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Let $G = (V,E)$ be a graph with $V = \{1, \dots , n\}$. Consider the function $f : V \to V$ defined by $$f(v) = \min\{u\mid u \in V, d_G(v, u) \leq 2\}$$ Prove that if there are $vw \in E(G)$ such that $f(v) \neq f(w)$ then $G$ contains the graph $P_4$ as an induced subgraph

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    What about the graph with two nodes and no edges?2012-11-02
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    @Hagen: It doesn’t satisfy the hypothesis.2012-11-02

2 Answers 2

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Suppose that $vw\in E$ is such that $f(v)\ne f(w)$. Since $x\ne y$, we may assume without loss of generality that $x

Now $d_G(v,w)=1\le 2$, so $x\le w$ and $y\le v$. Moreover, $d_G(w,w)=0\le 2$, so $y\le w$. Since $x

HINT: If $w=z$, then $d_G(x,w)\le 2$.

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    Draw the graph for this exercise, please2012-11-02
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    @Bunpentrutine-: It’s not really possible: there are infinitely many different graphs that satisfy the hypotheses of the theorem. At most you can sketch different configurations of the vertices $v,w,x,y,z$ and eliminate the impossible ones using the reasoning in my answer.2012-11-02
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    an example , please2012-11-02
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    @Bunpentrutine-: Can you find a graph that satisfies the hypotheses of the theorem? If so, you can draw it and pick out $v,w,x,y$, and $z$ for yourself. Draw several and do the same with each of them, and it **may** help you to see what’s going on with the argument, though I don’t guarantee it. Or is the problem that you can’t find an example of a graph satisfying the hypotheses of the theorem?2012-11-02
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    Prove that if G is P4-free then any two vertices u and v are in the same connected component if and only if f(u)=f(v)2012-11-04
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Assume $f(v)

Claim: $G$ has an induced subgraph as depicted below:

Induced $P_4$

If $\mathrm{dist}(v,f(v))<2$, or there are any other edges induced by the above subgraph, then $f(w) \leq f(v)$, giving a contradiction.

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    This is all the proof?2012-11-05