Download - he mo va mang noron
-
8/14/2019 he mo va mang noron
1/27
11
CHUONG 2.HE MO vA MANG NORON.
t)~.K.~ .TlJ N~
.~-E
.
;:
1'. -'"1"11 V3EN
O~OO908JNgay nay, ly thuy~t v t~p ma va logic ma, cling nhu ly thuy~t v m~ng
rOTondffkhong con la dium6i me v6i mQinguai nua. Chuang nay khong co thamvQngtrinh bay mQiv~n d lien quail d~n h~ma va m~ng naron, ma chi xin lieUlenm9t s6 khai ni~m ca bfmnh~t, co lien quail tI"\lCi~p d~n vi~c xiiy d\rngmo hinh h6trq ra quy~tdinh sau nay.I. H~ MO (FUZZY SYSTEM):
Trong vai th~p leYqua, cac h~ th6ng ma da:lam m9t CU9Chay th~ "ngo~n
-
8/14/2019 he mo va mang noron
2/27
-
8/14/2019 he mo va mang noron
3/27
13
-Phu dinh eua phu dinh:-A=A (2.7)- Lu~t DeMorgan:- --AuB=AnB (2.8)
(2.9)- --AnB=AuBAuA:;i:U va A nA:;i:0 (2.10)1.2.Cae tinh eht eua t~p mir:Cho 3 t~pma 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)
-
8/14/2019 he mo va mang noron
4/27
14
Au0=A (2.16)An0=0AuU=U (2.17)AnU=A1.3. Bi~u di~n t~p mir:Cho t~p vii tf\l U = {x, xl, X2,..., xn}.Cho t~pma 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
-
8/14/2019 he mo va mang noron
5/27
15
Khai ni~m bi~n ngon ngu d~ xuAt b6'i Zadeh, su d\mg nhu m(}tphuang ti~nd~ xApxi khi din hy sinh m(}tchut d(}chinh xac (khong cAnthi~t) d~ tranh duQ'cS\Tphue t~p qua mue luc mo tit m(}thi~n tUQ'llg,vAnd~ [9].
Biin mu (Fuzzy variable): mo tit b6'i b(}ba(X, U, R(X)) (2.21)X: ten eua bi~nU: t~p vu tr\lR(X): t~p mo con eua U, bi~u di~n m(}tgi6i h~ mo (fuzzy restriction) ap
d~ttrenX.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+ 1/80Bi~nngon ngu co thu b~c cao han bi~n mo: No lAycac bi~nmo lam gia tri
euaminh.MQtbi~nngon ngu mo tit b6'ib(}ni'im:
-
8/14/2019 he mo va mang noron
6/27
16
Lu~t G d sinh ra cae ten (nhan) cua cae thanh phAncua T(t6c dQ)hinhthanhb~ngtf\l'cgiac.Lu~tM dinh nghia nhu sau:M (Ch~) = T~p ma mo ta "mQtt6c dQkhoang du6i 40 km/h" v6i ham
thanh vien 11Ch~M (Vira) = T~p ma mo ta "mQtt6c dQgAll55 kmIh" v6i ham thanh vien
/lViraM (Nhanh) = T~p ma mo ta "mQtt6c dQkhoang tren 70 km/h" v6i ham
thinh vien IlNhanh
11
IIlch IlVira IlNhanhJ I-0.5
0 40 55 70 T6c dQ(km/h)
-
8/14/2019 he mo va mang noron
7/27
17
Kha BUNGBUNGRAtBUNGHoan toan BUNG
1 xmnh 2.3. Bi~nngon ngu "BUNG" dinh nghla b6'iBaldwin
Cae lu~t suy di~n:(A A (A =>B =>B(BA(A ~ B))~ A
(modus pollens) (2.23)(modus tollens)
A => B) A (B => C => (A => C) (tam do~ lu~)
KhaSAISAI
RAtSAIHoan tofmSAI
0 -0
-
8/14/2019 he mo va mang noron
8/27
18
v (A => B) ==v(A) => v(B) = Imin(a;,f3)/max{1-vpmin(vp w)} (2.27)ijNgoeii(2.27), con c6 nhi~u dinh nghla khac cua phep keGtheo c6 th dung
dtinhv(A=>B).2.3. L~p lu~n xfip xi:X, Y, z: bi~nma, nh~n gia tIt trong t~p vil tn,1D, V, WA, 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-1-X] leiphep chi~u cua quail h~ ma R len X.- Lu~t giao / rieng bi~t h6a (Conjunction / Particularization rule):N~uX 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)
-
8/14/2019 he mo va mang noron
9/27
19
v6i A 0 R chi k~t hqp fiax - mill cua t~p fib' A va quail h~ fib' R, dinhnghia nhu sau:
fl AoR(V)= maxmin(uA(u),I1R(u,v))u (2.35)- Modus pOllens tfmg quat:N~uX la A va (2.36)N~uX la B thi Y la CThi Y la A o(B EBC)v6i I1s(JJc(u,) = min(l,l-I1B(u) +l1c(v)) (2.37)- Nguyen ly fia rQng(Extensionprinciple):N~uX la A thi reX) la reA) (2.38)
.. ? ,3. Hf lJieu khien / Quyet iljnh mil (Fuzzy logic Control /DecisionSystem):
-
8/14/2019 he mo va mang noron
10/27
20
Trang h~ th6ng tren, v6i cac gia tri dfiuvao x, h~ IDase cho k~t qua dfiuray. N~uy la IDe>tanh de>ngdi~u khi~n cho IDe>thi~t bi, thi h~ th6ng tren la h~ di~ukhi~nIDa.Con khong, d6 la h~quy~tdinh IDa[9].
Be>IDah6a (Fuzzifier) chuy~n cac du li~u duQ'cdo luang ra thanh cac giatri ngonngu thich hQ'P.CO'sa lu~t IDa(Fuzzy rule base) gift nhung tri thuc v~n hanhti~n trinh cua cac chuyen gia trang lInh Vl,lCd6. De>ngcO' suy diSn (InferenceEngine) la c6t lai cua h~ th6ng. N6 c6 kha nang IDaphong vi~c ra quy~t dinh cuaconnguai b~ngcach l~p lu~nxApxi, d~tITd6, d~t duQ'cchi~n luQ'cdi~ukhi~n IDongmu6n.Be>khl'rIDa (Defuzzifier) se sinh ra cac quy~t dinh ho~c cac di~u khi~n "ra"titk~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
caegiatri ngon ngu tuO'ngung.
-
8/14/2019 he mo va mang noron
11/27
21
B' = A' 0 R = A' 0 (A -7B)Trong do, 0 la phep toan theo cong thuc (2.35).
(2.42)
Phep keo theo ma theo dinh nghia cua Mamdani co ham thanh vien nhusau:
Rc:a -7 b = a /\ bflA-7B(u,v) = flA (u) /\ JlB (v)Thi dl} minh hQa:Cho mQth~ lu~t g6m 2Iu~t. D~u vao la 2 bi~n Xlva X2. D~u ra la y.NEDxlla Alkvax21aA/ THIlla Bk,u(XI)= 5(Xl - input(i)) = {
I, XI= input(i)0, nguoclai
,u(X2)= 5(X2 - input(j)) = {I, X2= input(j)0, nguoclai
,uBk(Y) = m:x[ min[,u Af (input(i))"u A~(input(j))]]
(2.43)
(2.44 )
(2.45)
(2.46)
-
8/14/2019 he mo va mang noron
12/27
22
A21
II: InputU)IIII! A22I B2
mIll
Input(i) InputU)
mnh 2.5. Phuong phap suy di~nMamdani vai dAtivao rD.3.3. Khif mo-:
-
8/14/2019 he mo va mang noron
13/27
23
z' = Lflc(~). ~Lflc(z) (2.49)
Phuang phap miy giai h~n cho cae ham thanh vien co d~g d6i Xlmg,trongdo z hi gia trt z trung binh cua t~pma.
Jl1
0.90.4
a b z
IDnh 2.6. Phuang phap khir ma trung binh trQngs6. a(0.5) + b(0.9)z =0.5+0.9
-
8/14/2019 he mo va mang noron
14/27
-
8/14/2019 he mo va mang noron
15/27
25
Lap nh~p Lap An Lap xu.1tmnh 2.7.M~ngTImon3 lap
M6i nut trong lap nh~p nh~n gia tri cua biSn dQcl~p va chuy~nVaGm~ng.Cacnut trong lap Annh~n du li~utir nut nh~p, "t6ng trQnghoa", chuy~ntin hi~ucholapxuc1t.Cac nut t~i lap xu.1tcling t6ng trQnghoa tin hi~u nh~n tir lap An.M6i nuttrong lap xu.1ttuO'ngung vai mQtbiSn ph\! thuQc.M6i cling lien kSt trong m~ngnoronduco mQttrQngs6 rieng cua minh.
-
8/14/2019 he mo va mang noron
16/27
-
8/14/2019 he mo va mang noron
17/27
27
- s(u) la ham dan di~utang.- s(u) la ham lien t\lCva tran.MQi ham thoa 3 tlOOchAt tren du co th sir d\lng lam ham truyn trong
m~ng.Trong th\Ic t~, ham logistic g(u) thuang duQ'csir d\lng rQngrai.
1g(u) =~1+-;;-e
(2.50)
Ch~n tren
0 Ch~n du6i
Hinh 2.9. Ham logistic
-
8/14/2019 he mo va mang noron
18/27
28
a), , aj, ..., an:tn;mgs6 cling lien kt nut nh~p i va nut Andang xet.ao: trQng nguang nut Anu: t6ng tn;mgh6a cua nut Andang xet.
nU =ao + I aix;;=\ (2.51)y=g(u): t6ng trQng h6a duQ'cnen b~ng ham truyn: la kt xuftt cua nut An.2.3. Nut xuAt:
YI .1 bI I( \ zv=bo + I bjYj,Ym .m 1\ z=g(v)
-
8/14/2019 he mo va mang noron
19/27
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 xuAtcua nut Anj G=l..H) du
-
8/14/2019 he mo va mang noron
20/27
30
1 N K2LL(Zfm -tfm)2E = n=lk=lN.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~mcua no tuang ung v6i mQtdi~mtrong khong gian trQngs6. Chi~ucaDtren khong gian trQng s6 cua m6i di~m trong m~t 16ibi~u di~n sai s6 cua mohinh ung v6i cac trQngtuang ung do. f)i~m thfipnhfit tren m~t 16icho ta mo hinh cosai s6 it nhfit.
3.2. PhU'O'ngphap ghim gradient:G6m cac bu6c chinh sail:
BU'O'c1: chQnng~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~nmQtIlk nao do, cac gia tri cua bQtrQng
31
-
8/14/2019 he mo va mang noron
21/27
3.3. B~o ham rieng tU'O'ngng vOicae trQng s6:3.3.1. TrQng s6 nut xuit:Xet 1nut xu~t.
illnh 2.12. Nut xu~t va cac tn;mg s5 cua no.Gi:z: k~t xu~t thlJc cua nut xu~t (theo (2.54)).t: k~t xu~t dung cua nut
-
8/14/2019 he mo va mang noron
22/27
32
3.3.2. TrQng sAnut An:Xct 1 nut An.
y=g(u)
~u=ao+ L ajXjIDnh 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.
-
8/14/2019 he mo va mang noron
23/27
33
Qui t~c hQc delta la m(>ttrong nhung qui t~c nguyen thuy nh~t cua IantruynnguQ'c.Theo do, cac trQngs6 se duQ'cc~pnh~t sao cho hu6ng ma ham 16iE1\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~pm~u
N
(BE)-2:-m - n=1 8\vm n (2.60)
Thi:Wm= Wm-l + Cm (2.61)cm= - E dm (2.62)Trong do:
-
8/14/2019 he mo va mang noron
24/27
34
3.4.3. Qui t~c hQcthich nghi:Con duQ'c gQi 1echo bit "khoang thai gian m6i day" 1hQc thich nghi e tiOOtheo cong thuc:
{em-I + Ke =
m em-l x rjJ, d nfm > 0, dmfm ~ 0 (2.65)
Trong do:
35
-
8/14/2019 he mo va mang noron
25/27
3.5. Thu~t toaD hQcIan truy~n ngtfClc(Back-propagation Learning Rule)Algorithm BackPropagationDAuvela:
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: kt xuiit cua nut thu i trong lap q.KhOit~o:
- H~ sf>hQce (theo qui t~c delta - bar - delta).Emax(nguang 16i- max tolerable error).E=Ok=l (k=I..N: chi sf>d~ duy~t ht t~p m~u).
BU'Cyc: L~p v6i m~u k
-
8/14/2019 he mo va mang noron
26/27
36
Bmyc 2: Lan truy~n ti~nq=2L~p v6i q
-
8/14/2019 he mo va mang noron
27/27
Thu~t tmin BackPropagation tren day c~p nh~t trQng s6 khi da:duy~t quah~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