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Chapter 2: Problem 22

Show that if \\[ W(n) \leq \frac{(p-1)(p-2)}{2}+\frac{(n-p)(n-p-1)}{2} \\] then \\[ W(n) \leq \frac{n(n-1)}{2} \quad \text { for } \quad 1 \leq p \leq n \\] This result is used in the discussion of the worst-case time complexity analysis of Algorithm 2.6 (Quicksort).

Short Answer

Expert verified
The proof shows that if the original inequality \(W(n) \leq \frac{(p-1)(p-2)}{2}+\frac{(n-p)(n-p-1)}{2}\) holds true, then it will lead to \(W(n) \leq \frac{n(n-1)}{2}\), thereby verifying that \(W(n) \leq \frac{n(n-1)}{2}\) for \(1 \leq p \leq n\).

Step by step solution

01

Expand terms on the right side

First, expand the terms on the right side of the given inequality: \( \frac{(p-1)(p-2)}{2}+\frac{(n-p)(n-p-1)}{2} \), giving \( \frac{p^2-3p+2}{2}+\frac{n^2-2np+p^2-p}{2} \).
02

Simplify the equation

Simplify the equation by combining like terms, which yields: \( \frac{2p^2-3p+n^2-2np+2-p}{2} = \frac{2p^2-3p+n^2-2np+2-p}{2} = \frac{n^2-p^2+1}{2} \).
03

Substitute \(p=n/2\)

Replace p in the inequality with \(n/2\), which gives the right side as \( \frac{n^2 - (n/2)^2 + 1}{2} = \frac{n^2 - n^2/4 + 1}{2}\). When simplified, it gives \( W(n) \leq \frac{3n^2/4 + 1}{2} \).
04

Prove the required inequality

We can see that \( \frac{3n^2/4 + 1}{2} \leq \frac{n(n-1)}{2} \) because as \(n \geq 1\), \( \frac{3n^2}{4} + 1 \leq n^2 - n = n(n - 1) \). And both sides are divisible by 2, so the division by 2 will not affect the inequality. This proves that \(W(n) \leq \frac{n(n-1)}{2}\) if \(W(n) \leq \frac{3n^2/4 + 1}{2}\), thereby fulfilling the required condition.

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Most popular questions from this chapter

Write for the following problem a recursive algorithm whose worst-case time complexity is not worse than \(\Theta(n \lg n)\). Given a list of \(n\) distinct positive integers, partition the list into two sublists, cach of size \(n / 2,\) such that the difference between the sums of the integers in the two sublists is maximized. You may assume that \(n\) is a multiple of 2.

Use the divide-and-conquer approach to write a recursive algorithm that finds the maximum sum in any contiguous sublist of a given list of \(n\) real values. Analyze your algorithm, and show the results in order notation.

Write algorithms that perform the operations \(u \times 10^{m}\) \(u\) divide \(10^{n}\) \(u\) rem \(10 "\) where \(u\) represents a large integer, \(m\) is a nonnegative integer, divide returns the quotient in integer division, and rem returns the remainder. Analyze your algorithms, and show that these operations can be done in linear time.

Verify the following identity \\[ \sum_{p=1}^{n}[A(p-1)+A(n-p)]=2 \sum_{p=1}^{n} A(p-1) \\] This result is used in the discussion of the average-case time complexity analysis of Algorithm 2.6 (Quicksort).

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 \(T(n)\).
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