How do we change the positions of the elements of t matrix ?

2 views (last 30 days)
I need to use S as index to change positions of mateix A
S=[1 11 1 4 3 14 6 11 13 11 7 15 5 9 9 7]
C = 1:numel(ِِA);
S = unique(S,'stable');
S = [S C(~ismember(C,S))]
A =[111 222 30 4;
50 65 70 83;
10 27 39 40 ;
54 67 72 81 ]
I need to permutate positions of elements of matrix by bellow subistitute:
Substitute A(S(i) ) with A (S(M×N)−i+1).,where M and N is rows and cols of matrix
Here i = 1, 2,..., M × N/2
Note: this way for permutation from this paper:A novel plaintext-related image encryption scheme using hyper-chaotic system
I attached the part of permutation from algorithim in img
step1: I reshap matrix to vector
[row,col]=size(A);
len=row*col;
B=reshape(A,1,len);
A=B;
step2: Substitute A(S(i) ) with A (S(M×N)−i+1).,where M and N is rows and cols of matrix
Here i = 1, 2,..., M × N/2
for i =1:len/2
A(S(i))=A(S(len-i+1));
end
Ab=reshape(A,row,col);
AA=reshape(Ab,1,len);
I get this result after permutation A by S:
Ab =
81 222 30 222
50 30 70 50
67 40 72 40
70 67 72 81
Then i need to make reverse operation to get back the original matrix.
for i =1:len/2
AA(S(len-i+1))=AA(S(i));
end
Abb=reshape(AA,row,col);
I get this result
Abb =
81 222 30 222
50 30 70 50
67 40 72 40
70 67 72 81
but I couldn't get the original matrix "A"
A= 111 222 30 4
50 65 70 83
10 27 39 40
54 67 72 81
What should I do to get original matrix from permutated matrix?
Note: it is immportant to me use this method to permutate matrix: "Substitute A(S(i )) with A (S(M×N)−i+1)."
I get this result instead of original matrix A:
Abb =
81 222 30 222
50 30 70 50
67 40 72 40
70 67 72 81
--------------------------------------------my code-----------------------------
A =[111 222 30 4;50 65 70 83; 10 27 39 40 ; 54 67 72 81 ];% original matrix we need to permutate
S=[1 11 1 4 3 14 6 11 13 11 7 15 5 9 9 7];% index used to permutate matrix A
%{
Remove the repeated elements from matrix index S, and
then put the absent numbers at the end to generate a new sequence X.
%}
C = 1:numel(A);
S = unique(S,'stable');
S = [S C(~ismember(C,S))]
% start permutation
[row,col]=size(A);
len=row*col;
B=reshape(A,1,len);
A=B;
for i =1:8
A(S(i))=A(S(len-i+1));
end
Ab=reshape(A,row,col);
% To get original matrix
AA=reshape(Ab,1,len);
for i =1:8
%encrimg1(m1)=im1_reshap(ind1(m1));%permutation
AA(S(len-i+1))=AA(S(i));
end
Abb=reshape(AA,row,col);
  5 Comments
DGM
DGM on 1 Jan 2022
Edited: DGM on 1 Jan 2022
Well okay, that's one step, but I don't see how this is supposed to be reversible.
A = [111 222 30 4; 50 65 70 83; 10 27 39 40; 54 67 72 81];
S = [1 11 1 4 3 14 6 11 13 11 7 15 5 9 9 7];
n = numel(A);
C = 1:n;
S = unique(S,'stable');
X = [S C(~ismember(C,S))];
% forward transformation
% vectorization/loops aren't necessary here
k = 1:n/2; % assumes n is even
Ab = A;
Ab(X(k)) = Ab(X(n-k+1))
Ab = 4×4
81 222 30 222 50 30 70 50 67 40 72 40 70 67 72 81
% the mapping from A to Ab retains only half of the members of A
% making the process irreversible
numel(unique(A))
ans = 16
numel(unique(Ab))
ans = 8
% for clarity, these are the map indices of the subsets:
k
k = 1×8
1 2 3 4 5 6 7 8
n-k+1
ans = 1×8
16 15 14 13 12 11 10 9
I don't know what the rest of the book/paper suggests beforehand or afterward, but I don't see how this is supposed to be reversible unless we're both making incorrect assumptions about the process. What's the rest of the book say?
Arshub
Arshub on 1 Jan 2022
Edited: Arshub on 1 Jan 2022
the rest of the paper goes to diffusion phase after this permutation phase .
each phase indepentent each anther.

Sign in to comment.

Answers (0)

Products


Release

R2014a

Community Treasure Hunt

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

Start Hunting!