By Jeff Erickson

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**Extra resources for Algorithms**

**Example text**

One way to do this, at least when n is a power of two, is to split the pixel map into four n/2 × n/2 blocks, move each block to its proper position using a sequence of five blits, and then recursively rotate each block. Alternately, we could first recursively rotate the blocks and then blit them into place. (a) Prove that both versions of the algorithm are correct when n is a power of two. (b) Exactly how many blits does the algorithm perform when n is a power of two? (c) Describe how to modify the algorithm so that it works for arbitrary n, not just powers of two.

A) How many cells are there, as a function of n? Prove your answer is correct. (b) In the worst case, exactly how many cells can a horizontal line cross, as a function of n? Prove your answer is correct. Assume that n = 2k − 1 for some integer k. ] (c) Suppose we have n points stored in a kd-tree. Describe and analyze an algorithm that counts the number of points above a horizontal line (such as the dashed line in the figure) as quickly as possible. ] (d) Describe an analyze an efficient algorithm that counts, given a kd-tree storing n points, the number of points that lie inside a rectangle R with horizontal and vertical sides.

But there’s a degree of freedom we haven’t exploited—We get to choose the sample positions! Our conversion algorithm may be slow only because we’re trying to be too general. If we choose a set of sample positions with the right recursive structure, we can perform this conversion more quickly. 4 Divide and Conquer Any polynomial of degree n−1 can be expressed as a combination of two polynomials of degree (n/2)−1 as follows: p(x) = peven (x 2 ) + x · podd (x 2 ). The coefficients of peven are just the even-degree coefficients of p, and the coefficients of podd are just the odd-degree coefficients of p.

### Algorithms by Jeff Erickson

by Christopher

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