can anyone explain, how following program is working?(5th step)

 A should be a randomly created 5 x 5 matrix whose elements are all integers ranging from 0 to 7 in such a way that

 1) row 1 and column 1 have equal sums,
 2) row 2 has a sum of 7 and column 2 has a sum of 0,
 3) row 3 and column 3 have equal sums,
 4) row 4 has a sum of 0 and column 4 has a sum of 7,
 5) row 5 and column 5 have equal sums,
 6) its diagonal elements are all zeros, and
 7) all column and row sums are 7 or less,

function test

    A = zeros(4);

 for k = 1:7

   p = randperm(4); %  <-- This is the source of randomness

   for ix = 1:4

     A(p(ix),ix) = A(p(ix),ix) + 1; **..,please explain this step**
   end

 end

 A = [A(:,1),zeros(4,1),A(:,2:4)]; % Insert a column of four zeros

 A = [A(1:3,:);zeros(1,5);A(4,:)]; % Insert a row of five zeros

A(1,1) = 0; A(3,3) = 0; A(5,5) = 0; % Set non-zero diagonal elements to 0

 disp(A);

end

[Roger Stafford]

You've changed your question, Reshdev. That last step reduces the first, third, and fifth diagonal elements in A to zero. Since they are on the diagonal, that reduces each of the corresponding row and column sums by the same amounts and leaves them still equal as you requested.

Answer 1

[Roger Stafford]

The 'randperm' function is generating random permutations of the sequence, 1,2,3,4. For example suppose it generate p = 4,3,1,2. Then in the inner for-loop with ix = 1, we have p(ix) = 4. Hence

 A(p(ix),ix) = A(p(ix),ix) + 1

is doing this:

 A(4,1) = A(4,1) + 1;

It is adding 1 to the element of A in the fourth row of the first column. Then the next three steps are:

 A(3,2) = A(3,2) + 1
 A(1,3) = A(1,3) + 1
 A(2,4) = A(2,4) + 1

After those four steps you have

 A = [0 0 1 0;
      0 0 0 1;
      0 1 0 0;
      1 0 0 0]

Each row sum and each column sum is now 1. That is the property that any permutation, p, has. It would always leave you with equal row and column sums. When the outer loop is finished, all row and column sums are exactly 7.

Subsequent steps are of course meant to alter that situation.

[reshdev]

thank you, Roger

Answer 2

[Roger Stafford]

If you prefer vectorized solutions, I could just as well have put the solution in this form:

 [~,p] = sort(rand(4,7));
 A = accumarray([reshape(p+(p>3),[],1),repmat([1,3:5]',7,1)],1);
 A(1:12:end) = 0;