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does anyone know why this neural network is not giving the right target values????plz any helpppppp

Asked by abdulkader helwan on 11 Nov 2013
Latest activity Edited by Walter Roberson
on 4 Sep 2018
%a=imread('A.bmp');
%a=imresize(a,[6,5]);
%a=imresize(a,[1,30]);
inp=[0 0 1 0 0 0 0 1 1 0 0 1 0 1 0 0 1 0 1 0 0 1 1 1 1 1 0 0 0 1];
out=[1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0];
net= newff(inp,out,10)
tol = 0.001; % Error tolerance
%net.trainParam.show = 50;
net.trainParam.lr = 0.007;
net.trainParam.epochs = 500;
net.trainParam.alpha = 0.5;
%net.trainParam.goal = 1e-5;
[net,tr]=train(net,inp,out);
a=sim(net,inp)

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1 Answer

Answer by Greg Heath
on 13 Nov 2013
 Accepted Answer

% Weird data set:
plot(input,target,'o')
The data does not represent points from reasonable function. The target says" Regardless of the input, if the output is 0, your error will only be 3.33%".
Looking at an escalation of models (H=0:9) shows that the Linear model (H=0, NN with no hidden layer) is the best with a coefficient of determination (Wikipedia/R-squared) of only 0.0264. It is only slightly better than the constant output = 0 model.
close all, clear all, clc
N =30
input = [0 0 1 0 0 0 0 1 1 0 0 1 0 1 0 0 1 0 1 0 0 1 1 1 1 1 0 0 0 1];
target = [ 1, zeros( 1,N-1)];
% The simplest model is the constant output model that outputs the mean of the target regardless of the input.
y00 = (1/N)*ones(1,N); % 0.0333*ones(1,30)
e00 = target-y00; ; % [ 0.9667, -0.0333*ones(1,29)]
MSE00 = mse(target-y00) % 0.0322
% The next simplest is the linear model
W0 = target/[ ones(1,N) ; input ] % W0 = [ 0.0588 -0.0588 ]
y0 = W0 * [ ones(1,N) ; input ]; % y0 = 0.0588*(1-input)
e0 = target-y0; ; % [ 0.9412, -y0(2:end)]
MSE0 = mse(target-y0) % 0.0314 Not much better than constant model
R20 = 1 - MSE0/MSE00 % 0.0264 VERY BAD FIT
% Neural Model
[ I N ] = size(input) % [ 1 30 ]
[ O N ] = size(target) % [ 1 30 ]
Neq = N*O % 30 = No. of equations
% Nw = (I+1)*H+(H+1)*O = 1+3*H % No. of unknown weights
% No overfitting Neq > Nw==> H <= Hub (upper bound)
Hub = -1+ceil( (Neq-O) / ( I+O+1) ) % 9
Hmax = 9
dH = 1
Hmin = 0 % Linear Model
numH = length(Hmin:dH:Hmax) % 10
Ntrials = 10
rng(0)
j=0
for h=Hmin:dH:Hmax
j=j+1
if h==0
net = patternnet([]);
Nw = (I+1)*O
else
net = patternnet(h);
Nw = (I+1)*h+(h+1)*O
end
for i=1:Ntrials
net = configure(net, input,target);
net.divideFcn = 'dividetrain'; % Default 'dividerand' is much worse
[net tr y e ] = train(net,input,target);
R2(i,j) = 1 - mse(e)/MSE00;
end
end
R2 = R2 % 0.0264* ones(Ntrials,numH) Same as Linear
Hope this helps.
*Thank you for formally accepting my answer *
Greg

  2 Comments

Dear Professor
Could this loop be used also for Fittnet as well ??
Sure:
fitnet, timedelaynet, narnet, narxnet, ...

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