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Assignment 4 Solutions - COMP 251 Autumn 2013 Solutions to...

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COMP 251 Autumn 2013 Solutions to Assignment 4 Gheorghe Comanici Prakash Panangaden Question 1 [20 points] This problem is solved using a greedy algorithm, building the solution from east to west. We start with a list of all the houses. Find the westmost house, and place a base station 4 miles to its east . Remove all the houses covered by this station from the list of houses. Repeat the process until the problem has no more houses. Does our solution stay ahead of other solutions? Suppose somebody claims that some other solution S = { s 1 ,s 2 ,...,s k } is a better than our solution T = { t 1 ,t 2 ,...,t r } (where the stations are sorted from west to east). That is, k <r and both S and T cover all houses. Note that both s 1 and t 1 have to cover the west-most house h , but we chose t 1 as the east most point to cover h , therefore t 1 s 1 Assume t i s i . Then it must be the case that all houses covered { s 1 ,s 2 ,...,s i } will also be covered by { t 1 ,t 2 ,...,t i } , so if we add s i +1 to { t 1 ,t 2 ,...,t i } there won’t be any houses uncovered between t i and s i +1 . But we choose t i +1 as large as possible to cover these houses, so it must be the case that t i +1 s i +1 .
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