Unformatted text preview: Solutions to Quiz 4 Question 1
proc mean(a_1,...,a_n: integers) m:=a_1 for i:=2 to n m:=m+a_i end for m:=m/n end proc {m is the mean} worst case: n operations. Question 2
(a) Is 2n2  7n C3n2 , x > k for some k 1? Let k = 4, then both sides are always positive and definition is 2n2  7n 3Cn2 2n  7 3Cn. For n = 4, 1 12C works for C = 1. So C = 1, k = 4 works. (b) We know 2n2  7n O(3n2 ). Show 2n2  7n (3n2 ) 2n2  7n C3n2 , n > k. Set k = 4 and 2n  7 3Cn, k > 4. For n = 4, 2  3C 7/4. So C = 1/12, k = 4 works. Thus 2n2  7n (3n2 ). (c) No. By defination, 7n Cn2 , n > k. Thus 7n Cn2 and then 7 Cn, which cannot be true for any C and k. Question 3
(a) n comparisons by the first "if". O(n). (b) First pass, (n  1) comparisons. Second pass, (n  2), ... Thus comparisons = n1 k = (n1)n . O(n2 ) k=1 2 1 ...
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This note was uploaded on 06/07/2011 for the course COT 3100 taught by Professor Staff during the Summer '08 term at University of Florida.
 Summer '08
 Staff

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