t h p {Shape}, ta c "brick" xuthi n m t l n, v "sphere" hai ln. n cu i b c 4, tac t h p {Color} vi thu c tnh "green" v {Shape}v i thu c tnh "sphere" u c s l n xu t hi n l nnh t l 2. Thu t ton m c nh ch n tr ng h p th nh t xc nh lu t t h p l n nh t. Dng 3 v 4 c nh d u phn l p, ta c lu t d n nh sau:

Rule 1: IF color IS green THEN decision IS yes

70


Ti p t c th c hi n b c 4 n 8 cho cc m u cn l i (chanh d u) trong b ng con ny (t c dng 1 v 2). p dng tng

t nh trn, ta th y gi tr thu c tnh "medium" ca {Size}, "blue"c a "Color", "brick" v "sphere" ca {Shape} u xu t hi n m tl n. B i v s l n xu t hi n ny gi ng nhau, thu t gi i p d nglu t m c nh ch n tr ng h p u tin. Ta c thm lut sau:Rule 2: IF size IS medium THEN decision IS yes

nh d u cho dng 1 trong bng con th nh t. Ti p t c pd ng b c 4 n 8 trn dng cn l i (t c dng 2). Gi trthu ctnh "sphere" c a {Shape} xut hi n m t l n, ta c lu t th ba::Rule 3: IF shape IS sphere THEN decision IS yesDng 2 c nh d u. Nh v y, t t c cc dng trong b ng con 1 c nh d u, ta chuy n qua x l ti p b ng con 2.


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Thu c tnh "wedge" ca {Shape} xut hi n hai l n trong dng 1 v 2c a b ng con ny. nh d u cc dng ny v i lu t d n th t sau:

Rule 4: IF shape IS wedge THEN decision IS no

V i dng cn l i (t c dng 3) c a b ng con 2, ta c thu c tnh {Size}v i gi tr "large" c xut hi n trong b ng con 1. Do , theo thu t gi i,ta lo i b tr ng h p ny. Tng t nh v y cho gi tr "red" c a{Color} v "pillar" ca {Shape}. Khi , ILA tng j ln 1, v kh i t o cct h p 2 thu c tnh l {Size v Color}, {Size v Shape}, v {Color vShape}. Cc t h p th nh t v th ba tho mn i u ki n khng xu thi n trong b ng con 1 v i cc c p thu c tnh hi n c c a dng ny.Theo lu t m c nh, ta ch n lu t theo tr ng h p th nh t. nh d udng ny, ta c thm lu t d n th 5:

Rule 5: IF size IS large AND color IS red THEN decision IS no

72


Rule 1: IF color IS green THEN decision IS yesRule 2: IF size IS medium THEN decision IS yesRule 3: IF shape IS sphere THEN decision IS yesRule 4: IF shape IS wedge THEN decision IS noRule 5:IF size IS large AND color IS red THEN decision IS no

nh gi thu t gi i: S l ng cc lu t thu c xc nh m c thnh cng c a thu tgi i. y chnh l m c ch chnh c a cc bi ton phn l p thngqua m t t p m u h c. Ngoi ra, nh gi cc h h c quy n p lkh nng h th ng c th phn l p cc m u c a vo sau ny. Thu t gi i ILA c nh gi m nh hn hai thu t gi i v phngphp h c quy n p tr c y l ID3 v AQ).


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Chng 5: H th ng m cho cc bi n lin t c

5.1 Cc khi ni m1. T p r v hm c trng

- Ngn ng t nhin v logic m.- T p r (crisp set): Gi A l mt t p h p r, m t ph n t x

c th c xA ho c xA, C th s d ng hm mt khi ni m thu c v . N u xA, (x) = 1, ngu c l i n ux A, (x)=0. Hm c g i l hm c trng c at p h p A.

- T p m v hm thnh vin: Khc vi t p r, khi nimthu c v c m r ng nh m ph n nh m c x l

ph n t c a t p m A. M t t p m (fuzzy set): A cc trng b ng hm thnh vin v cho x l m t ph nt , A(x) ph n nh m c x thu c v A.

- M t t p m A trong tp v tr U c xc nh b i hm:A: U[0,1]

74

V d v t p m :- High- Young- S g n 7- T c nhanh

Bi u di n t p m :1. N u t p v tr U l r i r c v h u h n th t p m A trong U c bi u di n:

V d : Cho U={a, b, c, d}. Ta c th xc nh m t t p m A nh sau:

5.1 Cc khi ni m v Logic m

=U x

A

x

x A

)(

d

7.0+

c

0+

b

5.0+

a

3.0=A


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2. N u t p v tr U l lin t c th t p m A trong U c bi u di n:

V d : T p m A={S g n 2} c th xc nh hm thu c nh sau:

5.1 Cc khi ni m v Logic m (ti p)

=U x

A dx x

x A

)(

=

=

U x

A

dx x

A

e x2

2

2)-(x-

2)--(x

e

)(

76

Ghi ch: th hm thu c cho t p m A={S g n 2} c th xc nhcch khc nh sau:

5.1 Cc khi ni m v Logic m (ti p)

0 n u x < 1x-1 n u 1 x < 21 n u x = 2-x+3 n u 2 < x 30 n u x>3

(x)=


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5.2 Cc d ng c a hm thnh vin

Cc hm thnh vin c a t p m c 3 d ng c b n l: d ngtng, d ng gi m v d ng chungD ng S tng

Hnh 4.2. Hm S tngb) D ng S gi mc) D ng hnh chung

Hnh 4.3. Hm d ng chung

78

5.3 Bi n ngn ng (Linguistic Variable)

Logic m lin quan n l p lu n trn cc thu t ng m v m h trong ngn ng t nhin c a con ng i.Bi n nh n cc t trong ngn ng t nhin lm gi trg i l bi nngn ng .Bi n ngn ng c xc nh b i b b n (x, T, U, M): X l tn bin. V d : nhi t , tc , p su t, ... T l t p cc t (cc gi tr ngn ng ) m x c th nh n. V

d : x l tc th T c th l T={chm, v a, nhanh}

U l mi n gi tr m x c th nh n. V d , n u x l t c c a xe my th U=[0 .. 120 km/h] M l lu t ng ngh a, ng v i m i t tT v i m t t p m A.


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5.4 Bi n ngn ng (ti p)

Nng, l nhTh p, trung bnh, cao

Ch m, v a, nhanh

Nhi tcao

T c

Cc gi tr i n hnhBi n ngn ng

V d : Cho x l t c , T={ch m, v a, nhanh}, cc t ch m,v a, nhanh c xc nh b i cc t p m trong hnh sau:

Ch m V a Nhanh

30 50 60 70 120

80

5.5 Gia t

Gia t lm m h thm cc cu nh: r t, h i, c v , ...1. R t: R t(A)(x) = (A(x))2

V d : T p m g m nh ng ng i r t cao

2. Co gin / m t t: Co gin (A)(x) = (A(x))0.5V d : A l t p nh ng ng i t m th c th Co gin A l tp nh ng ng i thin vcao v th p trong nh ng ng i t m th c.

3. Nh n m nh/ th c s l:Nh n m nh(A)(x) = 2(A(x))2 n u 0 A(x) 0.5Nh n m nh(A)(x) = 1-2(1-A(x))2 n u 0.5 A(x) 1V d : Sau khi dng php ton ny v i t p m cao ta c t p nh ng ng ii th cs cao

4. M nh m / r t r t: R t r t(A)(x) = (A(x))nV d : Sau khi dng php ton ny v i t p m cao ta c t p nh ng ng i th cs cao


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5.6 Cc php ton trn t p m

Cho ba t p m A, B , C v i A(x),B (x),C(x)C=A B: C(x) = min(A(x),B (x))C=A B : C(x) = max(A(x),B (x))C= A : C(x) = 1-A(x)

Xt t p m cao v thp v chi u cao c a ng i:Cao = 0.0/1.5 + 0.2/1.55 + 0.5/1.60 + 0.8/1.65 + 1.0/1.70Th p = 1.0/1.5 + 0.8/1.55 + 0.5/1.60 + 0.2/1.65 + 0.0/1.70

Cao Th p (x) = 0.0/1.5 + 0.2/1.55 + 0.5/1.60 + 0.2/1.65 + 0.0/1.70Dng ch nh ng ng i t m th c: gi tr cao nh t gi a t p, th p nh t 2 bn

Cao Th p (x) = 1.0/1.5 + 0.8/1.55 + 0.5/1.60 + 0.8/1.65 + 1.0/1.70Dng ch nh ng ng i khng t m th c Cao (x) = 1.0/1.5 + 0.8/1.55 + 0.5/1.60 + 0.2/1.65 + 0.0/1.70Dng ch nh ng ng i khng cao hay t m th c hay th p.

82

5.7 Suy di n m

M nh m l m nh kh ng nh gi tr cho bi n ngn ng . is V d : Chi u cao l t m th c

Logic m s d ng t p m trong cc m nh m .IF X is A THEN Y is BV d : N u chi u cao l t m th c th tr ng l ng l

trung bnh N u A v B l tp m th H chuyn gia lu tr lin k t (A,B)trong ma tr n M (hay k hiu R). C th th hi n c A v B nh cc vector (A, B) thch hp v

t quan h ny vo ma tr n M.- Ma tr n lin k t m M nh x t p m A sang t p m B.

Hai k thu t suy di n thng d ng l; Suy di n Max-Min Suy di n c c i.


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5.7.1 Nhn ma tr n vector m

Cho A =(a1, a2, ..., an) v i ai = A(xi)Cho B =(b1, b2, ..., bp) v i b j = B(y j)Ma tr n Mn*p c xc nh nh sau: A o M =BTrong b j = Max { Min (ai, mij)} 1 i nV d : Cho A=(0.2, 0.4, 0.6, 1)

Cho0.1 0.6 0.80.6 0.8 0.60.8 0.6 0.50.0 0.5 0.5

b1 = Max{Min(0.2, 0.1), Min(0.4, 0.6), Min(0.6, 0.8), Min(1, 0.0)}= 0.6Tng t b2=0.6 v b3 = 0.5Nh v y, B=(0.6, 0.6, 0.5)

M =

84

5.7.2 Suy di n Max-Min- Khi bit ma tr n A v B hy to ma tr n M- Cho A hy tnh B- Suy di n Max-Min: mij = Min(ai , bj)V d : X: nhit , t p m A tr n X: nhit bnh th ng

Y: t c , t p m B trn Y: tc v a ph iGi s c lu t: IF nhi t bnh th ng THEN tc v a ph i.Gi s cc t p m th hi n b ng cc vector sau:Nhi t bnh th ng =(0/100, 0.5/125, 1/150, 0.5/175, 0/200)T c v a ph i = (0/10, 0.6/20, 1/30, 0.6/40, 0/50)Ma tr n M c t o ra nh sau:

0.0 0.0 0.0 0.0 0.00.0 0.5 0.5 0.5 0.00.0 0.6 1.0 0.6 0.00.0 0.5 0.5 0.5 0.00.0 0.0 0.0 0.0 0.0

Cho t p A = (0/100, 0.5/125, 0/150, 0/175, 0/200)Ta c b1= max{min(0,0), min(0.5,0), min(0,0), min(0,0), min(0,0)}=0B = (0/10, 0.5/20, 0.5/30, 0.5/40, 0/50)

M =


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85

5.7.3 Suy di n Tch c c i

- Dng php nhn t o cc thnh ph n c a ma tr n MSuy di n Tch c c i: mij = ai * bj

- Sau dng cch tnh Max-Min suy ra B t AV d : A = (0, 0.5, 1, 0.5, 0)

B = (0, 0.6, 1, 0.6, 0)Ma tr n M c t o ra nh sau:

0.0 0.0 0.0 0.0 0.00.0 0.3 0.5 0.3 0.00.0 0.6 1.0 0.6 0.00.0 0.3 0.5 0.3 0.00.0 0.0 0.0 0.0 0.0

Cho t p A = (0, 0.5, 0, 0, 0)

Ta c b1= max{min(0,0), min(0.5,0), min(0,0), min(0,0), min(0,0)}=0B = (0, 0.3, 0.5, 0.3, 0)

M =

86

5.8 NGUYN L XL CC BI TON M

Hnh 9.4. H th ng m

1. M ha: Chuy n i gi tr r u vo thnh vectori2. Xc nh cc lu t h p thnh v thu t ton xc nh gi

tr m3. Gi i m : Phng php i m tr ng tm

=

U A

U A

dx x

dx x x

x)(

)(

0


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87

5.8 NGUYN L XL BI TON M(ti p)

Hnh 9.4. H th ng m Bi ton 1: D li u Input l cc gi trr.V d : Xt bi ton m xc nh b i cc lu t sau:

Lu t 1:if x is A1and y is B1 Then z is C1Lu t 2 :if x is A2or y is B2Then z is C2Vo: tr x0, y0------------------------------------------------------------Ra : tr z0 tng ng

88

5.8 NGUYN L XL BI TON M(ti p)

ng v i t p m A1 ta c hm thnh vin A1(x)ng v i t p m A2 ta c hm thnh vin A2(x)ng v i t p m B1 ta c hm thnh vin B1(y)ng v i t p m B2 ta c hm thnh vin B2(x)ng v i t p m C1 ta c hm thnh vin C1 (x)ng v i t p m C2 ta c hm thnh vin C2 (x)


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89

V d : Gi i bi ton i u khi n t ng m cho h th ng b m n c l y n c t gi ng.

Trong khi h h t n c v trong gi ng c n c th my b m t ng b m.

0 0 0 N.t

B.H iLu B.V a 0 N.V a

B.Lu B.V a 0 N.Cao

H.C n H.L ng H. y

V i bi n ngn ng H c cc t p m h y (H. y), h l ng (H.L ng)v h c n (H.C n).

V i bi n ngn ng Gi ng c cc t p m nu c cao (N.Cao), nu c v a (N.V a), nu c t (N.t).V i bi n ngn ng k t lu n xc nh th i gian b m s c cc t p m b m v a (B.V a), b m lu (B.Lu), b m h i lu(B.H iLu).

90

V d (ti p)Trong x ch su c a H (0


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91

T b ng trn ta c cc lu t:Lu t 1: if x is H.L ng and y is N.Cao Then z is B.V a Lu t 2: if x is H.C n and y is N.Cao Then z is B.Lu Lu t 3: if x is H.L ng and y is N.V a Then z is B.V a Lu t 4: if x is H.C n and y is N.V a Then z is B.H i lu N u nh p tr Input x0 = 1 ( cao c a n c trong h ),

y0 = 3 ( cao c a n c trong gi ng)

V d (ti p)

92

V d (ti p)


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93

V d (ti p)

Cc Wig i l cc tr ng s c a lu t th iTheo l thuyt hm thnh vin c a k t lu n cho b i cng th c: C(z) = Wi K1i(Z) i = 1 N C(z) = W1.B.V a(z) + W2.B.Lu(z) + W3.B.V a(z) + W4.B.Hi Lu(z) C(z) = 3/10.B.Va(z) + 0.5.B.Lu(z) + 3/5.B.Va(z) + 0.5.B.HiLu(z)B c ti p theo l ta ph i gi i m t hm thnh vin c a k t lu n b ngcnh tnh tr ng tm c a hm C(z)Moment C(z) lv

V y Defuzzy(z) =16.06Do n u m c n c trong h v gi ng l 1m v 3m th thi gian c nbm l 16 pht v 06 giy.

U

C dz z)(

U

C dz z z )(

06.16)(

)(

0 ==

U C

U C

dz z

dz z z

z

94

Kh m : L y i m tr ng tm

=

=

=

=

=

==

n

i U U

n

i U

U

U

n

iU

U

n

iU

U A

U A

dx x

dx x x

dx x

dx x x

dx x

dx x x

x

i

i

i

i

1

1

1

10

)(

)(

)(

)(

)(

)(


48/50


49/50

97

4.6.8 NGUYN L XL CC BI TON M(ti p)

B i u khi n m :SISO: If (A is A1) Then (B is B1)

...If (A is An) Then (B is Bn)

MIMO: If (A1 is A11) and ... and (Am is Am1) Then (B1 is B11) and ... and (Bs is B1s)...If (A1 is An1) and ... and (Am is An1) Then (B1 is Bn1) and ... and (Bs is Bns)

MISO: If (A1 is A11) and ... and (Am is Am1) Then (B is B1)...

If (A1 is An1) and ... and (Am is An1) Then (B is Bn)SIMO: If (A is A1) Then (B1 is B11) and ... and (Bs is B1s)

...If (A is An) Then (B1 is Bn1) and ... and (Bs is Bns)

98

Bi t p: i u khi n m c n cBi ton i u khi n m c n c: Khng ph thu c vo l ng

n c ch y ra kh i bnh, c n ph i i u ch nh van cho l ngn c ch y vo bnh v a sao cho m c n c h trongbnh l lun khng i.

Gi s b i u khi n l con ng i, s c cc nguyn t c sau:

R1: N u m c n c l th p nhi u th van m c m to R2: N u m c n c l th p t th van m c m nh R3: N u m c n c l th van v tr ng .R4: N u m c n c l cao th van v tr ng .


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99

Bi t p: i u khi n m c n c(ti p)

Cc bi n ngn ng :+ x l m c n c:c 4 gi tr T={th p nhi u, th p t, , cao}+ y van: c 3 gi tr T={to, nh , ng}M c n c X [0 .. 3] (mt)

(x)

0.7

0.4

1 1.5 2 2.5 3

Th p nhi u Th p cao

Bi t p: i u khi n m c n c(ti p)

Van Y [0 .. 10] (cm)V i y l m c a van, T={to, nh , ng}

(y)

Cho m c n c cao x0= 2 m, h i m c a van y0 l baonhiu

1 10 y

ng nh to

bai giang he co so tri thuc

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