Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $G = (V,E)$ be a graph with $n$ vertices and minimum degree $\delta > 10$. Prove that there is a partition of $V$ into two disjoint subsets $A$ and $B$ so that $|A| \le O (\frac{n \ln \delta}{\delta})$ , and each vertex of $B$ has at least one neighbor in $A$ and at least one neighbor in $B$.

share|cite|improve this question
I don't understand what $|A|\leq O(\frac{n\ln\delta}\delta)$ means, as $n$ and $\delta$ are fixed (so I don't understand in what this gives a constraint about the cardinality of $A$. – Davide Giraudo Sep 27 '12 at 20:29
I think if $ T \le c n^k $ for some $c >0 $ we can write $T = O(n^k)$ or $T \le O(n^k)$. So it just asks us to prove the existence of some constant $c>0$ such that $|A| \le c \frac{n \ln \delta}{\delta}$. – Shubhodip Mondal Sep 28 '12 at 3:44
This is exercise 4 in section 1.6 in Alon and Spencer's "The Probabilistic Method". – Douglas S. Stones Sep 29 '12 at 6:34
Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. – Julian Kuelshammer Oct 15 '12 at 20:20
up vote 5 down vote accepted

Because this is a textbook problem, I try to put my thoughts related to the text. If we think about related problems, one can realize that $A$ is actually a dominating set: for every vertex in $B$, it has some neighbor in $A$. However, we require $B$ to have one more property: it contains no isolated vertices after $A$ is removed. Observe that, given a dominating set $D$ of $G$, if we move all the isolated vertices in $G|_{V-D}$ into $D$, then we have a dominating set with desired property satisfied.

This motivates a binomial sampling as follows: with a parameter $p \in (0,1)$, for each $v \in V$, $\Pr[v \in A_1] = p$. Let $A_2$ be the set of vertices not dominated by $A_1$ (that is, for each vertex $w \in A_2$, none of its neighbors is in $A_1$). As described in the text, $A_1 \cup A_2$ is a dominating set.

Now let $A_3$ be the set of vertices not in $A_1 \cup A_2$, but all its neighbors are in $A_1$ and $A_2$. If we move $A_3$ into the dominating set, then as discussed above, $A = A_1 \cup A_2 \cup A_3$ satisfies the requirement. We compute the expectation of $|A|$, which is ${\bf E}[|A_1|] + {\bf E}[|A_2|] + {\bf E}[|A_3|]$. The only difficulty lies in $|A_3|$. Consider $v \in A_3$, and one of its neighbor $w$. Then $\Pr[w \in A_1] = p$ and $\Pr[w \in A_2] < (1-p)^{\delta+1}$. Thus $\Pr[(w \in A_1) \vee (w \in A_2)] < p + (1-p)^{\delta+1}$. Therefore $\Pr[v \in A_3] < (1-p)(p + (1-p)^{1+\delta})^\delta$.

Therefore the expected size of $A$ is \begin{align*} n \Big( \big( p + (1-p)^{1+\delta} \big) + (1-p)\big( p + (1-p)^{1+\delta}\big)^{\delta} \Big) \end{align*} If we plug in $p = \ln(1+\delta)/(1+\delta)$, note that $p + (1-p)^{1+\delta} < 1$, so this is $O(n\ln\delta/\delta)$ as desired.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.