how to get sum of numbers appearing in triangle in a matrix

I want to find sum of numbers appearing in right triangular part from bottom right corner to top left corner in a 2-D matrix like

o = [1 2 3 4;5 6 7 8]
f = [1,2,3,4; 5,6,7,8; 2,3,4,5]

sum of o = 1+5+6+7+8 = 27

sum of f = 1+5+6+2+3+4+5 = 26

[Image Analyst]

It's not clear what you are doing. Do you only have short, wide arrays and you using a lower square triangle and then just copying the entire last row (like Stephen's solution does)? What if the array is taller than it is wide? Does that ever happen to you, and what elements do you sum if that is the case?

Answer 1

[Image Analyst]

You need to decide if the row and column are in the lower triangle and if it is, sum it up. This code works for both your (badly-named) "o" and your "f".

o = [1 2 3 4;5 6 7 8]
f = [1,2,3,4; 5,6,7,8; 2,3,4,5]

m = f % or set equal to f.
[rows, columns] = size(m)
deltaX = columns - 1;
deltaY = rows - 1;
slope = deltaY / deltaX
theSum = 0;
for row = 1 : rows
  for col = 1 : columns
    x = col;
    y = row;
    yOnSlope = slope*(x-1)+1;
    fprintf('row (y) = %d, col (x) = %d, yOnSlope = %.2f', y, x, yOnSlope);
    if y >= yOnSlope
      % It's in the lower triangle
      theSum = theSum + m(row, col);
      fprintf('<-- This is in the lower triangle');
    end
    fprintf('\n');
  end
end
% Echo to command window.
theSum

In the command window:

row (y) = 1, col (x) = 1, yOnSlope = 1.00<-- This is in the lower triangle
row (y) = 1, col (x) = 2, yOnSlope = 1.33
row (y) = 1, col (x) = 3, yOnSlope = 1.67
row (y) = 1, col (x) = 4, yOnSlope = 2.00
row (y) = 2, col (x) = 1, yOnSlope = 1.00<-- This is in the lower triangle
row (y) = 2, col (x) = 2, yOnSlope = 1.33<-- This is in the lower triangle
row (y) = 2, col (x) = 3, yOnSlope = 1.67<-- This is in the lower triangle
row (y) = 2, col (x) = 4, yOnSlope = 2.00<-- This is in the lower triangle
theSum =
    27

and for "f":

row (y) = 1, col (x) = 1, yOnSlope = 1.00<-- This is in the lower triangle
row (y) = 1, col (x) = 2, yOnSlope = 1.67
row (y) = 1, col (x) = 3, yOnSlope = 2.33
row (y) = 1, col (x) = 4, yOnSlope = 3.00
row (y) = 2, col (x) = 1, yOnSlope = 1.00<-- This is in the lower triangle
row (y) = 2, col (x) = 2, yOnSlope = 1.67<-- This is in the lower triangle
row (y) = 2, col (x) = 3, yOnSlope = 2.33
row (y) = 2, col (x) = 4, yOnSlope = 3.00
row (y) = 3, col (x) = 1, yOnSlope = 1.00<-- This is in the lower triangle
row (y) = 3, col (x) = 2, yOnSlope = 1.67<-- This is in the lower triangle
row (y) = 3, col (x) = 3, yOnSlope = 2.33<-- This is in the lower triangle
row (y) = 3, col (x) = 4, yOnSlope = 3.00<-- This is in the lower triangle
theSum =
    26

[Ayyaz Ahmed]

your code working correctly.Thanks a lot.but i could not under how you used slope method?can u explain it.

[Image Analyst]

I basically drew a line from one corner to the other. This line will not necessarily fall exactly at the center of the matrix element - the line will probably go above or below the exact center. So I just find out the y value (the row) that the line has when it's at that x (column). If the element row is below that, it's in the lower triangle, if the element row is above the line, it's in the upper triangle. For example if, say, in column 5 the line would have a value of 3.1234, then elements in column 5 that are in rows 1, 2, and 3 are above the line (because they're less than 3.1234), and rows 4, 5, 6, etc. are below the line (because they are greater than 3.1234).

Answer 2

[Stephen23]

The simplest solution is the best:

>> f = [1 2 3 4; 5 6 7 8; 2 3 4 5]
f =
     1     2     3     4
     5     6     7     8
     2     3     4     5
>> X = tril(f);
>> X(end,:) = f(end,:)
X =
     1     0     0     0
     5     6     0     0
     2     3     4     5
>> sum(X(:))
ans =
    26

[Image Analyst]

Plugging in o (instead of f) gives 15, not 27 like he wanted.

[Stephen23]

@Image Analyst: it must be an off day for you, do you really imagine that I did not test this?

>> f = [1 2 3 4; 5 6 7 8]
f =
     1     2     3     4
     5     6     7     8
>> X = tril(f);
>> X(end,:) = f(end,:)
X =
     1     0     0     0
     5     6     7     8
>> sum(X(:))
ans =
    27

[Image Analyst]

Sorry, you're right - I must have forgot to change one letter when I was changing your code from f to o. Sometimes, I post code off the top of my head without testing, though I know now you don't do that. +1 vote.