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Is minimum colors needed to assign colors to edges of complete graph $K_n$ so that every even simple cycle contains an odd number ($>1$) of colors much larger than $(\log n)^\beta$ or $n^{\frac{1}\beta}$ with some fixed $\beta>1$?

What is a good upper bound?

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  • $\begingroup$ one? [the rest of this comment is just filler to get to the required number of characters...] $\endgroup$ Commented Jul 24, 2015 at 16:14
  • $\begingroup$ You are right. I want to make number of colors assigned to edges of even cycles being not balanced. So one is the only trivial coloring that needs to be avoided. I missed it on first try. $\endgroup$
    – Turbo
    Commented Jul 24, 2015 at 21:44
  • $\begingroup$ Is it clear that such a coloring should exist for any number of colors? Every 4-cycle is required to use exactly 3 colors, which seems rather restrictive. $\endgroup$
    – Aravind
    Commented Jul 25, 2015 at 6:58
  • $\begingroup$ Good question. Answer below is 'if it exists, it satisfies some properties' which is sufficient for me. May be you can post a new question which would be very interesting. $\endgroup$
    – Turbo
    Commented Jul 25, 2015 at 7:04

2 Answers 2

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Yes.

Suppose that we did this with $k>1$ colors. For any pair of colors, the graph consisting of edges with these two colors obviously does not contain any even cycle. An even cycle free graph has at most $3(n-1)/2$ edges (see https://math.stackexchange.com/questions/438853/prove-that-the-maximum-number-of-edges-in-a-graph-with-no-even-cycles-is-floor3 ). Then add up these inequalities for any pair of colors, and observe that we counted every edge of $K_n$ exactly $k-1$ times.

$$ \binom{n}{2} (k-1) \leq \binom{k}{2}3(n-1)/2 $$ Thus ignoring constants we have that $$ n^2k \leq k^2 n $$ $$ n \leq k. $$ Note that $k\leq 3(n-1)/2$ also holds, as if there were more than this amount of colors, we could take an edge from any color, and the resulting graph would have an even cycle, obviously with an even number of colors. Thus roughly we have the bounds

$$ \frac{2}{3}n \leq k \leq \frac{3}{2} n. $$

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  • $\begingroup$ Should it be $n\leq k\leq 3n/2$? $\endgroup$
    – Turbo
    Commented Jul 24, 2015 at 23:11
  • $\begingroup$ No, it is actually $\frac{2}{3} \leq k \leq \frac{3}{2}$. At the first part of my answer I ignored constants. $\endgroup$ Commented Jul 24, 2015 at 23:14
  • $\begingroup$ Different question mathoverflow.net/questions/212276/… $\endgroup$
    – Turbo
    Commented Jul 25, 2015 at 1:34
  • $\begingroup$ Is Aravind's answer correct? $\endgroup$
    – Turbo
    Commented Jul 25, 2015 at 10:02
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For sufficiently large $n$ (say $n \geq 20$), there is no edge coloring of $K_n$ which ensures that every 4-cycle gets exactly 3 colors. In particular, the coloring asked in the question is not possible.

(i) There cannot be a monochromatic $P_3$ as this can be completed to a 4-cycle with at most 2 colors. Thus each color class is a star forest.

(ii) If va, vb, vc have the same color 1, then ab, bc and ca must have different colors other than 1.

(iii) From (ii), it is not possible for a vertex to have four edges incident on it of the same color (this would imply that the edges among those 4 neighbors of v are all distinctly colored).

(iv) It is not possible for a vertex to have more than one color repeating on edges incident on it.

(v) Thus for a vertex v except for 3 neighbors, all other edges out of v are distinctly colored. Now consider another vertex w and the edges from v,w to the remaining vertices. We easily get either a monochromatic 4-cycle or a 4-cycle with all distinct colors.

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  • $\begingroup$ What is term $P_3$? $\endgroup$
    – Turbo
    Commented Jul 25, 2015 at 11:05
  • $\begingroup$ I meant $P_4$ rather, path on 4 vertices. $\endgroup$
    – Aravind
    Commented Jul 25, 2015 at 19:39
  • $\begingroup$ please correct your answer, I do not understand the proof though. why dont you post as a question so others can figure out? $\endgroup$
    – Turbo
    Commented Jul 25, 2015 at 22:46

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