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Let $G$ be a connected simple graph not having $P_4$ or $C_3$ as an induced subgraph Prove that $G$ is a biclique.

I am looking for a proof that does not utilize induction.

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2 Answers2

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If $G$ has an odd cycle, let $C$ be one of minimal length $n$; clearly $n>3$. Let $v_1,v_2,v_3$, and $v_4$ be consecutive vertices of $C$; $G$ contains no triangle, so these vertices induce either a $P_4$ or a $C_4$. The former case is excluded by hypothesis. In the latter case we can replace the path $v_1v_2v_3v_4$ in $C$ by the edge $v_1v_4$ to get a cycle of length $n-2$, contradicting the minimality of $n$. Thus, $G$ cannot have an odd cycle and therefore must be bipartite.

Let the parts of $G$ be $V_0$ and $V_1$, and suppose that there are $v_0\in V_0$ and $v_1\in V_1$ that are not adjacent. $G$ is connected, so let $P$ be a path of minimal length from $v_0$ to $v_1$, say $u_1=v_0,u_2,\ldots,u_{2n}=v_1$, where $n\ge 2$. Minimality of $P$ and the hypothesis that $v_0$ is not adjacent to $v_1$ imply that $u_{2n-3}$ is not adjacent to $u_{2n}=v_1$, so $u_{2n-3},u_{2n-2},u_{2n-1}$, and $u_{2n}$ induce a copy of $P_4$. This is impossible, so $G$ must be a biclique.

Note that if we further require that $G$ not contain $P_4$ as a subgraph at all, we can show that $G$ is a star. Fix a vertex $v$ of maximal degree. If $\deg v=1$, $G$ is $K_{1,1}$. Otherwise, let $u$ and $w$ be distinct vertices adjacent to $v$. $G$ does not contain $C_3$, so $u$ and $w$ cannot be adjacent. Suppose that $u$ is adjacent to some vertex $x$ other than $v$; then $wvux$ is a copy of $P_4$, which is impossible. Thus, $G$ is a star with centre $v$.

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Brian M. Scott
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  • Wow, look who the op of the question you linked is. Now I'm feeling nostalgic. Thanks for your incredible support over the years. – Asinomás Feb 24 '15 at 05:04
  • @TheEmperorofIceCream: I’m embarrassed: I was concentrating so hard on whether the answers there were usable that I didn’t even notice until you pointed it out! You’re very welcome. – Brian M. Scott Feb 24 '15 at 05:10
  • @BrianM.Scott according to this proposition,a Bipartite graph must not contain an induced subgraph of path $P_4$,that means an even length cycle must also not contain path $P_4$ as induced subgraph but a Graph $G$ containg only cycle of length $6$ do contain induced subgraph $P_4$ ... i am confused !!! – sourav_anand Nov 19 '16 at 18:07
  • @sourav: You’ve misunderstood the proposition. It says that a connected simple graph that does not contain an induced $P_4$ or $C_3$ must be a complete bipartite graph; it doesn’t say anything about bipartite graphs in general. – Brian M. Scott Nov 19 '16 at 18:13
  • @BrianM.Scott a complete bipartite graph must be bipartite atleast ..!!! then how do a bipartite graph $C_6$ contains $P_4$ ..you can see the origin of my doubt here http://web.stanford.edu/class/cs103x/pset7_sol.pdf – sourav_anand Nov 19 '16 at 18:16
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    @sourav: The proposition only says something about graphs that do not contain induced $P_4$ or $C_3$. It doesn’t say anything about graphs, bipartite or not, that do contain one or both of these induced subgraphs. – Brian M. Scott Nov 19 '16 at 18:22
  • ok that was my doubt ..thank you for clearing my doubt @BrianM.Scott sir – sourav_anand Nov 19 '16 at 18:23
  • @sourav: You’re welcome. – Brian M. Scott Nov 19 '16 at 18:24
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The girth of $G$ is $4$ or $\infty$. Suppose it is not. Then take the smallest cycle $c_1,c_2,c_3,c_4,c_5\dots c_k$. Then the induced subgraph on $c_1,c_2,c_3,c_4$ is not a $P_4$. So there is an edge between two non consecutive vertices. A contradiction since this creates a smaller cycle.

If the girth is $\infty$ $G$ has no cycles and no $P_4$, since it is connected it is a star.

If the girth is $4$, since it has no odd cycles it is bipartite, also notice both parts have at least two vertices.Assume it is not a biclique,take the shortest geodesic of even length in $G$ of length $4$ or more, call it $v_1,v_2\dots v_{2n}$. then $v_1,v_2,v_{2n-1},v_{2n}$ induces a $P_4$.

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