BalancedK-satisfiability and biased randomK-satisfiability on trees

Abstract

We study and solve some variations of the random K-satisfiability (K-SAT) problem—balanced K-SAT and biased random K-SAT—on a regular tree, using techniques we have developed earlier. In both these problems as well as variations of these that we have looked at, we find that the transition from the satisfiable to the unsatisfiable regime obtained on the Bethe lattice matches the exact threshold for the same model on a random graph for K = 2 and is very close to the numerical value obtained for K = 3. For higher K, it deviates from the numerical estimates of the solvability threshold on random graphs but is very close to the dynamical one-step-replicasymmetry-breaking threshold as obtained from the first nontrivial fixed point of the survey propagation algorithm.

An alternate view of complexity in k-SAT problems

“…In this paper, we obtain instead the same expression for α s by calculating directly the fraction of literal assignments that have solutions. We demonstrate how to do this for both k-SAT as well as k-NAE-SAT, building on previous work [16,17], where we computed the fraction of satisfiable literal assignments exactly on trees. We outline a procedure, equivalent to performing this calculation now on a Bethe lattice, which gives us an easy way to derive an analog of the complexity for this whole class of problems.…”

“…The boundaries, or leaf-nodes, are assigned a fixed value 0 or 1 randomly, and have a degree = 1. Every node on this tree (except the boundary nodes) is the root of its subtree and considered as such, we can vary the literal assignment of the edges on this subtree and calculate the fraction of instances P 0 in which this node can take no value (since either value would violate a clause that it participates in), only one value (0 or 1) P 1 or both values P 2 [16,17]. Clearly P 0 + P 1 + P 2 = 1.…”

“…In general, for a tree, P 0 will depend on the level of a node (its distance from the leaves, with the leaves at level 0). For an infinite tree, the quantity P 0 eventually becomes independent of the level of the node (except for the central node which has d+1 subtrees) and the value it takes is given by the fixed point of the tree recursions [16,17]. These tree-recursions are written for the quantity Q; Q ≡ P1 2(1−P0) is the fraction of instances (taken only over all satisfiable instances on the sub-tree) in which the root can only take the one value not satisfying the clause connecting it to the node at the higher level, for any literal assignment.…”