when is insertion sort faster than merge sort?

Question

For a homework problem, I was told that insertion sort runs at 8n^2 and that merge sort runs at 64(n(lg n)). As part of the solution I was given, it said that insertion sort was

Answer 1

It came from this (algebraic) line of reasoning

steps_in_insertion_sort <= steps_in_merge_sort
8n^2 <= 64n lg n
n^2 <= 8n lg n
n <= 8 lg n

Then 43 works by trial and error, or by solving for the zero of the equation n - 8 lg n = 0.

For hacking by trial and error, note:

$ python
>>> 8 * log(43)/log(2)
43.41011803761678

Because "lg" means log-base-two.

(Aside: calculations like this do not really translate to any real-world indication that one algorithm will beat another. Seriously, exactly 43?)

Answer 2

This is the second exercise question in Introduction to Algorithms, 3rd edition by Cormen. The solution to this equation is not so straight forward for a newcomer to algorithms:

Insertion sort beats merge sort when 8n^2 < 64n lg n , n < 8 lg n , 2^ n/8 < n . This is true for 2 <= n <= 43 (found by using a calculator). Therefore, we can rewrite merge sort to use insertion sort for input of size 43 or less in order to improve the running time.