What is number of iterations deciding factor in the following code?

1 次查看(过去 30 天)
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...

采纳的回答

Walter Roberson
Walter Roberson 2016-12-30
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.

更多回答(1 个)

Image Analyst
Image Analyst 2016-12-30
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.
  5 个评论
Image Analyst
Image Analyst 2021-1-30
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

请先登录,再进行评论。

类别

Help CenterFile Exchange 中查找有关 Read, Write, and Modify Image 的更多信息

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by