Algorithmic Lovász local lemma
7 minute read
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Adapted from the note “CS167: Reading in Algorithms The Algorithmic Lovász Local Lemma” by Tim Roughgarden.
Warmup
Moser’s algorithm
We will describe the algorithm on the special case of exactly-$k$-CNFs (CNFs where each clause queries exactly $k$ variables), but it will be clear how to extend it to the following version:
Let $X_1, \ldots, X_n$ be independent random variables, and let $E_1, \ldots, E_m$ be events depending on subsets of those variables. Suppose that for each $j \in [m]$, $\Pr[E_j] \leq p$ and $E_j$ only shares variables with $\leq d$ of the other events. If $5pd \leq 1$, then $\Pr[\wedge_i \overline{E_i}] > 0$.
The algorithm goes as follows:
- pick some unsatisfied clause $C_j$;
- redraw all its variables at random;
- if this unsatisfies some other clause (or fails to satisfy $C_j$ itself), recurse on that clause;
- repeat until all clauses are satisfied.
It’s clear that this algorithm succeeds if it ever terminates, so we move directly to the runtime analysis.
Running time
Similar to Strings without repeats, we will:
- assume that the algorithm has run for $t$ steps (i.e. there were $t$ redraws);
- describe some (short) transcript of what the algorithm did during these $t$ steps;
- show that we can recover from it the value of all bits drawn so far;
- conclude from this “miraculous compression” that the algorithm is unlikely to run for $t$ steps.
Concretely, the transcript will include:
- the current values of $x_1, \ldots, x_n$ after $t$ steps (a string in $\zo^n$);
- which clauses were initially unsatisfied (a string in $\zo^m$);
- a description of the recursive branching structure of the $t$ steps (a string in $[d]^t$ indicating which neighbors were taken, and a string in $\zo^{2t}$ describing the sequence of recursions and backtrackings).
On the other hand, using the transcript, we can recover all $n + tk$ bits that were drawn while running the algorithm. Therefore, the probability that the algorithm ran for $t$ steps is at most
\[
\frac{2^n \cdot 2^m \cdot d^t \cdot 2^{2t}}{2^{n+tk}} = 2^m \cdot (4d/2^k)^t.
\]
Therefore, as long as $5pd \leq 1 \Leftrightarrow d \leq 2^{k}/5$, the algorithm has only a $2^m \cdot (4/5)^t$ chance of running for $t$ steps, which means that it will run in $O(m)$ steps with high probability.