# Back propagation algorithm of Neural Network : XOR training

105 views (last 30 days)

Show older comments

c=0;

wih = .1*ones(nh,ni+1);

who = .1*ones(no,nh+1);

while(c<3000)

c=c+1;

for i = 1:length(x(1,:))

for j = 1:nh

netj(j) = wih(j,1:end-1)*double(x(:,i))+wih(j,end)*1;

outj(j) = 1./(1+exp(-1*netj(j)));

end

% hidden to output layer

for k = 1:no

netk(k) = who(k,1:end-1)*outj'+who(k,end)*1;

outk(k) = 1./(1+exp(-1*netk(k)));

delk(k) = outk(k)*(1-outk(k))*(t(k,i)-outk(k));

end

% back proagation for j = 1:nh s=0; for k = 1:no s = s+who(k,j)*delk(k); end

delj(j) = outj(j)*(1-outj(j))*s;

s=0;

end

for k = 1:no

for l = 1:nh

who(k,l)=who(k,l)+.5*delk(k)*outj(l);

end

who(k,l+1)=who(k,l+1)+1*delk(k)*1;

end

for j = 1:nh

for ii = 1:ni

wih(j,ii)=wih(j,ii)+.5*delj(j)*double(x(ii,i));

end

wih(j,ii+1)=wih(j,ii+1)+1*delj(j)*1;

end

end

end

// The code above, I have written it to implement back propagation neural network, x is input , t is desired output, ni , nh, no number of input, hidden and output layer neuron. I am testing this for different functions like AND, OR, it works fine for these. But XOR is not working.

// Training x = [0 0 1 1; 0 1 0 1] // Training t = [0 1 1 0]

// who -> weight matrix from hidden to output layer

// wih -> weight matrix from input to hidden layer

// Can you help ?

##### 3 Comments

Greg Heath
on 24 Jan 2012

If you initialized weights randomly, you could see if it is

an initialization problem.

Have you noticed the loop accidentally included in the backpropagation comment?

Greg

### Accepted Answer

Greg Heath
on 25 Jan 2012

close all, clear all, clc

x = [0 0 1 1; 0 1 0 1]

t = [0 1 1 0]

[ni N] = size(x)

[no N] = size(t)

nh = 2

% wih = .1*ones(nh,ni+1);

% who = .1*ones(no,nh+1);

wih = 0.01*randn(nh,ni+1);

who = 0.01*randn(no,nh+1);

c = 0;

while(c < 3000)

c = c+1;

% %for i = 1:length(x(1,:))

for i = 1:N

for j = 1:nh

netj(j) = wih(j,1:end-1)*x(:,i)+wih(j,end);

% %outj(j) = 1./(1+exp(-netj(j)));

outj(j) = tansig(netj(j));

end

% hidden to output layer

for k = 1:no

netk(k) = who(k,1:end-1)*outj' + who(k,end);

outk(k) = 1./(1+exp(-netk(k)));

delk(k) = outk(k)*(1-outk(k))*(t(k,i)-outk(k));

end

% back propagation

for j = 1:nh

s=0;

for k = 1:no

s = s + who(k,j)*delk(k);

end

delj(j) = outj(j)*(1-outj(j))*s;

% %s=0;

end

for k = 1:no

for l = 1:nh

who(k,l) = who(k,l)+.5*delk(k)*outj(l);

end

who(k,l+1) = who(k,l+1)+1*delk(k)*1;

end

for j = 1:nh

for ii = 1:ni

wih(j,ii) = wih(j,ii)+.5*delj(j)*x(ii,i);

end

wih(j,ii+1) = wih(j,ii+1)+1*delj(j)*1;

end

end

end

h = tansig(wih*[x;ones(1,N)])

y = logsig(who*[h;ones(1,N)])

e = t-round(y)

Hope this helps.

Greg

##### 5 Comments

Shantanu Arya
on 10 Sep 2020

Greg Heath
on 10 Sep 2020

- I ALWAYS use the bipolar tanh in hidden layers. It ALWAYS works.
- What are x and t for 3 input XOR???

Greg

### More Answers (5)

Greg Heath
on 27 Jan 2012

It is well known that successful deterministic training depends on a lucky choice of initial weights. The most common approach is to use a loop and create Ntrial (e.g., 10 or more) nets from different random initial weights. Then choose the best net.

It is also well known that an odd bounded monotonically increasing activation function like TANSIG is the choice of preference for hidden layers because it does not restrict the polarity of the layer variables. It works even better when the input is shifted to have zero mean.

You can check the superiority of TANSIG and zero-mean yourself. You can also search the comp.ai.neural-nets FAQ and archives to find both agreement and numerical experiments.

For most real world problems the best choice for number of hidden nodes, H, is not known apriori. That is why I have posted many examples using a double loop: An outer loop over H and an inner loop over Ntrials random weight initializations. For examples, search the newsgroup using the keywords

heath clear Ntrials

Hope this helps.

Greg

##### 3 Comments

Imran Babar
on 6 May 2013

Havot Albeyboni
on 9 Dec 2020

Edited: Havot Albeyboni
on 9 Dec 2020

can anyone please explain this line ???

netj(j) = wih(j,1:end-1)*x(:,i)+wih(j,end);

shouldnt it be i.e ->> Y3=sigmoid(X1W13+ X2W23 - θ3 )

rgrds

Imran Babar
on 8 May 2013

Dear sir I want to use the same code for the following data set

Input dataset=[1 1 1 2;1 1 2 2;1 2 2 2; 2 2 2 2] Output=[5 6 7 8]

but it is always generating output as given below

1 1 1 1

I tried my best but unable to understand how may I get these results

##### 1 Comment

Greg Heath
on 10 May 2013

Your outputs are not within the range of logsig.

Either normalize your outputs to fit in {0,1)

or

change your output activation function (e.g., 'purelin')

Sohel Ahammed
on 4 Jul 2015

Ok. If i Want to test it, how i have to change. Ex: input : 1 0 expected output : 1 (From learing).

##### 0 Comments

dsmalenb
on 17 Oct 2018

##### 5 Comments

dsmalenb
on 17 Oct 2018

Greg Heath
on 9 Nov 2018

The 1 is a placeholder which is multiplied by a learned weight.

Hmm, I've been using that notation for decades and this is the 1st question re that that I can remember.

Greg

### See Also

### Categories

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!