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Set weight in neural network in [0 1]

3 vues (au cours des 30 derniers jours)
nguyen tien
nguyen tien le 28 Juil 2016
Commenté : nguyen tien le 30 Juil 2016
Hello everyone
my project is training parity 3 bit
my neural network model has 3 input, 6 neuron in hidden layer and 1 output.
I have a question for weight in neural network.
I want to set weight value in [-1 1] and bias in [-1 1], so how i can do it?.
this is my code
clear all; clc
n=input('learning rate, n = ');
emax=input('error max, emax = ')
fprintf('input:')
%x=[0 0 0;0 0 1;0 1 0;0 1 1;1 0 0;1 0 1;1 1 0;1 1 1]
x=[-1 -1 -1;-1 -1 1;-1 1 -1;-1 1 1;1 -1 -1;1 -1 1;1 1 -1;1 1 1]
fprintf('hidden weight:');
v=[0.3 -0.2 -0.1;-0.5 0 0.7;0.1 -0.5 0.1;0.2 0.3 -0.1;0.1 0.3 -0.7;0.5 -0.5 -0.4]
fprintf('ouput weight:')
w=[0.3 -0.1 0.6 -0.4 0.3 -0.5]
fprintf('desire output:')
d=[0 1 1 0 1 0 0 1]
%d=[-1 1 1 -1 1 -1 -1 1]
b1=[0.3 -0.4 0.6 -0.5 0.6 -0.3]
b2=0.8
k=0;
e=10;
%v11=[]
while ((e > emax) && (k<50000))
e=0;
k=k+1;
for i = 1:8
fprintf('epoch: k=%i\n',k);
fprintf('Input : i=%i\n',i);
for j=1:6
net_h(j)=(dot(x(i,:),v(j,:))+ b1(j));
z(j)=logsig(net_h(j));
end
disp('Input: x(1,2)=')
disp(x(i,:))
net_h=[net_h(1) net_h(2) net_h(3) net_h(4) net_h(5) net_h(6)]
z=[z(1) z(2) z(3) z(4) z(5) z(6)]
y=(dot(z,w)+b2);
e=e+1/2*(d(i)-y)^2;
fprintf('Output: y=%f\n',y);
fprintf('Desire: d=%f\n',d(i));
fprintf('Error: e=%f\n',e);
disp('update weight');
delta_o=(d(i)-y)*1;
wtr=w;
b2tr=b2
w=(w+n*delta_o*z)
b2=(b2+n*delta_o)
% w2=w/4
% b22=b2/4
for m=1:6
delta_h(m)=delta_o*wtr(m)*z(m)*(1-z(m));
v(m,1)=(v(m,1)+n*delta_h(m)*x(i,1));
v(m,2)=(v(m,2)+n*delta_h(m)*x(i,2));
v(m,3)=(v(m,2)+n*delta_h(m)*x(i,3));
b1(m)=(b1(m)+n*delta_h(m));
% v1(m,1)=v(m,1)/4;
% v1(m,2)=v(m,2)/4;
% v1(m,3)=v(m,3)/4;
% b11=b1/4;
end
% w2
% b22
% v1
% b11
w
v
end
end
hope your help,
thanks.

Réponse acceptée

Greg Heath
Greg Heath le 28 Juil 2016
1. Forget about controlling weight ranges. You have several more serious problems:
2. With an I-H-O = 3-6-1 node topology, the number of unknown weights is
Nw = (I+1)*H+(H+1)*O = (3+1)*6+(6+1)*1 = 31
whereas with N = 8 data points and O = 1 output node,
the number of training equations is no larger than
Ntrneq = Ntrn*1 <= N*1 = 8
2. Using a validation subset to mitigate overtraining
an overfit net would result in BOTH Nval and Ntrn
being inadequately small.
3. Other well known remedies (which can be combined)
a. Regularization: Add a constant times the sum of
weight magnitudes to the performance function
b. Add simulated noisy data about each original
data point within a sphere of radius of 0.5.
4. However, before getting all fancy, just try to
reduce the number of unknown weights by reducing the
number of hidden nodes.
Hope this helps,
*Thank you for formally accepting my answer*
Greg
  2 commentaires
nguyen tien
nguyen tien le 28 Juil 2016
Thanks for helping
actually, i run it, with learning rate = 0.1 and emax = 0.1. I got the result as i desired. and i also reduce the number of neuron, i used 4 neuron in hidden layer. i get weight v in input layer and w in hidden layer. but it include values <1 and values > 1 . so i want to control weight in range [-1 1] .i want this because i must control weight in circuit.
and i dont know this expression, can you give a document about it? Nw = (I+1)*H+(H+1)*O = (3+1)*6+(6+1)*1 = 31
thanks.
nguyen tien
nguyen tien le 30 Juil 2016
i see why have 31. but reduce neuron in hidden layer is not effect to weight value in range [-1 1].
Hope your helps.

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