Is there anyway to retrieve the original matrix from randomised matrix ? I used randperm function to randomised the code

CLC;
Clear all;
Close all;
A=[ 1 2 3;2 3 4;3 4 5];
[M,N] = size(A);
F= randperm(N);
B=A;
for i=1:M
  if(i<=M)
    B(i,F)=A(i,:);
    i=i+1;
   end
end
disp(B);

[Walter Roberson]

Under what conditions could your if be false

[Nippulati Uday]

Maybe when M variable is greater than M. Orelse my if condition can never be false

[Walter Roberson]

When can M be greater than M? Should one of those be N?

[KSSV]

Wont B(:,F) gives back A?

[Nippulati Uday]

No,I'm getting the right randomised sequence but I don't know how to retrieve the original Matrix from this randomised Matrix.plz help me with a decoding code

[Walter Roberson]

When you go to do the recovery, what values are you permitted to use as input?

[Nippulati Uday]

I can only use B matrix(randomised matrix) for recovery

Answer 1

[Walter Roberson]

You cannot do that.

You are permuting the columns, so you do not have to worry about the full matrix size: if you could work out any one row then you could apply that to all of the rows.

So consider one row of input. You permute it and know the result. To get back you would need to permute it back. But consider the examples:

A1=4 6 5, F1 = 3 1 2, B = 5 4 6
A2=6 4 5, F2 = 3 2 1, B = 5 4 6

You have the same B result for two different A, and therefore you cannot know whether to apply the inverse of F1 to get A1, or to apply the inverse of F2 to get A2.

Your sample A has the property that the first row is in increasing order. If we were permitted to rely on that then it would be easy to solve.