Nested for Loop needed

Hello,

Could anyone enlighten me as to how to produce the for loops required to perform the following:

2 x 3

2 x 4

2 x 5

3 x 4

3 x 5

4 x 5

The numbers actually refer to the rows of a matrix.

I need a loop which starts with the second row, and multiplies it sequentially with every row beneath it, then moves on to the row below and does the same.

Could anyone explain how this would be done for a matrix with 5 rows as above? I'm sure I could generalise it to n rows, but I have been struggling with this for some time now.

Kind regards,

Tom

[Tom]

I am getting mighty confused. All I have worked out is that if I label the first terms as j, and the second terms as i, and there are n rows. Then

j = 2:(n-1)

i = (j+1): n

And the sequence is:

take j, then iterate through i

iterate j, then iterate through i

repeat

[Tom]

I think I have finally solved it:

for j = 2:(n-1)

    for i = (j+1):n

    dotp(count1,:) = sum(x(j,:).*x(i,:));

    count1 = count1 + 1;

    end
end

[Image Analyst]

Hmmm... I think the more straightforward way that most people would do it is to use the prod() function, like I did in my answer below.

Answer 1

[Image Analyst]

Try this:

clc;
workspace;  % Make sure the workspace panel is showing.
m = magic(5)
outMatrix = m; % Initialize.
for row = 2 : size(m, 1)
  newRow = prod(m(row:end, :), 1);
  outMatrix(row, :) = newRow;
end
% Print to command window:
outMatrix

In the command window:

m =
  17    24     1     8    15
  23     5     7    14    16
   4     6    13    20    22
  10    12    19    21     3
  11    18    25     2     9
outMatrix =
        17          24           1           8          15
     10120        6480       43225       11760        9504
       440        1296        6175         840         594
       110         216         475          42          27
        11          18          25           2           9

[Tom]

Thanks for that. I can see what you are saying and I should probably get prod involved. I don't think your solution is final though, because for a matrix with 5 rows, the solution should be (the product of) 3 rows. Whereas her you have an output containing 5 rows.

[Image Analyst]

OK, well you can ignore the first row if you want. But then you said "starts with the second row, and multiplies it sequentially with every row beneath it, then moves on to the row below and does the same" so you'd have new rows like this

newRow2 = row2 .* row3 .* row4 .* row5
newRow3 = row3 .* row4 .* row5
newRow4 = row4 .* row5
newRow5 = row5

That's 4 rows in the new matrix. So I'm not sure how you're getting only 3 rows, unless you just ignore row 5 because it's just there by itself.