0

For the following pseudocode; I think that the number of flops is 2n^3. However, I am unsure that this is correct as the for loops make me doubt it. (Note: aij and xij represent entries for matrices A and X respectively)

for =1:
  for =1:
    for =:
      =+*    
    end
  end
end
anonymouseeeee
  • 3
  • 4

1 Answers1

0

This isn't really a matlab question. Here is a brute force way of working it out.

The inner equation has two flops, so the k loop has 2(n-j+1) flops.

To do the rest, it helps to know the sum of q from 1 to p is p(p+1)/2 and q^2 from 1 to p is p(p+1)(2p+1)/6.

The j loop is of 2(n-j+1) for j from 1 to i, so it has 2i(n+1)-i(i+1)=2i(n+1/2)-i^2 flops.

The overall, or i, loop is a sum of 2i(n+1/2)-i^2, giving n(n+1)(n+1/2) - n(n+1)(2n+1)/6 = n(n+1)(2n+1)/3.

You can see that this is the same as the sum from 1 to n of 2i^2.

A way to check this, e.g. in matlab, is to set some n, and put f=0 at the start and replace the inner equation with f=f+2;, then the result will be f=n(n+1)(2n+1)/3.

Ramashalanka
  • 8,564
  • 1
  • 35
  • 46
  • How come the k loop doesn't have 2(n-j) flops instead? – anonymouseeeee Oct 12 '20 at 20:29
  • The distance from j to n is n-j steps, but the number of steps including the first and the last step is n-j+1. For example from 1 to n has n-1+1=n steps, i.e. 1,2,...,n, not n-1 steps. From 2 to 4 has 4-2+1=3 steps, i.e. 2,3,4, not 4-2=2 steps. – Ramashalanka Oct 13 '20 at 01:39