# Thread: How to calculate Big Oh with figurates

1. I need to figure out how to calculate Big Oh (that is, O(N), etc.) values for statements inside double nested loops. Anyone know how? This relates to figurate numbers (http://en.wikipedia.org/wiki/Figurate_numbers)

this pertains to calculating worst-case efficiency through finding formulas for the number of total executions of each line of code of a sorting algorithm.

ex.--
-----------------
for (i = o; i < n; i++)
{
for (j = i; j < n; j++)
{
for(k=0; k<= 1; k++)
{
//statement
}
}
}

-----------------
--I know that the Big Oh value for the outermost loop is 2N+3 through manual
calculation.
--I know that the second or middle loop Big Oh value is (n^2+n)/2 via the figurate
--I know what the third or inner-most loop Big Oh value is-- [(n(n+1)(n+2)/3!)+1]
and that I need to multiply this equation by one representing the number of
executions of the statement within the third or innermost loop, but what is the
formula for this statement?

Can anyone help point me in the right direction?  2.

3. I think you have a serious questions and thats why no one in this forum give you a solution. In this case it looks like you need an expert in this AI science and you can found perfect and compatible answer in my reference link below.   