By Jeff Erickson
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N − 1]) STOOGESORT(A[0 .. m − 1]) (b) Would STOOGESORT still sort correctly if we replaced m = 2n/3 with m = 2n/3 ? Justify your answer. 14 Lecture 1: Recursion [Fa’10] Algorithms (c) State a recurrence (including the base case(s)) for the number of comparisons executed by STOOGESORT. (d) Solve the recurrence, and prove that your solution is correct. ] (e) Prove that the number of swaps executed by STOOGESORT is at most n 2 . 9. Describe an algorithm to compute the median of an array A[1 .. 5] of distinct numbers using at most 6 comparisons.
Thus, we can partition the cost function into three parts as follows: r−1 Cost(T, f [1 .. n]) = f [i] · #nodes between left(vr ) and vi i=1 n + f [i] i=1 n + f [i] · #nodes between right(vr ) and vi i=r+1 Now the first and third summations look exactly like our original expression (*) for Cost(T, f [1 .. n]). Simple substitution gives us our recursive definition for Cost: n Cost(T, f [1 .. n]) = Cost(left(T ), f [1 .. r − 1]) + f [i] + Cost(right(T ), f [r + 1 .. n]) i=1 The base case for this recurrence is, as usual, n = 0; the cost of performing no searches in the empty tree is zero.
N−1 2(n−1) 3(n−1) 1 ωn ωn ωn ··· ··· ··· ··· .. ··· 1 ωn−1 n 2(n−1) ωn 3(n−1) ωn .. . 2 ω(n−1) n To invert the discrete Fourier transform, converting sample values back to coefficients, we just have to multiply P ∗ by the inverse matrix V −1 . The following amazing fact implies that this is almost the same as multiplying by V itself: Claim: V −1 = V /n Proof: Let W = V /n. We just have to show that M = V W is the identity matrix. We can compute a single entry in M as follows: n−1 m jk = n−1 v jl · w l k = l=0 j−k If j = k, then ωn ωnjl lk · ωn /n = l=0 1 n−1 n ωnjl−lk l=0 = ω0n = 1, so m jk = 1 n−1 n l=0 7 1= n n = 1, 1 n−1 j−k l = (ω ) n l=0 n Lecture 2: Fast Fourier Transforms [Fa’10] Algorithms and if j = k, we have a geometric series j−k n−1 (ωnj−k )l m jk = l=0 = (ωn )n − 1 j−k ωn −1 = (ωnn ) j−k − 1 j−k ωn −1 = 1 j−k − 1 j−k ωn −1 = 0.