Merge Sort

Merge sort is based on the divide-and-conquer paradigm. Its worst-case running time has a lower order of growth than insertion sort. Since we are dealing with subproblems, we state each subproblem as sorting a subarray A[p .. r]. Initially, p = 1 and r = n, but these values change as we recurse through subproblems.

To sort A[p .. r]:

1. Divide Step

If a given array A has zero or one element, simply return; it is already sorted. Otherwise, split A[p .. r] into two subarrays A[p .. q] and A[q + 1 .. r], each containing about half of the elements of A[p .. r]. That is, q is the halfway point of A[p .. r].

2. Conquer Step

Conquer by recursively sorting the two subarrays A[p .. q] and A[q + 1 .. r].

3. Combine Step

Combine the elements back in A[p .. r] by merging the two sorted subarrays A[p .. q] and A[q + 1 .. r] into a sorted sequence. To accomplish this step, we will define a procedure MERGE (A, p, q, r).

Note that the recursion bottoms out when the subarray has just one element, so that it is trivially sorted.


Algorithm: Merge Sort

To sort the entire sequence A[1 .. n], make the initial call  to the procedure MERGE-SORT (A, 1, n).

MERGE-SORT (A, p, r)

1.     IF p < r                                                    // Check for base case
2.         THEN q = FLOOR[(p + r)/2]                 // Divide step
3.                 MERGE (A, p, q)                          // Conquer step.
4.                 MERGE (A, q + 1, r)                     // Conquer step.
5.                 MERGE (A, p, q, r)                       // Conquer step.

Example: Bottom-up view of the above procedure for n = 8.



What remains is the MERGE procedure. The following is the input and output of the MERGE procedure.

INPUT: Array A and indices p, q, r such that pq ≤ r and subarray A[p .. q] is sorted and subarray A[q + 1 .. r] is sorted. By restrictions on p, q, r, neither subarray is empty.

OUTPUT: The two subarrays are merged into a single sorted subarray in A[p .. r].

We implement it so that it takes Θ(n) time, where n = rp + 1, which is the number of elements being merged.


Idea Behind Linear Time Merging

Think of two piles of cards, Each pile is sorted and placed face-up on a table with the smallest cards on top. We will merge these into a single sorted pile, face-down on the table.

A basic step:

Each basic step should take constant time, since we check just the two top cards. There are at most n basic steps, since each basic step removes one card from the input piles, and we started with n cards in the input piles. Therefore, this procedure should take Θ(n) time.

Now the question is do we actually need to check whether a pile is empty before each basic step?
The answer is no, we do not. Put on the bottom of each input pile a special sentinel card. It contains a special value that we use to simplify the code. We use ∞, since that's guaranteed to lose to any other value. The only way that ∞ cannot lose is when both piles have ∞ exposed as their top cards. But when that happens, all the nonsentinel cards have already been placed into the output pile. We know in advance that there are exactly rp + 1 nonsentinel cards so stop once we have performed rp + 1 basic steps. Never a need to check for sentinels, since they will always lose. Rather than even counting basic steps, just fill up the output array from index p up through and including index r .

The pseudocode of the MERGE procedure is as follow:

MERGE (A, p, q, r )

1.      n1qp + 1
2.      n2rq
3.      Create arrays L[1 . . n1 + 1] and R[1 . . n2 + 1]
4.      FOR i ← 1 TO n1
5.            DO L[i] ← A[p + i − 1]
6.      FOR j ← 1 TO n2
7.            DO R[j] ← A[q + j ]
8.      L[n1 + 1] ← ∞
9.      R[n2 + 1] ← ∞
10.    i ← 1
11.    j ← 1
12.    FOR kp TO r
13.         DO IF L[i ] ≤ R[ j]
14.                THEN A[k] ← L[i]
15.                        ii + 1
16.                ELSE A[k] ← R[j]
17.                        jj + 1


Example [from CLRS-Figure 2.3]: A call of MERGE(A, 9, 12, 16). Read the following figure row by row. That is how we have done in the class.

The operation of lines 10 through 17 in the call MERGE (A, 9, 12, 6).


Running Time

The first two for loops (that is, the loop in line 4 and the loop in line 6) take Θ(n1 + n2) = Θ(n) time. The last for loop (that is, the loop in line 12) makes n iterations, each taking constant time, for Θ(n) time. Therefore, the total running time is Θ(n).


Analyzing Merge Sort

For simplicity, assume that n is a power of 2 so that each divide step yields two subproblems, both of size exactly n/2.

The base case occurs when n = 1.

When n ≥ 2, time for merge sort steps:

Summed together they give a function that is linear in n, which is Θ(n). Therefore, the recurrence for merge sort running time is

merge sort recurrence


Solving the Merge Sort Recurrence

By the master theorem in CLRS-Chapter 4 (page 73), we can show that this recurrence has the solution

T(n) = Θ(n lg n).

Reminder: lg n stands for log2 n.

Compared to insertion sort [Θ(n2) worst-case time], merge sort is faster. Trading a factor of n for a factor of lg n is a good deal. On small inputs, insertion sort may be faster. But for large enough inputs, merge sort will always be faster, because its running time grows more slowly than insertion sorts.


Recursion Tree

We can understand how to solve the merge-sort recurrence without the master theorem. There is a drawing of recursion tree on page 35 in CLRS, which shows successive expansions of the recurrence.

The following figure (Figure 2.5b in CLRS) shows that for the original problem, we have a cost of cn, plus the two subproblems, each costing T (n/2).

Construction of recursion tree


The following figure (Figure 2.5c in CLRS) shows that for each of the size-n/2 subproblems, we have a cost of cn/2, plus two subproblems, each costing T (n/4).

Construction of recursion tree


The following figure (Figure: 2.5d in CLRS) tells to continue expanding until the problem sizes get down to 1.

Construction of the recursion tree


In the above recursion tree, each level has cost cn.

The height of this recursion tree is lg n and there are lg n + 1 levels.


Mathematical Induction

We use induction on the size of a given subproblem n.

Base case: n = 1

Implies that there is 1 level, and lg 1 + 1 = 0 + 1 = 1.

Inductive Step

Our inductive hypothesis is that a tree for a problem size of 2i has lg 2i + 1 = i +1 levels. Because we assume that the problem size is a power of 2, the next problem size up after 2i is 2i + 1. A tree for a problem size of 2i + 1 has one more level than the size-2i tree implying i + 2 levels. Since lg 2i + 1 = i + 2, we are done with the inductive argument.

Total cost is sum of costs at each level of the tree. Since we have lg n +1 levels, each costing cn, the total cost is

cn lg n + cn.

Ignore low-order term of cn and constant coefÞcient c, and we have,

Θ(n lg n)

which is the desired result.




void mergeSort(int numbers[], int temp[], int array_size)

        m_sort(numbers, temp, 0, array_size - 1);


void m_sort(int numbers[], int temp[], int left, int right)

        int mid;

        if (right > left)


            mid = (right + left) / 2;

            m_sort(numbers, temp, left, mid);

            m_sort(numbers, temp, mid+1, right);

            merge(numbers, temp, left, mid+1, right);



void merge(int numbers[], int temp[], int left, int mid, int right)


            int i, left_end, num_elements, tmp_pos;

            left_end = mid - 1;

            tmp_pos = left;

            num_elements = right - left + 1;

while ((left <= left_end) && (mid <= right))


                if (numbers[left] <= numbers[mid])


                        temp[tmp_pos] = numbers[left];

                        tmp_pos = tmp_pos + 1;

                        left = left +1;




                        temp[tmp_pos] = numbers[mid];

                        tmp_pos = tmp_pos + 1;

                        mid = mid + 1;



        while (left <= left_end)


                        temp[tmp_pos] = numbers[left];

                        left = left + 1;

                        tmp_pos = tmp_pos + 1;


                while (mid <= right)


                        temp[tmp_pos] = numbers[mid];

                        mid = mid + 1;

                        tmp_pos = tmp_pos + 1;


                for (i = 0; i <= num_elements; i++)


                        numbers[right] = temp[right];

                        right = right - 1;



Update: January 14, 2010.