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Chapter 1: Problem 29

Give a \(\Theta(n \text { lg } n)\) algorithm that computes the reminder when \(x^{n}\) is divided by \(p .\) For simplicity, you may assume that \(n\) is a power of 2 That is, \(n=2^{k}\) for some positive integer \(k\)

Short Answer

Expert verified
By repeatedly squaring the base and halving the exponent, you can compute the remainder when \(x^n\) is divided by \(p\) where \(n = 2^k\) in \(O(\log n)\) time complexity.

Step by step solution

01

Initialization

Start with \(base = x\), \(exponent = n\) and \(result = 1\).
02

Exponentiation

While \(exponent > 0\), perform the following steps: If \(exponent\) is odd, then multiply \(result\) by \(base\) and compute the remainder when dividing by \(p\). If \(exponent\) is even, square the \(base\) and compute the remainder when dividing by \(p\). Then halve the \(exponent\) .
03

Final result

After the loop finishes, \(result\) will hold the value of \(x^n \mod p\).

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Most popular questions from this chapter

Show directly that \(f(n)=n^{2}+3 n^{3} \in \Theta\left(n^{3}\right)\). That is, use the definitions of \(O\) and \(\Omega\) to show that \(f(n)\) is in both \(O\left(n^{3)} \text { and } \Omega\left(n^{3)}\right.\right.\)

Algorithm A performs \(10 n^{2}\) basic operations, and algorithm \(\mathrm{B}\) performs 300 In \(n\) basic operations. For what value of \(n\) does algorithm B start to show its better performance?

Group the following function by complexity category. $$\begin{aligned}&n \ln n \quad(\lg n)^{2} \quad 5 n^{2}+7 n \quad n^{5 / 2}\\\&n ! \quad 2^{n !} \quad 4^{n} \quad n^{n} \quad n^{n}+\ln n\\\&5^{\lg n} \lg (n !) \quad(\lg n) ! \quad \sqrt{n} \quad e^{n} \quad 8 n+12 \quad 10^{n}+n^{20}\end{aligned}$$

Show that the function \(f(n)=\left|n^{2} \sin n\right|\) is in neither \(O(n)\) nor \(\Omega(n)\)

Under what circumstances, when a searching operation is needed, would sequential Search (Algorithm 1.1) not be appropriate?
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