What is number of iterations deciding factor in the following code?
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Veera Kalyani S
le 30 Déc 2016
Commenté : Ammy
le 31 Jan 2021
clc;
clearvars;
close all;
workspace;
fontSize = 33;
grayImage = imread('cameraman.tif');
subplot(1,2,1);
imshow(grayImage);
[rows, columns, numberOfColorChannels] = size(grayImage);
N=rows;
T=1.4938*N+40.8689;
disp(T);
t=0;
T2=ceil(T);
disp(T2);
c = T
iscram= grayImage;
while t<T2
for i= 1 : rows
for j= 1 : columns
r = mod(i+j,N)+1;
c = mod(i+(2*j)+1,N)+1;
imscram(i,j)=iscram(r,c);
end
end
t=t+1;
fprintf('t = %f, T2 = %f\n', t, T2);
end
subplot(1,2,2);
imshow(imscram);
(1) why should we use linear approximation of arnolds cat map, T=1.4938N+40.8689?
(2) which denotes the number of iterations in the above code?
(3) Even if i change t<700 or t<4, the output i got is the same..and also there is no change/improvement in NPCR and UACI value...
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Walter Roberson
le 30 Déc 2016
After your fprintf() you need to add
iscram = imscram;
Otherwise you are just repeating the same work over and over again instead of doing multiple iterations of the transform.
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Image Analyst
le 30 Déc 2016
That code is from someone else's post and it's wrong. I redid that code and a "fixed" version is attached. I don't use that T number he computed. It doesn't seem needed at all.
From Wikipedia, it seems that the code only works on square images. I tried to generalize it for rectangular images and always got the image wrapped around vertically. It never reconstituted itself. So maybe you'll have to use other scrambling methods.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/176352/image.jpeg)
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Image Analyst
le 30 Jan 2021
Make sure the imwrite() is not commented out like I had it in the demo:
% Save the image, if desired.
filename = sprintf('Arnold Cat Iteration %d.png', iteration);
% imwrite(currentScrambledImage, filename); % Uncomment this if you want to save the images.
fprintf('Saved image file %s to disk.\n', filename);
So take away the % and you should see Arnold Cat Iteration 67.png
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