Coupling Different Integer Encodings for SAT

Abstract

Boolean satisfiability (SAT) solvers have dramatically improved their performance in the last twenty years, enabling them to solve large and complex problems. More recently MaxSAT solvers have appeared that efficiently solve optimisation problems based on SAT. This means that SAT solvers have become a competitive technology for tackling discrete optimisation problems.

A challenge in using SAT solvers for discrete optimisation is the many choices of encoding a problem into SAT. When encoding integer variables appearing in discrete optimisation problems, SAT must choose an encoding for each variable. Typical approaches fix a common encoding for all variables. However, different constraints are much more effective when encoded with a particular encoding choice. This inevitably leads to models where variables have different variable encodings. These models must then be able to couple encodings, either by using multiple encoding of single variables and channelling between the representations, or by encoding constraints using a mix of representations for the variables involved. In this paper we show how using mixed encodings of integers and coupled encodings of constraints can lead to better (Max)SAT models of discrete optimisation problems.

A Proofs

Theorem 1

Unit propagation on the clauses of Eqs. (4) and (6) is propagation complete for constraint $x \in D(x) \wedge x = y \wedge y \in D(y)$ and $\mathcal {L}= DIRECT (x) \cup BINARY (y)$. Hence, it enforces domain consistency on x.

Proof

First we show that unit propagation enforces the exactly-one constraint. If we have $[\![x = d ]\!]$ and $[\![x = d' ]\!]$ true simultaneously, then Eq. (4) will force one binary variable in y to take two values, thus failing. If all variables $[\![x = d' ]\!]$ are set false except one $[\![x = d ]\!]$, then Eq. (6) will set the correct bits of y in the remaining solution (via setting the negation to false). The forward direction will then propagate $[\![x= d ]\!]$.

Next we show that the unit propagation is propagation complete with respect to the binary variables and missing values in the original domain. Suppose $S \subseteq D(x)$ are the remaining values in the domain. Then $\lnot [\![x = d ]\!]$ is set for $d \in D_i(x) - S$. Suppose that there is no support for $[\![\mathsf {bit}(y, k) ]\!]$ in S. Then unit propagation of Eq. (6) will set $\lnot [\![\mathsf {bit}(y, k) ]\!]$.

Finally, we show that unit propagation is propagation complete on the encoding variables. Given the subset of the current assignment literals $L \subseteq \phi$ restricted to the direct encoding variables for x and binary encoding variables of y. Let $S \subset D(x)$ be the set of possible domain values remaining, i.e. $S = \{ d ~|~ d \in D(x), \lnot [\![x = d ]\!] \not \in L \}$. Clearly the direct encoding variables are consistent with this by definition, and if S is a singleton, by the first argument $[\![x = d ]\!]$ is set true. We now consider each bit variable $[\![\mathsf {bit}(y, k) ]\!]$: if $[\![\mathsf {bit}(y, k) ]\!] \in L$ then by Eq. (4) each value in $d \in S$ has $\mathsf {bit}(d, k)$ is true, hence it is supported; similarly if $\lnot [\![\mathsf {bit}(y, k) ]\!] \in L$. Finally, if variable $[\![\mathsf {bit}(y, k) ]\!]$ does not appear in L, then there must be values in S with both truth values, otherwise one of the equations in Eq. (6) would have propagated.

Theorem 2

Unit propagation on the clauses of Eqs. (2) and (7) is propagation complete for constraint $x \in lb(x)..ub(x) \wedge x = y$ and language $\mathcal {L}= ORDER (x) \cup BINARY (y)$. Hence, it enforces the initial bounds on x. Note that these equations do not enforce (nor can the encoding variables represent) domain consistency on the domain of x.

Proof

Given the subset of the current assignment literals $L \subseteq \phi$ restricted to the order encoding variables for x and binary encoding variables for x, we show there is support for each possible value defined by L. Since each bound or bit is supported or propagated, propagation completeness follows. The order literals in L define a range l..u given by $l = \max \{ i ~|~ [\![x \ge i ]\!] \in L \}$ and $u = \min \{ i-1 ~|~ \lnot [\![x \ge i ]\!] \in L \}$. The order consistency Eq. (2) enforces that $\lnot [\![x \ge i ]\!] \in L$ for $i > u$ and $[\![x \ge i ]\!] \in L$ for $i < l$. We show l and u are supported. Suppose that $[\![\mathsf {bit}(x, b) ]\!] \ne \mathsf {bit}(l, b) \in L$ for some bit b. Then there is some stable segment for this bit including l, say $r_l..r_u$. Then the clause $[\![\mathsf {bit}(x, b) ]\!] \vee [\![x < r_l ]\!] \vee [\![x > r_u ]\!]$ will fire enforcing $[\![x > r_u ]\!]$ and hence l cannot be the lower bound. A similar argument applies to the upper bound.

We now show each possible value for each binary encoding variable $[\![\mathsf {bit}(x, b) ]\!]$ is supported in the range l..u. Suppose to the contrary that w.l.o.g. $[\![\mathsf {bit}(x, b) ]\!]$ is not true for any $x \in l..u$. Then there is a stable segment $r_l..r_u$ at least as large as l..u for $\lnot [\![\mathsf {bit}(x, b) ]\!]$. Then the clause $\lnot [\![\mathsf {bit}(x, b) ]\!] \vee [\![x < r_l ]\!] \vee [\![x > r_u ]\!]$ will propagate $\lnot [\![\mathsf {bit}(x, b) ]\!]$, which then must be in L. The same reasoning holds for the negation.

The channelling also enforces the outer bounds on the binary encoding variables. Clearly $d_1 \le l \le u \le d_m$ so the arguments for support of binary encoding variables automatically take into account the initial bounds.

Theorem 3

Unit propagation on the clauses of Eqs. (2) and (8) and clauses for lexicographic encoding of $y \ge lb(x)$ and together with the clause $\lnot [\![x \ge ub(y)+1 ]\!]$ is propagation complete for $x \le y$ for the language $\mathcal {L}= ORDER(x) \cup BINARY(y)$.

Proof

Suppose $\mathcal {D}\wedge x \le y \rightarrow [\![x \le d ]\!]$ we show that unit propagation will enforce this. The initial case given by $y \le ub(x)$ is enforced by the last clause. Let d be the maximum value y can take given $\mathcal {D}$, then we know that $\lnot [\![\mathsf {bit}(y, k) ]\!] \in \mathcal {D}$ for all $k \in \mathcal {B}(y), \mathsf {bit}(d, k) = 0$ otherwise y could take a larger value. The clause Eq. (8) for $[\![x \ge d+1 ]\!]$ has right-hand side $\bigvee _{k \in \mathcal {B}(y), \lnot \mathsf {bit}(d, k)}[\![\mathsf {bit}(y, k) ]\!]$ and hence propagates $\lnot [\![x \ge d+1 ]\!]$ as required.

Let Y be the set of all possibly values that y can take given the $\mathcal {D}$. Suppose $\mathcal {D}\wedge x \le y \rightarrow [\![\mathsf {bit}(y, k) ]\!]$ for some $k \in \mathcal {B}(y)$. We show that unit propagation will enforce this. If x still sits at its initial lower bound this will be forced by the encoding of $y \ge lb(x)$. Otherwise, this propagation was caused by $[\![x \ge d ]\!]$ in the current domain, where $d > \min (Y)$

So clearly (A) $\mathsf {bit}(v, k) = 1$ for all $v \in Y \cap d..\max (y)$. Let (B) $d'= \max \{ d'' ~|~ d'' \in Y, \mathsf {bit}(d'', k) = 0\}$. Clearly such a $d'$ exists otherwise we would already have propagated $[\![\mathsf {bit}(y, k) ]\!]$, and, since $d' < d$, $[\![x \ge d'+1 ]\!]$ is propagated by Eq. (2). We claim the clause 8 for $[\![x \ge d'+1 ]\!]$ will propagate $[\![\mathsf {bit}(y, k) ]\!]$. Suppose to the contrary then there is another bit $k'$ where $\mathsf {bit}(d', k') = 0$ where $\lnot [\![\mathsf {bit}(y, k') ]\!] \not \in \mathcal {D}$. Then $d' + 2^{k'} \in Y$ and either $d' + 2^{k'} \ge d$ contradicting (A), or $d' + 2^{k'} < d$ contradicting (B).

While it is not possible for $\mathcal {D}\wedge x \le y \rightarrow \lnot [\![\mathsf {bit}(y, k) ]\!]$ since a lower bound can never force a negative bit (although this can happen subsequently from the clauses for constraint $y \le ub(y)$).