By Nadia Creignou, Daniel Le Berre
This booklet constitutes the refereed lawsuits of the nineteenth overseas convention on idea and purposes of Satisfiability checking out, SAT 2016, held in Bordeaux, France, in July 2016.
The 31 common papers, five device papers awarded including three invited talks have been conscientiously reviewed and chosen from 70 submissions. The papers deal with various points of SAT, together with complexity, satisfiability fixing, satisfiability purposes, satisfiability modulop conception, past SAT, quantified Boolean formulation, and dependency QBF.
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Extra resources for Theory and Applications of Satisfiability Testing – SAT 2016: 19th International Conference, Bordeaux, France, July 5-8, 2016, Proceedings
Example text
For every linear code C with a k log(n)×n parity check matrix there is a CNF formula F in variable sets X and Z such that – the solution set of F projected to X is exactly C, – F has size O(kn3 log(n)2 ), and – F has modular pathwidth at most 2k − 1. Proof. It remains only to show the size bound on F . Note that we have n variables xj and kn log(n) variables zi,j . Moreover, we have kn log(n) constraints Ri . Each of those has 2 log(n) + 1 variables, so it can be encoded by O(n2 ) clauses with O(log(n)) variables each.
In the second case, j < n, and therefore i = i and j = j + 1. In this case, we set Rk = Rk for k = j + 1. We define Rj+1 as follows. Rj+1 = {(q c , a, rb ) | (q c , b, rb ) ∈ Rk , a ∈ π −1 (Ni,j+1 ), (c, a) ∈ H, (b, a) ∈ V }. w w w 1 2 n Then we have that q0 −−→ q1w1 −−→ q2w2 ... −−→ qnwn is an accepting path of A if and only if for each a ∈ Σ with (wj , a) ∈ H and (wj+1 , a) ∈ V , the path w wj w a w w 1 n j+1 q0 −−→ ... −−→ qj j − → qj+1 ... −−→ qnwn is an accepting path of A . wn |w ∈ L(A ), a ∈ Σ, (wj , a) ∈ H, (wj+1 , a) ∈ V }.
If the last automaton Am,n (N, π, V, H) accepts the empty language, then the picture N has no (π, V, H)-solution. On the other hand, if this language is not empty, then we still need to construct a solution. Let Ai = Ai,n (N, π, V, H) be the automaton accepting the boundary set ∂i,n (N, π, V, H). Let γ : Σ × Σ → Σ be a projection which sets γ(a, b) = a for each pair (a, b) ∈ Σ × Σ. In other words, γ erases the second coordinate of each pair (a, b) ∈ Σ × Σ. an . Also, for a string w ∈ Σ n , let A(w) be the minimum LDFA that accepts w, and no other string.