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Unformatted text preview: CSE 101 Midterm Solutions Chris Calabro July 23, 2009 1 Recurrence relation Suppose T (1) = 1 and for n > 1, T ( n ) = 4 T ( n 2 ) + 1. Claim 1. If n is a power of 2 , then T ( n ) = 4 3 n 2 1 3 . Proof. We use induction on n . If n = 1, then T (1) = 1 = 4 3 · 1 2 1 3 . If n > 1, then by the induction hypothesis, T ( n ) = 4 4 3 n 2 2 1 3 + 1 = 4 3 n 2 1 3 . One could also use the Master theorem to find the asymptotic growth rate of T , but that is not as specific as the above. 2 Median of 2 sorted lists Let X = ( X 1 ,...,X n ) ,Y = ( Y 1 ,...,Y n ) be 2 sorted lists of n elements each. For the purposes of this problem, the median of a list of size 2 n is the n th smallest element. The problem is to find the median of X ∪ Y . Let’s generalize the problem a little to finding the k th smallest element of 2 lists, X of size m and Y of size n . If m = 0, then this is Y k . If n = 0, then this is X k . But otherwise, m,n ≥ 1. If X d m 2 e > Y d n 2 e , then swap X,Y (and m,n ). So without loss of generality,)....
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This note was uploaded on 03/16/2010 for the course CSE 101 taught by Professor Staff during the Winter '08 term at UCSD.
 Winter '08
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