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

RRR 12-2007 Page 21Proof of Theorem 3. Let U = {1,...,|U|}, and each hyperedge H in an interval hypergraphH is given by I H = [L H : R H ], where L H ≤ R H . Let X = {[L 1 : R 1 ],...,[L k : R k ]}be a sub-hypergraph of H, i.e., X ⊆ H. If X is a minimal cover, then L i ≠ L j clearly holdsfor all i and j with i ≠ j, since no interval in X contains another. Moreover, we have thefollowing characterization.Proposition 9 Let X = {[L 1 : R 1 ],...,[L k : R k ]} be a sub-hypergraph of H ⊆ 2 U , such thatL 1 < L 2 < ... < L k . Then X is a minimal cover if and only if (i) L i+1 ≤ R i + 1 < L i+2 , fori = 1,...,k − 2, (ii) L k ≤ R k−1 + 1 < R k + 1, and (iii) [L 1 : R k ] = U.Proof. Suppose that X ⊆ H be a minimal cover. Then the minimality of X impliesR k−1 < R k . Note also that, for all i = 1,...,k − 1, we have L i+1 ≤ R i + 1, for otherwise thepoint R i + 1 cannot be covered by any interval in X. Furthermore, for i = 1,...,k − 2, wehave L i+2 > R i + 1, for otherwise, [L i : R i ] ∪ [L i+2 : R i+2 ] ⊇ [L i+1 : R i+1 ], contradicting theminimality of X. Since X is a cover of U, (i) and (ii) imply [L 1 : R k ] = U.Conversely, if X = {[L 1 : R 1 ],...,[L k : R k ]} satisfies properties (i)-(iii) stated in theproposition, then it is not difficult to see that X is a minimal cover of U.□Let F be the family of minimal collections of intervals in H that cover U, and let F ′be the family of all collections X = {[L 1 : R 1 ],...,[L k : R k ]} of intervals in H satisfyingproperties (i)-(ii) of Proposition 9 such that L 1 = 1. By definition, we have F ⊆ F ′ , and anyX ∈ F ′ is a minimal cover of [1 : R k ] ⊆ U (= {1,...,|U|}). In our backtracking procedure,we shall always choose S 1 from F ′ .Given two disjoint subsets S 1 = {[L 1 : R 1 ],...,[L k : R k ]} ∈ F ′ and S 2 ⊆ H, any intervalI = [L k+1 : R k+1 ] ∈ H\(S 1 ∪S 2 ) satisfies S 1 ∪{I} ∈ F ′ if and only if R k−1 +1 < L k+1 ≤ R k +1and R k+1 > R k . Furthermore, by Proposition 9, the check whether there is an X ∈ F suchthat X ⊇ S 1 ∪ {I} and X ∩ S 2 = ∅, can be performed in O(|H|) by checking if⋃{H ∈ H \ (S1 ∪ S 2 ) | L H > R k + 1} ⊇ [R k+1 + 1 : maxU]. (9)Since the depth of the backtracking tree is at most |U| by Proposition 9, the theorem follows.□3.2 Enumerating minimal cut conjunctionsLet T = (V,E) be a directed tree with a vertex set V and an arc set E, and let P = {(s i ,t i ) |s i ,t i ∈ V, for i = 1,...,n} be a collection of directed paths in T, defined by source-sinkpairs. To apply the generation algorithm described in the previous section, we let W = E,and for each X ⊆ E, we let π be the property that the graph (V,E\X) has no path between