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Given a graph G on six vertices, show that either G and its complement G' together contain at least two triangles.

Advanced question: Given G with m + 6 vertices, show G and G' contain together at least m+2 triangles.

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    Question for student, needs answer soon and I have run out of time – 2017-02-02
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    I'm gonna guess $G'$ is the complement of $G$? – 2017-02-02
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    @victoria urgency is not a concern for Stack Exchange. For the reasons why, see [this question (on a different site)](http://meta.stackoverflow.com/questions/326569/under-what-circumstances-may-i-add-urgent-or-other-similar-phrases-to-my-quest). – 2017-02-02
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    See answer to [this question](http://math.stackexchange.com/questions/101995/lower-bound-for-monochromatic-triangles-in-k-n). – 2017-02-02
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    Actually, it's true that $G$ and $G'$ must contain at least two triangles ***BETWEEN THEM***, but it's ***NOT TRUE*** that one oir the other must contain two triangles. Take a $6$-cycle $v_0v_1v_2v_3v_4v_5v_0$ and add the edge $v_0v_2;$ the resulting graph contains just ***ONE*** triangle, and its complement also contains just ***ONE*** triangle. Perhaps the student has miscopied the assignment. – 2017-02-02
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    Whoops, I misread the question. Thanks, @bof. – 2017-02-02
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    My error. I tried to summarize the question as much as possible and was not accurate. Sorry, I have been ill and am burned out. – 2017-02-02
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    @victoria Please edit your question to make the problem statement correct, so as not to confuse people readint this question and answer in the future. – 2017-02-02

1 Answers 1

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Color $K_n$ in 2 colors. I'll give a lower bound on the number of monochromatic triangles. This will give both of your results.

Let $A$ be the number of bichromatic angles, that is, the number of pairs of intersecting edges of opposite color. Then by summing over each vertex, one finds $$A\le n\left\lfloor \frac{n-1}2\right\rfloor\cdot\left\lceil \frac{n-1}2\right\rceil.$$

Every triangle has either 2 bichromatic angles, either the triangle is monochromatic. Hence if $d$ is the number of monochromatic triangles we find $$A=2\left( {n \choose 3}-d \right).$$

Combining this with the above inequality we find that the number of monochromatic triangles is $\Omega(n^3)$. The explicit inequality gives for example 2 triangles if $n=6$. Your other claim that there are at least $n$ triangles is not right when $n\le 9$, but it is when $n\ge 10$. I'll let that to you using the obtained inequality.

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    Thank you for your help. Sorry I am not a graph theorist, just a tutor trying to help out a student. A few clarifications requested, please. Is the colouring random? What is Omega (n^3) ? (Notation ?) – 2017-02-02
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    OK I worked through this to clarify it for the student and got it pretty well worked out. We colour an edge blue if it's in G and red if not, check. Got it. Thanks very much for all your trouble. – 2017-02-02
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    "Your other claims can also be verified"? You mean, the OP's claim that for $m+6$ vertices there are at least $m+6$ monochromatic triangles? I don't believe that's true for very small values of $m;$ offhand I only see how to prove that for $m\ge4.$ – 2017-02-02
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    @victoria Indeed, color an edge blue if it's in G and red otherwise. This is a standard thing because coloring problems are more concenient in graph theory. The Omega is Landau's asymptotic notation, it means there exists a constant such that for n large enough the number of triangles is $>Cn^3$. This is thus a much stronger result than yours (which is $\Omega(n)$. – 2017-02-03
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    @bof I agree this indeed does not satisfy the OP's needs for $m<4$, thanks for noticing, I'll edit the answer. – 2017-02-03
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    And @bof the fact that the OP's result cannot be verified for m<4 is because in that case the result is incorrect.. – 2017-02-03
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    By the way, note that the OP's question does not actually ask (when $n=6$) for two monochromatic triangles; it asks us to prove the stronger (and false) assertion "that either $G$ or its complement $G'$ contains at least two triangles", i.e., either two blue triangles or two red triangles. Perhaps a translation error. – 2017-02-03
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    Thanks again; we were able to work things out. I think I mistyped; think it was m + 1 or m + 2 triangles on m + 6 vertices. My apologies, really burned out. You are all a great help and your efforts are appreciated. How do I upvote this answer? – 2017-02-03