e u w E U must contain u orw otherwise v e has no neighbor in U ThusU is a

E u w e u must contain u orw otherwise v e has no

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e = { u , w } ∈ E , U must contain u or w (otherwise v e has no neighbor in U ). Thus U is a vertex cover in G . COMP 6651 / Fall 2013 , Dr. B. Jaumard 35
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. . . . . . . . . . . . . Definitions . . . . . . . . . Problem Reduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . NP-Completeness Proofs . Readings and References COMP 6651 / Fall 2013 , Dr. B. Jaumard 36
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. . . . . . . . . . . . . Definitions . . . . . . . . . Problem Reduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . NP-Completeness Proofs . Readings and References Dominating Set (5/5) Let U be a vertex cover of G . We show that it is also a dominating set in G . For v V \ U , there are two possibilities. If v V then, since it is not an isolated vertex, there is an edge { u , v } ∈ E . Since U is a vertex cover and v ̸∈ U , we have u U , so v has a neighbor in U . v = v e for some edge e = { u , w } ∈ E . Since U is a vertex cover, u U or w U , so v has a neighbor in U . This reduction can be performed in polynomial time , since it requires only the construction of G from G . COMP 6651 / Fall 2013 , Dr. B. Jaumard 37
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. . . . . . . . . . . . . Definitions . . . . . . . . . Problem Reduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . NP-Completeness Proofs . Readings and References 3SAT (1/5) . 3SAT . . . . . Given a Boolean expression in CNF such that each clause contains exactly three variables, determine whether it is satisfiable . Theorem . . . . . 3SAT is NP-complete COMP 6651 / Fall 2013 , Dr. B. Jaumard 38
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. . . . . . . . . . . . . Definitions . . . . . . . . . Problem Reduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . NP-Completeness Proofs . Readings and References 3SAT (1/5) . 3SAT . . . . . Given a Boolean expression in CNF such that each clause contains exactly three variables, determine whether it is satisfiable . Theorem . . . . . 3SAT is NP-complete COMP 6651 / Fall 2013 , Dr. B. Jaumard 39
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. . . . . . . . . . . . . Definitions . . . . . . . . . Problem Reduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . NP-Completeness Proofs . Readings and References 3SAT (2/5) A solution to 3SAT can be used to solve the regular SAT. Firstly, 3SAT clearly belongs to NP: We can guess a truth assignment and verify that it satisfies the expression in polynomial time. E arbitrary instance of SAT. Let us show that we can replace each clause of E with several clauses, each of which has exactly three variables C = x 1 x 2 ... x k arbitrary clause of E such that k 4 (positive literal for convenience of notation) Introduce new variables y 1 , y 2 , ..., y k - 3 C = ( x 1 x 2 y 1 ) ( x 3 y 1 y 2 ) ( x 4 y 2 y 3 ) ... ( x k - 1 x k y k - 3 ) COMP 6651 / Fall 2013 , Dr. B. Jaumard 40
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. . . . . . . . . . . . . Definitions . . . . . . . . . Problem Reduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . NP-Completeness Proofs . Readings and References 3SAT (3/5) C = x 1 x 2 ... x k arbitrary clause of E such that k 4 (positive literal for convenience of notation) C = ( x 1 x 2 y 1 ) ( x 3 y 1 y 2 ) ( x 4 y 2 y 3 ) ... ( x k - 1 x k y k - 3 ) Claim : C is satisfiable if and only if C is satisfiable. If C is satisfiable , then one of the x i ’s must be set to 1. In that case, we can set the values of the y i ’s in C such that all clauses in C are satisfied as well. Example: If x 3 = 1, then we set y 1 = 1 (takes care of first clause), y 2 = 0 (takes care of second clause) and the rest of the y i s to 0. In general: if x i = 1, set y 1 , y 2 , ..., y i - 2 to be 1, and the rest to be 0, which satisfies C .
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  • Fall '09
  • Computational complexity theory, NP-complete problems, NP-complete, Professor B. Jaumard

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