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Let vi denote the node that stores A[i], and let r be the index of the root node. Ignoring constant factors, the cost of searching for A[i] is the number of nodes on the path from the root vr to vi . Thus, the total cost of performing all the binary searches is given by the following expression: n Cost(T, f [1 .. n]) = f [i] · #nodes between vr and vi i=1 Every search path includes the root node vr . If i < r, then all other nodes on the search path to vi are in the left subtree; similarly, if i > r, all other nodes on the search path to vi are in the right subtree.

A1   y1  x 1n−1          x 2n−1    a2  =  y2  .  .   .  ..   .   .  .   .   .  n−1 x n−1 an−1 yn−1 x n−1 2 ··· x n−1 a0 y0 Given this formulation, we can clearly transform any coefficient vector a into the corresponding sample vector y in O(n2 ) time. 3 But we can speed this up by implicitly hard-coding the sample positions into the algorithm, To convert from samples to coefficients, we can simply multiply the sample vector by the inverse of V , again in O(n2 ) time.

6 Lecture 3: Backtracking [Fa’10] Algorithms This recursive definition can be translated mechanically into a recursive algorithm, whose running time T (n) satisfies the recurrence n T (n) = Θ(n) + T (k − 1) + T (n − k) . k=1 n The Θ(n) term comes from computing the total number of searches i=1 f [i]. Yeah, that’s one ugly recurrence, but it’s actually easier to solve than it looks. To transform it into a more familiar form, we regroup and collect identical terms, subtract the recurrence for T (n − 1) to get rid of the summation, and then regroup again.

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