Solving Sparse Instances of Max SAT via Width Reduction and Greedy Restriction

Abstract

We present a moderately exponential time and polynomial space algorithm for sparse instances of Max SAT. Our algorithms run in time of the form $O\left (2^{(1-\mu (c))n}\right )$ for instances with n variables and cn clauses. Our deterministic and randomized algorithm achieve $\mu (c) = {\Omega }\left (\frac {1}{c^{2}\log ^{2} c}\right )$ and $\mu (c) = {\Omega }\left (\frac {1}{c \log ^{3} c}\right )$ respectively. Previously, an exponential space deterministic algorithm with $\mu (c) = {\Omega }\left (\frac {1}{c\log c}\right )$ was shown by Dantsin and Wolpert [SAT 2006] and a polynomial space deterministic algorithm with $\mu (c) = {\Omega }\left (\frac {1}{2^{O(c)}}\right )$ was shown by Kulikov and Kutzkov [CSR [2007]]. Our algorithms have three new features. They can handle instances with (1) weights and (2) hard constraints, and also (3) they can solve counting versions of Max SAT. Our deterministic algorithm is based on the combination of two techniques, width reduction of Schuler and greedy restriction of Santhanam. Our randomized algorithm uses random restriction instead of greedy restriction.

Appendix: A Proof of Lemma 2

In this appendix, we give the precise proof of Lemma 2 to make this paper self-contained. We stress that all the proofs are due to Chen et al. [7] except that we consider Max formula SAT and $\widetilde {L}(\cdot )$ instead of De Morgan formula SAT and L(⋅).

Lemma 5 (Restatement of Lemma 2)

Let Φ be an instance of Max formula SAT on n variables. For any k≥4, we have

$$\begin{array}{@{}rcl@{}} \Pr\left[\widetilde{L}({\Phi}_{n-k}) \geq 2\cdot \widetilde{L}({\Phi})\cdot\left(\frac{k}{n}\right)^{3/2}\right]<2^{-k}. \end{array} $$

To prove this lemma, we need some definitions and lemmas.

Definition 1

A sequence of random variables X ₀, X ₁,…,X _n is a supermartingale with respect to a sequence of random variables R ₁, R ₂,…,R _n if $\mathbf {E}\left [X_{i} | R_{i-1}, \ldots , R_{1}\right ] \leq X_{i-1}$, for 1≤i ≤ n.

Chen et al. proved the following lemma.

Lemma 6 (Lemma 4.2 in [7])

Let $\{X_{i}\}^{n}_{i=0}$ be a supermartingale with respect to $\{R_{i}\}^{n}_{i=0}$ . Let Y _i =X _i −X _i−1 . If, for every 1 ≤ i ≤ n, the random variable Y _i (conditioned on R _i−1 ,…,R ₁ ) assumes two values with equal probability, and there exists a constant c _i ≥0 such that Y _i ≤c _i , then, for any λ, we have,

$$\begin{array}{@{}rcl@{}} \Pr\left[X_{n} - X_{0} \geq \lambda\right] \leq \exp\left(-\frac{\lambda^{2}}{2{\sum}^{n}_{i=1} {c_{i}^{2}}}\right). \end{array} $$

Now, we prove the counterpart of Lemma 4.4 in [7] with respect to $\widetilde {L}(\cdot )$

Lemma 7 (Lemma 4.4 in [ 7 ] with respect to $\widetilde {L}(\cdot )$)

Let Φ be an instance of Max formula SAT on n variables, and let $x=\arg \max _{x \in \text {var}({\Phi })}\widetilde {\text {freq}}_{\Phi }(x)$ . Then $\max \{\widetilde {L}({\Phi }[x=0]), \widetilde {L}({\Phi }[x=1])\} \leq \widetilde {L}({\Phi }) \cdot \left (1-\frac {1}{n}\right )$ , and $\mathbf {E}_{a \in \{0,1\}}\left [\widetilde {L}\left ({\Phi }[x=a]\right )\right ] \leq \widetilde {L}({\Phi }) \cdot \left (1-\frac {1}{n}\right )^{3/2}$.

Proof

First note that in any ϕ _i such that L(ϕ _i ) ≥ 2 and x ∈ var(ϕ _i ), for each leaf labeled with x or $\overline {x}$, its sibling subformula does not contain x or $\overline {x}$ according to the procedure of Simplify. Since $\widetilde {\text {freq}}(x)\ge \frac {\widetilde {L}({\Phi })}{n}$, after assigning 0 or 1 to x, the size of Φ decreases at least $\frac {\widetilde {L}({\Phi })}{n}$. Furthermore, when we set x to be 0 or 1 uniformly at random, then with probability 1/2, we can remove the sibling of each x or $\overline {x}$. Such a sibling subformula does not contain x or $\overline {x}$ and has at least one literal, thus,

$$\begin{array}{@{}rcl@{}} && \mathbf{E}_{a \in \{0,1\}}\left[\widetilde{L}\left({\Phi}[x=a]\right)\right] \leq \widetilde{L}({\Phi}) - \frac{\widetilde{L}({\Phi})}{n} - \frac{1}{2}\cdot\frac{\widetilde{L}({\Phi})}{n}\\ &=& \widetilde{L}({\Phi})\left(1-\frac{3}{2n}\right) \leq \widetilde{L}({\Phi}) \cdot \left(1-\frac{1}{n}\right)^{3/2}. \end{array} $$

□

Let R _i be the random value assigned to the restricted variables in step i. Set $\widetilde {L}_{i} := \widetilde {L}({\Phi }_{i})$ and $\ell _{i} := \log \widetilde {L}_{i}$. Define a sequence of random variables {Z _i } as follows:

$$\begin{array}{@{}rcl@{}} Z_{i} = \ell_{i}-\ell_{i-1}-\frac{3}{2}\log\left(1-\frac{1}{n-i+1}\right). \end{array} $$

Note that, given R ₁ ,…,R _i−1, the random variable Z _i assumes two values with equal probability.

Lemma 8 (Lemma 4.5 in [ 7 ] with respect to $\widetilde {L}(\cdot )$)

Let X ₀ =0 and $X_{i} = {\sum }^{i}_{j=1}Z_{j}$ . Then the sequence {X _i } is a supermartingale with respect to {R _i }, and, for each Z _i , we have $Z_{i} \leq c_{i} := -\frac {1}{2}\log (1-\frac {1}{n-i+1})$.

Proof

By Lemma 7, $\ell _{i} \leq \ell _{i-1} + \log \left (1-\frac {1}{n-i+1}\right )$. Then, $Z_{i} = \ell _{i}-\ell _{i-1}-\frac {3}{2}\log \left (1-\frac {1}{n-i+1}\right ) \leq -\frac {1}{2}\log \left (1-\frac {1}{n-i+1}\right ) = c_{i}$. By Jensen’s inequality, $\mathbf {E}[\ell _{i} | R_{i-1},\ldots , R_{1}] \leq \log \mathbf {E}[\widetilde {L}_{i} | R_{i-1}, \ldots , R_{i}]$, which, by Lemma 7, is at most $\log \left (\widetilde {L}_{i-1}\cdot \left (1-\frac {1}{n-i+1}\right )^{3/2}\right ) = \ell _{i-1}+\frac {3}{2}\log \left (1-\frac {1}{n-i+1}\right )$; this implies E[Z _i |R _i−1,…,R _i ] ≤ 0, and so {X _i } is indeed a supermartingale. □

Now, we ready to prove the proof of Lemma 5.

Proof of Lemma 5

Let λ be arbitrary, and let c _i ’s be as defined in Lemma 8. By Lemma 6 and Lemma 8, we get

$$\Pr\left[\sum\limits_{j=1}^{i} Z_{j} \geq \lambda \right] \leq \exp\left(-\frac{\lambda^{2}}{{2\sum}^{i}_{j=1}{c^{2}_{j}}}\right). $$

It is easy to show that ${\sum }^{i}_{j=1} Z_{j} = \ell _{i}-\ell _{0}-\frac {3}{2}\log \frac {n-i}{n}$ from the definition of Z _j . Hence,

$$\Pr\left[\sum\limits_{j=1}^{i} Z_{j} \geq \lambda \right] = \Pr\left[\ell_{i}-\ell_{0}-\frac{3}{2}\log\left(\frac{n-i}{n}\right)\geq \lambda\right]=\Pr\left[\widetilde{L}_{i}\geq e^{\lambda} \widetilde{L}_{0} \left(\frac{n-i}{n}\right)^{3/2}\right]. $$

For each 1 ≤ j ≤ i, we have $c_{j} = -\frac {1}{2}\log \left (1-\frac {1}{n-j+1}\right ) \leq \frac {1}{2}\cdot \frac {1}{n-j+1}$, using the inequality $\log (1+x) \leq x$. Thus, ${\sum }^{i}_{j=1} {c^{2}_{j}}$ is at most

$$\frac{1}{4}\sum\limits_{j=1}^{i}\left(\frac{1}{n-j+1}\right)^{2} \leq \frac{1}{4}\sum\limits_{j=1}^{i}\left(\frac{1}{n-j}-\frac{1}{n-j+1}\right)=\frac{1}{4}\left(\frac{1}{n-i}-\frac{1}{n}\right)\leq\frac{1}{4}\cdot\frac{1}{n-i}. $$

Taking i = n−k, we get

$$\Pr\left[\widetilde{L}_{n-k} \geq e^{\lambda}\widetilde{L}_{0}\left(\frac{k}{n}\right)^{3/2}\right] \leq \exp\left(-\frac{\lambda^{2}}{2{\sum}^{n-k}_{j=1}{c^{2}_{j}}}\right) \leq e^{-2\lambda^{2}k}. $$

Choosing $\lambda = \ln 2$ concludes the proof. □