Ps1sol - CS161 Design and Analysis of Algorithms Summer 2004 Problem Set#1 Solutions General Notes Regrade Policy If you believe an error has been

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CS161: Design and Analysis of Algorithms Summer 2004 Problem Set #1 Solutions General Notes Regrade Policy: If you believe an error has been made in the grading of your problem set, you may resubmit it for a regrade. If the error consists of more than an error in addition of points, please include with your problem set a detailed explanation of which problems you think you deserve more points on and why. We reserve the right to regrade your entire problem set, so your Fnal grade may either increase or decrease. If you use pseudocode in your assignment, please either actually use psuedocode or include a written explanation of your code. The TA does not want to parse through C or JAVA code with no comments. If you fax in your problem set, please make sure you write clearly and darkly. Some problem sets were very di±cult to read. Also, make sure that you leave enough margins so that none of your work is cut o². The version of the Master Theorem that was given in lecture is slightly di²erent from that in the book. We gave case 2 as f ( n ) = Θ( n log b a log k n ) = T ( n ) = Θ( n log b a log k +1 n ) for k 0 On exams, using the Master Theorem is normally quicker than other methods. But, remember that cases 1 and 3 only apply when f ( n ) is polynomially smaller or larger, which is di²erent from asymptotically smaller or larger. 1. [16 points] Ordering By Asymptotic Growth Rates Throughout this problem, you do not need to give any formal proofs of why one function is Ω, Θ, etc. .. of another function, but please explain any nontrivial conclusions. (a) [10 points] Do problem 3-3(a) on page 58 of CLRS. Rank the following functions by order of growth; that is, Fnd an arrangement g 1 , g 2 , . . . , g 30 of the functions satisfying g 1 = Ω( g 2 ) , g 2 = Ω( g 3 ) , . . . , g 29 = Ω( g 30 ). Partition your list into equivalence classes such that f ( n ) and g ( n ) are in the same class if and only if f ( n ) = Θ( g ( n )). lg(lg * n ) 2 lg * n ( 2) lg n n 2 n ! (lg n )! ( 3 2 ) n n 3 lg 2 n lg( n !) 2 2 n n 1 / lg n lnln n lg * n n · 2 n n lg lg n ln n 1 2 lg n (lg n ) lg n e n 4 lg n ( n + 1)! lg n lg * (lg n ) 2 2 lg n n 2 n n lg n 2 2 n +1
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Problem Set #1 Solutions 2 Answer: Most of the ranking is fairly straightforward. Several identities are helpful: n lg lg n = (lg n ) lg n n 2 = 4 lg n n = 2 lg n 2 2 lg n = n 2 / lg n 1 = n 1 / lg n lg * (lg n ) = lg * n - 1 for n > 1 In addition, asymptotic bounds for Stirling’s formula are helpful in ranking the expressions with factorials: n ! = Θ( n n +1 / 2 e - n ) lg( n !) = Θ( n lg n ) (lg n )! = Θ((lg n ) lg n +1 / 2 e - lg n ) Each term gives a diFerent equivalence class, where the > symbol means ω .
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Ps1sol - CS161 Design and Analysis of Algorithms Summer 2004 Problem Set#1 Solutions General Notes Regrade Policy If you believe an error has been

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