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Chapter 2: Problem 12

Show that the recurrence equation for the worst-case time complexity for Mergesort (Algorithms \(2.2 \text { and } 2.4)\) is given by \\[ W(n)=W\left(\left[\frac{n}{2}\right\rfloor\right)+W\left(\left\lceil\frac{n}{2}\right]\right)+n-1 \\] when \(n\) is not restricted to being a power of 2

Short Answer

Expert verified
The given recurrence equation \(W(n) = W(\left\lfloor\frac{n}{2}\right\rfloor) + W(\left\lceil\frac{n}{2}\right\rceil) + n - 1\) truly reflects the worst-case time complexity of the MergeSort algorithm. It accounts for the complexity of sorting each half and the process of merging them which takes at most \(n - 1\) comparisons.

Step by step solution

01

Understand MergeSort Algorithm

MergeSort is a recursive divide-and-conquer algorithm that splits a list into two halves, sorts them and merges the sorted halves. Its time complexity is given by the time it takes to split the array (which is constant regardless of the array's size since we are just finding the midpoint) plus the time to sort each half (each of which is a subproblem half the size of the original) plus the time to merge the two halves. This gives rise to a recurrence equation that represents the time complexity of MergeSort.
02

Explicit illustration of recurrence equation

We want to illustrate that the recurrence equation \(W(n) = W(\left\lfloor\frac{n}{2}\right\rfloor) + W(\left\lceil\frac{n}{2}\right\rceil) + n - 1\), where \(\lfloor\frac{n}{2}\rfloor\), denotes the floor of \(n/2\), and \(\lceil\frac{n}{2}\rceil\) denotes the ceiling of \(n/2\), represents the worst-case time complexity of MergeSort. MergeSort will divide the array into 2 parts of size \(\lfloor\frac{n}{2}\rfloor\) and \(\lceil\frac{n}{2}\rceil\). The worst-case time complexity for sorting each of these halves is denoted by \(W(\left\lfloor\frac{n}{2}\right\rfloor)\), and \(W(\left\lceil\frac{n}{2}\rceil)\) respectively. The process of merging these halves requires at most \(n - 1\) comparisons.
03

Validating the Recurrence Equation and Wrap Up

Each recursive call to MergeSort on a list of size \(n\) will end up in base cases of lists of size 1. Once those base cases are reached, the algorithm will begin merging each pair of sorted lists, using a maximum of \(n - 1\) comparisons per merge, and proceed until it reaches the top of the recursion tree where the initially divided list is now sorted. This is why we see the \(n - 1\) operation in our recurrence equation. This reflects the worst-case number of comparisons needed to merge the sorted lists at each level of recursion.

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Most popular questions from this chapter

Suppose that on a particular computer it takes \(12 n^{2}\), \(\mu\) s to decompose and recombine an instance of size \(n\) in the case of Algorithm 2.8 (Strassen). Note that this time includes the time it takes to do all the additions and subtractions. If it takes \(n^{3} \mu\) s to multiply two \(n \times n\) matrices using the standard algorithm, determine thresholds at which we should call the standard algorithm instead of dividing the instance further. Is there a unique optimal threshold?

Use Mergesort (Algorithms 2.2 and 2.4 ) to sort the following list. Show the actions step by step. \\[ 123 \quad 34 \quad 189 \quad 56 \quad 150 \quad 12 \quad 9 \quad 240 \\]

Suppose that, in a divide-and-conquer algorithm, we always divide an in stance of size \(n\) of a problem into 10 subinstances of size \(n / 3,\) and the dividing and combining steps take a time in \(\Theta\left(n^{2}\right)\). Write a recumence equation for the running time \(T(n),\) and solve the equation for \(T(n)\).

When a divide-and-conquer algorithm divides an instance of size \(n\) of a problem into subinstances each of size \(n / c\), the recurrence relation is typically given by \\[ \begin{array}{l} T(n)=a T\left(\frac{n}{c}\right)+g(n) \quad \text { for } n>1 \\ T(1)=d \end{array} \\] where \(g(n)\) is the cost of the dividing and combining processes, and \(d\) is a constant. Let \(n=c^{k}\) (a) Show that \\[ T\left(c^{k}\right)=d \times d^{k}+\sum_{j=1}^{k}\left[a^{k-j} \times g\left(c^{j}\right)\right] \\] (b) Solve the recurrence relation given that \(g(n) \in \Theta(n)\)

Suppose that, in a divide-and-conquer algorithm, we always divide an in stance of size \(n\) of a problem into \(n\) subinstances of size \(n / 3,\) and the dividing and combining steps take lincar time. Write a recurrence equation for the running time \(T(n),\) and solve this recurrence equation for \(T(n) .\) Show your solution in order notation.
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