5.1 A First Recurrence: The Mergesort Algorithm

Divide and conquer algorithms are a class of algorithmic techniques in which one breaks the input into several parts, solves the problem in each part recursively,and then combines the solutions to these subproblems into an overall solution. Analyzing these problems usually involves solving recurrence relation that bounds the running time recursively in terms of the running time on smaller instances. Mergesort uses the same template as other divide and conquer algorithms to sort a given list of items:

• Break up problem of size n into two equal parts of size ½n
• Solve two parts recursively
• Combine two solutions into overall solution

So,let's T(n) denote the Mergesort's worst-case running times on input instances of size n.

• Supposing that n is even, the algorithm spends O(n) time to divide the input into two pieces of size n/2 each
• It then spends T(n/2) to solve each one since T(n/2) is the worst-case running time for an input of size n/2
• Finally, it spends O(n) time to combine the solutions from the 2 recursive calls
• Thus, the running time satisfies the following recurrence relation :
  • T(n) ≤ 2T(n/2) + cn for n >2
  • And T(2) ≤ c

Thus, 2 is the base case of the recurrence relation in Mergesort algorithm. We assume n is even. To solve for the overall running time of the algorithm, we need to solve the recurrences involved in the algorithm.

Approaches to Solving Recurrences

There are two general approaches to solving recurrences:

• To Unroll the Recurrence : Account for the running time across the first few levels, and identify a pattern that can be continued as the recursion expands. Then we sum up the running times over all levels of recursion to get the overall running time.
• To substitute a Solution into the Mergesort Recurrence : Start with a guess for the solution, substitute it into the recurrence relation, and then check that it works. This approach is formally justified by induction on n.

Analyze the first few levels:

At the first level of recursion, we have a single problem of size n. It takes at most cn + time spent in all subsequent recursive calls.

At the next level, we have two problems of size n/2. Each takes at most cn/2 + time spent on subsequent recursive calls, for a total of at most cn.

A the third level, we have 4 problems each of size n/4, each taking time at most cn/4, for a total of at most cn.

Identifying the pattern :