HW_Collection - Fall Semester 2008 CS300 Algorithms...

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Unformatted text preview: Fall Semester 2008 CS300 Algorithms Homework #1 (due 10/06) 1. Take the following list of functions and arrange them in ascending order of growth rate. That is, if function g ( n ) immediately follows function f ( n ) in your list, then it should be the case that f ( n ) is O ( g ( n )). n 2 . 5 , √ 2 n, ( n + 10) , 10 n , 100 n , n 2 log n 2. Assume you have functions f and g such that f ( n ) is O ( g ( n )). For each of the following statements, decide whether you think it is true or false and give a proof or counterexample . (a) 2 f ( n ) is O (2 g ( n ) ). (b) f ( n ) 2 is O ( g ( n ) 2 ). 3. Denote the Euclidean distance between two points p and q by dist( p, q ). In the plane we have dist( p, q ) = q ( p x- q x ) 2 + ( p y- q y ) 2 Let P := { p 1 , p 2 , ··· , p n } be a set of n distinct points in the plane; these points are the sites. We define the Voronoi diagram of P as the subdivision of the plane into n cells, one for each site in P , with the property that a point q lies in the cell corresponding to a site p i if and only if dist( q, p i ) < dist( q, p j ) for each p j ∈ P with j = i . (See Fig. 1). The lower bound on time complexity for a sorting problem is Ω( n log n ), where n is the Figure 1: Voronoi diagram for a set of points P input size. Show that the lower bound of time complexity of computing Voronoi diagrams is Ω( n log n ) from the lower bound for the sorting problem. 1 Fall Semester 2008 CS300 Algorithms Solution #1 1. O ( √ 2 n ) ⊂ O ( n + 10) ⊂ O ( n 2 log n ) ⊂ O ( n 2 . 5 ) ⊂ O (10 n ) ⊂ O (100 n ) 2. (a) False; Take f ( n ) = 2 n and g ( n ) = n . (b) True; Since f ( n ) = O ( g ( n )), there exist some c ∈ R + and some n ∈ N such that f ( n ) ≤ c g ( n ) for all n ≥ n . By noticing that f ( n ) and g ( n ) are functions with nonnegative real values, f ( n ) 2 ≤ c g ( n ) f ( n ) ≤ c 2 g ( n ) 2 . Therefore, there always exist c = c 2 and n such that f ( n ) 2 ≤ cg ( n ) 2 for all n ≥ n . 3. Let us denote by V computing the Voronoi diagram of a given set P and by S sorting problem. To show that the lower bound of V is Ω( n log n ), you need to prove that S can be transformed to V in O ( n ). To do that, you have to find a solution of V after modifying the input of S into that of V and transform it to the adequate solution of S . Since the input of S is [ x 1 , x 2 , ··· , x n ] and the one of V is [( x 1 , y 1 ) , ( x 2 , y 2 ) , ··· ( x n , y n )], you can map each x i to ( x i , 0) in O ( n ). After computing V with this transformed input, we are given the Voronoi diagram which consists of n cells. You can find the leftmost cell in O ( n ). Then by visiting a cell next to the current one and reporting x-coordinate of the site of the cell, you can obtain the solution of S , [ x i 1 , x i 2 , ··· , x i n ] in O ( n ). Therefore S ∝ O ( n ) V and the lower bound L V of V is the following: L V = Ω( n log n )- O ( n ) = Ω( n log n ) [ x 1 ,x 2 , · · · x n ] ( x i 1...
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This note was uploaded on 02/04/2010 for the course COMPUTER S cs300 taught by Professor Unkown during the Spring '08 term at Korea Advanced Institute of Science and Technology.

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HW_Collection - Fall Semester 2008 CS300 Algorithms...

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