By Jeff Erickson
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Extra info for Algorithms
N], x): j←1 for i ← 1 to n if A[i] > x B[ j] ← A[i]; j ← j + 1 return B[1 .. j] LIS(A[1 .. n]): if n = 0 return 0 else max ← LIS(prev, A[2 .. n]) L ← 1 + LIS(A, FILTER(A[2 .. n], A)) if L > max max ← L return max The FILTER subroutine clearly runs in O(n) time, so the running time of LIS satisfies the recurrence T (n) ≤ 2T (n − 1) + O(n), which solves to T (n) ≤ O(2n ) by the annihilator method. This upper bound pessimistically assumes that FILTER never actually removes any elements; indeed, if the input sequence is sorted in increasing order, this assumption is correct.
We can improve the cubic running time by observing that both conversion problems boil down to computing the product of a matrix and a vector. The explanation will be slightly simpler if we assume the polynomial has degree n − 1, so that n is the number of coefficients or samples. Fix a sequence j x 0 , x 1 , . . , x n−1 of sample positions, and let V be the n × n matrix where vi j = x i (indexing rows and columns from 0 to n − 1): 1 x0 x 02 · · · x 0n−1 1 x1 x 12 · · · x 1n−1 x2 x 22 · · · x 2n−1 V = 1 .
Your algorithm should return the root and the depth of this subtree. 17 Lecture 1: Recursion [Fa’10] Algorithms The largest complete subtree of this binary tree has depth 2. 18. Consider the following classical recursive algorithm for computing the factorial n! of a non-negative integer n: FACTORIAL(n): if n = 0 return 0 else return n · FACTORIAL(n − 1) (a) How many multiplications does this algorithm perform? (b) How many bits are required to write n! in binary? Express your answer in the form Θ( f (n)), for some familiar function f (n).
Algorithms by Jeff Erickson