Thursday, January 17, 2008

Lecture 2: Algorithms and gaps for Set-Cover and Coverage

Questions/etc. about the lecture can go here.

I will discuss integrality gap instances for Set-Cover in a later post. I may also clarify the randomized rounding analysis for Max-Coverage. (Or -- it would be nice if someone else in class wanted to post something on this :)

For now, I'll quickly give you the detailed probabilistic analysis of randomized rounding for Set-Cover.

Recall that if we repeat the randomized rounding step $t$ times, the expected overall cost is at most $t \cdot$ Opt. (I think I said it exactly equals $t \cdot$ Opt in class but this is not quite right because you could pick the same set more than once in different rounding stages -- and you only have to pay for it once.) We will select $t = \ln n + C \ln \ln n$ for some large constant $C$.

We now observe that from Markov's inequality, the probability our output costs more than $(1 + 2/\ln n) \cdot t \cdot$ Opt is at most $1/(1 + 2/\ln n)$ which is about $1 - 2/\ln n$. Rigorously, it's less than $1 - 1/\ln n$. Multiplying out the $(1 + 1/\ln n) \cdot t$, we conclude:

(*) Pr[costs at most $(\ln n + O(\ln \ln n)) \cdot$ Opt] $\geq 1/\ln n$.

Now on the other hand, as we saw, after our $t$ rounds, the probability a particular $e$ is uncovered is at most $\exp(-t) = 1/(n \ln^C$ $n)$. Union-bounding over all elements, we get:

(**) Pr[solution is invalid] $\leq 1/\ln^C$ $n$.

Taking $C$ large (even $C = 2$ is okay, I guess) and combining (*) and (**) we conclude:

(***) Pr[valid solution of cost at most $(\ln n + O(\ln \ln n)) \cdot$ Opt] $\geq 1 / \ln n - 1/\ln^2$ $n \geq \Omega(1/\ln n)$.

We may now use Problem 1a on Homework 1.

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