Consider the problem of computing N!=1·2·3···N.

(a) If Nis an role="math" localid="1658397956489" n-bit number, how many bits long is N!, approximately ( inΘ(·)form)?

(b) Give an algorithm to compute N!and analyze its running time.

Short Answer

Expert verified

The running time complexity of the given algorithm can be analyzed as it runs for n iteration and that is multiplied and the final time complexity O(NlogN).

Step by step solution

01

Introduction

The factorial can be defined as the product of all the numbers between 1 and the given number including the given number. It can be represented as an integer with an exclamation mark.

02

Calculation of bits long with N!

We have,

N!=1·2·3···N.

It is known that,

Log2(N!)=Θ(Nlog2N)

So, N-factorial can be written as,

N!=Θ(Nlog2N)

It can be written as,Θ(n2n)

Given that, N is a n-bit number, then

N!isΘ(n2n)bitslong.

03

Algorithm to compute N!

Normal algorithm flow is shown below,

Factorial(N)If(N==0)Return1ElseReturnNfactorial(N-1)

The running time complexity of the given algorithm can be analyzing as it runs for n iteration and that are multiplied and the final time complexity O(NlogN).

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