GPT 6 Astra Ultra attempt on Math-5742
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Erdős Problem 570
Statement and scope.
For fixed $k\geq3$, is \[ R(C_k,H)\leq2m+\left\lfloor\frac{k-1}{2}\right\rfloor \tag{1} \] true for every graph $H$ with $m$ edges and no isolated vertices, once $m$ is sufficiently large depending on $k$? The assertion is known to be true. We prove the full sharp matching case for every $k$, and the triangle-versus-tree case. The extension to arbitrary $H$ is unresolved in this attempt.A construction showing that the proposed bound is sharp.
Let $t=\lfloor(k-1)/2\rfloor$ and $H=mK_2$. On $2m-1+t$ vertices, take a set $A$ of size $2m-1$. Colour edges within $A$ blue and every other edge red. The blue graph has matching number $m-1$. In a red cycle, no two successive vertices can both belong to $A$. Thus the number from $A$ is at most the number from its complement, and every red cycle has length at most $2t<k$. Consequently \[ R(C_k,mK_2)\geq2m+t. \tag{2} \]Matching upper bound via the stated matching theorem.
We use the Tutte--Berge formula as an explicit external input: for a finite graph $G$ on $N$ vertices with maximum matching size $\nu$, some vertex set $S$ satisfies \[ 2\nu=N-o(G-S)+|S|, \] where $o$ counts the connected components of odd order.Consider any colouring of $K_N$, with $N=2m+t$ and $m\geq k$. If its blue graph has no matching of size $m$, the formula gives \[ o(G-S)\geq t+2+|S|. \] Since $o(G-S)\leq N-|S|$, this also implies $|S|\leq m-1$. Hence $G-S$ has at least $m+t+1\geq k$ vertices and at least $t+2$ components. Edges joining distinct blue components are red.
If there are at least $k$ components, choose one vertex from each of $k$ of them to obtain a red clique. Otherwise, choose $k$ vertices meeting every component, first choosing one from each and then adding arbitrary available vertices. The complete multipartite red subgraph on these vertices has no part of size larger than \[ k-(t+2)+1=k-t-1=\lfloor k/2\rfloor. \] Its minimum degree is therefore at least $\lceil k/2\rceil$. It has a Hamilton cycle by the elementary argument below. This gives a red $C_k$ and proves, together with $2$, \[ R(C_k,mK_2)=2m+\left\lfloor\frac{k-1}{2}\right\rfloor \qquad(m\geq k). \tag{3} \]
The Hamilton-cycle step, proved explicitly.
Any graph on $k\geq3$ vertices of minimum degree at least $k/2$ is connected, since two components would each need at least $k/2+1$ vertices. Take a longest path $v_1,\ldots,v_a$. All neighbours of its endpoints lie on the path. Among the indices $1\leq i<a$, the conditions $v_1v_{i+1}\in E$ and $v_iv_a\in E$ hold together for some $i$: the two index sets have total size at least $k\geq a$, exceeding the available $a-1$ indices. These two edges and the path form a cycle on its $a$ vertices. If $a<k$, connectivity supplies an edge from this cycle to an outside vertex, producing a longer path. Hence $a=k$.Another full case: a triangle versus a tree.
For every tree $T$ on $m+1$ vertices, \[ R(C_3,T)=2m+1. \tag{4} \] Indeed, a graph of minimum degree at least $m$ contains every such tree: root the tree and embed its vertices one at a time after their parents. When embedding a child, at most $m-1$ other already used vertices can occupy neighbours of its parent's image, so an unused neighbour exists.It follows that a graph containing no $T$ is $(m-1)$-degenerate: every nonempty subgraph must have a vertex of degree at most $m-1$. Repeated deletion and reverse greedy colouring make it $m$-colourable. If a blue graph on $2m+1$ vertices has no $T$, one of these independent colour classes has at least three vertices, giving a red triangle. This proves the upper bound in $4$. For the lower bound, take two blue cliques of size $m$, with all cross edges red. Blue has no connected graph on $m+1$ vertices, and the red bipartite graph has no triangle. Since a tree has $m$ edges, $4$ is exactly $1$ for this case.
Remaining gap and attribution.
The matching argument cannot simply be applied to an arbitrary target: a large matching alone need not contain its adjacent edges. The tree argument uses a triangle obstruction and does not supply a cycle of a prescribed larger length. These are genuine limitations of the proofs given here.The matching proof is reconstructed from Proposition 9 of Cambie, Freschi, Morawski, Petrova and Pokrovskiy, Ramsey number of a cycle versus a graph of a given size, https://arxiv.org/abs/2601.10238. Their Theorem 3 settles the previously missing odd $k\geq7$ cases for arbitrary targets, with earlier results covering the other cycle lengths. That general induction is not proved here. The Tutte--Berge formula is the only external input used in $3$; the other displayed constructions and the tree case are proved in full. Current statement: https://www.erdosproblems.com/570, checked 24 September 2026. No novelty is claimed.