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,[])

[ANURAG]

clc

clear all

format short

%To solve the LPP by Simplex Method

%Min z=x1-3x2+2x3

%Subject to 3x1-x2+2x3<=7

%-2x1+4x2<=12

%-4x1+3x2+8x3<=10

%x1,x2,x3>=0

%First to change the objective function from minimization to maximization

%Max z=-x1+3x2-2x3

%To input parameters

C=[-1 3 -2]

info=[3 -1 2;-2 4 0; -4 3 8]

b=[7; 12; 10]

NOVariables=size(info,2)

s=eye(size(info,1))

A=[info s b]

Cost=zeros(1,size(A,2))

Cost(1:NOVariables)=C

BV=NOVariables+1:size(A,2)-1

%To calculate Z-Row(Zj-Cj)

ZRow=Cost(BV)*A-Cost

%To print the table

ZjCj=[ZRow; A]

SimpTable=array2table(ZjCj)

SimpTable.Properties.VariableNames(1:size(ZjCj,2))={'x_1','x_2','x_3','s_1','s_2','s_3','Sol'}

%Simplex Table Starts

Run=true;

while Run

if any(ZRow<0) %To check any negative value is there

fprintf('The current BFS is not optimal \n')

fprintf('\n=========The next iteration continues========\n')

disp('Old Basic Variable (BV)=')

disp(BV)

% To find entering Variable

ZC=ZRow(1:end-1)

[EnterCol, Pvt_Col]=min(ZC)

fprintf('The most negative element in Z-Row is %d Corresponding to Column %d \n', EnterCol, Pvt_Col)

fprintf('Entering Variable is %d \n', Pvt_Col)

%To find the leaving variable

sol=A(:,end)

Column=A(:,Pvt_Col)

if all(Column<=0)

error('LPP has unbounded solution. All entries <= 0 in column %d', Pvt_Col)

else

% To check minimum ratio is with positive entering column entries

for i=1:size(Column,1)

if Column(i)>0

ratio(i)=sol(i)./Column(i)

else

ratio(i)=inf

end

end

%To finding minimum ratio

[MinRatio, Pvt_Row]=min(ratio)

fprintf('Minimum ratio corresponding to pivot row is %d \n', Pvt_Row)

fprintf('Leaving Variable is %d \n', BV(Pvt_Row))

end

BV(Pvt_Row)=Pvt_Col

disp('New Basic Variables (BV) =')

disp(BV)

%Pivot Key

Pvt_Key=A(Pvt_Row,Pvt_Col)

%Update Table for next iteration

A(Pvt_Row,:)=A(Pvt_Row,:)./Pvt_Key

for i=1:size(A,1)

if i~=Pvt_Row

A(i,:)=A(i,:)-A(i,Pvt_Col).*A(Pvt_Row,:)

end

ZRow=ZRow-ZRow(Pvt_Col).*A(Pvt_Row,:)

%To print the updated table

ZjCj=[ZRow;A]

SimpTable=array2table(ZjCj)

SimpTable.Properties.VariableNames(1:size(ZjCj,2))={'x_1','x_2','x_3','s_1','s_2','s_3','Sol'}

BFS=zeros(1,size(A,2))

BFS(BV)=A(:,end)

BFS(end)=sum(BFS.*Cost)

CurrentBFS=array2table(BFS)

CurrentBFS.Properties.VariableNames(1:size(CurrentBFS,2))={'x_1','x_2','x_3','s_1','s_2','s_3','Sol'}

end

else

Run=false

fprintf('The current BFS is optimal and Optimality is reached \n')

end

end

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.