Back propagation algorithm of Neural Network : XOR training

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Ashikur
Ashikur am 22 Jan. 2012
Kommentiert: sultana albanhar am 15 Apr. 2022
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 Kommentare
Greg Heath
Greg Heath am 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
sultana albanhar
sultana albanhar am 15 Apr. 2022
how this code work XOR?

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Akzeptierte Antwort

Greg Heath
Greg Heath am 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 Kommentare
Shantanu Arya
Shantanu Arya am 10 Sep. 2020
Greg Heath, Why Doesn't this code work for 3 input XOR ?? If I replace the X and Y with 3 inputs then the error does not converge to 0 !!
Greg Heath
Greg Heath am 10 Sep. 2020
  1. I ALWAYS use the bipolar tanh in hidden layers. It ALWAYS works.
  2. What are x and t for 3 input XOR???
Greg

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Weitere Antworten (5)

Greg Heath
Greg Heath am 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 Kommentare
Imran Babar
Imran Babar am 6 Mai 2013
Thank you very much a nice example of MLP BP NN and very easy to understand. Though I am a novice in this field but I am now clear in the programming idea
Havot Albeyboni
Havot Albeyboni am 9 Dez. 2020
Bearbeitet: Havot Albeyboni am 9 Dez. 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

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Imran Babar
Imran Babar am 8 Mai 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 Kommentar
Greg Heath
Greg Heath am 10 Mai 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')

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Sohel Ahammed
Sohel Ahammed am 4 Jul. 2015
Ok. If i Want to test it, how i have to change. Ex: input : 1 0 expected output : 1 (From learing).

dsmalenb
dsmalenb am 17 Okt. 2018
Am I missing something here but I don't see any bias neurons. Maybe this is why you are getting some inputs to work and others not?
  5 Kommentare
dsmalenb
dsmalenb am 17 Okt. 2018
I'm sorry but that statement does not make much sense to me. Biases are added to shift the values within the activation function. Multiplying by 1 does nothing.
Greg Heath
Greg Heath am 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

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sultana albanhar
sultana albanhar am 14 Apr. 2022

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