Pixel Shuffling using Chaotic Tinkerbell map(https:​//en.wikip​edia.org/w​iki/Tinker​bell_map) and Henon Map(https:​//en.wikip​edia.org/w​iki/H%C3%A​9non_map)

I have tried to do double shuffling of pixels in an image using Tinkerbell mal and Henon Map.The code is shown below. It is showing an error "Index exceeds matrix dimensions".Please help me in this regard

clear all
close all
clc
in_img=double(imread('cameraman.tif'));
in_img=double(in_img)/255;
subplot(221)
imshow(in_img)
x=in_img;
[m,n]=size(in_img);
%Shuffling using Tinkerbel Map
% a=0.9,b=-0.6013,c=2.0,d=0.50;
a=0.3;b=0.6000;c=2.0;d=0.27;
for i=1:m
    for j=1:n
    %Shuffling using Tinkerbel Map
     r = [round(abs((i^2)-(j^2)+(a*i)+(b*j))),round(abs((2*i*j)+(c*i)+(d*j)))]
    ggg(i,j)=in_img(r(1)+1,r(2)+1);
    end
end
 Tinkerbel_Shuffled=ggg;
 subplot(222)
 imshow(Tinkerbel_Shuffled,[])

    %Shuffling using Henon map Map
   [m,n]=size(in_img);
   a = 1.4;b=0.3;
   for i=1:m
      for j=1:n
      %Shuffling using Henon map Map
      r = [round(abs(1-(a*(i^2))+j)),round(abs(b*i))];
      ggg(i,j)=Tinkerbel_Shuffled(r(1)+1,r(2)+1);
      end
  end
  subplot(223) 
  imshow(Double_Shuffled,[])

Answer 1

[Geoff Hayes]

Renjith - the problem is with the r

 r = [round(abs((i^2)-(j^2)+(a*i)+(b*j))),round(abs((2*i*j)+(c*i)+(d*j)))]

There is no guarantee that the either component of this array will be valid indices into the image

 ggg(i,j)=in_img(r(1)+1,r(2)+1);

What you can do is to find the modulus m and n for the first and second component of r respectively.

 r = mod([round(abs((i^2)-(j^2)+(a*i)+(b*j))),round(abs((2*i*j)+(c*i)+(d*j)))], [m,n]);

The same would need to be done for the Henon Mapping

 r = mod([round(abs(1-(a*(i^2))+j)),round(abs(b*i))],[m n]);

Though I don't know how this will impact the encryption...

Try the above and see what happens!

[Renjith V Ravi]

Yes, I have tried it and obtained the output.But now the problem is with inverse operation.THe inverse process is not occurring properly.

clear all
clc
g=imread('cameraman.tif');
% g=double(g)/255;
subplot(231)
imshow(g)
original=g;
[m,n]=size(g);
mo=m;
num = 12;
K=10.^10;
% K=19;
%....................Shuffling.......................
disp('........................Started Shuffling.....................')
%Tinkerbell shuffling
a=0.3;b=0.6000;c=2.0;d=0.27;
for k= 1:num
for i=1:m
  for j=1:n
    r = mod([round(abs((i^2)-(j^2)+(a*i)+(b*j))),round(abs((2*i*j)+(c*i)+(d*j)))], [m,n]);
%     r = mod([(i+j),(j+K*sin(((i+1)*n)/2*pi))],mo);
%     r = [((i-1)+(K*sin(j-1))),((j-1)+i)];
%     r = uint8(r);
    ggg(i,j)=g(r(1)+1,r(2)+1);
%       ggg(i,j)=g(r(1),r(2));
    end
end
  g=ggg;
end
 subplot(232)
 x=ggg;
 imshow(ggg,[]);
 title('Tinkerbell shuffled')
 %henon Shuffling
[m,n]=size(x);
 a = 1.4;b=0.3;
% num_iter=input('Enter the round for Arnold shuffling');
num_iter = 5;
%     a=1.4;
%     b=0.3;
 for k=1:num_iter
for i=1:m
    for j=1:m
        r = mod([round(abs(1-(a*(i^2))+j)),round(abs(b*i))],[m n]);
         xxx(i,j)=x(r(1)+1,r(2)+1);
     end
 end
  x=xxx;
 end
 subplot(233)
 imshow(xxx);
 title('henon Shuffled Image')
 save xxx x
 disp('........................Shuffling process completed succesfully.....................')
 %...................Inverse Shufling.............................
 disp('........................Started Inverse Shuffling.....................')
 %Henon Inverse Shuffling
 for k=1:num_iter
for i=1:m
    for j=1:n
r = mod([round(abs(1-(a*(i^2))+j)),round(abs(b*i))],[m n]);
x1(r(1)+1,r(2)+1)=xxx(i,j);
     end
 end
  xxx=x1;
 end
 subplot(234)
 imshow(x1);
 title('henon Inverse Shuffled')
%Tinkerbell inverse shuffling
a=0.3;b=0.6000;c=2.0;d=0.27;
for k=1:num
for i=1:m
  for j=1:n
       r = mod([round(abs((i^2)-(j^2)+(a*i)+(b*j))),round(abs((2*i*j)+(c*i)+(d*j)))], [m,n]);
        g1(r(1)+1,r(2)+1)=x1(i,j); 
     end
 end
  xxx=g1;
end
 subplot(224)
 imshow(g1)
 title('Tinkerbell_inverse_shuffled')
  disp('........................Inverse Shuffling process completed succesfully.....................')
 psnr = psnr(g1,original)
 ssim=ssim(g1,original)

[Walter Roberson]

You created a new Question about this, and I have replied there.