Approximate Inference for Clusters in Solution Spaces

depiction of a solution space with several clusters Figure 2: Left: recursive counting by partitioning the solution space. Right: fragmentation of solution clusters. away the number of clusters in which x1 takes both values, 0 and 1. Thus, #clusters(F) = #clusters(F)|x1=0 + #clusters(F)|x1=1− #clusters(F)|x!=0&x1=1. The tricky part is to compute #clusters(F)|x1=v1 for any v1, as this no longer necessarily equals the number of clusters in the simplified formula F |x1=v1 . The reason is possible cluster fragmentation—a single cluster of F could fragment into two or more clusters if we restrict the value of x1 to v1, as depicted in the right pane of Fig. . Thus, one way to prove that Z(−1) is exact on a given domain (e.g., 2-SAT) is to show that there is no cluster fragmentation in that domain, and this indeed forms the main combinatorial argument of our proof of exactness. The second problem that may arise but only in domains with three or more values (e.g., 3-COL) is what we call simultaneous valuation—even if a variable takes three different values, say {R,G,B}, within a cluster, there may not be any valuation of the remaining n− 1 variables for which it can take simultaneously take each subset of values, say, {R,G} in the cluster. Thus, when computing the number of clusters in the simplified formula F |x1∈{R,G}, which is one term we need for applying inclusion-exclusion, we will incorrectly get a zero count from this cluster even when x1 takes both values R and G in this cluster. Therefore, one way to prove exactness of Z(−1) on domains like 3-COL involves showing that we can circumvent the simultaneous valuation issue in that domain by, e.g., requiring that the input graph has a triangle in each of its connected components.