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

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Arshub 2021-12-31

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https://ww2.mathworks.cn/matlabcentral/answers/1620340-how-do-we-change-the-positions-of-the-elements-of-t-matrix

编辑： Arshub 2022-1-1

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;
65 70 83; 
27 39 40 ; 
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);

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DGM 2021-12-31

编辑：DGM 2022-1-1

在 MATLAB Online 中打开

Is the matrix Ab even correct? The matrix Ab is a combination of a subset of the elements of A. You can't reconstruct A by permutation if you don't retain all the members of A, so I'm going to assume Ab is wrong.

This is probably where the problem lies, but I can't really guess what the intent is here.

C = [1 3 4 5 6 7 9 11 13 14 15 1 7 9 11 11];
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.
% This code doesn't do what the above description says it does
% Nothing here changes S, and C now has more duplicates than it did before
for i = 1:numel(C)
    idx = find(S == C(i));
    C = [C S(idx(2:end))];
end
C
C = 1×28
     1     3     4     5     6     7     9    11    13    14    15     1     7     9    11    11     1     7     9    11    11     1     7     9    11    11    11    11
numel(C)
ans = 28
numel(unique(C))
ans = 11

If the goal is to do what the comment describes, then it's unclear why C does not span the indices of A:

% this does what the description says is supposed to happen
% S has duplicates removed, missing indices appended
C = 1:numel(A);
S = unique(S,'stable');
S = [S C(~ismember(C,S))]
S = 1×16
     1    11     4     3    14     6    13     7    15     5     9     2     8    10    12    16

DGM 2022-1-1

编辑：DGM 2022-1-1

在 MATLAB Online 中打开

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 2022-1-1

编辑：Arshub 2022-1-1

the rest of the paper goes to diffusion phase after this permutation phase .

each phase indepentent each anther.

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

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

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