math in it

8/8/2019 Math in It

1/42

I HC QUC GIA THNH PH H CH MINH

CHNG TRNH O TO THC S CNTT QUA MNG

CHUYN PHNG PHP TON TRONG TIN HC

KHA LUN

LOGIC M, S M V H M

GING VIN HNG DNGS,TSKH. HONG KIM

HC VIN THC HIN

NGC ANH CH0301002 BI TRN QUANG V CH0301088

2005

8/8/2019 Math in It

2/42

Logic m, s m v h m

2

Li ni u

Con ngi giao tip bng ngn ng t nhin, m bn cht ca ngn ng t nhin lm h v khng chnh xc. Tuy vy, trong hu ht tnh hung, con ngi vn hiunhng iu m ngi khc mun ni vi mnh. Kh nng hiu v s dng ng ngnng t nhin, thc cht l hiu v x l ng thng tin khng chnh xc cha trong, c th coi l thc o mc hiu bit, thng minh ca con ngi. Con ngicng lun m c my tnh, ngi bn, ngi gip vic c lc ca mnh, ngy cngthng minh v hiu bit hn. V vy, nhu cu lm cho my tnh hiu v x l cnhng thng tin khng chnh xc, xp x, ng chng l mt nhu cu bc thit.

Logic m ra i cung cp mt cng c hu hiu nghin cu v xy dng cch thng c kh nng x l thng tin khng chnh xc. Nh c logic m m con ngixy dng c nhng h iu khin c tnh linh ng rt cao. Chng c th hot ngtt ngay trong iu kin c nhiu nhiu hoc nhng tnh hung cha c hc trc.

Nh c logic m m con ngi xy dng c nhng h chuyn gia c kh nng suylun nh nhng chuyn gia hng u v c kh nng t hon thin thng qua vic thunhn tri thc mi.

Ngy nay logic m c phm vi ng dng rng ri trn th gii, t nhng h thngcao cp phc tp nh nhng h d bo, nhn dng, robos, v tinh, du thuyn, my

bay, n nhng dng hng ngy nh my git, my iu ho khng kh, mychp hnh t ng, Nhng trung tm ln v l thuyt cng nh ng dng ca logicm hin nay l M, Nht, v Chu u.

Vit Nam, vic nghin cu v l thuyt cng nh ng dng ca logic m clch s gn hai thp k v thu c nhng thnh tu to ln. Tuy vy vn cn thit

phi pht trin hn na c v chiu su ln chiu rng.

Bi thu hoch ny ca nhm hc vin Ngc Anh v Bi Trn Quang V l ktqu tm hiu v logic m, phng php xy dng mt h iu khin m in hnh vminh ho l thuyt bng mt h m n gin iu khin my git t ng.

Chng em xin chn thnh cm n GS.TSKH Hong Kim, ging vin mn hcPhng php ton trong tin hc, truyn t nhng kin thc qu bu v cng nghtri thc v c bit l v logic m, gip cho chng em vit bi thu hoch ny. Xinchn thnh cm n ban c vn hc tp v ban qun tr Chng trnh o to thc sCng ngh thng tin qua mng ca i Hc Quc Gia Thnh ph H Ch Minh toiu kin v ti liu tham kho.

8/8/2019 Math in It

3/42

Logic m, s m v h m

3

Mc lcLi ni u ................................ ................................ ................................ ........ 2Mc lc ................................ ................................ ................................ ............. 3

CHNG I. LOGIC M ................................ ................................ ...................... 5TP M ................................ ................................ ................................ ............ 5

Khi nim tp m ................................ ................................ .......................... 5Cc dng hm thuc tiu biu ................................ ................................ ........ 6

Nhm hm n iu ................................ ................................ ................... 6Nhm hm hnh chung ................................ ................................ ............. 6

Cc khi nim lin quan ................................ ................................ ................. 7Cc php ton trn tp m................................ ................................ .............. 7Cc php ton m rng ................................ ................................ .................. 8

S M ................................ ................................ ................................ ............ 10nh ngha ................................ ................................ ................................ ... 11Cc php ton ................................ ................................ .............................. 11

Nguyn l suy rng ca Zadeh ................................ ................................ ..... 11LOGIC M ................................ ................................ ................................ ..... 12

Bin ngn ng ................................ ................................ ............................. 12Mnh m ................................ ................................ ................................ . 13Cc php ton mnh m ................................ ................................ .......... 13Php ton ko theo m lut if-then m thng dng ................................ .... 13Lut modus-ponens tng qut ................................ ................................ ...... 14

CHNG II. H M ................................ ................................ ......................... 16KIN TRC CA H M TNG QUT ................................ ...................... 16C S LUT M ................................ ................................ .......................... 17B SUY DIN M ................................ ................................ ......................... 17

Trng hp mt u vo v mt lut ................................ ........................... 18

Trng hp hai u vo v mt lut ................................ ............................. 18Trng hp nhiu u vo v nhiu lut ................................ ...................... 19B M HO ................................ ................................ ................................ .. 19

M ho n tr ................................ ................................ ............................. 19M ho Gaus ................................ ................................ ............................... 19M ho tam gic ................................ ................................ .......................... 19

B GII M ................................ ................................ ................................ .. 20Phng php ly max ................................ ................................ .................. 20Phng php ly trng tm ................................ ................................ .......... 20Phng php ly trung bnh tm ................................ ................................ .. 20

H M L MT H XP X VN NNG ................................ ................... 20SO SNH H M VI MNG NRON ................................ ....................... 21

GII THIU MT S H M TRONG THC T ................................ ....... 22CHNG III. PHNG PHP THIT K H IU KHIN M T TP DLIU VO V RA ................................ ................................ ................................ .. 24

T VN ................................ ................................ ................................ . 24THIT K H IU KIN M BNG BNG D LIU VO .................... 24

CHNG IV. MINH HO H M: H IU KHIN MY BM NC TNG ................................ ................................ ................................ ..................... 26

CHNG TRNH MINH HO H M IU KHIN MY GIT ............ 29Cc bin ngn ng ................................ ................................ ....................... 29

8/8/2019 Math in It

4/42

Logic m, s m v h m

4

Cc gi tr ngn ng ................................ ................................ ..................... 30Cc lut m ................................ ................................ ................................ . 30Hng dn s dng chng trnh ................................ ................................ . 30Kt qu chy chng trnh ................................ ................................ ........... 34

Thut ng ................................ ................................ ................................ ........ 41Ti liu tham kho ................................ ................................ ........................... 42

8/8/2019 Math in It

5/42

Logic m, s m v h m

5

CHNG I.LOGIC M

TP M

Khi nim tp m

Mt tp hp trong mt khng gian no , theo khi nim c in s chia khnggian thnh 2 phn r rng. Mt phn t bt k trong khng gian s thuc hoc khngthuc vo tp cho. Tp hp nh vy cn c gi l tp r. L thuyt tp hp cin l nn tng cho nhiu ngnh khoa hc, chng t vai tr quan trng ca mnh.

Nhng nhng yu cu pht sinh trong khoa hc cng nh cuc sng cho thy rngl thuyt tp hp c in cn phi c m rng.

Ta xt tp hp nhng ngi tr. Ta thy rng ngi di 26 tui th r rng l trv ngi trn 60 tui th r rng l khng tr. Nhng nhng ngi c tui t 26 n

60 th c thuc tp hp nhng ngi tr hay khng? Nu p dng khi nim tp hpc in th ta phi nh ra mt ranh gii r rng v mang tnh cht p t chng hn l45 xc nh tp hp nhng ngi tr. V trong thc t th c mt ranh gii m ngn cch nhng ngi tr v nhng ngi khng tr l nhng ngi trung nin.

Nh vy, nhng ngi trung nin l nhng ngi c mt tr no . Nu coi tr ca ngi di 26 tui l hon ton ng tc l c gi tr l 1 v coi tr cangi trn 60 tui l hon ton sai tc l c gi tr l 0, th tr ca ngi trungnin s c gi tr p no tho 0 < p < 1.

Nh vy nhu cu m rng khi nim tp hp v l thuyt tp hp l hon ton tnhin. Cc cng trnh nghin cu v l thuyt tp m v logic m c L.Zadehcng b u tin nm 1965, v sau lin tc pht trin mnh m.

nh ngha: Cho khng gian nn U, tp A U c gi l tp m nu A cxc nh bi hm

AQ :X->[0,1].

AQ c gi l hm thuc, hm lin thuc hay hm thnh vin (membershipfunction)

Vi xX th AQ (x) c gi l mc thuc ca x vo A.Nh vy ta c th coi tp r l mt trng hp c bit ca tp m, trong hm

thuc ch nhn 2 gi tr 0 v 1.

K hiu tp m, ta c cc dng k hiu sau: Lit k phn t: gi s U={a,b,c,d} ta co th xc nh mt tp m

A=dcba02.03.01.0

A = _ aUxxx A |)(,Q

A = Ux

A

x

x)(Qtrong trng hp U l khng gian ri rc

A =U

Axx /)(Q trong trng hp U l khng gian lin tc

8/8/2019 Math in It

6/42

Logic m, s m v h m

6

Lu l cc k hiu v khng phi l cc php tnh tng hay tch phn, mch l k hiu biu th tp hp m.

V d. Tp m A l tp s gn 2 xc nh bi hm thuc2)2( ! xA eQ ta c th

k hiu: A = _ aUxxx |)2(, 2 hoc A = g

g

xx /)2( 2

Cc dng hm thuc tiu biu

Theo l thuyt th hm thuc c th l mt hm bt k tho AQ :X->[0,1]. Nhngtrong thc t th c cc dng hm thuc sau y l quan trng v c tnh ng dng caohn c.

Nhm hm n iuNhm ny gm n iu tng v n iu gim. V d tp hp ngi gi c hm

thuc n iu tng theo tui trong khi tp hp ngi tr c hm thuc n iu

gim theo tui. Ta xt thm v d minh ho sau: Cho tp v tr E = Tc =a_ 120,100,80,50,20 n v l km/h. Xt tp m F=Tc nhanh xc nh bi hmthuc nhanhQ nh th

Nh vy tc di 20km/h c coi l khng nhanh. Tc cng cao th thuc ca n vo tp F cng cao. Khi tc l 100km/h tr ln th thuc l 1.

Nhm hm hnh chungNhm hm ny c th dng hnh chung, bao gm dng hm tam gic, hm hnh

thang, gauss.Xt v d cng vi tp v tr E trn, xt tp m F=Tc trung bnh xc nh

bi hm thuc

ee

ee

ue

!

1005050/)100(

502030/)20(

100200

xkhix

xkhix

xxkhi

trungbnhQ

1

0.85

0.5

10020 50 80

E

nhanhQ

120

1

0.4

10020 50 80

E

trungbnhQ

120

8/8/2019 Math in It

7/42

Logic m, s m v h m

7

Cc khi nim lin quan

Gi s A l tp m trn v tr U, c hm thuc AQ th ta c cc khi nim sau: Gi ca A, k hiu supp(A) l mt tp r bao gm tt c cc phn t x

U sao choA

Q (x) > 0

Nhn ca A l mt tp r bao gm tt c cc phn t x U sao cho AQ (x)= 1

Bin ca A l mt tp r bao gm tt c cc phn t x U sao cho 0< AQ (x) < 1

cao ca A, k hiu height(A) l cn trn ng caAQ (x).

height(A)= )(sup xA

Ux

Q

Tp m A c gi l tp m chun tc (normal fuzzy set) nuheight(A)=1. Tc l tp m chun tc c nhn khc rng.

Cc php ton trn tp m

Gi s A v B l cc tp m trn v tr U th ta c cc nh ngha sau:

Quan h bao hm

A c gi l bng B khi v ch khi xU, AQ (x) = BQ (x) .

A c gi l tp con ca B, k hiu A B khi v ch khi xU, AQ (x) e BQ (x)

Phn b

Phn b m ca tp m A l tp m A vi hm thuc c xc nh bi:A

Q (x) = 1 - AQ (x) (1)

HpHp ca tp m A v tp m B l tp m AB vi hm thuc c xc nh bi:

BAQ (x) = max( AQ (x), BQ (x)) (2)

GiaoGiao ca tp m A v tp m B l tp m AB vi hm thuc c xc nh bi:

BAQ (x) = min( AQ (x), BQ (x)) (3)

Tch ccGi s 1A , 2A , , nA l cc tp m trn cc v tr 1U , 2U , , nU tng ng. Tch

-cc ca 1A , 2A , , nA l tp m A = 1A v 2A v v nA trn khng gian tch

1U v 2U v v nU vi hm thuc c xc nh bi:

AQ ( 1x , 2x , , nx ) = min( 1AQ ( 1x ), 2AQ ( 2x ), , nAQ ( nx ))

1x 1U , 2x 2U , , nx nU (4)

8/8/2019 Math in It

8/42

Logic m, s m v h m

8

Php chiu

Gi s A l tp m trn khng gian tch 1U v 2U . Hnh chiu ca A trn 1U l tp

m 1A vi hm thuc c xc nh bi:

1AQ (x) =

2

maxUy

AQ (x, y) (5)

nh ngha trn c th m rng cho trng hp khng gian tch n chiu

Mrng hnh tr

Gi s 1A l tp m trn v tr 1U . M rng hnh tr ca 1A trn khng gian tch

1U v 2U l tp m A vi hm thuc c xc nh bi:

AQ (x, y) = 1AQ (x) (6)

Cc php ton m rng

Ngoi cc php ton chun: phn b, hp, giao c cp trn cn c nhiucch m rng php ton trn tp m khc c tnh tng qut ha cao hn.

Phn b mGi s xt hm C:[0,1] -> [0,1] cho bi cng thc C(a) = 1 a, a [0,1]. Khi

hm thuc ca phn b chun tr thnhA

Q (x) = C( AQ (x)). Nu tng qut ho tnh

cht ca hm C th ta s c tng qut ho nh ngha ca phn b m. T ta cnh ngha:

Phn b mca tp m A l tp m A vi hm thuc c xc nh biA

Q (x) =

C( AQ (x)), trong C l mt hm s tho cc iu kin sau:i. Tin C1 (iu kin bin): C(0) = 1, C(1) = 0

ii. Tin C2 (n iu gim): a, b [0,1]. Nu a < b th C(a) u C(b)

Hm C tho cc iu kin trn c gi l hm phn b.Ta thy rng hm thuc ca phn b chun l mt hm c bit trong h cc hm

phn b.Vd:

Hm phn b Sugeno C(a) =a

a

P

1

1trong P l tham s tho P > -1. Hm b

chun l trng hp c bit ca hm Sugeno khi P = 0.

Hm phn b Yager C(a) = wwa1

)1( trong w l tham s tho w > 0. Hm bchun l trng hp c bit ca hm Yager khi w = 1.

Hp m ccphpton S-normPhp ton max trong cng thc hm hp m chun c th c tng qut ho

thnh cc hm S-norm:Mt hm s S: [0,1]x[0,1] -> [0,1] c gi l mt S-norm nu tho cc iu kin

sau:i. Tin S1 (iu kin bin): S(0,a) = a, a[0,1]

ii. Tin S2 (giao hon): S(a,b) = S(b,a), a,b[0,1]iii. Tin S3 (kt hp): S(S(a,b),c) = S(a,S(b,c)), a,b,c[0,1]

8/8/2019 Math in It

9/42

Logic m, s m v h m

9

iv. Tin S4 (n iu tng): Nu a e b v ce d th S(a,c)eS(b,d), a,b,c,d[0,1]

S-norm cn c gi l co-norm hoc T-i chun.

Hp ca tp m A v tp m B l tp m AB vi hm thuc c xc nh bi:

BAQ (x) = S( AQ (x), BQ (x))trong S l mt S-norm

Ngoi hm max, ta c mt s hm S -norm quan trng sau y:

Tng Drastic :

""

!

!

!

0,01

0

0

baif

aifb

bifa

ba

Tng chn: ),1min( baba !

Tng i s: abbaba !

Php hp Yager:

-

! www

wbabaS

1

)(,1min),(

Trong w l tham s tho w > 0

Giao m ccphpton T-normTa c nh ngha hm T-norm l tng qut ho ca hm min:Mt hm s T: [0,1]x[0,1] -> [0,1] c gi l mt T-norm nu tho cc iu kin:

i. Tin T1 (iu kin bin): T(1,a) = a, a[0,1]ii. Tin T2 (giao hon): T(a,b) = T(b,a), a,b[0,1]

iii. Tin T3 (kt hp): T(T(a,b),c) = T(a,T(b,c)), a,b,c[0,1]iv. Tin T4 (n iu tng): Nu ae b v ce d th T(a,c)eT(b,d),

a,b,c,d[0,1]

T-norm cn c gi l T-chun hoc chun tam gic.

Giao ca tp m A v tp m B l tp m AB vi hm thuc c xc nh nhsau:

BAQ (x) = T( AQ (x), BQ (x))

Trong T l mt T-norm.

Ngoi hm min, ta c mt s hm T -norm quan trng sau y: Tch Drastic:

!

!

!

1,10

1

1

baif

aifb

bifa

ba

Tch chn: )1,0max( ! baba Tch i s: abba !. Php giao Yager:

8/8/2019 Math in It

10/42

Logic m, s m v h m

10

-

! www

wbabaT

1

))1()1((,1min1),(

Trong w l tham s tho w>0

nh l: Vi mi T-norm bt k T v S-norm bt k S ta c:

ab e T(a,b) e min(a,b) e max(a,b) e S(a,b) e ab

Tch -cc m

Tch -cc ca tp m 1A , 2A , , nA trn cc v tr 1U , 2U , , nU tng ng l

tp m A = 1A v 2A v v nA trn khng gian tch 1U v 2U v v nU vi hm

thuc c xc nh nh sau:

AQ ( 1x , 2x , , nx ) = 1AQ (x) T 2AQ (x) T T nAQ (x)

1x 1U , 2x 2U , , nx nU Trong T l mt T-norm bt k.

Ta thy y l nh ngha m rng cho tch -cc chun khi thay th hm min bngmt T-norm bt k.

Quan h mCho U v V l cc v tr. Khi mt quan h m hai ngi R gia U v V l mt

tp m trong tch -cc UxV. Nh vy ta c th xc nh hm thuc cho quan h mtheo cch tnh hm thuc cho tch -cc m.

Khi U = V ta ni R l quan h trn U.

Tng qut mt quan h m R gia cc tp 1U , 2U , , nU l tp m A = 1A v

2A v v

nA trn khng gian tch

1U v

2U v v

nU . Trong

iA

iU , i = 1..n

Hp ca cc quan h mHp ca quan h m R t U n V v quan h m Z t V n W l quan h m

RoZ t U n W c hm thuc xc nh bi

RoSQ (u,w) =Vv

max _ T( RQ (u,v), ZQ (v,w)) a

Trong T l mt T-norm bt k.Cc hm thuc quan trng sau c dng rng ri xc nh hp ca cc quan h

m : Hm hp max-min:

RoSQ (u,w) = Vvmax _ min( RQ (u,v), ZQ (v,w)) a Hm hp max-tch (hay max-prod):

RoSQ (u,w) =Vv

max _R

Q (u,v) .Z

Q (v,w) a

S MMt lp tp m quan trng c nhiu ng dng thc t l s m

8/8/2019 Math in It

11/42

Logic m, s m v h m

11

nhngha

Tp m M trn ng thng thc R l tp s m nu:a) M l chun ho, tc l c im x sao cho Q M(x) = 1

b) ng vi mi a E R, tp mc {x: M(x) u E } l on ng

Ngi ta thng dng cc s m tam gic, hnh thang v dng Gauss

Cc php ton

a) Cng:[a,b] + [d,e] = [a+d, b+e]

b) Tr:[a,b] - [d,e] = [a-e, b-d]c) Nhn:[a,b] * [d,e] = [min(ad,ae, bd, be), max(ad,ae, bd, be)]d) Chia:[a,b] / [d,e] = [min(a/d,a/e, b/d, b/e), max(a/d,a/e, b/d, b/e)]

Nguyn l suyrngcaZadeh

lm vic vi cc h thng c nhiu bin vo, nguyn l suy rng ca Zadeh lrt quan trng

nh ngha: Cho Ai l tp m vi cc hm thuc Q Ai trn khng gian nn Xi,(i=1..n). Khi tch A1xA2x..An l tp m trn X=X1xX2x..Xn vi hm thuc:

Q A(x)=min{ Q Ai(xi); i=1..n} Trong x=(x1,x2,..xn)

Gi s mi bin u vo xi ly gi tr l Ai(i=1..n). Hm f:X->Y chuyn cc gi tru vo l Ai thnh gi tr u ra B. Khi B l tp m trn Y vi hm thuc xcnh bi:

Q B(x)=max{min(Q Ai(xi)); i=1..n : xf 1 (y)} nu f 1 (y) { J

Q B(x)=0 nu f 1 (y) =J

Trong f 1 (y) = {x X : f(x)=y}

Ta c th p dng nguyn l suy rng cho nh ngha suy rng ca php cng nhmt hm 2 bin m. Tng t cho cc php ton tr, nhn, chia.

T cc php ton c bn ngi ta xy dng nn s hc m. C nhiu cch xydng mt s hc m. Sau y l s hc m da trn khi nim E -cuts (lt ct alpha).E -cuts ca s m l khong ng thc vi mi 0< E = 0 bB, cC th A.(B.C)=A.B+A.C6. O A-A; 1A/A7. Nu A E v B F th:

8/8/2019 Math in It

12/42

Logic m, s m v h m

12

a. A+B E+Fb. A-B E-Fc. A.B E.Fd. A/B E/F

LOGIC M

Bin ngn ng

Ta xt mt bin nhn gi tr trong mt min gi tr no , chng hn nhit c th nhn gi tr s l 1QC, 2 QC, l cc gi tr chnh xc. Khi , vi mt gi trc th gn vo bin s gip chng ta xc nh c tnh cht, quy m ca bin. Ngoira chng ta cn bit c nhng thng tin khc lin quan n bin . V d chng tahiu l khng nn chm tay trn vo vt c nhit l 80QC tr ln. Nhng trongthc t th chng ta thng ni khng nn chm vo vt c nhit cao ch t khini khng nn chm vo vt c nhit l 80QC tr ln. Thc t l li khuyn uth c ch hn bi v nu nhn c li khuyn sau th ta d b ng nhn l c th

chm tay vo vt c nhit l 79QC trong khi vt c nhit 80 QC tr ln thkhng. Nhng vn t ra l nu nghe theo li khuyn u th ta c th xc nh rl nhit bng bao nhiu th c th chm tay vo? Cu tr li l tu vo kin catng ngi. Vi nhit l 60 QC th c ngi cho l cao trong khi ngi khc thkhng. Tuy cc kin l khc nhau nhng c mt iu chc chn l khi gi tr ca

bin nhit cng tng th cng d dng c chp nhn l cao. Nh vy nu xthm caoQ nhn bin nhit v tr v t l kin ng l cao th caoQ s l hm

thuc ca tp m nhit cao trn v tr nhit

Bin nhit c th nhn gi tr cao l mt gi tr ca ngn ng t nhin nn nc gi l mt bin ngn ng (linguistic variable)

Khi nim bin ngn ng c Zadeh a ra nm 1973 nh sau: Mt bin ngn ng c xc nh bi b (x, T, U, M) trong : x l tn bin. V d nhit , tc , m, T l tp cc t l cc gi tr ngn ng t nhin m x c th nhn. V d x

l tc th T c th l {chm, trung bnh, nhanh} U l min cc gi tr vt l m x c th nhn V d x l tc th U c

th l {0km/h,1km/h, 150km/h} M l lut ng ngha, ng mi t trong T vi mt tp m At trong U

1

0.9

10050 80

Nhit

caoQ

120

8/8/2019 Math in It

13/42

Logic m, s m v h m

13

T nh ngha trn chng ta c th ni rng bin ngn ng l bin c th nhn gitr l cc tp m trn mt v tr no .

Mnh m

Trong logic c in (logic v t cp mt), mt mnh phn t P(x) l mt phtbiu c dng x l P trong x l mt i tng trong mt v tr U no tho tnh

cht P. V d x l s chn th U l tp cc s nguyn v P l tnh cht chia ht cho 2. Nh vy ta c th ng nht mt mnh phn t x l P vi mt tp (r) A =_ xU | P(x) a.

T ta c:P(x) = P (x)

Trong P l hm c trng ca tp A ( xA P (x) = 1). Gi tr chn l ca P(x)ch nhn mt trong hai gi tr 1 v 0 (true v false) tng ng vi s kin x thuc Ahoc khng

Trong trng hp P l mt tnh cht m chng hn nh s ln th ta s c mtmnh logic m phn t. Khi tp hp cc phn t trong v tr U tho P l mttp m B c hm thuc

BQ sao cho:

P(x) = BQ (x)Lc ny P(x) c th nhn cc gi tr tu trong [0,1]. V ta thy c th ng nht

cc hm thuc vi cc mnh logic m.

Cc php ton mnh m

Trong logic c in, t cc mnh phn t v cc php ton (AND), (OR), (NOT) ta c th lp nn cc mnh phc. Ta c: P(x) = 1 P(x)P(x) Q(y) = min(P(x), Q(y))P(x) Q(y)=max(P(x), Q(y))P(x)=>Q(y) = P(x) Q(y) = max(1-P(x), Q(y))P(x)=>Q(y) = P(x) (P(x) Q(y)) = max(1-P(x), min(P(x), Q(y)))

Nh vy, ta s c m rng mt cch t nhin t logic c in sang logic m viquy tc tng qut ho dng hm b m cho php ph nh, hm T-norm cho phpgiao v S-norm cho php hp. S m rng ny da trn s tng quan gia mnh logic m vi hm m v cc php ton trn tp m. Ta c:

AQ (x) = C( AQ (x))

AQ (x) BQ (y) = T( AQ (x), BQ (y))

AQ (x) BQ (y) = S( AQ (x), BQ (y))

AQ (x) => BQ (y) = S(C( AQ (x)), BQ (y)) (1)

AQ (x) => BQ (y) = S( C( AQ (x)), T( AQ (x), BQ (y)) ) (2)

Trong C l hm b m (hay ph nh m), T l hm T-norm, S l hm S-norm.Cc hm ny trnh by trong phn php ton trn tp m.

Php ton ko theo m lut if-then m thng dng

Cc php ton ko theo c vai tr quan trng trong logic m. Chng to nn cclut m thc hin cc php suy din trong tt c cc h m. Do mt mnh mtng ng vi mt tp m nn ta c th dng hm thuc thay cho cc mnh .

Sau y l mt s php ko theo quan trng c s dng rng ri:

8/8/2019 Math in It

14/42

Logic m, s m v h m

14

Php ko theo Dienes RescherNu p dng cng thc (1) vi S-norm max v C l hm b chun cho ta c php

ko theo Dienes Rescher

AQ (x) => BQ (y) = max(1- AQ (x), BQ (y))

Php ko theo LukasiewiczNu p dng cng thc (1) vi S-norm l hm hp Yager vi w=1 v C l hm b

chun cho ta c php ko theo Lukasiewicz:

AQ (x) => BQ (y) = min(1, 1- AQ (x)+ BQ (y))

Php ko theo ZadehNu p dng cng thc (2) vi S-norm l max, T-norm min hoc tch v C l hm

b chun cho ta c php ko theo Zadeh:

AQ (x) => BQ (y) = max( 1- AQ (x), min( AQ (x), BQ (y))) (a)

AQ (x) => BQ (y) = max( 1- AQ (x), AQ (x). BQ (y)) (b)

Ko theo MamdaniTa c th coi mnh AQ (x) => BQ (y) xc nh mt quan h 2 ngi R UxV.

Trong U l khng gian nn ca x (v tr cha x), V l khng gian nn ca y (v trcha y). Khi gi tr chn l ca mnh AQ (x) => BQ (y) l gi tr hm thuc cacp (x,y) vo R. Theo cng thc xc nh hm thuc ca quan h m ta c

AQ (x) => BQ (y) = T( AQ (x), BQ (y))Trong T l mt T-norm. Khi chn T l min hoc tch ta c cc php ko

theo Mamdani:

AQ (x) => BQ (y) = min( AQ (x), BQ (y)) (a)

AQ (x) => BQ (y) = AQ (x). BQ (y) (b)

Lut modus-ponens tng qut

Tng t logic c in, trong logic m cng c lut modus-ponens nh sau:GT1 (lut) : if x l A then y l BGT2 (s kin) : x l A--------------------------------------------------------KL : y l B

Trong A, B, A, B l cc bin ngn ng (c ngha l cc tp m).

Cng thc tnh kt lun ca lut modus-ponens nh sau:'BQ (y) = sup

x

T(R

Q (x,y), 'AQ (x)) (*)

Trong T l mt hm T-norm v R l quan h hai ngi xc nh bi php kotheo. Cch tnh

RQ (x,y), chnh l cch tnh gi tr chn l ca php ko theo trnh by

phn trc. Nh vy tu theo cch chn cch tnh lut ko theo khc nhau m ta ccch tnh kt qu ca lut modus-ponens khc nhau.

V d: Gi s quan h gia nhit v p sut cho bi lut sau:

8/8/2019 Math in It

15/42

Logic m, s m v h m

15

Nu nhit l cao th p sutl ln.Nhit nhn cc gi tr trong U = {30, 35, 40, 45}Ap sut nhn cc gi tr trong V = {50,55,60,65}Ta c cc tp m xc nh bi cc bin ngn ng nhit v p sut nh sau:

A = nhit cao =45

1

40

9.0

35

3.0

30

0

B = p sut ln =651

601

555.0

500

p dng lut ko theo Mamdani tch ta c quan h m sau (gi tr dng i, ct j lgi tr hm thuc ca cp nhit i v p sut j vo quan h)

R=

6560555045

40

35

30

115.00

9.09.045.00

3.03.015.00

0000

-

By gi, gi s ta bit s kin nhit l trung bnh vA = nhit trung bnh =

451.0

408.0

351

306.0

p dng cng thc (*) ta suy ra B =65

8.0

60

8.0

55

45.0

50

0

8/8/2019 Math in It

16/42

Logic m, s m v h m

16

CHNG II.H M

KIN TRC CA H M TNG QUT

Mt h m tiu biu c kin trc nh hnh v

Thnh phn trung tm ca h m l c s lut m (fuzzy rule base). C s lut mbao gm cc lut m if-then biu din tri thc ca chuyn gia trong lnh vc no .Trong trng hp mt h iu khin m c th th c s lut m chnh l tri thc vkinh nghim ca cc chuyn gia trong vic iu khin khi cha p dng h m.

Thnh phn quan trng k tip l b suy din m (fuzzy inference engine). Nhimv ca b phn ny l kt hp cc lut trong c slut m,p dng vo tp m uvo theo cc phng php suy din m xc nh tp m u ra.

D liu u vo ca h iu khin m l cc tn hiu do cc b phn cm bin mitrng cung cp sau khi s ho nn c tnh cht r (khi nim r y c ngha l

cc tn hiu khng phi l cc tp m, ch khng c ngha l cc tn hiu khng cnhiu). V vy cn phi c b m ho (fuzzier) chuyn cc d liu s u vothnh cc tp m b suy din m c th thao tc c.

D liu u ra ca b suy din m dng cc tp m s c b gii m(defuzzier) chuyn thnh tn hiu s trc khi truyn n cc c quan chp hnh nhtay my, cng tc, van iu khin,

Do cc d liu u vo v u ra c s ho nn ta ch cn xem xt cc h mlm vic vi cc bin s. Trng hp tng qut, h m nhn mt vector n chiu uvo v cho ra mt vector m chiu u ra. H m nh th c gi l h m nhiuu vo nhiu u ra (MIMO). Nu m bng 1, ta c h h m nhiu u vo mt

u ra (MISO). Mt h m nhiu u vo nhiu u ra c th phn tch thnh nhiuh nhiu u vo mt u ra. Do ta ch cn tm hiu k v h m nhiu u vo mt u ra vi cc bin s. Khi ch ni v h m nhiu - mt th ta s ngm hiu lmt h m nhiu u vo mt u ra vi cc bin s

C slut m

B suydin m

B mho

B giim

u vo (s)u vo (tp

m)

Tham kholut m

u ra (tpm)

u ra (s)

8/8/2019 Math in It

17/42

Logic m, s m v h m

17

K hiu RVRUU nn

ii !

!

,1

, trong iU l min xc nh ca cc bin vo

i, i=1..n v V l min gi tr ca bin ra y, ta c m hnh h m nhiu u vo mtu ra nh hnh v

Cc mc k tip s m t k hn v cc b phn chc nng ca h m.

C SLUT MC s lut m ca h m n u vo mt u ra gm m lut if-then m c dng:

If x1 l Ak1 v x2 l Ak2 v v xn l Akn then y l Bk , k=1..m (1)

Trong k l ch s ca lut (lut th k trong tp lut), xi l cc bin u vo, Akil cc tp m trn Ui (i=1..n), y l bin u ra v Bk l tp m trn V (k=1..m)

Cc lut m dng (1) c gi l cc lut if-then m chun tc. Cc lut m khngchun tc c th bin i a v dng chun tc tng ng.

C nhiu phng php xc nh cc lut m a vo c s lut m. Ccphng php thng dng l nh cc chuyn gia trong lnh vc p dng, hoc t quanst, thc nghim thng k c c cc tp d liu mu u vo v ra tng ng, t dng cc k thut khai m d liu rt ra cc lut.

B SUY DIN M

Chng ta s nghin cu phng php thit k b suy din trong trng hp c slut m gm m lut if-then m chun tc, nhiu u vo v mt u ra (MISO).

Cc lut if-then c th c p dng bng cc cng thc tng qut nh trnh bytrong chng logic m nhng trong thc t th thng c tnh bng cng thcMamdani max-min hoc max-tch (max-prod) . Chng ta s xem xt k kin trc bsuy din m s dng phng php suy din max-min. Khi chuyn qua phng phpsuy din max-tch th ch cn thay min bng php nhn trong cc cng thc.

Cho A, A, B ln lt l cc tp m trn v tr X, X, Y. Lut ifA then B c thhin nh mt quan h m R=AvB trn XvY. Khi tp m B suy ra t A c xcnh bi:

'BQ (y) = max {min [ 'AQ (x), RQ (x,y)]} (*)

H mnhiu u vo mt u ra

1Ux

2Ux

nUx

Vy

8/8/2019 Math in It

18/42

Logic m, s m v h m

18

Trng hp mt u vo v mt lut

Ta c 'BQ (y) = maxx

{min [ 'AQ (x), RQ (x,y)]}

= maxx

{min [ 'AQ (x), AQ (x), BQ (y)]}

= min {

maxx

(min [ AQ (x), AQ (x)]), BQ (y)}

= min {maxx

AA 'Q (x), BQ (y)}

= min { h AA' , BQ (y)}

Trong h AA ' l cao ca tp m AA

Trng hp hai u vo v mt luty l trng hp lut c pht biu Nu x l A v y l B th z

l C.

Lut: Nu x l A v y l B th z l CS kin: x l A v y l B-------------------------------

Kt lun: z l C

Lut m vi iu kin c 2 mnh nh trn c th biu din dng AxB => C.Suy lun tng t trng hp mt u vo v mt lut ta c:

'cQ (z) = min { h AxBxBA '' , CQ (z)}

M A x B A x B = (A A) x (B B) nn h AxBxBA '' = min {h AA ' ,h BB ' }Vy 'cQ (z) = min { h AA ' ,h BB ' , CQ (z)}

Suy rng ra cho trng hp nhiu u vo Ai, i=1..n v mt lut

Lut: Nu x1 l A1 v x2 l A2 v... v xn l An th z l CS kin: x1 l A1 v x2 l A2 v... v xn l An

-------------------------------Kt lun: z l C

'cQ (z) = min { (ni ..1

min! h AiiA ' ), CQ (z)}

Minh ha:

B

BA A

h

x

Q

y

Q

8/8/2019 Math in It

19/42

Logic m, s m v h m

19

Trng hp nhiu u vo v nhiu lut

Trong trng hp nhiu u vo v nhiu lut, ta tnh kt qu u ra cho tng lutsau kt qu ca h s l cc php giao hoc hp cc kt qu ring ty theo bncht ca h l hi hay tuyn cc lut.

Nu trong mt lut c dng Nu x l A hoc y l B th z l C ta tch thnh 2 lutring bit Nu x l A th z l C v Nu y l B th z l C tnh.

B M HOM ho l qu trnh bin i mt vector x=(x1,x2,,xn) U nR thnh mt tp

m A trn U. A s l u vo cho b suy din m. M ho phi tho cc tiu chunsau:

im d liu x phi c thuc cao vo A Vector x thu nhn t mi trng ngoi c th sai lch do nhiu nn A phi

phn nh c tnh gn ng ca d liu thc Hiu qu tnh ton: n gin cho cc tnh ton trong b suy din.

Sau y l mt s phng php m ho thng dng

M ho n trMi im d liu x c xem nh mt tp m n tr tc l tp m A c hm

thuc xc nh nh sau:

'AQ (u)=

{

!

xuif

xuif

0

1

M ho Gaus

Mi gi tr xi c biu din thnh mt s m Ai. Tp A l tch -cc ca ccAi

iA'Q ( iu ) =

2

i

ii

a

xu

e

M ho tam gic

Mi gi tr xi c biu din thnh mt s m Ai. Tp A l tch -cc ca ccAi

iA'Q ( iu ) =

e

iii

iiii

ii

bxuif

bxuifb

xu

||0

||||

1

h1

A A

x

Q

C

C

z

Q

B B

y

Q

h2

8/8/2019 Math in It

20/42

Logic m, s m v h m

20

B GII MGii m (hay cn gi l kh m) l qu trnh xc nh mt im y t mt tp m

trn B trn V. (B l u ra ca b suy din m ). Gii m phi tho cc tiu chunsau:

im y l i din tt nht cho B. Trc quan y l im c thuc caonht vo B v trung tm tp gi ca B.

Hiu qu tnh ton nhanh Tnh lin tc. Khi B thay i t th y cng thay i t

Sau y l mt s phng php gii m thng dng

Phng php ly max

Phng php ny chn y l im c thuc cao nht vo B

Xc nh tp r H=

!

)()(| '' sup vyVy BVv

BQQ

Sau c th chn y trong H nh sau: y bt k y l im cc bin (ln nht hoc nh nht) y l trung im ca H

Phng php ly trng tm

Phng php ny chn y l im trng tm ca tp B

y =

V

B

V

B

dvv

dvvv

)(

)(

'

'

Q

Q

P

hng php ly trung bnh tmV B thng l hp hoc giao ca m tp m thnh phn do vy ta c th tnh gnng gi tr y l bnh qun c trng s ca tm m tp m thnh phn. Gi s x i v h i

l tm v cao ca tp m thnh phn B i ta c:

y =

!

!

m

ii

m

iii

h

hx

1

1

.

Phng php ny c ng dng nhiu nht v kt qu u ra y c xt n nhhng ca tt c cc lut tng t nh phng php trng tm nhng phc tp tnhton t hn.

H ML MT H XP X VN NNG

Cc h m c ng dng thnh cng trong rt nhiu lnh vc, c bit l trongiu khin cc qu trnh cng nghip v trong h chuyn gia. Mt cu hi t ra l tisao h m li c phm vi ng dng rng ln v hiu qu nh th? Mt cch trc quanchng ta c th gii thch rng l v cc h m c th s dng tri thc ca cc

8/8/2019 Math in It

21/42

Logic m, s m v h m

21

chuyn gia c pht biu trong ngn ng t nhin nn c bn cht m. L do na lv cc s liu thu nhn c t mi trng l xp x, khng chnh xc nn cng c bncht m.

V mt l thuyt, nh l quan trng sau y c chng minh, l nn tngvng chc cho h m nhiu u vo mt u ra (MISO). M h m nhiu u vo

mt u ra c th coi nh n v cu thnh ca h m nhiu u vo nhiu u ra(MIMO).

Mt h m nhiu u vo mt u ra xc nh mt hm thc n bin y = F(x), ngvi mi vector u vo nn Rxxxx ! ,...,, 21 vi gi tr u ra Ry . Ta gi l F ls hm c trng ca h m ny.

nh l: Gi s U l tp compact trong nR , RV , f l hm f: U -> V. Nu f lintc th tn ti mt h m sao cho hm F c trng ca n xp x f vi chnh xc tu cho trc. Tc l

I

)()(sup xfxFUx

Chng minh nh l trn c th tham kho trong cc sch sau:

Lee C. S. G. and Lin C. T., Neural Fuzzy Systems, A neuro Fuzzy Synergismto Intelligent System, Prentice Hall, 1996

Wang L. X. A course in Fuzzy Systems and Control, Prentice Hall, 1997

SO SNH H M VI MNG NRON

H m v mng nron (neural network) u l cc h xp x vn nng v u lnhng h m phng hot ng ca no ngi. Nhng trong khi mng nron m

phng kin trc sinh l ca no th h m li m phng c ch tm l ca no. C thhn, mng nron m phng cc t bo thn kinh v mng li cng tc gia chng.Trong khi h m li m phng c ch suy lun xp x, ng chng ca no da vocc lut m nu-th. iu th v l c hai cch tip cn ny u thu c nhng ktqu to ln v chng b tr cho nhau, gip xy dng nhng h thng thng minh ckin trc lai phc tp ngy cng mnh.

8/8/2019 Math in It

22/42

Logic m, s m v h m

22

Hnh v di y tm lc s so snh gia mng nron v h m

GII THIU MT S H M TRONG THC T

ng dng ca logic m trong thc t rt phong ph v a dng. Chng ta c thphn chia cc ng dng thnh cc dng chnh sau y:

1. Cc h iu khin2. Cc h chuyn gia3. Cc h nhn dng4. Cc h m phng, gi lp5. Cc h chn on trong y t (l mt dng h chuyn gia c bit)

Sau y l mt s h thng ng dng logic m thnh cng trn th gii v trongnc:

ABVBL h chuyn gia chn on ph khoa. Suy din da trn logic m v d liu c

biu din l cc s m v bin ngn ng

CADIAG-2L h chuyn gia chn on y khoa tng qut. Kt hp thng k d liu v logic

m.

Mng nron

H m

HcB nhTr tu

Tr nhS biu din

Hm thuc

Trng s

Tip cn ngn ng

Tip cn s hc

Cc c tnh sinh l ca no ngi

Cc c tnh tm l ca no ngi

8/8/2019 Math in It

23/42

Logic m, s m v h m

23

CLINAIDL h c s tri thc cho chn on v n thuc. H dng lut xp x kt hp l

thuyt tp m

DIABETO-IIIL h chuyn gia chn on v iu tr bnh tiu ng.

MILORDL h chuyn gia chn on y khoa tng qut. H cho php biu din tri thc

khng chc chn nh ngha bi chuyn gia.

NEUMONIAL h chuyn gia chn on v iu tr bnh vim phi. H ny l ng dng c th

ca MILORD

8/8/2019 Math in It

24/42

Logic m, s m v h m

24

CHNG III.PHNG PHP THIT K H IUKHIN M T TP DLIU VO V RA

T VN Mt h thng iu kin m c mc ch m phng suy ngh ca con ngi khi iu

khin mt i tng no . Nhn chung, hiu bit ca con ngi gm 2 loi: hiu bit r (conscious knowledge) v hiu bit cha r (subconscious knowledge). Hiubit r l hiu bit c th din t bng ngn ng, cng thc, thut ton, chnhl tri gic. Cn hiu bit cha r l hiu bit m con ngi bit cch p dng, thchin ng nhng khng din t chnh xc c. Chng hn kin thc x l bng camt cu th gii. Anh ta hu nh khng th gii thch c l vi mt ng bngkh, lm th no a bng vo li mc d l iu m anh ta thc hin thngxuyn. Hiu bit cha r thng c tch ly t kinh nghim v bn cht l cmgic. a c kin thc hiu bit cha r vo h iu khin m th ta cn phi

lng ha cc iu kin u vo v u ra to thnh cc tp d liu vo-ra. Sau thit k h thng da trn c s tp d liu . y cng l mt trong nhng phng

php thu nhn tri thc t d liu th.

C nhiu phng php xc nh h iu kin m t tp d liu vo-ra nhng chngta ch tm hiu phng php lp bng d liu vo

THIT K H IU KIN M BNG BNG DLIU VO

Gi s ta c tp cc cp d liu vo-ra (xi, yi) i=1..N.

Trong xi [a1, b1] x x [ak, bk] Rk

v yi [c1,c2] R

Cc bc xy dng mt h iu kin m nh sau:

Bc 1: Xc nh ttc cc bin vo v ra

Bc 2: Xc nh min gitr bin vo v ra v cc hm thuc ca chng- Phn chia min gi tr ca cc bin sao cho c ngha v ph hp thc t. Mi

min s c i din bi mt tp m.- Vi mi tp m trn, xc nh hm thuc ca chng. Cc hm dng tam gic

v hnh thang thng c chn

Bc 3: Xc nh cc lutm

- Xt tng cp d liu vo-ra to ra tng lut ring bit. Vi mi cp d liuvo-ra (x,y) ta cho rng c mt lut A=>B nu QA

(x) l gi tr ln nht trong

cc gi tr hm thuc ca cc tp m u vo i vi x v QB

(y) l gi tr ln

nht trong cc gi tr hm thuc ca cc tp m u ra i vi y.- Xc nh trng s ca tng lut. Nu c 2 lut mu thun th chn lut c

trng s cao hn- Cc lut chn c a vo bng lut. V d: bng lut ca mt h iu

kin my bm nc

8/8/2019 Math in It

25/42

Logic m, s m v h m

25

H.y H.Lng H.Cn N.Cao 0 B.Va B.Lu N.Va 0 B.Va B.HiLu

N.t 0 0 0

Bc 4: Chnphngphp suy din

tnh hm thuc u ra th c nhiu phng php nh trnh by chnglogic m, nhng v l do d ci t v tc tnh ton nhanh nn phng php max-min, max-prod, sum-min, sum-prod thng c chn.

Bc 5: Chnphngphp giimPhng php trng tm hoc trung bnh tm thng c chn v l do d ci t

v tc tnh ton nhanh.

Bc 6: Tiu ha h lutv th nghim m hnhH thng c cho chy th v so snh vi cc kt qu c c bi chuyn gia.

Nu kt qu cha ph hp th cn hiu chnh cc hm thuc v cc lut cng nh

phng php suy din v gii m

8/8/2019 Math in It

26/42

Logic m, s m v h m

26

CHNG IV.MINH HO H M: H IU KHINMY BM NC T NG

minh ha cho l thuyt logic m v h iu kin m, chng ta s cng xem xt

mt h iu khin m iu khin my bm nc t ng. y l v d minh ha hthng m trong Gio trnh Cng ngh tri thc v ng dng ca GS.TSKH HongKim.

Vn : iu khin my bm t ng bm nc t ging vo h. Thi gian bms c h thng t ng tnh ton cn c trn mc nc ca ging v h. M hnh

bi ton nh hnh v:

Bc 1: Xc nh bin ngn ng

Vi bin ngn ng H (mc nc h) c cc tp m h y (H.y), h lng(H.Lng) v h cn (H.Cn)

Vi bin ngn ng Ging (mc nc ging) c cc tp m nc cao (N.Cao), ncva (N.Va) v nc t (N.t)

Vi bin ngn ng Bm (thi gian bm) c cc tp m bm lu (B.Lu), bm hilu (B.Hi Lu) v bm va (B.Va)

Bng quan h gia cc bin ngn ng:

H.y H.Lng H.CnN.Cao 0 B.Va B.LuN.Va 0 B.Va B.HiLuN.t 0 0 0

Bc 2: Xc nh hm thuc ca cc bin ngn ngHm thuc ca H nc:

BmGing

10 m

0 m

y

H

2 m

0 m

x

8/8/2019 Math in It

27/42

Logic m, s m v h m

27

H.y(x) = x/2 0

8/8/2019 Math in It

28/42

Logic m, s m v h m

28

Bc 3: Xc nh cc lutm Lut 1 (r1): ifxis H.Lng and y is N.Cao then z is B.Va Lut 2 (r3): ifxis H.Cn and y is N.Cao then z is B.Lu Lut 3 (r3): ifxis H.Lng and y is N.Va then z is B.Va Lut 4 (r4): ifxis H.Cn and y is N.Va then z is B.HiLu

Bc 4: Chnphngphp suy din m v giimVi mt cp gi tr u vo (x,y) ta tnh trng s ca tng lut i vi (x,y). Trng

s ca lut r chnh l thuc ca gi tr u vo (x,y) i vi gi thit ca r. T tatnh c gi tr u ra cho bi r. Gi tr u ra ca h l tng hp ca u ra ca ttc cc lut trong h. chnh l tp m u ra. Gii m ta c gi tr r u ra v lgi tr tn hiu iu khin ca h m.

V d gi tr u vo l x=1 (mc nc h) v y=3 (mc nc ging)Tnh cc trng s ca cc lut: k hiu wi l trng s ca lut riQ H.Lng(x) = 1 v QN.Cao(y) = 3/10 = 0.3 => w1 = min(1, 0.3) = 0.3Q H.Cn(x) = 0.5 v QN.Cao(y) = 3/10 = 0.3 => w2 = min(0.5, 0.3) = 0.3Q H.Lng(x) = 1 v QN.Va(y) = 3/5 = 0.6 => w3 = min(1, 0.6) = 0.6Q H.Cn(x) = 0.5 v QN.Va(y) = 3/5 = 0.6 => w4 = min(0.5, 0.6) = 0.5

Hm thuc ca kt lun:

QC

(z) = )(1

zKli

n

iiw Qv

!

= w1.B.Va(z) + w2.B.Lu(z) + w3.B.Va(z)+ w4.B.HiLu(z)= 0.3.B.Va(z) + 0.3.B.Lu(z) + 0.6.B.Va(z)+ 0.5.B.HiLu(z)= 0.9.B.Va(z) + 0.3.B.Lu(z) + 0.5.B.HiLu(z)

H.y

1

2

H.Cn

1

2N.Cao

1

10

N.t

1

10

N.Va

1

105

H.Lng

1

21

B.Lu

1

30

B.Va

1

3015

B.HiLu

1

3020

8/8/2019 Math in It

29/42

Logic m, s m v h m

29

=

!!

!!

!!

3020)20/2(5.030/3.0)15/2(9.0

201520/5.030/3.0)15/2(9.0

15020/5.030/3.015/9.0

zzzz

zzzz

zzzz

=

!!

!!

!!

3020075.08.2 2015025.08.1

150095.0

zz

zz

zz

Gii m:

F )(zcQ =

!!

!!

!!

30200375.08.2

20150125.08.1

1500475.0

2

2

2

zzz

zzz

zz

)(zz cQ =

!!

!!

!!

3020075.08.2

2015025.08.1

150095.0

2

2

2

zzz

zzz

zz

F )(zz cQ =

!!

!!

!!

3020025.04.1

20153/025.09.0

1503/095.0

32

32

3

zzz

zzz

zz

30

0

)( dzzz cQ = 15

0

)( dzzz cQ + 20

15

)( dzzz cQ + 30

20

)( dzzz cQ

= [ 33/095.0 z ]15

0+ [ 32 3/025.09.0 zz ]

20

15+ [ 32 025.04.1 zz ]

30

15

= 450.833

30

0 )( dzzcQ =

15

0 )( dzzcQ +

20

15 )( dzzcQ +

30

20 )( dzzcQ

=[ 20475.0 z ]15

0+ [ 20125.08.1 zz ]

20

15+ [ 20375.08.2 zz ]

30

15

= 26.75Vy deffuzzy(z) = 450.833 / 26.75 = 16.8536

Vy vi mc nc trong h l 1m v ging l 3m th cn bm 16.845=16 pht 51giy

CHNG TRNH MINH HO H M IU KHIN MY GIT

Chng trnh ny m phng h m iu khin my git theo cc lut c bn sau:a) Qun o cng nhiu th cng cn nhiu bt git v my chy cng lu

b) Qun o cng bn th cng cn nhiu bt git v my chy cng lu

M hnh ho bi ton theo cc yu t di y:

Ccbinngnng

C 4 bin ngn ng:1. khi lng qun o

8/8/2019 Math in It

30/42

Logic m, s m v h m

30

2. bn3. thi gian chy my4. khi lng bt git

Ccgitrngnng

1. khi lng qun o 0-6kg (t, va, kh nhiu, nhiu)

2. bn 0-1 (t, va, kh nhiu, nhiu)3. thi gian chy my 0-120 (t, va, kh lu, lu)4. khi lng bt git 0-1kg (t, va, kh nhiu, nhiu)

Cc lutm

biu din lut dng If A and B then C and D ta tch thnh 2 lut lIf A and B then C v If A and B then D theo nguyn l phn r h MIMO

thnh MISO

Biu din cc lut trong file d liu theo quy c gi tr ngn ng cui cng chnhl kt lun ca lut:

# Fuzzy washer rule baserule_1=klqa_it; doban_it; tgcm_itrule_2=klqa_it; doban_it; botgiat_it

rule_3=klqa_it; doban_vua; tgcm_itrule_4=klqa_it; doban_vua; botgiat_it

rule_5=klqa_it; doban_nhieu; tgcm_vuarule_6=klqa_it; doban_nhieu; botgiat_vua

rule_7=klqa_vua; doban_it; tgcm_itrule_8=klqa_vua; doban_it; botgiat_it

rule_9=klqa_vua; doban_vua; tgcm_vua

rule_10=klqa_vua; doban_vua; botgiat_vua

rule_11=klqa_vua; doban_nhieu; tgcm_laurule_12=klqa_vua; doban_nhieu; botgiat_nhieurule_13=klqa_nhieu; doban_it; tgcm_vuarule_14=klqa_nhieu; doban_it; botgiat_vua

rule_15=klqa_nhieu; doban_vua; tgcm_laurule_16=klqa_nhieu; doban_vua; botgiat_nhieurule_17=klqa_nhieu; doban_nhieu; tgcm_laurule_18=klqa_nhieu; doban_nhieu; botgiat_nhieu

Hngdn sdngchngtrnhChng trnh vit bng Java 1.4 chy chng trnh, cn thc hin cc bc sau:

a) copy ton b th mc MathInIT trong CD vo a cng, gi s l thmc C:\MathInIT

b) nhp p file C:\MathInIT\run.bat khi ng chng trnh. Ta sthy mn hnh chnh xut hin nh sau:

8/8/2019 Math in It

31/42

Logic m, s m v h m

31

c) chn menu File/Open hoc kch nt Open chn file d liu m hnh

my git (file cha khai bo cc bin ngn ng v cc lut m)

d) Chn tab Input nhp gi tr cc bin vo (khi lng qun o v bn) sau nhn nt Compute tnh gi tr u ra

8/8/2019 Math in It

32/42

Logic m, s m v h m

32

e) Chn tab Solve xem li gii

8/8/2019 Math in It

33/42

Logic m, s m v h m

33

f) Chn tab Graph xem biu ca cc hm thuc ca cc bin ngnng c khai bo trong file m hnh

8/8/2019 Math in It

34/42

8/8/2019 Math in It

35/42

Logic m, s m v h m

35

0.0 if 0.5

8/8/2019 Math in It

36/42

Logic m, s m v h m

36

Input values:

klqa = 1.3

doban = 0.4

**********************************************

Weight of rules:

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

Membership value of klqa_it = 0.5666709999999999

Membership value of doban_it = 0.19999999999999996

Min(0.5666709999999999, 0.19999999999999996) = 0.19999999999999996Weight of rule rule_1 = 0.19999999999999996

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


Membership value of doban_vua = 0.8

Min(0.5666709999999999, 0.8) = 0.5666709999999999

Weight of rule rule_3 = 0.5666709999999999

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


Membership value of doban_nhieu = 0.0


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

Membership value of klqa_vua = 0.433329



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




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




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

Membership value of klqa_nhieu = 0.0



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

8/8/2019 Math in It

37/42

Logic m, s m v h m

37




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




**********************************************

Membership function of the output MC(z):

MC(z) =0.19999999999999996 + 0.025(z) if 0.0

8/8/2019 Math in It

38/42

Logic m, s m v h m

38

IF klqa_vua AND klqa_vua THEN botgiat_vuaFuzzy rule: rule_12IF klqa_vua AND klqa_vua THEN botgiat_nhieuFuzzy rule: rule_14IF klqa_nhieu AND klqa_nhieu THEN botgiat_vuaFuzzy rule: rule_16IF klqa_nhieu AND klqa_nhieu THEN botgiat_nhieuFuzzy rule: rule_18IF klqa_nhieu AND klqa_nhieu THEN botgiat_nhieu

**********************************************

Input values:

klqa = 1.3

doban = 0.4

****************************************** ****

Weight of rules:

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




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




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




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




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




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




8/8/2019 Math in It

39/42

Logic m, s m v h m

39

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




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




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




**********************************************

Membership function of the output MC(t):

MC(t) =0.19999999999999996 + 0.009999999999999998(t) if 0.0

8/8/2019 Math in It

40/42

Logic m, s m v h m

40

klqa = 1.3 kgdoban = 0.4

then the output values are:

tgcm = 51.23 'botgiat = 94.87 g

8/8/2019 Math in It

41/42

Logic m, s m v h m

41

Thut ng

approximate reasoning suy lun xp x

center of gravity trng tm

Confidence tin cy

Conjuntive php hi

Core nhn ca tp m, l tp r gm ccphn t c gi tr hm thuc = 1

crisp set tp r

defineable set tp c th xc nh

Defuzzier b gii m

Disjunctive php tuyn

elementary set tp c bn, tp c s, v tr trong tp m cnh ngha

Fuzzier b m ho

fuzzy inference engine b suy din m

fuzzy logic logic m

fuzzy rule base c s lut m

fuzzy set tp m

max-min phng php tnh lut modus-ponens m dngT-norm min v S-norm max

max-prod phng php tnh lut modus-ponens m dngT-norm tch v S-norm max

membership function hm thuc, hm thnh vin

MIMO(Multi Input Multi Output) h m nhiu u vo v nhiu u ra

MISO(Multi Input Single Output) h m nhiu u vo v mt u ra

precise set tp chnh xc

product fuzzy conjunction giao m tch

rule aggregation kt hp lut

SISO(Single Input Single Output) h m mt u vo v mt u ra

S-norm S-chun hay T-i chun, l hm tng qutha t hm max

Support gi ca tp m, l tp r gm cc phn tc gi tr hm thuc > 0

T-norm T-chun, l hm tng qut ha t hm min

weighted sum tng c trng s

8/8/2019 Math in It

42/42

Logic m, s m v h m

Ti liu tham kho

[1] GS, TSKH. Hong Kim, gio trnh Phng php ton trong tin hc, HQGTp. HCM 2005

[2] GS, TSKH. Hong Kim, gio trnh Cng ngh tri thc v ng dng, HQGTp. HCM 2004

[1] PGS, TS. Nguyn Trng Thun, iu khin logic & ng dng, Tp 1, NXBKhoa Hc v K Thut, 2000

[2] Nguyn Hong Phng, Nadipuram R. Prasad, L Linh Phong, Nhp mn trtu tnh ton, NXB Khoa Hc v K Thut, 2002

[3] Nguyn Hong Phng, Bi Cng Cng, Nguyn Don Phc, Phan XunMinh, Chu Vn H, H m v ng dng, NXB Khoa Hc v K Thut, 1998

[4] Bi Cng Cng, Nguyn Don Phc, H m, mng nron v ng dng,NXB Khoa Hc v K Thut, 2001

[5] TS. inh Mnh Tng, Tr tu nhn to , NXB Khoa Hc v K Thut, 2002

[6] Trung Tun, Hchuyn gia , NXB Gio Dc, 1999

math in it

Documents