# 191014K02 | Day 2 Lecture 1

## 191014K02: Day 2 Lecture 1

### Vertex Cover

The naive algorithm by brute force — examining all possible subsets — is O(n^k \cdot m) in damages. Can we do better?

The answer turns out to be yes: we can improve from O(n^k \cdot m) to deterministic 2^k \cdot n^{O(1)} time, which is fixed-parameter tractable in k.

Having said that, we will begin with a very elegant randomized algorithm for Vertex Cover, which essentially picks an edge at random and then, one of its endpoints at random, for as long as it can.

Here are few claims about cute algorithm:

`ALG`

always runs in polynomial^{1}time.- S is always a vertex cover.
- \operatorname{Pr}[S is an optimal vertex cover] \geqslant 1/2^k.

The first two claims follow quite directly from the operations of the algorithm and the definition of a vertex cover.

What about the third? Well: let `OPT`

be some fixed optimal vertex cover. Suppose |`OPT`

| \leqslant k. Initially, note that S \subseteq `OPT`

. In each round, \operatorname{Pr}[s \in S] \geqslant 1/2, since S \cap \{u,v\} \neq \varnothing by definition. If s \in `OPT`

in *every* round of the algorithm, then S = `OPT`

, which is awesome: and said awesomeness manifests with probability 1/2^k.

Bonus: repeat the algorithm and retain the smallest solution to get an overall constant success probability:

1-\left(1-\frac{1}{2^k}\right)^{2^k} \geqslant 1-1 / e.

**Approximation.** Do we expect `ALG`

to be a reasonable approximation? It turns out: yes!

In particular: we will show that the size of the vertex cover output by `ALG`

is at most twice |`OPT`

| in expectation.

For a graph G, define X_G to be the radom variable returning the size of the set S output by the algorithm.

For integers k,n define: X_{n,k}=\max_G E[X_G],

where the \max is taken over all graphs with \leqslant n vertices^{2} and vertex cover of size \leqslant k.

Now let’s analyze the number X_{n,k}. Let G^\star be the “worst-case graph” that bears witness to the \max in the definition of X_{n,k}. Run the first step of `ALG`

on G^\star. Suppose we choose to pick {\color{indianred}s} in this step.
\begin{aligned}
X_{n,k}=E[X_{G^\star}] & = {\color{indianred}1}+\left(\frac{1}{2}+\varepsilon \right) E\left[{\color{darkseagreen}X_{G^\star-s} \mid s \in \text{OPT}}\right]+\left(\frac{1}{2} - \varepsilon\right) E\left[{\color{palevioletred}X_{G^\star-s} \mid s \notin \text {OPT}}\right] \\
& = 1 + \left(\frac{1}{2} + \varepsilon \right){\color{darkseagreen}X_{n, k-1}}+ \left(\frac{1}{2} - \varepsilon \right){\color{palevioletred}X_{n-1, k}}\\
& \leqslant 1 + \left(\frac{1}{2} + \varepsilon \right){\color{darkseagreen}X_{n, k-1}}+ \left(\frac{1}{2} - \varepsilon \right){\color{dodgerblue}X_{n, k}}\\
& = 1 + \frac{1}{2} X_{n, k-1} + \varepsilon \cdot X_{n, k-1} + \frac{1}{2} X_{n, k} - \varepsilon X_{n, k}\\
& \leqslant 1 + \frac{1}{2} X_{n, k-1} + \varepsilon \cdot {\color{dodgerblue}X_{n, k}} + \frac{1}{2} X_{n, k} - \varepsilon X_{n, k}\\
& = 1 + \frac{1}{2} X_{n, k-1} + {\color{indianred}\varepsilon \cdot X_{n, k}} + \frac{1}{2} X_{n, k} - {\color{indianred}\varepsilon X_{n, k}}\\
& = 1 + \frac{1}{2} X_{n, k-1} + \frac{1}{2} X_{n, k}
\end{aligned}

Note that:

- X_{n,k} \geqslant X_{n-1,k} and X_{n,k} \geqslant X_{n,k-1}.
- \operatorname{Pr}[s \in \text{OPT}] \geqslant \frac{1}{2}, in particular we let \operatorname{Pr}[s \in \text{OPT}] = \frac{1}{2} + \varepsilon.
- \operatorname{Pr}[s \notin \text{OPT}] = 1 - \operatorname{Pr}[s \in \text{OPT}] = \frac{1}{2} - \varepsilon.
- {\color{darkseagreen}G^\star-s} is a graph on at most {\color{darkseagreen}n} vertices with a vertex cover of size {\color{darkseagreen}\leqslant k-1}
- {\color{palevioletred}G^\star-s} is a graph on at most {\color{palevioletred}n} vertices with a vertex cover of size {\color{palevioletred}\leqslant k}

Rearranging terms, we get:

\frac{1}{2} X_{n,k} \leqslant 1 + \frac{1}{2} X_{n,k-1} \equiv X_{n,k} \leqslant 2 + X_{n,k-1}

Expanding the recurrence, we have: X_{n,k} \leqslant 2k, as claimed earlier.

### Feedback Vertex Set

Now we turn to a problem similar to vertex cover, except that we are “killing cycles” instead of “killing edges”.

If we try to mimic the cute algorithm from before, we might easily be in trouble: note that the driving observation — that an edge has at least one of its endpoints in the solution with a reasonable enough probability — can fail spectacularly for FVS:

One thing about this example is the large number of pendant vertices sticking out prominently, and these clearly contribute to the badness of the situation. Happily, it turns out that we can get rid of these:

Consider graphs with no pendant vertices and fix an optimal FVS S. Is it true that a reasonable fraction of edges are guaranteed to be incident to S? Well… not yet:

However, continuing our approach of conquering-by-observing-patterns-in-counterexamples, note that the example above has an *abundance* of vertices that have degree two. Can we get rid of them? Well, isolated and pendant vertices were relatively easy because they don’t even belong to cycles, but *that* is not quite true for vertices of degree two. Can we still hope to shake them off?

One thing about a degree two vertex u is that if there is a cycle that contains u it *must* contain both its neighbors^{5}. So you might suggest that we can ditch u and just work with its neighbors instead. This intuition, it turns out, can indeed be formalized:

Let us also get rid of self-loops (because we can):

Now let’s apply Lemmas 1—3 exhaustively, which is to say we keep applying them until none of them are applicable (as opposed to applying them until we feel exhausted ). Once we are truly stuck, we have a graph H that is: (a) a graph whose *minimum* degree is three; and (b) equivalent to the original G in the sense that any minimum FVS for H can be extended to a minimum FVS of G by some time travel: just play back the applications of Lemmas 1—3 in reverse order.

Recall that all this work was to serve our hope for having a cute algorithm for FVS as well. Let’s check in on how we are doing on that front: consider graphs whose minimum degree is three and fix an optimal FVS S. Is it true that a reasonable fraction of edges are guaranteed to be incident to S? Or can we come up with examples to demonstrate otherwise?

This is a good place to pause and ponder: play around with examples to see if you can come up with bad cases as before. If you find yourself struggling, it would be for a good reason: we actually now *do* have the property we were after! Here’s the main claim that we want to make.

We argue this as follows: call an edge good if it has at least one of its endpoints in S, and bad otherwise.

We will demonstrate that the number of good edges is at least the number of bad edges: this implies the desired claim.

**The bad edges.**Let X := G \setminus S. The bad edges are precisely E(G[X]).**The good edges.**Every edge that has one of its endpoints in X and the other in S is a good edge. Recall that G has minimum degree three, because of which:- for every leaf in G[X], we have at least
*two*good edges, and - for vertex that is degree two in G[X], we have at least one good edge.

- for every leaf in G[X], we have at least

So at this point, it is enough to show that twice the number of leaves and degree two vertices is at least |E(G[X])| = |X|-1. But this is quite intuitive if we simple stare at the following mildly stronger claim:

2 \cdot (\text{\# leaves}) + \text{\# deg-} 2 \text{vertices} \geqslant |X|

which is equivalent to:

2 \cdot ({\color{darkseagreen}\text{\# leaves}}) + {\color{palevioletred}\text{\# deg-}2 \text{vertices}} \geqslant ({\color{darkseagreen}\text{\# leaves}}) + {\color{palevioletred}\text{\# deg-}2 \text{ vertices}} + \text{\# deg-}(\geqslant 3) \text{ vertices}.

After cancelations, we have:

(\text{\# leaves}) \geqslant \text{\# deg-}(\geqslant 3) \text{ vertices}.

Note that this is true! Intuitively, the inequality is suggesting every branching vertex in a tree pays for at least one leaf — this can be formalized by induction on n.

All this was leading up to cute randomized algorithm v2.0 — i.e, adapted for FVS as follows:

This follows from Lemmas 1—3 and induction on n.

This follows from Lemmas 1—3, the key lemma, and induction on n.

### Exact Algorithms

`Coming Soon.`

## Footnotes

(even linear)↩︎

We could have also considered X_k = \max_G E[X_G], where the max is over all graphs whose vertex cover is at most k: but there are infinitely many graphs that have vertex covers of size at most k, and it is not

*immediate*that the max is well-defined, so we restrict ourselves to graphs on n vertices that have vertex covers of size at most k.↩︎We allow for more than one edge between a fixed pair of vertices and self-loops.↩︎

A graph without cycles that is not necessarily connected.↩︎

Except when the cycle

*only*contains u, i.e, u has a self-loop.↩︎