Why is insertion sort Θ(n^2) in the average case?

Question

Insertion sort has a runtime that is Ω(n) (when the input is sorted) and O(n²) (when the input is reverse sorted). On average, it runs in Θ(n²

Answer 1

For fun I wrote a program which ran through all data combinations for a vector of size n counting comparisons and found that the best case is n-1 (all sorted) and the worst is (n*(n-1))/2.

Some results for different n:

  n min     ave     max ave/(min+max) ave/max

  2   1     1         1        0.5000
  3   2     2.667     3        0.5334
  4   3     4.917     6        0.5463
  5   4     7.717    10        0.5512
  6   5    11.050    15        0.5525
  7   6    14.907    21        0.5521
  8   7    19.282    28        0.5509
  9   8    24.171    36        0.5493
 10   9    29.571    45        0.5476
 11  10    35.480    55        0.5458
 12  11    41.897    66        0.5441

It seems the average value follows min closer than it does max.

EDIT: some additional values

 13  12    48.820    78        0.5424        
 14  13    56.248    91        0.5408

EDIT: value for 15

 15  14    64.182   105        0.5393

EDIT: selected higher values

 16  15    72.619   120        -       0.6052
 32  31   275.942   496        -       0.5563
 64  63  1034.772  1953        -       0.5294
128 127  4186.567  8128        -       0.5151
256 255 16569.876 32640        -       0.5077

I recently wrote a program to compute the average number of comparisons for insertion sort for higher values of n. From these I have drawn the conclusion that as n approaches infinity the average case approaches the worst case divided by two.