Question

Given a list of complexities:

How do you order then in their Big O order?

I think the answer is below?

Question now is how does log(n!) become n log(n). Also I don’t know if I got the n! and (n-1)! right. Is it possible that c^n can be bigger than n!? When c > n?

In general how do I visualize such Big O problem … it took me quite long to do this … compared to coding so far … Any resources, videos MIT Open Courseware resources, something with explaination

Answer 1

You might want to see how the functions grow. Here’s a quick plot from Wolfram Alpha:

link

In general, n^n grows much faster than c^n for any n greater than some n_0 (because n will overtake c at some point, even if c is extremely large). log grows much slower than quadratic or exponential, and slightly faster than linear.

For O(log(n!)) = O(nlogn), I believe there was something called Stirling’s Approximation. It boils down to seeing that O(n!) = O(n^n) as n! = n*(n-1)*(n-2)*...*2*1, so n^n = n*n*n*...*n is an upper bound. It can be proven that is it a lower bound as well, but you don’t need that.

Since log(n^n) = nlogn by log rules, O(log(n!) = O(log(n^n)) = O(nlogn).