merge sort

(Latest Revision: 09/19/2005)
Merge Sort



If you do a merge sort of 2^k things 'from the bottom up' it goes
like this:

You are merging                         The step requires

2^k-1 pairs of 2⁰=1-element lists
                                     2^k moves and 2^k-1 compares
2^k-2 pairs of 2¹=2-element lists
                                     2^k moves and from 2*2^k-2 to 3*2^k-2 compares
2^k-3 pairs of 2²=4-element lists
                                     2^k moves and from 4*2^k-3 to 7*2^k-3 compares
2^k-4 pairs of 2³=8-element lists
                                     2^k moves and from 8*2^k-4 to 15*2^k-4 compares
.
.
.
                                     2^k moves and from 2^j*2^k-j-1 to (2^j-1)*2^k-j compares
2^k-j-1 pairs of 2^j-element lists
                                     2^k moves and from 2^j*2^k-j-1 to (2^j+1-1)*2^k-(j+1) compares
2^k-j-2 pairs of 2^j+1-element lists
.
.
.
2⁰=1pair of 2^k-1-element lists
                                     2^k moves and from 2^k-1*2⁰ to (2^k-1)*2⁰ compares
one 2^k-element lists

Thus the total number of moves required by the sort is k*2^k =
(logN)*N, where N=2^k is the size of the list.

The total number of compares required could be as little as k*2^k-1.
That is the sum of the lower bounds on each "phase".  

The maximum number of compares is 
Σ(2^j-1)2^k-j, where the index j ranges from 1 to k.

That is the same as Σ2^k - Σ2^k-j, (in both sums: j=1 to k);

which is the same as k*2^k - Σ2ⁱ, (i=0 to k-1)

which is the same as k*2^k - (2^k-1) = (k-1)*2^k + 1.

It's pretty easy to see that both the min and max numbers of compares are O(N*logN).

Example: Imagine we are merge-sorting a list of 8 objects.  The max comparison "tree" looks like this:

slot #     1   2   3   4   5   6   7   8
slots    |   |   |   |   |   |   |   |   |
             1       1       1       1
                 3               3
                         7

The figure denotes that we could require 1 compare for each of the initial 4
pairs of 1-element lists, 3 compares for each of the subsequent 2 pairs of
2-element lists, and finally 7 compares for the final pair of 4-element lists.
This would be a total of 4+6+7=17 compares.  The number 17 is also given by
the formula (k-1)*2^k + 1, where k=3: (2)*2³ + 1 =
2*8+1=17.