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Chapter 2: Problem 21

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).

Short Answer

Expert verified
Therefore, this verifies the given identity \[ \sum_{p=1}^{n}[A(p-1)+A(n-p)]=2 \sum_{p=1}^{n} A(p-1) \]

Step by step solution

01

Understand and Write Down the Problem

We are given to prove the identity: \[ \sum_{p=1}^{n}[A(p-1)+A(n-p)]=2 \sum_{p=1}^{n} A(p-1) \] The aim is to verify that both sides of this equation are indeed equal.
02

Simplify the Left Hand Side

Apply the distributive property of addition over summation \[ \sum_{p=1}^{n}[A(p-1)+A(n-p)] = \sum_{p=1}^{n}A(p-1) + \sum_{p=1}^{n}A(n-p) \] Next, make the change of index by setting \( q = n-p \). Then, as \( p \) goes from \( 1 \) to \( n \), \( q \) will go from \( n-1 \) to \( 0 \). Thus, \[ \sum_{p=1}^{n}A(n-p) = \sum_{q=n-1}^{0}A(q) = \sum_{q=0}^{n-1} A(q)\] Hence, the left side of our original equation can be written as: \[ \sum_{p=1}^{n}A(p-1) + \sum_{q=0}^{n-1} A(q)\]
03

Simplify Right Hand Side

The right hand side of the equation is \(2 \sum_{p=1}^{n} A(p-1)\). Removing the constant 2 out of the sum, we will be left with \(2 \sum_{p=1}^{n} A(p-1)\)
04

Compare Both Sides

Now the left hand side becomes: \[ \sum_{p=1}^{n}A(p-1) + \sum_{q=0}^{n-1} A(q)=\sum_{p=0}^{n-1}A(p) + \sum_{p=0}^{n-1}A(p)=2 \sum_{p=0}^{n-1}A(p)\] Comparing this to the right hand side: \(2 \sum_{p=1}^{n} A(p-1)\), it's clear that they are the same.

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Most popular questions from this chapter

Use Binary Search (Algorithm 2.1 ) to search for the integer 120 in the following list (array) of integers. Show the actions step by step. $$\begin{array}{lllllllll} 12 & 34 & 37 & 45 & 57 & 82 & 99 & 120 & 134 \end{array}$$

How many multiplications would be performed in finding the product of two \(64 \times 64\) matrices using Strassen's Method (Algorithm 2.8)?

Show that the worst-case time complexity for Binary Scarch (Algorithm 2.1) is given by \\[ W(n)=\lfloor\lg n\rfloor+1 \\] when \(n\) is not restricted to being a power of \(2 .\) Hint: First show that the recurrence equation for \(W(n)\) is given by \\[ \begin{array}{l} W(n)=1+W\left(\left[\frac{n}{2}\right]\right) \quad \text { for } n>1 \\ W(1)=1 \end{array} \\] To do this, consider even and odd values of \(n\) separately. Then use induction to solve the recurrence equation.

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.

Implement both the standard algorithm and Strassen's Algorithm on your computer to multiply two \(n \times n\) matrices \(\left(n=2^{h}\right) .\) Find the lower bound for \(n\) that justifies application of Strassen's Algorithm with its overhead.
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