HW7_Solution

# HW7_Solution - HW7 SOLUTION 1 Chapter 8 Problem 2 First to...

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Unformatted text preview: HW7 SOLUTION 1. Chapter 8 Problem 2: First, to see that the problem is in NP , we notice that given a customer set X with size k, we can verify that if these customers is diverse in time O ( kn ). To show NP-hardness, we reduce from Set Packing problem. Given a set U of n elements, a collection S 1 ,...,S m of subsets of U and a number k ,for each set S i = { p 1 ,...,p j } , we have a customer i purchasing p 1 ,...,p j If there exist a collection i 1 ,...,i k of those S i with the property that no two of them intersect, then X = { i 1 ,...,i k } is the diverse set of { 1 ,...,m } since S i j T S i j = ∅ if i j 6 = i j ,i j ,i j ∈ X Conversely, if X is a solution to the Diverse Set problem, then for any i,j ∈ X,i 6 = j , by the definition of diversity, S i T S j = ∅ , thus X is a collection of S 1 ,...,S m with the property that no two of them intersect. This proves that the reduction is correct and it’s obvious that it is done in polynomial time. Problem 4 : For each process i = 1 ,...n , we let R i ⊆ { 1 ,...,m } denote the set of resource that process i requests. The question is then : is there a collection of k of these sets R i that are pairwise disjoint? (a) The Resource Reservation problem is in NP because given a set of k process, we merely have to check that their corresponding sets R i have pairwise empty intersections, which clearly works in polynomial time. Notice that this immediately implies that all special cases in parts ( b )- ( d ) are also in NP We will show in part ( d ) that that special case is NP - hard already, so the general problem is also NP - hard (b) when k = 2, all we need to do is to find if there are two sets with empty...
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HW7_Solution - HW7 SOLUTION 1 Chapter 8 Problem 2 First to...

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