randperm and sorting array

as example :

A = [8 9 0]

Perm1 = randperm(length(A));
B= A(:,Perm1);
B_eks=B

[val_sort, id_sort] = sort(Perm1);
 A_eks = B_eks(:,id_sort);

i want it for looping, so A_eks(:,:,1) = [8 9 0],;A_eks(:,:,2) = [8 9 0],A_eks(:,:,2) = [8 9 0],A_eks(:,:,4) = [8 9 0], i have try like this:

for i=1:4
Perm1(:,:,i) = randperm(length(A));
B(:,:,i)= A(:,Perm1(:,:,i));
B_eks(:,:,i)=B(:,:,i)
end

for k=1:4
[val_sort, id_sort] = sort(Perm1(:,:,k));
 A_eks(:,:,k) = B_eks(:,id_sort);
end
but it didn't work like i want, what should i do ?

[KSSV]

A = [8 9 0]
Perm1 = randperm(length(A));
B= A(:,Perm1);
B_eks=B
[val_sort, id_sort] = sort(Perm1);
 A_eks = B_eks(:,id_sort);

This piece of code gives you back A again....why to do? :(

[KL]

Maybe your example with just 3 elements (as KSSV says, they are already sorted!) is not sufficient, probably you need to explain your input and expected output better.

[KL]

@cvklpstunc: permuted indices are being sorted, which makes the permutation itself redundant. A_eks will all be the same as A.

[maharani meidy]

@KSSV @KL: yes, i need it to be A again, because i use it for extraction from watermarked video. the watermark is apply to every frame, so a_eks needs to be the same with A. is it answer ur question ?

@cvklpstunc : i want my data permutated in random position for it secure. is there easier way for it?

thanks all ^^

[KSSV]

If you want it to be same..keep it same....why to run all the stuff? Using randperm is good if you want to permute array randomly.

[maharani meidy]

@KSSV : well.. i want it to be more secure than just keep it the same...^^

Answer 1

[KL]

Change

 A_eks(:,:,k) = B_eks(:,id_sort);

to

A_eks(:,:,k) = B_eks(:,id_sort,k);

[maharani meidy]

thank you ^^