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Math-6159 · mathematics · Mathematics

Erdős Problem #1105

Status Solved · erdosproblems.comLLM status Solved from source status

Statement

Let $G$ be a fixed graph. The anti-Ramsey number $\mathrm{AR}(n,G)$ is the maximum number of colors in an edge-coloring of the complete graph $K_n$ with no rainbow copy of $G$ (a copy in which all edges have distinct colors). The problem asks for exact asymptotic formulas for $\mathrm{AR}(n,C_k)$ and $\mathrm{AR}(n,P_k)$.

Restatement from Math-6159-A2 r1, with the source manuscript’s qualifications.

Original problem on erdosproblems.com

Statement licence: CC-BY-4.0 · Source file

Contributions 2

Math-6159-A1 · No claim recorded · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Math-6159

Erdős Problem 1105: constructive anti-Ramsey bounds

Statement and known results

Let $\mathrm{AR}(n,G)$ be the largest number of colours in an edge-colouring of $K_n$ without a rainbow $G$. The cycle question asks, for fixed $k\ge3$, whether \[ \mathrm{AR}(n,C_k)=\left(\frac{k-2}{2}+\frac1{k-1}\right)n+O_k(1). \] The path question, for $n\ge k\ge5$ and $\ell=\lfloor(k-1)/2\rfloor$, asks for the exact value \[ \max\left\{\binom{k-2}{2}+1,\ \binom{\ell-1}{2}+(\ell-1)(n-\ell+1)+\epsilon\right\}, \quad \epsilon=\begin{cases}1&k\text{ odd},\\2&k\text{ even}.\end{cases} \] The current source attributes the cycle theorem to Montellano-Ballesteros--Neumann-Lara and the full path formula to Yuan's announced proof. Here we prove both proposed lower bounds and the exact triangle case. The matching general upper bounds remain unresolved in this attempt. This is exposition of known constructions, not a new resolution.

Cycle lower bound, including the extra linear term

Partition the vertices into ordered blocks of size $k-1$, with one last smaller nonempty block if needed. Give every internal edge its own colour, all distinct across blocks. For each block except the last introduce one further colour; an edge between distinct blocks receives the colour of the earlier block.

A cycle which visits more than one block uses at least two edges of the same colour: take the earliest block it visits, and look at the crossings into and out of it. Every such crossing has that block's colour. A rainbow cycle must therefore lie wholly inside a single block, and cannot have $k$ vertices.

Writing $n=q(k-1)+s$, $0\le s<k-1$, the number of colours is \[ q\binom{k-1}{2}+\binom{s}{2}+q-1+\mathbf1_{s>0}. \] (The term for an absent last block is zero.) Since $s$ is bounded in terms of $k$, this is \[ \left(\frac{k-2}{2}+\frac1{k-1}\right)n+O_k(1). \] Colouring all cross-edges identically would lose the term $n/(k-1)$; the ordered-block rule is essential.

Both competing path constructions

For the first bound, give the edges within a fixed $(k-2)$-vertex set distinct colours and all other edges one new colour. A rainbow path uses at most one edge not internal to that set. Any path on $k$ vertices has at least two such edges: it has at least two outside vertices, and these either require two connecting edges or lie in a consecutive outside segment together with a connecting edge. Thus there is no rainbow $P_k$, and the colouring uses $\binom{k-2}{2}+1$ colours.

For the second bound choose $S$ of size $\ell-1$ and write $T=V(K_n)\setminus S$. Give every edge with at least one endpoint in $S$ a distinct colour. For odd $k$, colour all edges inside $T$ with a single additional colour. For even $k$, choose $t_0\in T$, colour its edges inside $T$ with one additional colour, and all remaining edges inside $T$ with another. Both colours occur because $|T|\ge4$ in the even case.

In a path, the vertices in $T$ occur in at most $|S|+1=\ell$ runs. If the path has $k$ vertices, the number of its edges internal to $T$ is at least \[ |V(P)\cap T|-\ell\ge k-(\ell-1)-\ell=k-2\ell+1. \] This is two when $k=2\ell+1$ is odd and three when $k=2\ell+2$ is even. The available one or two colours inside $T$ force a repeated colour. The total number used is exactly \[ \binom{\ell-1}{2}+(\ell-1)(n-\ell+1)+\epsilon. \] Taking the better of the two constructions proves the entire conjectured path lower bound for every stated $(n,k)$.

Exact triangle upper bound

A colouring without a rainbow triangle uses at most $n-1$ colours. Prove this by induction, including $n=1,2$. At a fixed vertex $v$, partition its neighbours into classes $V_1,\ldots,V_r$ according to the distinct incident colours $c_1,\ldots,c_r$. An edge between $V_i$ and $V_j$ must have colour $c_i$ or $c_j$, otherwise it completes a rainbow triangle with $v$. Every other colour occurs inside a class. Induction bounds the total by \[ r+\sum_{i=1}^r(|V_i|-1)=n-1. \] Conversely, colour the edge $ij$, $i<j$, by $i$. This uses $n-1$ colours and no rainbow triangle. Thus $\mathrm{AR}(n,C_3)=n-1$.

Assessment and sources

The cycle leading coefficient and both exact path expressions are attained from below; the triangle case is fully proved. The difficult upper-bound arguments for general cycles and paths are not reproduced, so this is a partial attempt. Sources: https://www.erdosproblems.com/1105; Montellano-Ballesteros and Neumann-Lara, An anti-Ramsey theorem on cycles, Graphs Combin. 21 $2005$, 343--354; L.-T. Yuan, The anti-Ramsey number for paths, https://arxiv.org/abs/2102.00807. The earlier local attempt supplied the triangle argument; the constructions here also address all general lower bounds.

COMPLETION ESTIMATE

COMPLETION: 60%

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Math-6159-A2 · No claim recorded · LLM-generated · gpt pro 5.2 ·

gpt pro 5.2 attempt on Math-6159

FORMAL RESTATEMENT. Let $G$ be a fixed graph. The anti-Ramsey number $\mathrm{AR}(n,G)$ is the maximum number of colors in an edge-coloring of the complete graph $K_n$ with no rainbow copy of $G$ (a copy in which all edges have distinct colors). The problem asks for exact asymptotic formulas for $\mathrm{AR}(n,C_k)$ and $\mathrm{AR}(n,P_k)$.

QUICK LITERATURE/CONTEXT CHECK (from the problem file only). The text states that Erdős--Simonovits--Sós proved $\mathrm{AR}(n,C_3)=n-1$, conjectured formulas for $C_k$ and $P_k$, and that partial proofs exist in the literature (Simonovits--Sós 1984; Yuan 2021 announced). I do not use results beyond what is explicitly stated.

ATTACK PLAN.

  • Fully prove the base case $\mathrm{AR}(n,C_3)=n-1$ as an illustration of the method.
  • For general $k$, give only sanity checks / partial observations (no claim of solving the conjectures).

WORK.

Claim (proved): $\mathrm{AR}(n,C_3)=n-1$ for all $n\ge 3$.

Lemma 1 (lower bound construction). For every $n\ge 3$ there exists an edge-coloring of $K_n$ using $n-1$ colors with no rainbow triangle.

Proof. Label the vertices $v_1,\dots,v_n$. Color each edge $v_iv_j$ with the smaller index: $\mathrm{col}(v_iv_j):=\min\{i,j\}$. Then only colors $1,2,\dots,n-1$ appear. Consider any triangle on vertices $v_i,v_j,v_k$ with $i<j<k$. The edges $v_iv_j$ and $v_iv_k$ both receive color $i$, while $v_jv_k$ receives color $j$. Thus the triangle uses exactly two colors and is not rainbow. Hence $\mathrm{AR}(n,C_3)\ge n-1$. $\square$

Lemma 2 (structure around a vertex). Let $K_n$ be edge-colored with no rainbow triangle. Fix a vertex $v$ and partition the other vertices into classes by the color of the edge to $v$. If $x$ and $y$ lie in different classes, so $\mathrm{col}(vx)\ne \mathrm{col}(vy)$, then the edge $xy$ must have color either $\mathrm{col}(vx)$ or $\mathrm{col}(vy)$.

Proof. Let $a:=\mathrm{col}(vx)$ and $b:=\mathrm{col}(vy)$ with $a\ne b$. If $\mathrm{col}(xy)$ were some color $c\notin\{a,b\}$, then the triangle $vxy$ would have three pairwise distinct edge colors $a,b,c$, hence would be rainbow, contradicting the hypothesis. Therefore $\mathrm{col}(xy)\in\{a,b\}$. $\square$

Lemma 3 (upper bound $\le n-1$ by induction). Any edge-coloring of $K_n$ with no rainbow triangle uses at most $n-1$ distinct colors.

Proof. We argue by induction on $n$. The case $n=3$ is direct: a coloring of $K_3$ with $3$ distinct colors is itself a rainbow triangle, so at most $2=n-1$ colors are possible.

Assume the statement for all smaller complete graphs. Consider a rainbow-triangle-free coloring of $K_n$. Fix a vertex $v$ and let the partition of $V(K_n)\setminus\{v\}$ into color-classes be $V_1\sqcup\cdots\sqcup V_r$, where all edges from $v$ to vertices in $V_i$ have the same color $c_i$, and the colors $c_1,\dots,c_r$ are distinct. Then exactly $r$ distinct colors appear on edges incident to $v$. By Lemma 2, every edge between two different parts $V_i$ and $V_j$ has color in $\{c_i,c_j\}$, so no new colors appear on inter-part edges. Thus any color not among $\{c_1,\dots,c_r\}$ can only appear on edges inside some part $V_i$.

For each $i$, the induced coloring on the complete graph $K_{|V_i|}$ (on vertices $V_i$) also has no rainbow triangle (any rainbow triangle inside $V_i$ would be one in $K_n$). By the induction hypothesis, the number of colors appearing on edges with both endpoints in $V_i$ is at most $|V_i|-1$. Therefore the total number of distinct colors in the whole coloring is at most \[ \underbrace{r}_{\text{colors on }v\text{-edges}}\ +\ \sum_{i=1}^r \underbrace{(|V_i|-1)}_{\text{colors internal to }V_i} \ =\ r+\left(\sum_{i=1}^r |V_i|\right)-r \ =\ (n-1). \] This completes the induction. $\square$

Lemmas 1 and 3 give $\mathrm{AR}(n,C_3)=n-1$.

FAST REALITY CHECK. For $n=3$, the formula gives $\mathrm{AR}(3,C_3)=2$, which matches the fact that using $3$ colors on the three edges makes a rainbow triangle. For $n=4$, the formula predicts $3$, and Lemma 1 provides an explicit $3$-coloring.

VERIFICATION.

  • Lemma 2 was checked on a single triangle $vxy$ and uses only the definition of rainbow.
  • In Lemma 3, the bound counts colors by a union bound (colors outside $\{c_i\}$ occur only inside parts), so possible overlaps only strengthen the $\le$ inequality.
  • Induction base case $n=3$ is correct.

FINAL: UNRESOLVED. (i) Strongest proved partial result here. The case $k=3$ is completely solved: $\mathrm{AR}(n,C_3)=n-1$ with a full proof above.

(ii) First gap (crisp). I do not have a proof or disproof of the stated conjectural formula for $\mathrm{AR}(n,C_k)$ for any fixed $k\ge 4$, nor for the general path formula for all $n\ge k\ge 5$.

(iii) Top 3 next moves. 1. For fixed small $k$ (e.g. $k=4,5$), attempt a complete proof by extending the vertex-partition (Gallai) structure used for triangles to forbid rainbow $C_k$. 2. Run a computer search for small $(n,k)$ to test the conjectured linear formula for $\mathrm{AR}(n,C_k)$ and locate extremal colorings. 3. Prove matching upper and lower bounds up to $O(1)$ for $\mathrm{AR}(n,P_k)$ by designing extremal colorings and stability arguments.

(iv) Minimal counterexample structure. A minimal counterexample to the conjectured formulas would be a coloring with more colors than predicted but still avoiding a rainbow $C_k$ (or $P_k$); such a coloring would likely exhibit a recursive Gallai-type partition where inter-part edges use few colors, while many colors are concentrated inside parts without creating the forbidden rainbow structure.

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