What can be a reverse permutation formula of n*m matrix

The first row remain as it is, but from the second to the end it changed as shown below: Given A as original matrix, B the result matrix. the first row at B remain as it was in A, but at second row, the first element is obtained After add the first element at row above and it's last element in corresponding row, such as B[2,1]=A[1,1]+A[2,4] which is 9=5+4. But the second element at row two is B[2,2]=A[2,1]+A[1,2] which is 13=10+3. So 13=11+2,4=3+1, again for next row B[3,1]=8+10, B[3,2]= 6+11 etc..

A=[4 3 2 1;10 11 3 5; 6 4 7 8; 9 10 40 5]

B=[ 4 3 2 1; 9 13 13 4; 18 16 7 12; 11 23 17 48].

I need the reverse formula to its original matrix.. Assume Matrix A is of size 256*256 or 512*512

[Matt J]

I think you have incorrect entries in B(3:4,2). I think the real B matrix should be,

 B =

     4     3     2     1
     9    13    13     4
    18    17     7    12
    11    13    17    48

Answer 1

[Matt J]

Assuming A and B are always square,

n=length(A);

E=speye(n);
S=circshift(E,[1,0]);

T=kron(S,E)+kron(E,S);
T(1:n,1:n)=E;
T(1:n,n+1:end)=0;

 fun=@(Z) reshape(Z.',[],1);
ifun=@(Z) reshape(Z,n,n).';


A=ifun(T\fun(B))

[dani elias]

Thank you for quick replying. I have trying to re-write the codes but it did not work.

Is it possible to use Control structures such as for loop and if statements ?.

My target is to get Matrix B and return that matrix to its original form as it was. Assume original matrix was A.

matrix A should be square matrix.

[Matt J]

Why are you trying to rewrite the code? Does the version I gave you not work?

[dani elias]

Sorry,it was mistake during coding. The solution you gave, works perfectly. Thank you so much. And it's performance is awesome,"very fast" . Can I have simple codes as this one for A to B?

[Matt J]

B=ifun(T*fun(A));

[dani elias]

Thank you. I do appreciate.

Answer 2

[Matt J]

This version is also quite fast - maybe even faster than my other, fully vectorized answer.

n=length(A);

%% Forward
tic;
B=A;
B(2:end,:)=A(1:end-1,:)+circshift(A(2:end,:),[0,1]);
toc

%% Inverse
A=B;
tic
for i=2:n
 
  A(i,:)=circshift(A(i,:)-A(i-1,:),[0,-1]);  
      
end

toc

[dani elias]

Thank you. I do appreciate your contributions.