[Computational Complexity] A Kolmogorov Complexity Proof of the Lovász...
- Robin Moser gave one of the best STOC talks ever in his award-winning paper A Constructive Proof of the Lovász Local Lemma. Not only an amazing result but also an incredibly inventive simple proof that he came up with while preparing the talk.
Here is a sketch of Moser's proof written as a Kolmogorov complexity argument. This is not the full Lovász Local Lemma but captures the main principle.
Theorem: Suppose we have a k-CNF formula φ with n variables and m clauses and each clause shares a variable with at most r other clauses. There is a constant d such that if r < 2k-d then φ is satisfiable. Moreover we can find that assignment in time polynomial in m and n.
Here's the algorithm:
Pick a random assignment of φ
While there is an unsatisfiable clause C
Replace the variables of C with new random variables
While there is clause D that shares a variable with C that is not satisfied
Assume Fix(C) always terminates. Every clause that was satisfied before we called Fix(C) will still remain sastisfied and C will also now be satisfied. So Solve makes at most m calls to Fix.
We need to show all the Fix(C) terminate. Suppose the algorithm makes at least s Fix calls including all the recursive ones. We will show s is bounded and thus the algorithm terminates.
Fix a Kolmogorov random string x of length n+sk (random relative to φ, k, s, r, m and n) and assume the algorithm uses the first n bits as the initial assignment and k bits each to replace the variables in each Fix call.
If we know which clause is being fixed, we know the clause is violated so we know all the bits of this clause and thus we learn k bits of x. We then replace those bits with another part of x.
So we can describe x by the list of clauses we fix plus the remaining n bits of the final assignment. We can describe the C such that Fix(C) is called by Solve by m log m bits and the remaining fixed clauses by log r + O(1) bits because either it is one of r clauses that intersects the previous clause or we indicate the end of a recursive call (keeping track of the recursion stack).
So we must have m log m + s(log r+O(1))+n ≥ n+sk or s(k-log r-O(1)) ≤ m log m.
To show s is bounded, we need k-log r-O(1) to be positive or r<2k-d for some constant d.
Note this in fact shows s = O(m log m) so the algorithm runs in polynomial time. We choose the x randomnly which with high probability will be Kolmogorovly random. QED
In the talk, Moser gives the bound r<2k-3 and in follow-up work with Gábor Tardos shows r<2k/e (e = 2.71…) which is the best that comes out of the original Lovász Local Lemma.
Posted By Lance to Computational Complexity at 6/02/2009 03:46:00 AM