Unit Propagation by Means of Coordinate-Wise Minimization

Abstract

We present a novel theoretical result concerning the applicability of coordinate-wise minimization on the dual problem of linear programming (LP) relaxation of weighted partial Max-SAT that shows that every fixed point of this procedure defines a feasible primal solution. In addition, this primal solution corresponds to the result of a certain propagation rule applicable to weighted Max-SAT. Moreover, we analyze the particular case of LP relaxation of SAT and observe that coordinate-wise minimization on the dual problem resembles unit propagation and also has the same time complexity as a naive unit propagation algorithm. We compare our theoretical results with max-sum diffusion which is a coordinate-wise minimization algorithm that is used to optimize the dual of the LP relaxation of the Max-Sum problem and can in fact perform a different kind of constraint propagation, namely deciding whether a given constraint satisfaction problem (CSP) has non-empty arc consistency closure.

A Proof of Theorem 2

All of the following propositions assume that y is an ILM of (7) with $\mathcal {H} = \emptyset$, $c \in \mathcal {S}$ is an arbitrary clause, and $i \in [n]$ is also arbitrary.

Proposition 2

If $S_{ci} = 1$ and $S_i^Ty > 0$, then $w_c \ge y_c$.

Proof

Proof by contradiction – assume $w_c < y_c$. From $S_i^Ty > 0$, it follows from definition of $k'_{ci}$ that $S_{ci}y_c + k_{ci}' > 0$, thus $y_c > -S_{ci}k_{ci}'$. Since there are two breakpoints ($-S_{ci}k_{ci}'$ and $w_c$) that are strictly lower than $y_c$ and $w_c > 0$, $y_c$ is not in the relative interior of optimizers by reasoning in Sect. 2.1.    $\square$

Proposition 3

If $S_{ci} \ne 0$ and $S_i^Ty=0$, then $w_c \ge y_c$.

Proof

Following the reasoning in Sect. 2.1, $y_c$ should be in the relative interior of (9) for two smallest breakpoints $b_1,b_2$. Since $y_c$ is equal to breakpoint $b_j = -S_{ci}k'_{ci}$, it must hold that $b_j = b_2 = b_1$ (or $b_j = b_2 = 0 \ge b_1$, which trivially leads to $y_c = 0 < w_c$), otherwise $y_c$ is not in the relative interior of optimizers. Since $w_c>0$ is also a breakpoint, it must be greater or equal to the smallest breakpoint $b_1$, this yields $w_c \ge b_1 = b_j = -S_{ci}k'_{ci} = y_c.$ $\square$

Proposition 4

If $S_{ci} = -1$, $|c| = 1$, and $S_i^Ty \ge 0$, then $w_c \le y_c$.

Proof

Function (7a) restricted to $y_c$ has two breakpoints, $-S_{ci}k_{ci}'$ and $w_c$. Since $S_i^Ty = -y_c + k_{ci}' \ge 0$, it holds that $y_c \le -S_{ci}k_{ci}'$. By the reasoning in Sect. 2.1, it can either hold that $w_c< y_c < -S_{ci}k_{ci}'$ or $w_c = y_c = -S_{ci}k_{ci}'$, hence $w_c \le y_c$.     $\square$

Proposition 5

If $S_{ci} = -1$ and $S_i^Ty < 0$, then $w_c \ge y_c$.

Proof

Similar to Proposition 2. It holds that $y_c > -S_{ci}k_{ci}'$, which follows from $S_{ci} = -1$ and $S_i^Ty < 0$. Assuming $w_c < y_c$ leads to a contradiction.    $\square$

Proposition 6

If $S_{ci} = 1$, $|c| = 1$, and $S_i^Ty \le 0$, then $w_c \le y_c$.

Proof

Similarly to Proposition 4, function (7a) restricted to $y_c$ has two breakpoints, $-S_{ci}k_{ci}'$ and $w_c$. Since $S_i^Ty = y_c + k_{ci}' \le 0$, it holds that $y_c \le -S_{ci}k_{ci}'$. By the reasoning in Sect. 2.1, it can either hold that $w_c< y_c < -S_{ci}k_{ci}'$ or $w_c = y_c = -S_{ci}k_{ci}'$, hence $w_c \le y_c$.    $\square$

Proof

(Theorem 2). We will prove the two parts of Theorem 2 separately, both by contradiction. For the first part, assume that $S_i^Ty \ge 0$ (which should be contradictory because it results in $x_i = 1$ or $x_i = \tfrac{1}{2}$). It holds that

$$\begin{aligned} \sum _{\begin{array}{c} c' \in \mathcal {S} \\ |c'| = 1 \\ S_{c'i} = -1 \end{array}} y_{c'} \ge \sum _{\begin{array}{c} c' \in \mathcal {S} \\ |c'| = 1 \\ S_{c'i} = -1 \end{array}} w_{c'} > \sum _{\begin{array}{c} c \in \mathcal {S} \\ S_{ci} = 1 \end{array}} w_c \ge \sum _{\begin{array}{c} c \in \mathcal {S} \\ S_{ci} = 1 \end{array}} y_c \end{aligned}$$

(17)

where inequality on the left follows from Proposition 4, inequality in the middle is equivalent to the assumption in Theorem 2 and inequality on the right follows from Propositions 2 and 3. This results in

$$\begin{aligned} \sum \limits _{\begin{array}{c} c' \in \mathcal {S} \\ S_{c'i} = -1 \end{array}} y_{c'} = \sum \limits _{\begin{array}{c} c' \in \mathcal {S} \\ |c'|> 1 \\ S_{c'i} = -1 \end{array}} y_{c'} + \sum _{\begin{array}{c} c' \in \mathcal {S} \\ |c'| = 1 \\ S_{c'i} = -1 \end{array}} y_{c'} \ge \sum _{\begin{array}{c} c' \in \mathcal {S} \\ |c'| = 1 \\ S_{c'i} = -1 \end{array}} y_{c'} > \sum _{\begin{array}{c} c \in \mathcal {S} \\ S_{ci} = 1 \end{array}} y_c \end{aligned}$$

(18)

where inequality on the right follows from (17) and inequality in the middle follows from non-negativity of $y_c$, hence $\sum \limits _{\begin{array}{c} c' \in \mathcal {S} \\ S_{c'i} = -1 \end{array}} y_{c'} > \sum \limits _{\begin{array}{c} c \in \mathcal {S} \\ S_{ci} = 1 \end{array}} y_c$, i.e., $S_i^Ty < 0$, which is contradictory with $S_i^Ty \ge 0$.

For the second part, assume that $S_i^Ty \le 0$ (which should be contradictory, because it results in $x_i = 0$ or $x_i = \tfrac{1}{2}$). Similarly as in the previous part, it holds that

$$\begin{aligned} \sum _{\begin{array}{c} c' \in \mathcal {S} \\ |c'| = 1 \\ S_{c'i} = 1 \end{array}} y_{c'} \ge \sum _{\begin{array}{c} c' \in \mathcal {S} \\ |c'| = 1 \\ S_{c'i} = 1 \end{array}} w_{c'} > \sum _{\begin{array}{c} c \in \mathcal {S} \\ S_{ci} = -1 \end{array}} w_c \ge \sum _{\begin{array}{c} c \in \mathcal {S} \\ S_{ci} = -1 \end{array}} y_c \end{aligned}$$

(19)

where the left inequality is given by Proposition 6, the middle inequality follows from the condition in Theorem 2, and the inequality on the right follows from Propositions 3 and 5. We proceed analogously as in the first part and obtain that $S_i^Ty > 0$, which is contradictory.    $\square$