Unsatisfiable (k,(4*2^k/k))-CNF formulas

arXiv:0810.1904v1 [cs.DM] 10 Oct 2008 Unsatisfiable CNF-formulas Heidi Gebauer ∗ November 17, 2018 Abstract A Boolean formula in a conjunctive normal form is called a (k, s)-formula if every clause contains exactly k variables and every variable occurs in at most s clauses. We show that there are unsatisfiable (k, 4 · 2k k )-CNF formulas. 1 A better bound for unsatisfiable formulas Theorem 1.1 For every sufficiently large k there is an unsatisfiable (k, 4 · 2k k )-CNF. Note that due to Kratochv´ıl, Savick´y and Tuza [2] every (k, 2k ek)-CNF is satisfiable. So our result shows that this bound is tight up to a factor 4e. Proof: We consider the class C of hypergraphs G whose vertices can be arranged in a binary tree TG such that every hyperedge of G is a path of TG. For positive integers k, s ≥1 we denote by a (k, s)-tree a k-uniform hypergraph G ∈C such that • every full branch of TG contains a hyperedge of G and • every vertex of TG belongs to at most s hyperedges of G When there is no danger of confusion we write G for TG. The following lemma is the core of our proof. Lemma 1.2 For every sufficently large k there is a (k, 2 · 2k k )-tree G. We first show that Lemma 1.2 implies Theorem 1.1. Suppose that there is a (k, 2 · 2k k )-tree G and let G′ be a copy of G. Let H be the hypergraph obtained by generating a new root v and attaching G as a left subtree and G′ as a right subtree. Note that H is a (k, 2 · 2k k )-tree as well. Let (x1, x′ 1), (x2, x′ 2), . . . , (xr, x′ r) denote the pairs of siblings of H. We set x′ i := ¯xi for every i, i = 1, . . . , r (i.e. each non-root vertex represents a literal x ∈{x1, ¯x1, x2, ¯x2, . . . , xr, ¯xr}). Let E(H) ∗Institute of Theoretical Computer Science, ETH Zurich, CH-8092 Switzerland. Email: gebauerh@inf.ethz.ch. 1 denote the set of hyperedges of H. Then for every hyperedge {y1, y2, . . . , yn} ∈E(H) we form the clause C{y1,y2,...,yn} = (y1 ∨y2 ∨. . . ∨yn) and set F := V e∈E(H) Ce. Note that every variable xi of F occurs in at most 2 · ∆(F) clauses with ∆(F) denoting the maximum degree a variable in F. Indeed, the number of occurrences of the variable xi is bounded by the number of occurrences of the literal xi plus the number of occurrences of the literal ¯xi, which is at most 2∆F. So F is a (k, 2 · 2k k )-CNF. It remains to show that F is not satisfiable. Let α be an assignment to {x1, . . . , xr}. Observation 1.3 Note that there is (at least) one full branch bfull of H such that all literals along bfull are set to FALSE by α. By assumption bfull contains a hyperedge h. But α does not satisfy the clause Ch, implying that α does not satisfy F. Since α was chosen arbitrarily, F is not satisfiable. □ It remains to prove our key lemma. Proof of Lemma 1.2: We need some notation first. The vertex set and the hyperedge set of a hypergraph H are denoted by V (H) and E(H), respectively. By a slight abuse of notation we consider E(H) as a multiset, i.e. every hyperedge e can have a multiplicity greater than 1. By a bottom hyperedge of a tree TH we denote a hyperedge covering a leaf of TH. Let d = 2k k . For simplicity we assume that k is a power of 2, implying that d is power of 2 as well. To construct the required hypergraph G we establish first a (not necessarily k-uniform) hyper- graph H and then successively modify its hyperedges and TH. The following lemma is about the first step. Lemma 1.4 There is a hypergraph H ∈C with maximum degree 2d such that every full branch of TH has 2i bottom hyperedges of size log d + 1 −i for every i with 0 ≤i ≤log d. Proof of Lemma 1.4: Let T be a binary tree with log d + 1 levels. In order to construct the desired hypergraph H we proceed for each vertex v of T as follows. For each leaf descendant w of v we let the path from v to w be a hyperedge of multiplicity 2l(v) where l(v) denotes the level of v. Figure 1 shows an illustration. The construction yields that each full branch of TH has 2i bottom hyperedges a b c Figure 1: An illustration of H for d = 4. The hyperedge {a, b, c} has multiplicity 1, {b, c} has multiplicity 2 and {c} has multiplicity 4. 2 of size log d + 1 −i for every i with 0 ≤i ≤log d. So it remains to show that d(v) ≤2d for every vertex of v ∈V (T). Note that every vertex v has 2log d−l(v) leaf descendants in TH, implying that v is the start node of 2log d−l(v) · 2l(v) ≤d hyperedges. So the degree of the root is at most d ≤2d. We then apply induction. Suppose that d(u) ≤2d for all nodes u with l(u) ≤i −1 for some i with 1 ≤i ≤log d and let v be a vertex on level i. By construction exactly half of the hyperedges containing the ancestor of v also contain v itself. Hence v occurs in at most 1 2 · 2d = d hyperedges as non-start node. Together with the fact that v is the start node of at most d hyperedges this implies that d(v) ≤d + d ≤2d. □ The next lemma deals with the second step of the construction of the required hypergraph G. Lemma 1.5 There is a hypergraph H′ ∈C with maximum degree 2d such that each full branch of TH′ has 2i botto

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