how to count number of iterations.?

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Nimisha
Nimisha on 16 Dec 2014
Commented: Adam on 16 Dec 2014
I = imread('Coins.jpg');
Id = im2double(I); % I is a uint8 grayscale image
T = 0.5*(min(Id(:)) + max(Id(:)));
deltaT = 0.01; % convergence criterion
done = false;
while done
g = Id >= T;
Tnext = 0.5*(mean(Id(g)) + mean(Id(g)));
done = abs(T - Tnext) < deltaT;
T = Tnext;
end
i want to count number of iteration for this program.! as well i want to store value of T at each iteration.!
  2 Comments
Nimisha
Nimisha on 16 Dec 2014
I = imread('Coins.jpg');
Id = im2double(I); % I is a uint8 grayscale image
done = false;
count = 0
T = zeros(1,100); % Try to presize to something big enough to hold all iterations if you can estimate it
while done
g = Id >= T;
Tnext = 0.5*(mean(Id(g)) + mean(Id(g)));
count = count + 1;
T(count) = Tnext
end
Why we changed T.? Cant we do with previous T.?
And above is giving results as 0
Adam
Adam on 16 Dec 2014
I edited my answer below. I didn't notice the initialisation of T higher up and the >= test against it in the while loop.

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Accepted Answer

Adam
Adam on 16 Dec 2014
Edited: Adam on 16 Dec 2014
count = 0
newT = zeros(1,100); % Try to presize to something big enough to hold all iterations if you can estimate it
while done
g = Id >= T;
Tnext = 0.5*(mean(Id(g)) + mean(Id(g)));
count = count + 1;
newT(count) = Tnext
done = abs(T - Tnext) < deltaT;
end
Presize newT if you can come up with a sensible bounded estimate, otherwise the warning you get is likely irrelevant anyway. Your code will be a little slower, but unlikely to matter in this case.
  6 Comments
Nimisha
Nimisha on 16 Dec 2014
Ya, This ~ sign works. Thank You very Much :)
Adam
Adam on 16 Dec 2014
No problem :)
Would have been a bit quicker if I absorbed your whole original post first time, but I was trying to just scan it for the parts relevant to the question which turned out to be a mistake!

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