John Michell answered . 2024-04-21 08:12:19

 Nw = (I+1)*H+(H+1)*O

exceed the number of training equations

 Ntrneq = Ntrn*O

This will occur as long as H <= Hub where Hub is the upperbound

 Hub = -1+ceil( (Ntrneq-O) / (I+O+1) )

Based on Ntrneq and Hub I decide on a set of numH candidate values for H

 0 <= Hmin:dH:Hmax <= Hmax

 numH = numel(Hmin:dH:Hmax)

and the number of weight initializations for each value of H, e.g.,

 Ntrials = 10

If the training target is ttrn = t(indtrn), the mean-square-error of a naïve constant output net (independent of the input) is

 MSEtrn00 = mean(var(ttrn',1));

Then using MSEtrn00 as a normalization reference, I use the following double loop format

 rng(0)
 j=0
 for h = Hmin:dH:Hmax
     j=j+1
     net = ...
     net.divideFcn = 'dividetrain';
          ... 
     for i = 1:Ntrials
         net = configure(net,x,t);
         ...
         [ net tr y e ] = train(net,x,t);
          ...
         R2(i,j) = 1-mse(e)/MSEtrn00;
      end
 end

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