Erdős Problem #612

declared status 'open'. Formalized: no. Let $G$ be a connected graph with $n$ vertices, minimum degree $d$, and diameter $D$. Show if that $G$ contains no $K_{2r}$ and $(r-1)(3r+2)\mid d$ then$$D\leq \frac{2(r-1)(3r+2)}{2r^2-1}\frac{n}{d}+O(1),$$and if $G$ contains no $K_{2r+1}$ and $3r-1 \mid d$ then$$D\leq \frac{3r-1}{r}\frac{n}{d}+O(1).$$ Current best: It is known (see [EPPT89] for example) that any connected graph on $n$ vertices with minimum degree $d$ has diameter$$D\leq 3\frac{n}{d+1}+O(1).$$This was disproven for the case of $K_{2r}$-free graphs with $r\geq 2$ by Czabarka, Singgih, and Sz\'{e}kely [CSS21], who constructed arbitrarily large connected graphs on $n$ vertices which contain no $K_{2r}$ and have minimum degree $d$, and diameter$$\frac{6r-5}{(2r-1)d+2r-3}n+O(1),$$which contradicts the above conjecture for each fixed $r$ as $d\to \infty$. Prize: no. Tags: graph theory.

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