Distributed algorithms for the Lovász local lemma and graph coloring

Abstract

The Lovász local lemma (LLL), introduced by Erdős and Lovász in 1975, is a powerful tool of the probabilistic method that allows one to prove that a set of n “bad” events do not happen with non-zero probability, provided that the events have limited dependence. However, the LLL itself does not suggest how to find a point avoiding all bad events. Since the works of Alon (Random Struct Algorithms 2(4):367–378, 1991) and Beck (Random Struct Algorithms 2(4):343–365, 1991) there has been a sustained effort to find a constructive proof (i.e. an algorithm) for the LLL or weaker versions of it. In a major breakthrough Moser and Tardos (J ACM 57(2):11, 2010) showed that a point avoiding all bad events can be found efficiently. They also proposed a distributed/parallel version of their algorithm that requires $O(\log ^2 n)$ rounds of communication in a distributed network. In this paper we provide two new distributed algorithms for the LLL that improve on both the efficiency and simplicity of the Moser–Tardos algorithm. For clarity we express our results in terms of the symmetric LLL though both algorithms deal with the asymmetric version as well. Let p bound the probability of any bad event and d be the maximum degree in the dependency graph of the bad events. When $epd^2 < 1$ we give a truly simple LLL algorithm running in $O(\log _{1/epd^2} n)$ rounds. Under the weaker condition $ep(d+1) < 1$, we give a slightly slower algorithm running in $O(\log ^2 d\cdot \log _{1/ep(d+1)} n)$ rounds. Furthermore, we give an algorithm that runs in sublogarithmic rounds under the condition $p\cdot f(d) < 1$, where f(d) is an exponential function of d. Although the conditions of the LLL are locally verifiable, we prove that any distributed LLL algorithm requires ${\varOmega }(\log ^* n)$ rounds. In many graph coloring problems the existence of a valid coloring is established by one or more applications of the LLL. Using our LLL algorithms, we give logarithmic-time distributed algorithms for frugal coloring, defective coloring, coloring girth-4 (triangle-free) and girth-5 graphs, edge coloring, and list coloring.

Appendix: Tools

Lemma 18

(Chernoff Bound) Let $X_1,\dots , X_n$ be indicator variables such that $\Pr (X_i=1) = p$. Let $X = \sum _{i=1}^{n} X_i$. Then, for $\delta > 0$:

$$\begin{aligned}&\Pr (X\ge (1+\delta ){\text {E}}[X])< \left[ \frac{e^{\delta }}{(1+\delta )^{(1+\delta )}} \right] ^{{\text {E}}[X]}\\&\quad \Pr (X\le (1-\delta ){\text {E}}[X]) < \left[ \frac{e^{\delta }}{(1-\delta )^{(1-\delta )}} \right] ^{{\text {E}}[X]} \end{aligned}$$

The two bounds above imply that for $0< \delta < 1$, we have:

$$\begin{aligned}&\Pr (X\ge (1+\delta ){\text {E}}[X])< e^{-\delta ^2 {\text {E}}[X] / 3} \\&\quad \Pr (X\le (1-\delta ){\text {E}}[X]) < e^{-\delta ^2 {\text {E}}[X] / 2}. \end{aligned}$$

Lemma 19

Let $\mathcal {E}_1, \ldots , \mathcal {E}_n$ be (likely) events and $X_1, \ldots , X_n$ be indicator variables such that for each $1 \le i \le n$ and $X = \sum _{i=1}^{n} X_i$,

$$\begin{aligned} \max _{\varvec{X}_{i-1}}\Pr (X_i \mid \varvec{X}_{i-1}, \mathcal {E}_1,\ldots \mathcal {E}_i) \le p \end{aligned}$$

where $\varvec{X}_i$ denotes the shorthand for $(X_1,\ldots ,X_i)$.⁴ Then for $\delta > 0$:

$$\begin{aligned} \Pr \left( (X >(1+\delta )np) \cap \left( \bigcap _{i} \mathcal {E}_i\right) \right)&\le \left[ \frac{e^{\delta }}{(1+\delta )^{(1+\delta )}} \right] ^{np} \end{aligned}$$

and thus by the union bound,

$$\begin{aligned} \Pr (X >(1+\delta )np)&\le \left[ \frac{e^{\delta }}{(1+\delta )^{(1+\delta )}} \right] ^{np} + \sum _{i} \Pr (\overline{\mathcal {E}_i}). \end{aligned}$$

Proof

For now let us treat $\mathcal {E}_i$ as 0/1 random variables and let $\mathcal {E} = \prod _i \mathcal {E}_i$. For any $t>0$,

$$\begin{aligned}&\Pr \left( (X >(1+\delta )np) \cap \left( \bigcap _{i} \mathcal {E}_i\right) \right) \end{aligned}$$

(6)

$$\begin{aligned}&\quad = \Pr \left( \left( \prod _{i=1}^{n} \mathcal {E}_i \right) \cdot \exp (tX) > \exp (t(1+\delta )np)\right) \nonumber \\&\quad \le \frac{{\text {E}}\left[ \left( \prod _{i=1}^{n} \mathcal {E}_i\right) \cdot \exp (tX)\right] }{\exp (t(1+\delta )np)} \nonumber \\&\quad = \frac{{\text {E}}\left[ \left( \prod _{i=1}^{n} \mathcal {E}_i \cdot \exp (tX_i) \right) \right] }{\exp (t(1+\delta )np)} \end{aligned}$$

(7)

We will show by induction that

$$\begin{aligned} {\text {E}}\left[ \left( \prod _{i=1}^{k} \mathcal {E}_i \exp (tX_i) \right) \right] \le (1+p(e^t - 1))^k \end{aligned}$$

When $k = 0$, it is trivial that ${\text {E}}[\mathcal {E}] \le 1$.

$$\begin{aligned}&{\text {E}}\left[ \left( \prod _{i=1}^{k} \mathcal {E}_i \exp (tX_i) \right) \right] \\&\quad \le {\text {E}}\left[ \left( \prod _{i=1}^{k-1} \mathcal {E}_i \exp (tX_i) \right) \right. \\&\qquad \left. \cdot {\text {E}}\left[ \mathcal {E}_k \exp (tX_k) \mid \varvec{X_{i-1}}, \mathcal {E}_1, \ldots , \mathcal {E}_{k-1}\right] \right] \\&\quad = {\text {E}}\left[ \left( \prod _{i=1}^{k-1} \mathcal {E}_i \exp (tX_i) \right) \right. \\&\qquad \left. \cdot \Pr (\mathcal {E}_k) \cdot {\text {E}}\left[ \exp (tX_k) \mid \varvec{X_{i-1}}, \mathcal {E}_1, \ldots , \mathcal {E}_{k}\right] \right] \\&\quad \le {\text {E}}\left[ \left( \prod _{i=1}^{k-1} \mathcal {E}_i \exp (tX_i) \right) \right. \\&\qquad \left. \cdot {\text {E}}\left[ \exp (tX_k) \mid \varvec{X_{i-1}}, \mathcal {E}_1, \ldots , \mathcal {E}_{k}\right] \right] \\&\quad = {\text {E}}\left[ \left( \prod _{i=1}^{k-1} \mathcal {E}_i \exp (tX_i) \right) \right. \\&\qquad \left. \cdot (1+\Pr (X_k \mid \varvec{X}_{i-1},\mathcal {E}_1, \ldots , \mathcal {E}_k)(e^{t} - 1)) \right] \\&\quad \le {\text {E}}\left[ \left( \prod _{i=1}^{k-1} \mathcal {E}_i \exp (tX_i) \right) \cdot (1+p(e^{t} - 1) ) \right] \\&\quad = {\text {E}}\left[ \left( \prod _{i=1}^{k-1} \mathcal {E}_i \exp (tX_i) \right) \right] \cdot (1+p(e^{t} - 1)) \\&\quad \le (1+p(e^{t} - 1))^{k} \end{aligned}$$

Therefore, by (6),

$$\begin{aligned}&\Pr \left( (X >(1+\delta )np) \cap \left( \bigcap _{i} \mathcal {E}_i\right) \right) \\&\quad = \frac{{\text {E}}[\mathcal {E} \cdot \prod _{i=1}^{n} \exp (tX_i) ]}{\exp (t(1+\delta )np)} \\&\quad \le \frac{(1+p(e^t-1))^{n}}{\exp (t(1+\delta )np)} \\&\quad \le \frac{\exp (np(e^t - 1))}{\exp (t(1+\delta )np)} \\&\quad = \left[ \frac{\exp (\delta ) }{(1+\delta )^{1+\delta }} \right] ^{np}. \end{aligned}$$

The last equality follows from the standard derivation of Chernoff Bound by choosing $t = \ln (1+\delta )$. $\square$

Corollary 5

Suppose that for any $\delta > 0$,

$$\begin{aligned} \Pr \left( (X >(1+\delta )np) \cap \left( \bigcap _{i} \mathcal {E}_i\right) \right) \le \left[ \frac{e^{\delta }}{(1+\delta )^{(1+\delta )}} \right] ^{np} \end{aligned}$$

then for any $M \ge np$ and $0< \delta < 1$,

$$\begin{aligned} \Pr \left( (X > np + \delta M) \cap \left( \bigcap _{i} \mathcal {E}_i\right) \right)&\le \left[ \frac{e^{\delta }}{(1+\delta )^{(1+\delta )}} \right] ^{M} \\&\le e^{-\delta ^2M/3} \end{aligned}$$

Proof

Without loss of generality, assume $M = tnp$ for some $t \ge 1$, we have

$$\begin{aligned}&\Pr \left( (X > np + \delta M) \cap \left( \bigcap _{i} \mathcal {E}_i\right) \right) \\&\quad \le \left[ \frac{e^{t \delta }}{(1+t \delta )^{(1+t \delta )}} \right] ^{np} \\&\quad = \left[ \frac{e^{\delta }}{(1+t \delta )^{(1+t \delta )/t}} \right] ^{M} \\&\quad \le \left[ \frac{e^{\delta }}{(1+\delta )^{(1+\delta )}} \right] ^{M} \qquad \qquad \qquad \qquad \qquad \quad \qquad \quad \quad \qquad (*) \\&\quad \le e^{-\delta ^2 M/ 3} \quad \quad \qquad \frac{e^{\delta }}{(1+\delta )^{(1+\delta )}} \le e^{-\delta ^2 / 3} \hbox { for } 0< \delta < 1 \end{aligned}$$

Inequality (*) follows if $(1+t \delta )^{(1+t \delta )/t} \ge (1+\delta )^{(1+\delta )}$, or equivalently, $((1+t \delta )/t) \ln (1+t \delta ) \ge (1+\delta ) \ln (1+\delta )$. Letting $f(t) = ((1+t \delta )/t )\ln (1+t \delta ) - (1+\delta ) \ln (1+\delta )$, we have $f'(t) = \frac{1}{t^2}\left( \delta t - \ln (1+\delta t) \right) \ge 0$ for $t > 0$. Since $f(1) = 0$ and $f'(t) \ge 0$ for $t > 0$, we must have $f(t) \ge 0$ for $t\ge 1$. $\square$