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Math-677 · mathematics · Geometry · Mathematics

Erdős Problem #103

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Statement

Verbatim problem statement (from file). "Let $h(n)$ count the number of incongruent sets of $n$ points in $\R^2$ which minimise the diameter subject to the constraint that $d(x,y)\ge 1$ for all points $x\ne y$. Is it true that $h(n)\to\infty$? It is not even known whether $h(n)\ge 2$ for all large $n$."

Definitions and conventions. For $n\ge 1$, consider the class $\mathcal{C}_n$ of all $n$-point sets $A\subset\R^2$ with pairwise distances at least $1$. Define the minimal achievable diameter \[\Delta_n := \inf\{\diam(A): A\in\mathcal{C}_n\}. \] Call $A\in\mathcal{C}_n$ diameter-minimising if $\diam(A)=\Delta_n$. Two sets are congruent if one can be mapped to the other by an isometry of $\R^2$. Define $h(n)$ as the number of congruence classes of diameter-minimising sets in $\mathcal{C}_n$.

Question. Does $h(n)\to\infty$ as $n\to\infty$?

Restatement from Math-677-A2 r1, with the source manuscript’s qualifications.

Original problem on erdosproblems.com

Statement licence: CC-BY-4.0 · Source file

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Math-677-A1 · No claim recorded · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Math-677

Erdős Problem 103: minimum-diameter packings and movable points

1. Statement and scope

Let $\Delta(n)$ be the minimum diameter of an $n$-point planar set whose distinct points have mutual distances at least one. Let $h(n)$ be the number of its optimal configurations up to congruence, allowing reflections and ignoring labels. Does $h(n)\to\infty$? This remains unresolved. We prove existence and size bounds, determine the first four orders, and give a sufficient geometric condition for infinitely many optimal configurations at a fixed order.

2. Existence and the diameter scale

The case $n=1$ has $\Delta(1)=0$. For $n\ge2$, a subset of a $q\times q$ unit grid, with $q=\lceil\sqrt n\rceil$, gives \[ \Delta(n)\le\sqrt2(q-1).\tag{1} \] Choose a sequence of admissible configurations whose diameters tend to the infimum, label their points, and translate each first point to the origin. Discarding finitely many terms, every point lies in a common bounded closed disk, by $1$ and an extra margin of one. A convergent subsequence in the finite product of these disks has a limit configuration. The inequalities $|p_i-p_j|\ge1$ are closed, so the limiting points remain distinct and admissible. Diameter is the maximum of finitely many continuous distance functions, hence is continuous. The limiting set attains $\Delta(n)$.

The open radius-$1/2$ disks about any admissible points are disjoint and lie inside a radius-$(D+1/2)$ disk about one of the points, where $D$ is the diameter. Area comparison gives \[ \Delta(n)\ge(\sqrt n-1)/2.\tag{2} \] Thus $\Delta(n)=\Theta(\sqrt n)$ as $n\to\infty$. This order-of-magnitude statement does not count optima. Also, every optimum for $n\ge2$ has a pair at distance exactly one: otherwise dilation by the reciprocal of its minimum distance would strictly decrease its positive diameter while preserving admissibility.

3. Four small orders, with the equality case checked

For $n=1,2,3$ the unique optima up to congruence are respectively one point, a unit segment, and a unit equilateral triangle. Indeed, for $n=2,3$ the diameter is at least one, and attaining one forces every distance to equal one.

For $n=4$, a unit square has diameter $\sqrt2$. We show it is the unique optimum. If the four points form a convex quadrilateral, at least one interior angle is at least $90$ degrees. The two incident sides both have length at least one, so the cosine rule shows that the diagonal joining their other endpoints has length at least $\sqrt2$. If the diameter is exactly $\sqrt2$, no interior angle can exceed $90$ degrees, since then the same inequality is strict. The four angles sum to $360$ degrees, so all are right angles. The rectangle has side lengths at least one and diagonal at most $\sqrt2$, forcing both sides to equal one.

In the remaining cases, if one point lies strictly inside the triangle formed by the other three, the three angular gaps around it are less than $180$ degrees and sum to $360$ degrees. One is at least $120$ degrees. Its two rays end at points of distance at least one from the interior point, and the cosine rule then gives a mutual distance at least $\sqrt3>\sqrt2$. If a point lies on an edge of the triangle, the edge's two parts each have length at least one, so the diameter is at least two. A collinear set of four points has diameter at least three. These cases exhaust the nonconvex configurations. Therefore \[ \Delta(4)=\sqrt2,\qquad h(1)=h(2)=h(3)=h(4)=1.\tag{3} \]

4. A movable-point criterion for infinitely many optima

Suppose an optimal configuration $P$ of order $n\ge3$ contains a point $v$ satisfying \[ 1<|v-q|<\Delta(n)\qquad\text{for every }q\in P\setminus\{v\}.\tag{4} \] Such a point is sometimes called a rattler; condition $4$, rather than terminology, is what is used. Since all its incident distances are strictly below the diameter, a pair of the other, fixed points attains $\Delta(n)$. There are finitely many strict inequalities in $4$, so a sufficiently small open disk about $v$ consists entirely of positions satisfying them. Moving $v$ within that disk preserves minimum separation and leaves the fixed diameter pair unchanged. Every resulting configuration is therefore optimal.

These motions yield infinitely many incongruent sets, not just infinitely many coordinate descriptions. Fix two distinct stationary points $a,b\in P\setminus\{v\}$. In any one congruence class of $n$-point sets, the set $S$ of pairwise distance values is finite. If a moved configuration belongs to that class, its moving point $x$ has $|x-a|,|x-b|\in S$. For each pair of radii, the corresponding two circles have at most two intersections. Thus this congruence class accounts for at most $2|S|^2$ possible moving positions. An open disk cannot be covered by finitely many finite sets, so it contains positions in infinitely many distinct congruence classes; in fact there are continuum many.

A useful contrapositive is that if an optimal configuration belongs to an order with finite $h(n)$, every one of its points must be incident to a pair at distance one or to a pair at distance $\Delta(n)$. This is a necessary condition for finiteness, not a proof of rigidity or a sufficient condition for uniqueness.

5. Remaining gap

No argument above proves that large optimal configurations contain a point satisfying $4$, or otherwise produces many optimal shapes. The elementary uniqueness at small orders does not bear on the eventual question. In logical terms, failure of $h(n)\to\infty$ would give a bounded subsequence of values; it would not by itself imply eventual uniqueness or even an eventual uniform bound.

Source: https://www.erdosproblems.com/103. The existence, packing bounds, equality cases and movable-point criterion are proved in this attempt.

Completion rate estimate: 25%.

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Math-677-A2 · No claim recorded · LLM-generated · gpt pro 5.2 ·

gpt pro 5.2 attempt on Math-677

Erdős Problem #103

FORMAL RESTATEMENT

Verbatim problem statement (from file). "Let $h(n)$ count the number of incongruent sets of $n$ points in $\R^2$ which minimise the diameter subject to the constraint that $d(x,y)\ge 1$ for all points $x\ne y$. Is it true that $h(n)\to\infty$? It is not even known whether $h(n)\ge 2$ for all large $n$."

Definitions and conventions. For $n\ge 1$, consider the class $\mathcal{C}_n$ of all $n$-point sets $A\subset\R^2$ with pairwise distances at least $1$. Define the minimal achievable diameter \[\Delta_n := \inf\{\diam(A): A\in\mathcal{C}_n\}. \] Call $A\in\mathcal{C}_n$ diameter-minimising if $\diam(A)=\Delta_n$. Two sets are congruent if one can be mapped to the other by an isometry of $\R^2$. Define $h(n)$ as the number of congruence classes of diameter-minimising sets in $\mathcal{C}_n$.

Question. Does $h(n)\to\infty$ as $n\to\infty$?

QUICK LITERATURE/CONTEXT CHECK

No external browsing was used. The file provides no explicit constructions or bounds for $h(n)$ beyond the fact that it is not known whether $h(n)\ge 2$ for all large $n$. Below I prove: (i) $\Delta_n$ is achieved (a minimiser exists), and (ii) crude asymptotics $\Delta_n=\Theta(\sqrt{n})$.

ATTACK PLAN

  • Existence/compactness track: Show the infimum diameter $\Delta_n$ is achieved by some configuration (so $h(n)\ge 1$ is well-defined).
  • Quantitative track: Establish rough bounds for $\Delta_n$ using packing arguments and explicit grid constructions.
  • Multiplicity track: To show $h(n)\to\infty$, one would need to build many noncongruent minimisers or prove that minimisers are not eventually unique; absent such a construction, aim for partial statements (e.g., non-uniqueness for infinitely many $n$).

WORK

FAST REALITY CHECK (very small $n$ by hand).
  • $n=1$: $\Delta_1=0$, and (up to congruence) the minimiser is unique, so $h(1)=1$.
  • $n=2$: $\Delta_2=1$ (two points at distance 1), unique up to congruence, so $h(2)=1$.
  • $n=3$: $\Delta_3=1$ (equilateral triangle of side 1), unique up to congruence, so $h(3)=1$.
(I did not compute $h(n)$ for $n\ge 4$; continuous optimisation makes brute force nontrivial.)
Lemma (Existence of a diameter minimiser): For each integer $n\ge 1$, there exists a set $A_n\subset\R^2$ of $n$ points with pairwise distances at least $1$ and with $\diam(A_n)=\Delta_n$. In particular, $h(n)$ is well-defined and satisfies $h(n)\ge 1$ for all $n$.
Proof: Fix $n\ge 1$. Let $(A^{(m)})_{m\ge 1}$ be a sequence in $\mathcal{C}_n$ such that \[\diam(A^{(m)})\to \Delta_n\quad\text{as }m\to\infty. \] For each $m$, choose an ordering $A^{(m)}=\{a^{(m)}_1,\dots,a^{(m)}_n\}$. Apply a translation so that $a^{(m)}_1=0$ (the origin) for all $m$; this does not change pairwise distances or diameter. Let $D_m=\diam(A^{(m)})$. Then every point of $A^{(m)}$ lies in the closed disk $\overline{B}(0,D_m)$. Since $D_m$ is bounded (it converges), all points lie in a fixed compact disk $\overline{B}(0,R)$ for some $R$ and all large $m$.

By compactness of $\overline{B}(0,R)^n$, the sequence of ordered $n$-tuples $(a^{(m)}_1,\dots,a^{(m)}_n)$ has a convergent subsequence; relabel the subsequence as $(a^{(m)}_i)$ again, and let $a_i\in\overline{B}(0,R)$ be the limit of $a^{(m)}_i$ for each $i$. Set $A:=\{a_1,\dots,a_n\}$ (as a set; if some limits coincide we address this below).

Claim 1: The limit points are distinct and satisfy pairwise distance $\ge 1$. Indeed, for any $i\ne j$, the function $f(x,y)=\|x-y\|$ is continuous, so \[\|a_i-a_j\|=\lim_{m\to\infty}\|a^{(m)}_i-a^{(m)}_j\|\ge 1, \] since each $A^{(m)}\in\mathcal{C}_n$. In particular $\|a_i-a_j\|\ge 1$ implies $a_i\ne a_j$. Thus $A\in\mathcal{C}_n$.

Claim 2: $\diam(A)\le \liminf_{m\to\infty}\diam(A^{(m)})=\Delta_n$. For any $i,j$, by continuity we have \[\|a_i-a_j\|=\lim_{m\to\infty}\|a^{(m)}_i-a^{(m)}_j\|\le \liminf_{m\to\infty} D_m. \] Taking the maximum over $i,j$ gives $\diam(A)\le \liminf D_m=\Delta_n$. But by definition $\Delta_n\le \diam(A)$ for all $A\in\mathcal{C}_n$, hence $\diam(A)=\Delta_n$. So $A$ is a minimiser. ∎

Proposition (Crude asymptotics for the minimal diameter): There exist absolute constants $c,C>0$ such that for all $n\ge 1$, \[c\sqrt{n} \le \Delta_n \le C\sqrt{n}. \] For instance, one may take $c=\sqrt{2/\pi}/2$ and $C=2$.
Proof: Lower bound. Let $A\in\mathcal{C}_n$ be any $n$-point set with pairwise distances at least $1$, and let $D=\diam(A)$. Apply the packing Lemma from Problem #100 (which uses only the separation condition) to obtain $D\ge c_0\sqrt{n}-1$ for some absolute $c_0>0$. For $n\ge 4$ this implies $D\ge (c_0/2)\sqrt{n}$ by absorbing the $-1$ term; for $n<4$ one can adjust constants. Thus $\Delta_n\ge c\sqrt{n}$ for some absolute $c>0$.

Upper bound. Let $m=\lceil\sqrt{n}\rceil$. Consider the integer grid points \[G_m:=\{(i,j): i,j\in\{0,1,\dots,m-1\}\}\subset\R^2. \] This set has $m^2\ge n$ points with pairwise distances at least $1$ (because distinct integer lattice points have Euclidean distance at least $1$). Select any $n$ of these points; call the resulting set $A$. Then $A\in\mathcal{C}_n$. Its diameter is at most the diagonal of the $m\times m$ square: \[\diam(A)\le \sqrt{(m-1)^2+(m-1)^2}\le \sqrt{2}\,m \le 2\sqrt{n}, \] where the last inequality uses $m\le \sqrt{n}+1$ and a crude bound. Therefore $\Delta_n\le 2\sqrt{n}$. Combine to get $c\sqrt{n}\le \Delta_n\le C\sqrt{n}$. ∎

VERIFICATION

  • The existence proof uses only compactness and continuity; the constraint $d(x,y)\ge 1$ is closed, so it survives limits.
  • The proof checks that points do not collide in the limit because the distance lower bound persists.
  • The upper bound construction uses a square grid; pairwise distances among distinct lattice points are indeed $\ge 1$.
  • Constants are not optimised; only $\Theta(\sqrt{n})$ is claimed.

FINAL

UNRESOLVED.
  1. Strongest proved partial result here: A diameter minimiser exists for every $n$ (so $h(n)\ge 1$), and the minimal diameter satisfies $\Delta_n=\Theta(\sqrt{n})$.
  2. First gap (crisp): Construct, for infinitely many $n$, two noncongruent diameter-minimising configurations of $n$ points with pairwise distances $\ge 1$ (i.e. prove $h(n)\ge 2$ infinitely often), or else prove eventual uniqueness.
  3. Top 3 next moves (concrete):
    1. Compute $\Delta_n$ and enumerate minimisers for small $n$ (say $n\le 15$) using numerical optimisation + rigorous verification of optimality, to detect patterns of non-uniqueness.
    2. Prove a rigidity lemma: if $A$ minimises diameter, then a positive fraction of points must lie on the boundary of the minimal enclosing disk, limiting degrees of freedom (could imply finiteness or discreteness of minimisers).
    3. Produce explicit competing constructions with the same diameter: e.g. take two different subsets of a dense lattice in a nearly minimal enclosing disk and try to show both achieve the same $\Delta_n$.
  4. What a minimal counterexample would likely look like: If $h(n)$ does not tend to infinity, then beyond some $N$ there would be only boundedly many congruence classes of optimal configurations. A minimal counterexample to $h(n)\to\infty$ would thus be an infinite tail of $n$ for which the minimiser is unique (or comes from a small finite family) and varies continuously with $n$.

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