Generating vertices of polyhedra and related monotone ... - Rutcor

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$On the Compound Poisson Distribution. Acta Sci. Math. - Rutcor$

RRR 12-2007 Page 27hence, it is not NP-hard unless each problem from NP is quasi-polynomially solvable. Moreover,verifying each of the following dual identitiesD 1 ∨ ... ∨ D n ≡ C, C 1 ∧ ... ∧ C n ≡ D,for arbitrary monotone DNFs D,D 1 ,...,D n and CNFs C,C 1 ,...,C n is obviously equivalentto dualization too. In contrast, verifying each of identitiesD 1 ∧ ... ∧ D n ≡ C 0 , C 1 ∨ ... ∨ C n ≡ D 0 ,is NP-hard, since they generalize D ∧ C 0 ≡ C 0 and C ∨ D 0 ≡ D 0 , respectively.Finally, let us note that CD 0 is equal to a “dipole” CNF, where all clauses of C arepositive (contain no negations) and all clauses of D 0 = (x 1 ∨ y 1 )...(x k ∨ y k ) are negative(contains only negative literals). Hence, equivalence of claims (0) and (5) implies that SAT isNP-complete already for dipole CNFs, or in other words, that Monotone SAT is NP-complete,a fact that we already made use of in Section 2.3.4.Some statements from Proposition 11 can be naturally reformulated as (NP-hard) generationproblems. For example, (2) D 0 ∨ C ≡ D 0 , calls for enumerating all prime implicantsof a monotone Boolean expression D 0 ∨ C. One can immediately get k of them, x i y i fori = 1,...k, however, it is NP-hard to decide whether the obtained list is complete or it canbe extended. In the next sections we will develop this approach and derive corollaries forreliability theory and vertex enumeration.Now, let us consider two similar dual identitieswhere D 0 and C 0 are defined by (11).D 1 ∧ ... ∧ D n ≡ D 0 , C 1 ∨ ... ∨ C n ≡ C 0 (12)Proposition 12 It is coNP-complete to verify any of the identities of (12) already for monotonequadratic DNFs D 1 ,...,D n and CNFs C 1 ,...,C n .Proof. By duality, it will suffice to prove the first claim only. We will reduce it from SAT.Let C be a CNF (satisfying conditions (i), (ii), and (iii) of Remark 1) of k variables x 1 ,...,x k .Let us substitute y j for each x j and get a monotone CNF C = c 1 ∧ ... ∧ c n .Now for i = 1,...,n let us set in (12) D i = D 0 ∨ c i . Conventionally, we delete the termx j y j from D i if clauses c i contains x j or y j , for i = 1,...,n and j = 1,...,m. Then, byassumption (i), we have D 1 ∧ ... ∧ D n = D 0 if and only if the CNF C is not satisfiable. □Let us remark that the following, more general, identities
Page 28 RRR 12-2007D 1 ∧ ... ∧ D n ≡ D, C 1 ∨ ... ∨ C n ≡ C (13)can be reduced to dualization, and hence, verified in quasi-polynomial time if one of thefollowing conditions is satisfied.Case 1. All DNFs D i (CNFs C i ) are linear, or in other words, they are just clausesD i = c i = x 1 i ∨ ... ∨ x k ii (resp., terms C i = t i = x 1 i ...x k ii ) for i = 1,...,n. In this case (13)is reduced to the form D ≡ C which is exactly dualization.Case 2. The number of terms (resp., clauses) in each of the DNFs D 1 ,...,D n (resp.,CNFs C 1 ,...,C n ) is bounded by a constant. In this case each DNF D i (resp., CNF C i ) canbe efficiently dualized in polynomial time, and the size of each such dual is bounded by apolynomial.We can reformulate Proposition 12 as NP-hardness of the following generation problem:Product of Hypergraphs. To each monotone DNF D let us assign (as usual) a hypergraphH = (V,E) whose vertices are the variables and whose edges are the terms of D.Proposition 13 Given n hypergraphs H i = (V,E i ), the problem of generating all minimalsubsets of V which contain an edge of H i for each i = 1,...,n is NP-hard, even if all H i aregraphs.Proof. We can just translate the previous proof (of Proposition 12) in terms of graphs.Assign a vertex u j (resp., v j ) to each literal x j (resp., y j ) for j = 1,...,k, introduce a newvertex w, and set V = {w,u 1 ,v 1 ...,u k ,v k }. Then for i = 1,...,n, let us assign to quadraticDNF D i = D 0 ∨c i the graph G i = (V,E i ) such that E i = {(u j ,v j ) | j = 1,...,k} ∪ {(w,u j ) |x j ∈ c i } ∪ {(w,v j ) | y j ∈ c i }. By Proposition 12, the product of the obtained n graphsG 1 ,...,G n , or in other words, all minimal subsets of V which contain an edge from E i foreach i = 1,...,n, is NP-hard to generate.□Another corollary of Propositions 12 and 13 is also related to the sets of n graphs inwhich, however, edges, not vertices, are in common. Given n (directed) graphs G i = (V i ,E i )for i = 1,...,n, whose edges are labeled by the same indices E = {e 1 ,...,e m }, generate allminimal subsets of E such that the corresponding edges contain a (directed) cycle in G i foreach i = 1,...,n. It is easy to see that, by Propositions 12, 13, this problem is NP-hard.Let us remark that the problem is open for the case when n is bounded by a constant, inparticular, for n = 2.We can fix a pair of vertices s i ,t i ∈ V i for each i = 1,...,n and consider (directed) (s i ,t i )paths instead of (directed) cycles. The corresponding generation problem remains NP-hard.There are also interesting corollaries of Propositions 12 and 13 in reliability theory. Forinstance, given a directed graph G = (V,E) and n pairs of terminals s i ,t i ∈ V for i = 1,...,n,
Page 2 and 3: Rutcor Research ReportRRR 12-2007,
Page 4 and 5: RRR 12-2007 Page 3one can only hope
Page 6 and 7: RRR 12-2007 Page 5whether the verte
Page 8 and 9: RRR 12-2007 Page 7hypergraph of H,
Page 10 and 11: RRR 12-2007 Page 9with base P and a
Page 12 and 13: RRR 12-2007 Page 11Corollary 1 Let
Page 14 and 15: RRR 12-2007 Page 13Procedure GEN-B(
Page 16 and 17: RRR 12-2007 Page 152.3.3 The perfec
Page 18 and 19: RRR 12-2007 Page 17Question: Is the
Page 20 and 21: RRR 12-2007 Page 19Procedure GEN-C(
Page 22 and 23: RRR 12-2007 Page 21Proof of Theorem
Page 24 and 25: RRR 12-2007 Page 23i.e., go from th
Page 26 and 27: RRR 12-2007 Page 25can be extended.
Page 30 and 31: RRR 12-2007 Page 29consider the pro
Page 32 and 33: RRR 12-2007 Page 31(0) CNF C is sat
Page 34 and 35: RRR 12-2007 Page 33Proposition 16 G
Page 36 and 37: RRR 12-2007 Page 35[9] E. Boros, K.
Page 38 and 39: RRR 12-2007 Page 37[32] L. Khachiya

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