Conway thrackle constant

Description of constant

In topological graph theory, a thrackle is a drawing of a finite graph in the plane in which every pair of edges meets precisely once, either at a common endpoint or at a proper crossing. If $t(n)$ denotes the maximum number of edges in a thrackle on $n$ vertices, then Conway’s thrackle conjecture states that $t(n)=n$ for every $n\ge 3$. [FP2011-def-tn-conj]

For a finite graph $G$, write $V(G)$ for its vertex set and $E(G)$ for its edge set. We define

\[C_{78} \ :=\ \sup\left\{ \frac{\lvert E(G)\rvert}{\lvert V(G)\rvert} : G \text{ admits a thrackle drawing in the plane} \right\}.\]

Equivalently,

\[C_{78} \ =\ \sup_{n\ge 1}\frac{t(n)}{n}.\]

Thus Conway’s thrackle conjecture is equivalent to the assertion that $C_{78}=1$. [FP2011-def-tn-conj]

At present, the best established range is

\[1\ \le\ C_{78}\ \le\ 1.393.\]

The lower bound comes from thrackled cycles, and the upper bound from Xu’s current record bound $m\le 1.393(n-1)$ as recorded in recent survey-style sources. [FP2011-open-sharp] [KSTZ2025-current-best-special]

Known upper bounds

Bound Reference Comments
$2$ [CN2000] Lovász–Pach–Szegedy proved $m\le 2n-3$, hence $C_{78}\le 2$. [CN2000-ub-lps-cn]
$\frac{3}{2}$ [CN2000] Cairns–Nikolayevsky improved the linear bound to $m\le \frac{3}{2}(n-1)$, hence $C_{78}\le \frac{3}{2}$. [CN2000-ub-lps-cn]
$\frac{167}{117}$ [FP2011] Fulek–Pach proved $t(n)\le \frac{167}{117}n<1.428n$, so $C_{78}\le \frac{167}{117}$. [FP2011-def-tn-conj]
$1.4$ [FP2019] Previous record, attributed there to Goddyn–Xu: $m\le 1.4n$, hence $C_{78}\le 1.4$. [FP2019-ub-gx-fp]
$1.3984$ [FP2019] Fulek–Pach improved the bound to $m\le 1.3984n$, hence $C_{78}\le 1.3984$. [FP2019-ub-gx-fp]
$1.393$ [KSTZ2025] Current best upper bound: Xu’s $m\le 1.393(n-1)$ bound, as recorded in recent literature, gives $C_{78}\le 1.393$. [KSTZ2025-current-best-special]

Known lower bounds

Bound Reference Comments
$1$ [FP2011] Any cycle of length at least five can be drawn as a thrackle, and such a cycle has $\lvert E\rvert/\lvert V\rvert=1$. [FP2011-open-sharp]

References

Contribution notes

Prepared with assistance from ChatGPT 5.2 Pro.