i am completely stuck and i cant figure out what direction to head with this it would be much appreciated if someone would be able to give me a few hints for what direction to head for this ty

edit:

not looking for it to be solved just for a way that i can go about solving it.

You are looking for summation of all cells?

Also I am not familiar with the big oh notation, but I can look into it

This post was edited by cialda on Feb 5 2014 08:39pm

Yes looking for all of the cells, I understand the big oh notation the problem for me is figuring out how to sum all the cells up.

Well up to n you have
1+2+2+3+3+3.... = summation of I from 1 to n of n*n

For diagonal right of n, you have
N-1 cells at a value of n+1

Next
N-2 cells at a value of n+2

And so forth until 2n-1


This post was edited by cialda on Feb 5 2014 08:59pm

i understood the first part but i dont get what you are getting at for the second and thir

So once you start hitting the largest diagonal (n), the amount of cells in the diagonals start decreasing again, but the cell values are decreasing.

It should be in a form similar to summation of (n+k)(n-k) or the likes. For k upto n-1

I'll check it out in a bit when done watching tv

This post was edited by cialda on Feb 5 2014 08:58pm

you have 1*1, 2*2, ... n*n, (n+1)*(n-1), (n+2)*(n-2), ..., 1*(2n-1)

now if you take the first and the last and continue from there you have [1+(2n-1)]*1, [2+(2n-2)]*2, ... [(n-1)+(2n-{n-1})]*n-1, n*n

again you can simplify that to 2n*sum(1,...,n-1) + n*n = n * ( [2 * {(n-1) * n / 2}] + n ) = n^3

so should enable you to solve your problem

thank you very much