How to only perform the upper triangular part of matrix operations?

Hi all,

I'm trying to solve this problem, where I have to use a nested loop like this:

clear; clc;
demo = zeros(4, 4);
indj = 1;
for l = 1:2

      for j = 1:2
          indi = 1;

          for k = 1:2

              for i = 1:2

                  disp([indi, indj])
                  demo(indi, indj) = 1;
                  indi = indi + 1;

              end

          end
          indj = indj + 1;
      end
  end

In this code, i, j, k, l each has 2 operations, thus total number of operations is 2^4 = 16. A 4 by 4 matrix 'demo' is filled by these 16 operations (each operation is represented by 'demo(indi, indj) = 1;'), while indi and indj are used to indicate the coordinate of these operations. 'demo' is a 4 by 4 matrix made of 1s.

>> demo
demo =
     1     1     1     1
     1     1     1     1
     1     1     1     1
     1     1     1     1

Now in my problem, because 'demo' is symmetric along the main diagonal, I'm trying to only compute the upper triangular part of 'demo' to save cost, i.e. I want 'demo' to be like:

>> demo
demo =
     1     1     1     1
     0     1     1     1
     0     0     1     1
     0     0     0     1

while the form of using i, j, k, l cannot be changed. I tried the following (only change to k = l:2 and i = j:2):

clear; clc;
  demo = zeros(4, 4);
  indj = 1;
  for l = 1:2

        for j = 1:2
            indi = 1;

            for k = l:2

                for i = j:2

                    disp([indi, indj])
                    demo(indi, indj) = 1;
                    indi = indi + 1;

                end

            end
            indj = indj + 1;
        end
    end

It doesn't work. Any ideas?

Answer 1

[James Tursa]

If you can't change the looping, then maybe just insert an if-test:

                    if( indi <= indj )
                        demo(indi, indj) = 1;
                    end

[Xh Du]

Thanks! This works! Notice that this needs to be inserted in the first piece of code in my original question.