4 Let Set Double Cover Decision Problem Defined Input Set Y Set H Subsets Y Y U S Integer Q43806824
4. Let Set-Double-Cover be the decision problem defined as: Input: A set Y and a set H of subsets of Y with Y = U S, and an integer l. SEH Output: “YES” if there exists a D CH with |D <l such that every ye Y belongs to at least two sets in D; “NO” otherwise. Give a polynomial-time reduction from the decision problem Set-Cover (defined on page 26 in the notes from Lecture 12) to Set-Double-Cover. (2 marks) Example 2: The Minimum Set Cover Problem, cont. Example: An instance (X, F) of the Minimum Set Cover Problem, where X consists of 12 points and F = {S1, S2, S3, S4, S5, S6} E.g., {S1, S3, S4, S5} is a set cover of F of size 4. (Not a minimum set cover of F.) COMP3011 Lecture 12 25 / 46 Example 2: The Minimum Set Cover Problem, cont. Decision problem corresponding to the Minimum Set Cover Problem: Set-Cover Input: A set X and a set F of subsets of X with X= U S, and an integer k. SEF Output: “YES” if there exists a set cover of F of size at most k; “NO” otherwise. Theorem Set-Cover is NP-complete. Proof: • Set-Cover ENP because for any candidate set cover C of F, it can be checked in polynomial time that C CF, C<k, and that US contains all of X. SEC • To prove that each problem in N P can be reduced to Set-Cover, give a polynomial-time reduction from the NP-complete problem Vertex-Cover: For any instance (G= (V, E);j) of Vertex-Cover, create an instance (X;F; k) of Set-Cover by defining S(v) = {e e E: e contains v} for every ve V and taking X = E, F = {S(v):VEV} , and k=j. This takes polynomial time. Then it can be shown that (Exercise 35.3-2) G has a vertex cover of size <j if and only if F has a set cover of size <k. COMP3011 Lecture 12 26 / 46 Show transcribed image text 4. Let Set-Double-Cover be the decision problem defined as: Input: A set Y and a set H of subsets of Y with Y = U S, and an integer l. SEH Output: “YES” if there exists a D CH with |D
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