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I am doing an exercise related to maximizing Independent Set, I have $G = (V = \{v_1, . . . , v_n\}, E)$ as an undirected graph. This graph as $n!$ possible orderings for the vertices $V$.

If we pick any such orderings uniformly at random i.e. let $\sigma = (\sigma_1, . . . , \sigma_n)$ be such random ordering then consider :

Begin with $S = \emptyset$.

Then, at each step for $i = 1$ to $n$, if for all $u \in S$, $(u, v_{\sigma_i}) \notin E$, add $v_{\sigma_i}$ to $S$.

Suppose we denote by $d$ the maximum degree of a vertex in $V$. We need to prove that the proposed algorithm achieves an independent set with expected value of at least $1/d$ fraction of the optimal solution.

My Proposed incomplete solution:

We denote OPT solution as $S^*$ and $|S^*| = OPT$ the size of the max independent set.
Let $S$ be the solution and $|S|$ is its size. We can write $X_i = 1$ if $v_{\sigma_i}$ vertex is added to $S$ and $0$ if not. Let $X$ be the number of vertices added to $S$.

Consider any vertex $v \in V$. If $v$ is not added to the independent set $S$, it must be because one of its neighbors was added to the independent set $S$ before it. The probability that $v$ is added to the independent set $S$ is at most $1/(d+1)$, it’s an upper bound. We calculate the expected value of the algorithm: we denote $u$ as an arbitrary vertex of $V$ and $N(v)$ the neighbors of $v$.

$$\mathbb{E}[X] = \mathbb{E}[X_1] + ... + \mathbb{E}[X_n]$$

\begin{align} \mathbb{E}[X_i] & = Pr[v_{\sigma_i} \text{ is added to }S | u \notin N(v_{\sigma_i})] \times Pr[u \notin N(v_{\sigma_i})]\\ & \qquad + Pr[v_{\sigma_i}\text{ is added to }S | u \in N(v_{\sigma_i})] \times Pr[u \in N(v_{\sigma_i})]\\ & = 1\times Pr[u \notin N(v_{\sigma_i})] + 0 \times Pr[u \in N(v_{\sigma_i})]\\ & = Pr[u \notin N(v_{\sigma_i})] \\ & = 1 - Pr[u \in N(v_{\sigma_i})] \end{align} since the maximum degree of a vertex is $d$, we have $Pr[u \notin N(v_{\sigma_i})] \leqslant $ (something? I'm not sure what) for all $i$ from $1$ to $n$.

Here I got stuck on how to proceed further. If there are mistakes in this I would be happy if you identify and correct.

D.W.
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1 Answers1

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Let $Y_v$ be the random variable that indicates whether vertex $v$ is added to the independence set. The probability that $v$ appears before all vertices that are incident to $v$ in a random ordering is $\frac1{\deg(v)+1}$, since all orderings are equally likely and all $\deg(v)+1$ vertices involved are symmetric to one another. On this situation $v$ will be added to the independence set. ($v$ can be added as well on other situations since none of its neighbors might have been added when $v$ appears after some of its neighbors.) We have $$E[Y_v]\ge\frac1{\deg(v)+1}\ge\frac1{d+1}.$$

So the expected value of $Y$, which is the size of the independence set returned by the given random algorithm is $$E[Y]=\sum_vE[Y_v]\ge\frac{|V|}{d+1}.$$

On the other hand, we will see that the size of the maximum independence set is at most $\frac {d|V|}{d+1}$. We can assume $G$ is connected since it is enough to prove for each connected component. We assume $d\ge1$; otherwise $d=0$, the situation is trivial while the proposition is ill-defined with $\frac1d$.

Let $M$ be a maximum independence set. Since each vertex in $M$ is incident to a vertex not in $M$ (otherwise that vertex is an isolated vertex, meaning $G$ is not connected), we have $$|M| \le|\text{vertices that are incident to }V\setminus M |.$$ Since each vertex is incident to at most $d$ vertices, we have $$|\text{vertices that are incident to }V\setminus M|\le d|V\setminus M|.$$ Since $|V\setminus M|=|V|-|M|$, we get $|M|\le d(|V|-|M|)$, i.e., $$|M|\le \frac {d|V|}{d+1}.$$

So $$\frac{E[Y]}{|M|}\ge\frac{\frac{|V|}{d+1}}{\frac {d|V|}{d+1}}\ge\frac1d,$$ which is what we are asked to prove.

John L.
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