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

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abdulkader helwan on 11 Nov 2013
Edited: Walter Roberson on 4 Sep 2018
%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)

Greg Heath on 13 Nov 2013
% 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

Dear Professor
Could this loop be used also for Fittnet as well ??
Greg Heath on 14 Mar 2015
Sure:
fitnet, timedelaynet, narnet, narxnet, ...