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Unformatted text preview: CSE 200 HW 2 Solutions Russell Impagliazzo with Chris Calabro May 12, 2010 1 Restricted 3SAT Show that 3SAT remains NPcomplete when restricted to formulas where every variable appears in at most three clauses. Since we can verify that a formula has the above property in polynomial time, the restricted 3SAT is still in NP . We reduce from arbitrary 3SAT to the restricted 3SAT as follows. Let Φ be a 3 CNF in variables x 1 , ..x n and with clauses C 1 , ..C m . We construct Φ as follows: Our new set of variables contains a variable x i,j for each 1 ≤ i ≤ n and 1 ≤ j ≤ m . For each clause C j , we replace each variable x i in the clause by x i,j and each negated variable ¬ x i with ¬ x i,j . Call the result C j , and add it to Φ .. Note that x ← y is equivalent to y ∨ ¬ x . Thus, we can express x i, 1 ← x i, 2 ← ...x i,n ← x i, 1 as a set of m clauses of size 2. For each i , we add these m clauses to Φ . These two types of clauses are all the clauses in Φ . Note that each x i,j can only appear up to three times in Φ : once in C j , and once each in the implications between x i,j 1 and x i,j and between x i,j and x i,j +1 (operations in indices being done modulo m ). This gives the reduction. We now need to show that Φ is satisfiable if and only if Φ is. First, assume Φ is satisfiable, by assigning each x i value a i . Then consider the assignment to the variables x i,j that assigns each x i,j value a i . Since each clause C j is satisfied by ~a , and we assign each x i,j the value x i was assigned, each C j is satisfied. Since we give all x i,j the same value, each implication is satsified. Thus, Φ is satisfiable. If Φ is satisfiable by x i,j = a i,j , we must have all a i,j with the same i equal to satisfy the implication clauses. Define a i to be this common value of a i,j . Then the assignment giving x i value a i satisfies Φ, since a i,j satisfies each clause C j and we give the x i the same value as x i,j . Thus, the reduction exactly preserves satisfiability, and hence the restricted version is still NPcomplete. 2 Larger Independent Set The larger independent set problem is, given a graph G and an independent set I in G , is there an independent set of G larger than I ? This problem is in NP , 1 since an independent set larger than I would have size at most n , and could be verified in polynomialtime. To show that it is NPcomplete, we reduce from the Big Independent Set problem: Given G, k , is there an independent set I of size k in G ? Given G = ( V, E ) and k , we let V be V together with k 1 new nodes, w 1 , ..w k 1 . We let E be E together with v, w j for each v ∈ V and 1 ≤ j ≤ k 1....
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 Winter '12
 Edmonds
 Graph Theory, US standard clothing size, independent set

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