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Math-895-A2 · revision r1 · on Math-895

gpt pro 5.2 attempt on Math-895

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Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

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FORMAL RESTATEMENT. Fix $d\ge 1$. For a finite set $A\subseteq\mathbb{R}^d$ of size $n$, define its diameter \[\operatorname{diam}(A):=\max\{\|x-y\|: x,y\in A\}. \] Assume that for any two distinct unordered pairs $\{x,y\}\ne\{x',y'\}$ with $x\ne y$ and $x'\ne y'$, the corresponding distances satisfy \[\big|\|x-y\| - \|x'-y'\|\big|\ge 1. \] (Equivalently: the multiset of $\binom{n}{2}$ pairwise distances is $1$-separated.) The conjecture asks whether necessarily \[ \operatorname{diam}(A) \ge (1+o(1))\, n^2\quad\text{as } n\to\infty. \]

Potential ambiguity. The phrase "all pairwise distances differ by at least $1$" is interpreted above as a separation condition between distances of distinct pairs. This is consistent with the statement that the trivial diameter lower bound is $\binom n2$ (see Lemma 670.2).

QUICK LITERATURE/CONTEXT CHECK. The prompt states that the lower bound $\binom n2$ is trivial and that Erdős proved the $(1+o(1))n^2$ lower bound for $d=1$. Per project rules, I do not use external sources beyond this.

ATTACK PLAN.

  • Prove the "trivial" lower bound $\operatorname{diam}(A)\ge \binom n2$ carefully, including the necessary preliminary fact that the minimum distance is at least $1$ when $n\ge 3$.
  • Do small-$n$ checks and a computation in $d=1$ (Golomb ruler search) to see the gap between $\binom n2$ and the known/expected $\Theta(n^2)$ behavior.
  • No general improvement beyond $\binom n2$ is proved here for $d\ge 2$; conclude unresolved.

WORK.

Fast reality check (tiny $n$). For $n=2$, there is only one distance, so the separation condition is vacuous; the diameter can be any positive number. For $n=3$, one can realize distances $\{1,2,3\}$ by collinear points at $0,1,3$ in $\mathbb{R}$, giving diameter $3=\binom 32$. For $n=4$, collinear points at $0,1,4,6$ realize all distances $1,2,3,4,5,6$, giving diameter $6=\binom 42$. So the trivial bound can be tight for small $n$.

Lemma 670.1 (Minimum distance is at least $1$ for $n\ge 3$). Let $A\subseteq\mathbb{R}^d$ be a set of $n\ge 3$ points satisfying the distance-separation condition. Then the minimum pairwise distance in $A$ is at least $1$.

Proof. Let $\delta$ be the minimum distance between distinct points of $A$, and choose distinct points $P,Q\in A$ with $\|P-Q\|=\delta$. Since $n\ge 3$, choose a third point $R\in A\setminus\{P,Q\}$. The distances $\|P-R\|$ and $\|Q-R\|$ correspond to distinct unordered pairs from $\{P,Q\}$, so each must differ from $\delta$ by at least $1$. Also, by minimality of $\delta$, both are at least $\delta$. Therefore \[ \|P-R\|\ge \delta+1,\qquad \|Q-R\|\ge \delta+1. \] Moreover, since the unordered pairs $\{P,R\}$ and $\{Q,R\}$ are distinct, the separation condition gives $|\|P-R\|-\|Q-R\||\ge 1$. Thus one of these two distances is at least the other plus $1$. Without loss of generality assume $\|Q-R\|\ge \|P-R\|+1$. Then combining with $\|P-R\|\ge \delta+1$ yields \[\|Q-R\|\ge (\delta+1)+1 = \delta+2. \] Now apply the triangle inequality in $\mathbb{R}^d$ to the triangle $PQR$: \[ \|Q-R\| \le \|Q-P\| + \|P-R\| = \delta + \|P-R\|. \] Using $\|Q-R\|\ge \|P-R\|+1$ gives \[ \|P-R\|+1 \le \delta + \|P-R\|, \] so $1\le \delta$. Therefore the minimum distance $\delta\ge 1$. $\square$

Lemma 670.2 (Trivial diameter lower bound). Let $A\subseteq\mathbb{R}^d$ be a set of $n\ge 3$ points satisfying the distance-separation condition. Then \[ \operatorname{diam}(A)\ge \binom{n}{2}. \]

Proof. There are $N:=\binom{n}{2}$ unordered pairs of distinct points in $A$, hence $N$ pairwise distances. Let these distances be listed in increasing order: \[d_1<d_2<\cdots<d_N. \] The separation condition says that for any two distinct distances in this list, their difference is at least $1$. In particular, consecutive distances satisfy $d_{i+1}-d_i\ge 1$ for each $i$. Summing these inequalities yields \[ d_N - d_1 = \sum_{i=1}^{N-1} (d_{i+1}-d_i) \ge N-1. \] By Lemma 670.1, $d_1\ge 1$. Therefore $d_N\ge 1+(N-1)=N=\binom n2$. Since $d_N$ is one of the pairwise distances, it is at most the diameter, i.e. $d_N\le \operatorname{diam}(A)$. Hence $\operatorname{diam}(A)\ge \binom n2$. $\square$

Computation in $d=1$ (optimal integer Golomb rulers for $n\le 7$). In one dimension, taking integer points $0=x_1<\dots<x_n=L$ with all pairwise differences distinct automatically satisfies the "distance differences $\ge 1$" condition. I brute-forced the smallest possible $L$ for $n\le 7$ (optimal Golomb rulers), obtaining: 

n=2: L=1,  marks (0,1)
n=3: L=3,  marks (0,1,3)
n=4: L=6,  marks (0,1,4,6)
n=5: L=11, marks (0,1,4,9,11)
n=6: L=17, marks (0,1,4,10,12,17)
n=7: L=25, marks (0,1,4,10,18,23,25)
These illustrate that the diameter can equal $\binom n2$ for $n\le 4$ but must exceed it for $n\ge 5$ in $d=1$.

VERIFICATION.

  • Lemma 670.1: checked that it uses $n\ge 3$ essentially (to pick a third point $R$). For $n=2$ the claim is false since there is no constraint.
  • Lemma 670.2: verified that the ordering argument uses only the pairwise distance separation and Lemma 670.1.
  • Small-$n$ examples: $\{0,1,3\}$ gives distances $\{1,2,3\}$ and separation $\ge 1$; $\{0,1,4,6\}$ gives distances $\{1,2,3,4,5,6\}$.

FINAL. UNRESOLVED.

(i) Strongest proved partial results.

  • For any $d$ and any $n\ge 3$, the minimum pairwise distance is at least $1$ (Lemma 670.1).
  • Consequently, for any $d$ and any $n\ge 3$, the diameter satisfies $\operatorname{diam}(A)\ge \binom n2$ (Lemma 670.2), matching the "trivial" bound stated in the prompt.
  • In $d=1$, small-$n$ computations show the optimal diameter equals $\binom n2$ for $n\le 4$ and exceeds it for $n\ge 5$.

(ii) First gap (crisp statement). For fixed $d\ge 2$, prove or disprove the asymptotic lower bound \[ \operatorname{diam}(A) \ge (1+o(1))\,n^2 \] for every $n$-point set $A\subseteq\mathbb{R}^d$ whose pairwise distances are $1$-separated.

(iii) Top 3 next moves.

  1. Dimension reduction attempt: prove that any such $A$ contains a large subset whose points are "almost collinear" in a quantitative sense, reducing to the $d=1$ phenomenon.
  2. Geometric constraints on distance sets: exploit triangle inequalities across many triples to show that the $\binom n2$ distances cannot be packed into an interval of length $\approx \binom n2$, forcing diameter strictly larger by a linear factor.
  3. Construction search in $d\ge 2$: attempt to build explicit families in $\mathbb{R}^2$ or $\mathbb{R}^3$ with diameter close to the trivial lower bound $\binom n2$, which would refute the conjectured $(1+o(1))n^2$ bound.

(iv) What a minimal counterexample would likely look like. A counterexample (if the conjecture is false) would be a sequence of point sets $A_n\subset\mathbb{R}^d$ with $|A_n|=n$ such that:

  • the $\binom n2$ pairwise distances are all distinct and $1$-separated, but
  • the diameter satisfies $\operatorname{diam}(A_n) \le (c+o(1))n^2$ with some constant $c<1$ (ideally $c\approx 1/2$, matching the trivial lower bound up to lower order).
Such a construction would have to realize the distance multiset with unusually "tight packing" compatible with triangle inequalities in fixed dimension.