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I'm having some trouble with proving the following: Let $|S(v)|$ be the set of colours available to colour vertex v. The claim is that for every bipartite graph $G=(V,E)$, if $|S(v)| > log_2n$ for each v then it is possible to find a valid colouring of G.

Any hints?

Thanks.

nonuser
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Gabriel Gava
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  • What is $n$? Are you saying let $n = |V|$? Also, bipartite graphs are two colorable. So it is always possible to find a coloring using a BFS. I think your question needs a bit more context and clarification. – ml0105 Apr 10 '14 at 21:01
  • @ml0105 The bipartite graph might not be 2-colorable, if the sets of feasible colors do not contain the two same colors for all the vertices. – dtldarek Apr 10 '14 at 21:05
  • Got it. That was my confusion. Thanks for clarifying! – ml0105 Apr 10 '14 at 21:06
  • Yes, $n = |V|$, I forgot to define. And @dtldarek explained right about the 2-colorable part, thanks. – Gabriel Gava Apr 10 '14 at 21:09

1 Answers1

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Disclaimer:

This is a simplified proof from Nancy Eaton's On Two Short Proofs About List Coloring Graphs . It's rather simple, but I was unable to form hints in a way that would make the intention clear, so I present all the details. As this is homework, please make an additional effort to rework the proof again, only by yourself.

Also, should you need to hand it in, be sure to cite Nancy Eaton's paper as a reference.

Let $C$ be the set of all the colors, i.e. $C = \bigcup_{v \in V} S(v)$, and set $V_0, V_1$ to be the bipartition of the graph.

We will prove that there exists a function $f : C \to \{0,1\}$ such that $f^{-1}(0)$ and $f^{-1}(1)$ indicate proper colorings for the $V_0$ and $V_1$ parts of the bipartite graph respectively. To prove our claim it is enough to show that for all $v \in V_i$ we have

$$S(v) \cap f^{-1}(i) \neq \varnothing$$

The proof will be probabilistic, that is, consider a random function $f : C \to \{0,1\}$, i.e. one picked uniformly from ${\{0,1\}}^C$. We have that for any $c \in C$

$$\mathbb{P}\Big(f(c) = 0\Big) = \frac{1}{2}.$$

Now

\begin{align} \mathbb{P}\Big(\exists v \in V_i.\ S(v) \cap f^{-1}(i) = \varnothing\Big) &\leq \sum_{v \in V_i} \mathbb{P}\Big(S(v) \cap f^{-1}(i) = \varnothing\Big) \\ &\leq \sum_{v \in V_i} \left(\frac{1}{2}\right)^{|S(v)|} \\ & < \sum_{v \in V_i} \frac{1}{n} \end{align}

Summing it up for both $i = 0$ and $i = 1$ we get

$$\mathbb{P}(f \text{ fails for some } v \in V) < \sum_{v \in V} \frac{1}{n} = 1.$$

That means, that there is a non-zero probability that it won't fail, and as we are in discrete, finite probability space, there exists $f$ that satisfies all the constraints.

I hope this helps $\ddot\smile$

dtldarek
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  • Thank you very much @dtldarek, it was very well explained, I understood everything :) – Gabriel Gava Apr 10 '14 at 22:57
  • What do you mean $f^{-1}(i)$ indicate proper colorings of $V_i$? Do you mean any color that is in the pre-image of $i$ can color any vertex in $V_i$? How does that work? – H. Walter Dec 28 '20 at 09:45
  • Ok I think I get it, since $f$ is a function it cannot be that $f^{-1}(0)\cap f^{-1}(1)\ne\emptyset$, so colors of $V_0$ and colors of $V_1$ are disjoint sets. One thing though: the second inequality should be an equality I think – H. Walter Dec 28 '20 at 10:02