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Show that it is possible to color the edges of $K_n$ with at most $3 \sqrt n$ colors so that there are no monochromatic triangles.

(This was previous question and I get an explanation in the comments: Does this problem make a sense? I would expect at least, not at most. Where is my thinking wrong?)

Here is a solution: Actually, the problem is trivial. Proceed inductively.

Just split $K_n$ in to two parts with ${n\over 2}$ elements in both if $n$ is even or ${n-1\over 2}$ and ${n+1\over 2}$ elements if $n$ is odd. Color all edges between this parts with one color. In one part we will not use more than $3\sqrt{n+1\over 2}$ colors. So we have used $$3\sqrt{n+1\over 2} +1$$ colors and all we have to check if $$3\sqrt{n+1\over 2} +1\leq 3\sqrt{n}$$ which is true.


Since I found this here, problem 41 I wonder how to solve it with a probabilistic method?

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    $\begingroup$ With "at least" it would be trivial, you would give each edge a different color and voila. The point is how to do it using a small number of colors. $\endgroup$ – Michal Adamaszek Aug 13 at 11:09
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    $\begingroup$ Let $n=16$. You want to ask: is it possible to color the edges of $K_{16}$ with at least $12$ colors and have no monochromatic triangles. Sure, we color each edge with a different color and we used $120$ colors, which is at least $12$, so problem solved. If you ask can we do it with at most $12$ colors it becomes more interesting. $\endgroup$ – Michal Adamaszek Aug 13 at 11:20
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    $\begingroup$ But you can not color $K_{16}$ with two colors to not having monocolor triangle. Can you? @MichalAdamaszek $\endgroup$ – Aqua Aug 13 at 11:28
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    $\begingroup$ You should find a coloring with some number of colors less than $3\sqrt{n}$, not with every possible number of colors less than $3\sqrt{n}$. For example coloring $K_{16}$ with $4$ colors solves the problem for $n=16$. $\endgroup$ – Michal Adamaszek Aug 13 at 11:31
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    $\begingroup$ Here is a long formulation. Prove that for every integer $n$ there exists an integer $k$ such that $k\leq 3\sqrt{n}$ and such that the edges of $K_n$ can be colored with exactly $k$ colors and without monochomatic triangles. $\endgroup$ – Michal Adamaszek Aug 13 at 11:37
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Fix $3\sqrt{n}$ colors. For each edge, choose one of those $3\sqrt{n}$ colors uniformly at random to color it with, with the random choice being independent for each edge.

The probability a given three vertices form a monochromatic triangle is $(\frac{1}{3\sqrt{n}})^2 = \frac{1}{9n}$.

The number of triangles whose three edge colorings are not independent of a given triangle is $3n-8$ (the triangle itself and the at most $n-3$ triangles sharing a given edge of the given triangle).

It holds that $e\frac{1}{9n}(3n-7) \le 1$, so by Lemma II, there is a nonzero probability that there are no monochromatic triangles, as wanted.

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  • $\begingroup$ Can you tell me please, why is 2. senence true. Is not it on $^3$? $\endgroup$ – Aqua Aug 18 at 6:58
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    $\begingroup$ @Aqua If $k$ is the number of colours, the probability that a given triangle has all $3$ edges of a specified colour is $1/k^3$; the probability that all $3$ edges are the same colour (any colour) is $1/k^2$. (If I toss $3$ coins, the probability that they all come up heads is $1/8$, the probability that they all come up tails is $1/8$, but the probability that they all match is $1/4$.) $\endgroup$ – bof Aug 18 at 7:05
  • $\begingroup$ So we have event $A_i$ ...i-th triangle is monochromatic, and we want to see what is a probability that none of it occurs? @bof $\endgroup$ – Aqua Aug 18 at 7:24
  • $\begingroup$ This is very nice solution. And I would say this is pretty much straightforward proof, geedy. You just have to know LLL to check how much that greedines cost you. $\endgroup$ – Aqua Aug 18 at 7:36
  • $\begingroup$ Do you know a solution without LLL? $\endgroup$ – Aqua Aug 18 at 11:43

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