(Solved):Consider the problem of computing N! = 1 2 3 N. (a) If N is an n-bit number, how many bits long is N!, approximately (in () form)? (b) Give an algorithm to compute N! and analyze its running time. View Answer…

 

Question

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

(a) If N is an n-bit number, how many bits long is N!, approximately (in () form)?

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

 

EXPERT ANSWER

Part a:

Using Stirling’s approximation to the factorial, N! is approximately equal to {eq}sqrt{(2 pi N) N^N/e^N}

{/eq}, so:

{eq}log_2(N!) log_2((2 pi N)^{1/2}N^N / e^N)

{/eq}

= {eq}(1 + log_2 pi + log_2 N)/2

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