OPEN
This is open, and cannot be resolved with a finite computation.
Let $G$ be a graph given by $n$ points in $\mathbb{R}^2$, where any two distinct points are at least distance $1$ apart, and we draw an edge between two points if they are distance $1$ apart.
Let $g(n)$ be maximal such that any such graph always has an independent set on at least $g(n)$ vertices. Estimate $g(n)$, or perhaps $\lim \frac{g(n)}{n}$.
Such graphs are always planar. Erdős initially thought that $g(n)=n/3$, but Chung and Graham, and independently Pach, gave a construction that shows $g(n)\leq \frac{6}{19}n$. Pach and Toth
[PaTo96] improved this to $g(n)\leq \frac{5}{16}n$.
Pollack
[Po85] noted that the
Four colour theorem implies $g(n)\geq n/4$, since the graph is planar. Pollack reports that Pach observed that this in for unit distance graphs the four colour theorem can be proved by a simple induction.
This lower bound has been improved to $\frac{9}{35}n$ by Csizmadia
[Cs98] and then $\frac{8}{31}n$ by Swanepoel
[Sw02]. The current record bounds are therefore\[\frac{8}{31}n \approx 0.258n \leq g(n) \leq 0.3125n=\frac{5}{16}n.\]Pollack
[Po85] also reports a letter from Erdős which poses the more general problem of, given $n$ points in $\mathbb{R}^d$ with minimum distance $1$, let $g_d(n)$ be maximal such that there always exist at least $g_d(n)$ many points which have minimum distance $>1$. Is it true that $g_d(n) \gg n/d$ in general? The upper bound $g_d(n) \ll n/d$ is trivial, considering widely spaced unit simplices.
See
[1070] for the general estimate of independence number of unit distance graphs.
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This page was last edited 02 October 2025.
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