He Mo Va Mang Noron

11

CHUONG 2.

HE. MO vA MANG NORON . t)~.K~ .TlJ.

;:

N~ ~-E ' -'" .

1"11 V3EN 1 .

.

O~OO908J Ngay nay, ly thuy~t v€ t~p ma va logic ma, cling nhu ly thuy~t v€ m~ng rOTondffkhong con la di€u m6i me v6i mQinguai nua. Chuang nay khong co tham vQngtrinh bay mQiv~n d€ lien quail d~n h~ ma va m~ng naron, ma chi xin lieUlen m9t s6 khai ni~m ca bfm nh~t, co lien quail tI"\lCti~p d~n vi~c xiiy d\rng mo hinh h6 trq ra quy~tdinh sau nay.

I. H~ MO (FUZZY SYSTEM): Trong vai th~p leYqua, cac h~ th6ng ma da:lam m9t CU9Cthay th~ "ngo~n m\lc" cac ky thu~t thong thuang trong nhi€u 11nhV\lCkhac nhau cua khoa hQc va cong ngh~, d~c bi~t la trong di€u khi~n va nh~n d~ng. Logic ma dffduqc sir d\mg I

I .

h~t suc hi~u qua trong hang lo~t 11nhV\lC,ill cac san phfun tieu dung va d~n cac h~

-

,

thong cong nghi~p. Cling v6i ky thu~t ma do, cong ngh~ thong tin trong ho trq ra

I quy~tdinh va cac h~ chuyen gia da:tim l~i duqc suc s6ng cua minh, v6i kha nang I

l~p lu~n h~t suc hi~u qua tren m9t s6 luqng t6i thi~u cac lu~t duqc cling c~p b6i h~mO'.

1. T{ip mil (Fuzzy Set): T~p ma A trong t~p vu tr\l U duqc dinh nghla b6i mQtc~p thu tv: A = { (x, JlA(X))/ x E U}

!lA(.)duqc gQila ham thanh vien cua t~p ma A.

(2.1)

12

J.1A(X) la dQthuQc cua x trong t~p ma A. J.1.

A

1

0

x IDnh 2.1. Ham thanh vien cua t~p ma A

1.1. Cae phep tmiDireD t~p ma: Cho 2 t~p ma A, B trong t~p vii t11,1 U.

- Phepbli: Khi J.1A(X)E [0, 1] . Bli cua t~p ma A - ki hi~u A J.1A(X) = 1 - J.1A(X)

- co ham thanh vien:

'v'XEU

(2.2)

- Phep giao:

Giaocua 2 t~pma A va B - ki hi~u A n B - co ham thanh vien: J.1An B (x)

= mill

[J.1A(X), J.1B(X)] = J.1A(X)/\ J.1B(X)

'v'XEU

(2.3)

- Phep hQi: HQicua 2 t~p ma A va B - ki hi~u A u B - co ham thanh vien: J.1Au B (x) = max [J.1A(X),J.1B(X)]

=

J.1A(X) V J.1B(X)

'v'XEU

(2.4)

- B~ng nhau: Hai t~p ma A va B b~ng nhau khi va chi khi: J.1A(X) = J.1B(X)

'v'XEU

(2.5)

- T~p con: T~p ma A gQila con cua t~p ma B - ki hi~u A c B - khi va chi khi: IlA(X) :::;IlB(X)

'v'XEU

(2.6)

13

- Phu dinh eua phu dinh: -

A=A

(2.7)

- Lu~t DeMorgan:

-

--

AuB=AnB

(2.8)

-AnB=AuB

(2.9)

AuA:;i:U

va

A nA:;i:0

(2.10)

1.2. Cae tinh eh§t eua t~p mir: Cho 3 t~p ma A, B, C tren t~p vii tIVU. - Giao hmln:

AuB=BuA

(2.11)

AnB=BnA

- K~thqp: Au (B u C) = (A u B) u C

(2.12)

An (B n C) = (A n B) n C - Phfm ph6i:

Au (B n C) = (A u B) n (A u C)

(2.13)

An (B u C) = (A n B) u (A n C)

- B~e e~u: N~u A c B c C thi A c C

(2.14)

- Cae tinh chat khae: AuA=A AnA=A

(2.15)

14

Au0=A

(2.16)

An0=0

AuU=U

(2.17)

AnU=A 1.3. Bi~u di~n t~p mir: Cho t~p vii tf\l U = {x, xl, X2,..., xn}. Cho t~p ma A co ham thanh vien !lA. T~p ma A duQ'cbi~u di~n: A = { (Xl, !lA(XI)),(X2,!lA(X2)),..., (xn, !lA(Xn))}

(2.18)

N~u U f<Jir~c thi A co th~ vi~t: n

A= Ill/Xl + !l2/X2 +...

+ !In/Xn=

LJli/X;

(2.19)

;=1

Trong do, phep "+" la h<)icua cac phfu1ill. N~u U lien tvc thi A bi~u di~n:

A= LJlA(X)lx

(2.20)

2. Logic mil va l~p lu~n xdp xi (Fuzzy Logic - Approximate Reasoning) : 2.1. Bi~n ngon ngft (Linguistic Variables): Bi~n ngon ngil' la m<)tkhai ni~m quail trQngtrong logic ma va I~p lu~ xAp xi. No dong vai tro then ch6t trong nhi€u tmg dvng, d~c bi~t la trong cac h~ chuyen gia fia va di€u khi~n mO'. Bi~n ngon ngil' la m<)t bi~n ma gia tri cua no la tir (word) hay diu (sentence)trong ngon ngil'tv nhien. Thi d1}:"T6c d<)"la m<)tbi~n ngon ngil' n~u no lAycac gia tri: "ch~m", "nhanh", "rAtnhanh" ...

15

Khai ni~m bi~n ngon ngu d~ xuAt b6'i Zadeh, su d\mg nhu m(}tphuang ti~n d~ xApxi khi din hy sinh m(}t chut d(}chinh xac (khong cAn thi~t) d~ tranh duQ'c S\T phue t~p qua mue luc mo tit m(}thi~n tUQ'llg,vAn d~ [9]. Biin mu (Fuzzy variable): mo tit b6'i b(}ba (X, U, R(X))

(2.21)

X: ten eua bi~n U: t~p vu tr\l R(X): t~p mo con eua U, bi~u di~n m(}tgi6i h~ mo (fuzzy restriction) ap d~ttren X. Thi d\l: X = "Gia" U = {1O,20, ..., 80} R(X) = 0.1/20 + 0.2/30 + 0.4/40 + 0.5/50 + 0.8/60 + 1/70+ 1/80

Bi~n ngon ngu co thu b~c cao han bi~n mo: No lAycac bi~n mo lam gia tri euaminh. MQtbi~n ngon ngu mo tit b6'ib(}ni'im: (x, T(x), U, G, M)

(2.22)

x: ten cua bi~n T(x): t~p hQ'Pcae ten cua bi~n ngon ngu cua x ma gia tri cua no la cac bi~n mb'tren U. G: lu~t eu phap d~ sinh ra ten cua cac gia tri cua x

M: lu~t ngu nghla d~ k~t hQ'Pm6i gia tri cua x v6i nghla eua no. Thi d\l: x: bi~n ngon ngu "T6e d(}" - lAygia tri tren U = [0, 100]

T(x) = T(t6c d(})= {RAT Ch~m, Ch~m, Vila, Nhanh ...}

16

Lu~t G d€ sinh ra cae ten (nhan) cua cae thanh phAn cua T(t6c dQ) hinh thanh b~ngtf\l'cgiac. Lu~t M dinh nghia nhu sau: M (Ch~)

= T~p ma mo ta "mQt t6c dQ khoang du6i 40 km/h" v6i ham

thanh vien 11Ch~

M (Vira) = T~p ma mo ta "mQt t6c dQ gAll55 kmIh" v6i ham thanh vien /lVira

M (Nhanh) = T~p ma mo ta "mQt t6c dQ khoang tren 70 km/h" v6i ham thinh vi en IlNhanh

11

I

0.5

IlVira

Ilch

J

IlNhanh

I-

0 40

55

70

T6c dQ (km/h)

Hinh 2.2. Cac gia tri cua bi€n ligon ngu "T6c dQ" 2.2. Logic mO': Trong logic c6 di€n, chilli tri la mQt trong hai gia tri: DUNG hoi.ic SA!. Logic ma Ii ma rQng cua logic c6 di€n, trong do, cae chiin tri la cae gia tri cua bi€n ngon ngu DUNG [9].

17

Kha SAI

Kha BUNG

SAI

BUNG RAt BUNG

RAtSAI

Hoan toan BUNG

Hoan tofm SAI

-

0

x

1

0

mnh 2.3. Bi~n ngon ngu "BUNG" dinh nghla b6'iBaldwin Cae lu~t suy di~n: (A A (A => B» => B

(modus pollens)

(BA(A

(modus tollens)

~ B))~

A

«A => B) A (B => C» => (A => C)

(2.23)

(tam do~ lu~)

GQiyeA) va v(B) la cae chan tri ngon ngu cua m~nh d~ A va B, bi€u di~n du6'idl;tng: yea) = aI/vI + az/vz + ... + an/vn v(b) = ~l/WI + z/wz + ... + ~m/wm v6'i aj, ~j, Vi,Wi E [0, 1]

Taco: ,

V

(NOT A) == 1 -

V

(A) = al/(I-vI) + az/(l-vz) + ... + an/(l-vn)

v (A AND B) ==yeA) A v(B) = Lmin(ai'p)/min(vj' ij

v (A OR B)

==yeA)

v v(B) = Lmin(aj,p)/max(vi' ij

w) w)

(2.24) (2.25) (2.26)

18

v (A => B) ==v(A) => v(B) = Imin(a;,f3)/max{1-vpmin(vp ij

w)} (2.27)

Ngoeii(2.27), con c6 nhi~u dinh nghla khac cua phep keGtheo c6 th€ dung d€tinhv(A=>B). 2.3. L~p lu~n xfip xi: X, Y, z: bi~n ma, nh~n gia tIt trong t~p vil tn,1D, V, W A, B, c: vi tu ma - Lu~t chi~u (Projection rule of inference): N~u (X, Y) leiR thi X lei[R-1-X]

(2.28)

v6'i [R-1X] leiphep chi~u cua quail h~ ma R len X. - Lu~t giao / rieng bi~t h6a (Conjunction / Particularization rule): N~u X leiA veiX leiB thi X leiA n B

(2.29)

N~u (X, Y) leiA veiX leiB thi (X, Y) leiAn (BxV) N~u (X, Y) leiA vei(Y, Z) leiB thi (X, Y, Z) = (AxW) n (D x B)

(2.30)

- Lu~t tach rai (Disjunction / Cartesian product rule): N~u X leiA ho~c X leiB thi X leiA u B

(2.31)

N~u X leiA, Y leiB thi (X, Y) leiA x B

- Lu~t phu

dinh (Negation rule):

N~u NOT(X leiA) thi X lei A

(2.32)

- Lu~t k~ thua thu tv (Entailment rule ofinferrence): N~u X leiA veiA c B thi X leiB

(2.33)

- Lu~t k~t hgp (Conpositional rule of inference): N~u X leiA vei(X, Y) leiR thi Y leiA

0

R

(2.34)

19

v6i A

0

R chi k~t hqp fiax - mill cua t~p fib' A va quail h~ fib' R, dinh

nghia nhu sau:

(2.35)

fl AoR(V) = maxmin(uA(u),I1R(u,v)) u

- Modus pOllens tfmg quat:

N~u X la A va

(2.36)

N~u X la B thi Y la C Thi Y la A o(B EBC) (2.37)

v6i I1s(JJc(u, v) = min(l,l-I1B(u) +l1c(v))

- Nguyen ly fia rQng(Extension principle): N~u X la A thi reX) la reA) ..

(2.38) ,

?

3. Hf lJieu khien / Quyet iljnh mil (Fuzzy logic Control / DecisionSystem):

x

r-------------------

BQ Mb'hoa

fleX)

iI

FY Qng ca

I

1 fl(Y) I I I I

suy di€n

BQ khu Mb'

, I I I I I I I I I I I

Casa lu~t I:

I I I I I I I I

fib' J

mnh 2.4. H~ quy~t dinh dung logic fib'

y

20

Trang h~ th6ng tren, v6i cac gia tri dfiuvao x, h~ IDase cho k~t qua dfiura y. N~u y la IDe>thanh de>ngdi~u khi~n cho IDe>tthi~t bi, thi h~ th6ng tren la h~ di~u khi~nIDa.Con khong, d6 la h~ quy~t dinh IDa [9]. Be>IDa h6a (Fuzzifier) chuy~n cac du li~u duQ'cdo luang ra thanh cac gia tri ngon ngu thich hQ'P.CO'sa lu~t IDa(Fuzzy rule base) gift nhung tri thuc v~n hanh ti~n trinh cua cac chuyen gia trang lInh Vl,lCd6. De>ngcO' suy diSn (Inference Engine) la c6t lai cua h~ th6ng. N6 c6 kha nang IDaphong vi~c ra quy~t dinh cua con nguai b~ng cach l~p lu~n xApxi, d~ tITd6, d~t duQ'cchi~n luQ'cdi~u khi~n IDong mu6n.Be>khl'rIDa (Defuzzifier) se sinh ra cac quy~t dinh ho~c cac di~u khi~n "ra" tit k~tqua IDacling cApbai de>ngcO'suy diSn. 3.1. CO' sO' lu~t miY: Cac lu~t IDa trang h~ th6ng duQ'c bi~u diSn du6i d~ng:

R!:NED x la Ai, ... AND Yla Bi THI z=Ci (i=l, 2, ..., n)

(2.39)

v6i x, y, z la cac bi~n ngon ngu IData tr~ng thai cua h~ th6ng va A, B, C la caegia tri ngon ngu tuO'ngung. Me>tcach vi~t khac cua lu~t IDa: R!:NED x la Ai, ..., AND Yla Bi THI z = fi(x, ..., y)

(2.40)

v6'i f leihaID cua cac bi~n IDa ta tr~ng thai h~ th6ng. 3.2. DQng CO'sur di~n: Ap d\lng l~p lu~n IDa vao h~ th6ng, qui t~c IDodus pOllens tbng quat c6 th~ vi~t l~i nhu sail:

Gia thi~t 1: NED x la A THI y la B

(2.41)

Gia thi~t 2: x la A' K~t lu~n: y leiB' GQi quail h~ IDa Ria quail h~ d~ diSn ta phep keo theo R = A -7 B. Ta c6:

21

B' = A'

0

R = A' 0 (A

-7 B)

(2.42)

Trong do, 0 la phep toan theo cong thuc (2.35). Phep keo theo ma theo dinh nghia cua Mamdani co ham thanh vien nhu sau: Rc:a -7 b = a /\ b

(2.43)

flA-7B(u,v) = flA (u) /\ JlB (v)

Thi dl} minh hQa: Cho mQt h~ lu~t g6m 2Iu~t. D~u vao la 2 bi~n Xl va X2. D~u ra la y.

NEDxlla Alkva x21aA/ THIlla Bk ,u(XI)= 5(Xl - input(i)) =

I, { 0,

,u(X2)= 5(X2 - input(j)) =

XI = input(i) nguoclai

I,

X2 = input(j)

{ 0,

nguoclai

,uBk(Y) = m:x[ min[,u Af (input(i))"u

A~(input(j))]]

(2.44 )

(2.45) (2.46)

22

I I

: InputU)

I I I I I !

A21

I

B2

A22

mIll

Input(i)

InputU)

mnh 2.5. Phuong phap suy di~n Mamdani vai dAtivao rD. 3.3. Khif mo-: *

,

GQiz la gia tri cua z sail khi khu mer.C6 nhieu phuong phap khu merkhac nhau[15]:

-Nguyen ly thanh vien Ian nh~t (Max -

membership principle):

*

/lcCz) ~ /lc (z),

vai mQi z E Z

(2.47)

- PhuO'llgphap Centroid: z * = f,ue (z)

. zdz

f,ue (z)dz

- PhuO'llgphap trung binh trQng s6 (Weighted average method):

(2.48)

23

z' = Lflc(~). ~ Lflc(z)

(2.49)

Phuang phap miy giai h~n cho cae ham thanh vien co d~g d6i Xlmg,trong do z hi gia trt z trung binh cua t~p ma. Jl 1 0.9

0.4

a

b

z

IDnh 2.6. Phuang phap khir ma trung binh trQngs6

.

z = a(0.5) + b(0.9) 0.5+0.9

24

II. M~NG NORON (NEURAL NETWORK): M~ng naron nhan t~o la h~ th6ng duqc xay dvng nh~m dua vao su d\mg m<)tsf>nguyen 1;'t6 chuc tuang 1\1'nhu cua b<)nao con nguai trong xu 1;'thong tin. M~ngnaron hua h~n san sinh ra m<)tth€ h~ mai cua cac h~ th6ng xu 1;'thong tin. M~ngnaron nhan t~o lIng d\mg t6t trong cac Hnh vvc, nhu: d6i sanh va phan lap mfiu,xfipxi ham, t6i Uti,luqng tu hoa vector, gom c\lm du li~u... M~ng naron g6m m<)ts6 luqng 100 cac dan vi xu 1;'k€t n6i vai nhau ho~t d<)ngsong song va cfiu hinh nen ki€n truc cua h~ th6ng. Do mo ph6ng cach thuc ho~td<)ngcua cac naron th~n kinh, nen m~ng naron co kha nang hQc,tMt~o va t6ng quathoa tir du li~u da duqc hufin luy~n. Trong ph~n nay, chi xin trinh bay cac vfin dS cua m~ng naron vai thu~t giai IantruySnnguqc d~ xay dvng mo hinh phi tuy€n tren t~p du li~u.

1. Kiin true m(lng nO'ron va CO'chi ho{lt il{Jng [5J: Dan vi t~o nen m~ng naron la cae nut. Cae nut nay se n~m (] eae lap (layer) khaenhau. M6i lap eo m<)tnhi~m V\lrieng:

- Lap nh~p (input layer): nh~n du li~u d~u vao. Cae nut thu<)elap nh~p gQi la nut nh~p.

- Lap xufit (output layer): k€t xufit du li~u. Cae nut thu<)clap xufit gQi la nutxuk - Lap ~n (hidden layer): lap nay eo th~ eo ho~e khong, tuy lo~i m~ng. S6 luQ'Ilg lap ~n eua m<)tm~ng naron eling tuy theo nguai thi€t k€ m~ng. Cae nut thu<)e lap ~ngQila nut ~n.

25

Lap nh~p

Lap An

Lap xu.1t

mnh 2. 7.M~ng TImon3 lap M6i nut trong lap nh~p nh~n gia tri cua biSn dQcl~p va chuy~n VaGm~ng. Cacnut trong lap Annh~n du li~u tir nut nh~p, "t6ng trQnghoa", chuy~ntin hi~u cho lap xuc1t.Cac nut t~i lap xu.1tcling t6ng trQnghoa tin hi~u nh~n tir lap An.M6i nut trong lap xu.1ttuO'ngung vai mQt biSn ph\! thuQc. M6i cling lien kSt trong m~ng norond€u co mQttrQngs6 rieng cua minh. s6 luQ'ngbiSn dQc l~p, biSn ph\! thuQc ma m~ng TImoncho phep la tliy y. £>6ngthai, m~ng TImonlam vi~c duQ'ctren ca biSn dinh luQ'llgl~n cac biSn lap. Truac tien, gia tri cua biSn dQc l~p duQ'cchuy~n cho lap nh~p cua m~ng. T~icac nut nh~p kh6ng co xu ly naG, ma no chuy~n gia tri cua no cho cac nut An. M6i nut Antinh t6ng trQng hoa cua t.1tca du li~u nh~p b~ng cach cQng d6n t.1tca tichgiua gia tri nut Anvai trQng s6 cling lien kSt giua no vai nut nh~p. KS tiSp, mQt hamtruy€n duQ'cap d\!ng tren t6ng trQnghoa nay cling vai mQtnguang cua nut An do, cho ra gia tri th\fc cua nut An. (Ham truy€n la nen gia tri VaGmQt mi€n giai h~n.).No gui gia trt nay cho nut xu.1t.M6i nut xu.1tcling th\fc hi~n thao tac tuang t\f nhunut An,cho ra gia tri kSt xu.1tcua nut xu.1t.Gia tri cua cac nut xuc1tla gia tri cua cac biSn ph\! thuQc c~n xac dinh. M~ng nO'ronho~t dQng a hai tr~ng thai:

26

- HQc. - Anhx~. Trongtr~ngthai hQc,t~pdu li~umfiudua vao m~g baa g6m ca ghitri dAu vao lfingia tri k~t xuAtcua cac bi~n ph\! thuQc.Thong tin nh~p duQ'cdua vao m~ng. Cac du li~u duQ'ctinh toan va cho ra k~t qua xuAt.K~t qua xuAtnay cua m~ng duQ'c so scinhv6'i k~t xuAtthlJc duQ'ccho trong t~p mfiu d~ rut ra sai s6. Sai s6 nay duQ'c Iantruy~nnguQ'ctra l~i cac nut xuAt,nut ~n, d~ cac nut nay di~u chinh l~i cac trQng s6 cua minh. Qua trinh Ian truy~n theo hai chi~u nay duQ'cti~n hanh nhi~u lAn,cho d~nkhi sai s6 chApnh~n duQ'c:nghia la trQngs6 cac nut trong m~ng da dung: m~ng duQ'cluy~nxong. Khi m~ng da duQ'cluy~n xong, thi nguai ta co th~ dung m~g cho cac m\!c dich cua minh. Nghia la bay gia, dua du li~u cac bi~n dQc l~p vao m~ng, m~ng se chogia tri cac bi~n ph\! thuQccAnxac dinh.

2. cau true va ho{lt il{jngeua nut an, nut xuat va m{lng ndron [5J: 2.1.Ham truy~n:

/" /...

,.

,.

,.

,.

,.

,.

,.

,.

,.

,.

,.

,.

,.

,.

,.

,.

,.

,

-,' ,.

IDnh 2.8. D6 thi ham truy~n d~g S MQtham s(u) la mQtham truy~n d~ng S n~u thoa: -

s(u) la ham bi ch~n.

27

- s(u) la ham dan di~u tang. -

s(u) la ham lien t\lCva tran.

MQi ham thoa 3 tlOO chAt tren d€u co th€ sir d\lng lam ham truy€n trong

m~ng. Trong th\Ic t~, ham logistic g(u) thuang duQ'csir d\lng rQngrai. 1 g(u)

(2.50)

=~

1+-;;e Ch~n tren

0

Ch~n du6i

Hinh 2.9. Ham logistic 2.2. Nut in:

al

Xl I

II

GJ u=ao+ L ajxj

Xn

I

an

y=g(u)

Hinh 2.10. Nut Annh~n n gia tri OO~p Xt,

,Xj, ""xn:giatrinnutOO~p.

)

....

y

28

a),

, aj, ..., an:tn;mgs6 cling lien k€t nut nh~p i va nut Andang xet.

ao: trQng nguang nut An

u: t6ng tn;mgh6a cua nut Andang xet. n

U

= ao + I;=\ aix;

(2.51)

y=g(u): t6ng trQng h6a duQ'cnen b~ng ham truy€n: la k€t xuftt cua nut An. 2.3. Nut xuAt:

YI

.1

bI

I( v=bo +

,

.m

Ym

1\

I

bjYj

\

z

z=g(v)

IDnh 2.11. Nut xufttnh~ tin hi~u cua m nut An y),

, Yj,..., Ym:gia trj k€t xuftt cua m nut An.

b),. ..., bj, ..., bm:trQngs6 cling lien k€t nut Anj va nut xuftt dang xet. bo:trQngnguang nut xuftt v: t6ng trQngh6a cua nut xufttdang xet. z=g(v): t6ng trQng h6a duQ'cnen b~ng ham truy€n: la k€t xuftt cua nut xuftt. m

(2.52)

V = bo + IbjYj j=\

2.4. M~Dg Do-rOD:

Cho mQt m~ng naron 3 lap

- 1 lap

An.GQi:

29

I: s6 nut nh~p cua m~ng. H: s6 nut Ancua m~ng. 0: s6 nut xu~t cua m~ng. K~t xuAt cua nut Anj G=l..H) du<JctiOO:

" 1

U

J

= aoJ + ~ ax .

I)

(2.53)

I

i=1

Yj= g(Uj)

K~t xuAtcua nut xuAtk trong m~ng co 0 nut xuAt:(k=1..0) H Vk

= bOk

(2.54)

+ IbjkYj j=1

Zk

= g(Vk) ,

,

3. Lan truyen ngU'(1csai so- PhU'ung phtip giiim gradient [5J 3.1. M~t l&i:

HQc la qua trinh tim cac trQng s6 cua m~ng saDcho anh x~ cua m~ng kh6p nhAtv6i be>du li~u chua cac mfiucua ham dich. Sai s6 trung binh binh phuO'ng E thuang du<Jc sir d\mg d€ do luang SlJ trimg kh6p giua aOOx~ cfin xiiy dlJilg v6i ham dich cho tru6c (thong qua t~p mfiu).

GQi: K: s6 bi~n ph\! thue>c

N: kich thu6c t~p mfiu Zkn:k~t xuAtcua nut k cua m~ng v6i mfiun (k=1..K) (n=1..N) tkn:k~t xuAt dung theo t~p mfiu Thi: sai s6 trung binh binh phuO'ng tren toan be>t~p mfiu

30

1 E =

N

K

2 LL(Zfm n=lk=l

-tfm)2

N.K

(2.55)

v~ m~t hinh hQc,co th~ xem E nhu mQtm~t 16i.M~t 16ila mQtsieu ph~ng trong do m6i di~m cua no tuang ung v6i mQtdi~m trong khong gian trQngs6. Chi~u caDtren khong gian trQng s6 cua m6i di~m trong m~t 16ibi~u di~n sai s6 cua mo hinh ung v6i cac trQngtuang ung do. f)i~m thfip nhfit tren m~t 16icho ta mo hinh co sai s6 it nhfit. 3.2. PhU'O'ng phap ghim gradient:

G6m cac bu6c chinh sail: BU'O'c1: chQn ng~u nhien mQt di~m xo trong khong gian trQng s6. BU'O'c2: Tinh dQa6c cua m~t 16it~i xo. BU'O'c3: c~p nh~t cac trQng s6 theo hu6ng d6c nhfit cua m~t 16i. BU'O'c4: xem di~m nay nhu di~m Xom6i

L~p l~i bu6c 2 d~n bu6c 4 thi d~n mQt Ilk nao do, cac gia tri cua bQtrQng s6 se ti~p c~n duQ'cdi~m thfipnhfit trong m~t 16i. Khong th~ tinh toan duQ'cd~ng cua toan m~t 16i.Nhung co th~ tinh duQ'c chi~u caDva dQ d6c m~t 16it~i bfit kYdi~m nao trong khong gian trQng.Trong toan hQc,dQ d6c m~t 16i duQ'cxac dinh b~ng d~o ham rieng theo tUng trQng s6 trong m~ng naron. Khi da:bi~t duQ'c dQ d6c cua m~t 16i, se sir d\mg thong tin nay b~ng cac qui t~c hQc d~ c~p nh~t bQtrQng s6. Ph~n 3.3 se trinh bay v~ d~o ham rieng theo cac trQng s6 cua m~ng naron. Khi da: co d~o ham rieng, nghla la da:bi~t dQ d6c m~t 16i, ph~n 3.4 se trinh bay qui t~c hQc thich nghi d~ c~p nh~t bQtrQng s6 cua m~ng.

31

3.3. B~o ham rieng tU'O'ng«ng vOi cae trQng s6: 3.3.1. TrQng s6 nut xuit: Xet 1 nut xu~t.

illnh 2.12. Nut xu~t va cac tn;mg s5 cua no. G<;>i:

z: k~t xu~t thlJc cua nut xu~t (theo (2.54)).

t: k~t xu~t dung cua nut (2.56)

p=(z-t)z(l-z) bj: cac tr<;>ngs5 cua nut xu~t G=O..H) H: s5 nut An

Yj:k~t xu~t cua nut tlnj G=I..H) D~o ham rieng tuang lmg v6i cac tr<;>ng s5 la: BE

P

,j =0

Bbj = { PYj ,j:5, H

(2.57)

32

3.3.2. TrQng sA nut An: Xct 1 nut An.

~ u=ao+ L

ajXj

y=g(u)

IDnh 2.13. Nut Anva cac trQngs6 cua no. GQi:

q =(~Pkbk }(1- y)

(2.58)

v6i p theo (2.56) 0: s6 nut xufit cua m(;lng. I: s6 nut nh~p cua m(;lng D(;loham rieng tuang (mg v6i cac trQng s6 Ia:

BE

q ,i = 0 Ba; = { qx; ,i ~ I

(2.59)

3.4. Cac qui t~c hQc: Sau khi da tinh duQ'cde>d6c m~t I6i, ta sir d\lng cac k€t qua cua (2.57) va (2.59)theo nhi~u cach khac nhau d~ c~p nh~t be>trQngs6 cho m(;lng. 3.4.1. Qui t~c delta:

33

Qui t~c hQc delta la m(>ttrong nhung qui t~c nguyen thuy nh~t cua Ian truy€n nguQ'c.Theo do, cac trQng s6 se duQ'cc~p nh~t sao cho hu6ng ma ham 16iE 1\1txu6ng se d6c nhfit. GQi: Wm:trQng s6 cAn c~p nh~t (Ia aij d6i v6'i nut An, bjk d6i v6'i nut xu~t) t~i bu6'c thu m. dm: d(>l6n cua vectO'tfmg cac vectO' d~o ham rieng ham 16i.

N: kich thu6'c t~p m~u N

BE

(2.60)

d -2:m

- n=1( 8\vm ) n

Thi: Wm = Wm-l

+ Cm

(2.61)

cm = - E dm

(2.62)

Trong do:

E: tham s6 hQc (learning rate parameter): do ngmJi dung quy€t dinh,

:

I ~

I

,

,

thuemgduQ'cchQntheo phuO'ngphap thu-va-sai. Day la h~n che cua qui Hicdelta. ,

,

cm:bien thien trQng so W &bu6'c m

3.4.2. Qui t~c mo men: La m(>tcM ti€n cua qui t~c delta. Theo do, gia tri cua E se thay d6i thich hgpv6'itung bu6'c hQc. Nghia la: n€u cac bu6'c hQc trn6'c dang giam m~nh thi theo da - bu6'c t6'i cling se giam m~nh -+ tang h~ s6 hQc E d€ bi€n thien trQng Cmtang len - va nguQ'c

l~i. Khi do, bi€n thien trQng duQ'ctinh: cm = ~ Cm-l + (1-~) E dm

(0:::;~ < 1)

(2.63)

34

3.4.3. Qui t~c hQc thich nghi: Con duQ'c gQi 1<\phuO'ng philp delta - bar- delta, 1<\phuO'llg philp cai ti€n duQ'cxem 1<\hi~u qua ooAt cua phuO'llg philp delta.

Y tu&ng cua phuO'llg philp 1<\:m6i trQng sf>w se co m(>ttham

sf>hQc e khac

OOau.Khi c~p nh~t trQng sf>,n€u hu6'ng giam 16i hi~n h<\OOcling hu6'ng v6i bu6c tru6c, e 16'n,con n€u nguQ'c hu6'ng, e se 006. Hu6'ng giam 16i duQ'c xac dinh b6'i dAti cua dm (2.60). N€u dm duO'llg, 16i giam khi trQng sf>giam. N€u dmam, 16i giam khi trQng tang. PhuO'llg philp hQc thich nghi v~n d\mg khai ni~m hu6'ng 16i "vira m6i" giam OOusau: fm+)

= e fm+ (1- e) dm (0:::;e < 1)

(2.64)

Tham sf>echo bi€t "khoang thai gian m6i day" 1<\bao lau. Khi do, h~ sf>hQc thich nghi e tiOOtheo cong thuc:

=

e m

em-I

+K

{ em-l x rjJ

,

d nfm > 0

,

dmfm ~ 0

(2.65)

Trong do: K, : tham sf> thu(>c (0, 1)

Khi da:co emthi bi€n thien trQng trong (2.63) tiOOl~i theo cong thuc: cm = ~ cm-) - (1-~t) em dm

(2.66)

35

3.5. Thu~t toaD hQcIan truy~n ngtfClc(Back-propagation Learning Rule) Algorithm BackPropagation DAuvela:

T~pm~u luy~nv6i N m~u. ciiu hinh m~ng: m nut nh~p, n nut xuiit. sf>lap ~n: Q. GQi: qYi: k€t xuiit cua nut thu i trong lap q.

KhOi t~o: -

H~ sf>hQce (theo qui t~c delta - bar - delta). Emax(nguang 16i - max tolerable error).

E=O k=l (k=I..N: chi sf>d~ duy~t h€t t~p m~u). BU'Cyc 1: L~p v6i m~u k

+ q=l (lap nh~p) + X;(k):gia tri nh~p cua nut nh~p i (i=1..m) cua m~u k. + Truy€n diJ li~u nh~p cua m~u thu k VelOlap nh~p QYi= IYi =

x?) cho tiit ca cac nut i thuQclap q

+ dY) : k€t xuiit dung cua nut xuiitj cua mdu k

36

Bmyc 2: Lan truy~n ti~n

q=2 L~p v6i q<=Q L~p cho tUng nut i cua lOpq QYi=

t6ng trQnghoa cho nut thu i cua lOpq

(K~t qua cua bu6c llC1Y la n k~t xuAt cua n nut xuAt) B1ftYC3: Tinh sai s6 k~t xuAt

E =~ tcdY)_Qy)2 +E 2 j=l

Tinh sai s6 theo d~o ham rieng cac trQngs6 theo cong thuc (2.57) va (2.59) va cac cong thuc (2.56), (2.58). Bmyc 4: Lan truy~n nguQ'c sai s6 Tinh bi~n thien cac trQng s6 va cac h~ s6 lien quail theo cac cong thuc tuong lIng tIT(2.60) d~n (2.66) (tuy phuong phap hQc C\lth€ chQn).

Bmyc5: Quay l~i bu6c 11~p cho mdu k~ ti~p. Sang Bu6c 6 khi k>N. Bmyc6: N~u E<Emaxthi dUng (ho~c ap d\lng cac qui t~c ki€m tra cheo d€ dUng ti~n trinh luy~n) NguQ'c l~i thi: C~p nh~t cac trQng s6 theo cac bi~n thien trQng co trong Bmyc 4.

E=O. k=1. Quay l~i Bmyc 1. End BackPropagation

37

Thu~t tmin BackPropagation tren day c~p nh~t trQng s6 khi da:duy~t qua h~t t~p mati (chu khong phai qua tung mati). Cac bu6c va cong thuc trinh bay da: duQ'cchQnlQCphil hgp v6i mo hinh se xay d\lng sau nay.

---000--Tren day vua trinh bay m<)ts6 vftn d~ v~ h~ ma va mc;tngnaron, t~p trung vao nhftng chi ti~t se sir d\lng trong mo hinh sau nay. Trong chuang 3 va chuang 4, khi xay d\lng va thi~t k~ mo hinh, cac vftn de lien quail d~n thi~t k~ rieng cua mo hinh se duQ'ctrinh bay C\lth~.