503MidtermExamSolF08

503MidtermExamSolF08 - UML CS 91.503 Midterm Exam Fall,...

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UML CS 91.503 Midterm Exam Fall, 2008 1 of 10 MIDTERM EXAM SOLUTIONS Stats : to be determined later (with ?? points added) - Minimum: - Maximum: - Average: - Standard Deviation: 1: (5 points) Asymptotic Growth of Functions (a) (1 point) List the 4 functions below in nondecreasing asymptotic order of growth: () ( )( ) 2 3 lg 3 lg lg 32 lg n n n n n n n 1) n n 3 2) 2 lg n n 3) n n lg lg 3 4) 32 lg n smallest largest Rationale : 0 3 lim = n n n , so ( ) n n 3 is the smallest. ( ) ) lg lg ( lg lg 3 2 2 2 n n O n n n n = because ) lg lg ( lg 2 n n O n . ( ) n n 5 32 lg = ; this exponential function dominates the other 3 functions. ( ) n n n f n n O n f n n f n n f Ω Ω Θ 3 ) ( 4) ) ) lg (( ) ( 3) ) ( ) ( 2) ) ( ) ( ) 1 4 2 3 3 2 2 1 ( ) 32 lg n n n lg lg 3 3 n ( ) 2 lg n n 2 n ( ) n n 3 f 1 ( n ) f 2 ( n ) f 3 ( n ) f 4 ( n )
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UML CS 91.503 Midterm Exam Fall, 2008 2 of 10 b) (1 point) )) ( ( ) ( 2 1 n f n f Ο TRUE FALSE Proof : ) ( ) ( ) ( ) ( 2 1 2 1 n O n f n n f Θ [1] by the definition of the Θ operator. ) ( 3 2 n O n [2]. Applying transitivity to [1] and [2] yields ) ( ) ( 3 1 n O n f [3]. Now, via transpose symmetry from ) ( ) ( 3 2 n n f Ω we have )) ( ( 2 3 n f O n [4]. Applying transitivity to [3] and [4] yields )) ( ( ) ( 2 1 n f n f Ο . c) (1 point) )) ( ( ) ( 4 3 n f n f Ω TRUE FALSE Counter-example : n n f = ) ( 3 and 3 4 ) ( n n f = . d) (1 point) ) ( ) ( 3 3 n n f Ο TRUE FALSE Proof : ) ) lg (( ) ( 2 3 n n O n f combined transitively with ) ( ) lg ( 3 2 n O n n yields ) ( ) ( 3 3 n n f Ο . e) (1 point) )) ( ( ) ( 4 1 n f n f Ω TRUE FALSE Counter-example : 2 1 ) ( n n f = and 3 4 ) ( n n f = 2: (5 points) Recurrence Find a tight upper bound on the closed-form solution for the following recurrence: n n T n T + = ) 1 ( 3 ) ( where T(n) is constant for sufficiently small n . That is, find a function ) ( n g such that )) ( ( ) ( n g O n T . Solution : The Master Theorem does not apply here. A recursion tree can be used. The tree has 1 + n levels if T (0)=1. ) ( 3 i n i work is done at the i th level, except for the bottom level, where n n T 3 ) 0 ( 3 = work is done (thanks to Jane for pointing out the work at the bottom level). The total work is: = + = + = + = = = = = n n i i n i i n n i i n i i n n i i i n i n i n 3 3 3 3 3 3 3 )) ( 3 ( 1 0 1 0 1 0 1 0 1 0 (using closed-form solutions to the summations): ) 3 ( 4 3 2 3 4 7 3 4 3 3 3 ) 1 ( 2 ) 1 3 ( ) 1 ( n n n n n n O n n n n = + + + . Thus, ( ) n O n T 3 ) ( .
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UML CS 91.503 Midterm Exam Fall, 2008 3 of 10 3: (5 points) Analyze Pseudocode Mystery 1 has one argument: a positive integer value n . ) ( 1 n Mystery n n " with called Mystery1 " print = 1 if n return then 3 to 1 for i ) 4 / ( 2 do n Mystery return ) ( 2 n Mystery n n " with called Mystery2 " print = 1 if n return then ) 4 / ( 1 n Mystery return Derive a tight upper bound on Mystery 1 ’s worst-case asymptotic running time as a function of n . Solution : In the worst case, let n be a power of 4. ). 1 ( ) 16 / ( 3 )) 16 / ( ) 1 ( ( 3 ) 1 ( ) ( Θ + = + Θ + Θ = n T n T n T Case 1 of the Master Theorem applies, yielding ) ( ) ( 3 log 16 n n T Θ = . So, a tight upper bound is ) ( ) ( 3 log 16 n O n T = .
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503MidtermExamSolF08 - UML CS 91.503 Midterm Exam Fall,...

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