Asymptotic Complexity comparison

Question

Can anybody explain which one of them has highest asymptotic complexity and why,

10000000n vs 1.000001^n vs n^2

Answer 1

You can use standard domination rules from asymptotic analysis.

Domination rules tell you that when n -> +Inf, n = o(n^2). (Note the difference between the notations O(.) and o(.), the latter meaning f(n) = o(g(n)) iff there exists a sequence e(n) which converges to 0 as n -> +Inf such that f(n) = e(n)g(n). With f(n) = n, g(n) = n^2, you can see that f(n)/g(n) = 1/n -> 0 as n -> +Inf.)

Furthermore, you know that for any integer k and real x > 1, we have n^k/x^n -> 0 as n -> +Inf. x^n (exponential) complexity dominates n^k (polynomial) complexity.

Therefore, in order of increasing complexity, you have:

n << n^2 << 1.000001^n

Note:10000000n could be written O(n) with the loose written conventions used for asymptotic analysis in computer science. Recall that the complexity C(n) of an algorithm is O(n) (C(n) = O(n)) if and only if (iff) there exists an integer p >= 0 and K >= 0 such that for all n >= p the relation |C(n)| <= K.n holds.

Answer 2

When calculating asymptotic time complexity, you need to ignore all coefficients of n and just focus on its exponent.

The higher the exponent, the higher the time complexity.

In this case

We ignore the coefficients of n, leaving n^2, x^n and n.

However, we ignore the second one as it has an exponent of n. As n^2 is higher than n, the answer to your question is n^2.