感知器神经网络方案.doc

感知器神经网： clear all close all P=[0 0 1 1;0 1 0 1] T=[0 1 1 1] net=newp(minmax(P),1,hardlim,learnp) net=train(net,P,T) Y=sim(net,P) plotpv(P,T) plotpc(net.iw{1,1},net.b{1})性神经网络clear all close all P=[1.0 2.1 3 4] T=[2.0 4.01 5.9 8.0] lr=maxlinlr(P) net=newlin(minmax(P),1,0,lr) net.trainParam.epochs=300 net.trainParam.goal=0.05 net=train(net,P,T) Y=sim(net,P) clear all close all P=[1.1 2.2 3] T=[2.1 4.3 5.9] net=newlind(P,T) Y=sim(net,P) x=-4:.1:4 y=tansig(x) plot(x,y,^-r)Sigmoid As a reminder:?σ(x)=1/1+exp(?x) Its derivative:?ddxσ(x)=(1?σ(x))?σ(x) The problem here is?exp, which quickly goes to infinity, even though the result ofσ?is restricted to the interval?[0,?1]. The solution: The sigmoid can be expressed in terms of?tanh:?σ(x)=1/2(1+tanh(x/2)). Softmax Softmax, which is defined as?softmaxi(a)=exp(ai)/∑jexp(aj)?(where?a?is a vector), is a little more complicated. The key here is to expresssoftmax?in terms of the?logsumexp?function:?logsumexp(a)?=?log(∑?iexpai), for which good, non-overflowing implementations are usually available. Then, we have?softmax(a)?=?exp(a???logsumexp(a)). As a bonus: The vector of partial derivatives / the gradient of?softmax?is analogous to the sigmoid, i.e??/?aisoftmax(a)=(1?softmaxi(a))?softmaxi(a).