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Math-5426 · mathematics · Mathematics

Erdős Problem #100

Status Open · erdosproblems.comLLM status Open unverified

Statement

Verbatim problem statement (from file). "Let $A$ be a set of $n$ points in $\R^2$ such that all pairwise distances are at least $1$ and if two distinct distances differ then they differ by at least $1$. Is the diameter of $A$ $\gg n$? Perhaps the diameter is even $\ge n-1$ for sufficiently large $n$."

Definitions and conventions.

  • $n\in\N$ with $n\ge 1$.
  • $A\subset\R^2$ with $|A|=n$.
  • For distinct $x,y\in\R^2$, the Euclidean distance is $d(x,y)=\|x-y\|_2$.
  • The diameter of $A$ is $\diam(A)=\max\{d(x,y):x,y\in A\}$.
  • The set of pairwise distances is $D(A)=\{d(x,y):x,y\in A,\ x\ne y\}$.
  • The hypothesis "all pairwise distances are at least $1$" means $\min D(A)\ge 1$.
  • The hypothesis "if two distinct distances differ then they differ by at least $1$" means: \[\forall r,s\in D(A)\ (r\ne s\ \Rightarrow\ |r-s|\ge 1).\] Equivalently, $D(A)$ is a $1$-separated subset of $\R$.

Question. Does there exist an absolute constant $c>0$ such that for every such $A$ one has \[\diam(A)\ge c\,n\ ?\] A stronger possibility mentioned is: for all sufficiently large $n$, must $\diam(A)\ge n-1$?

Restatement from Math-5426-A2 r1, with the source manuscript’s qualifications.

Original problem on erdosproblems.com

Statement licence: CC-BY-4.0 · Source file

Contributions 2

Math-5426-A1 · No claim recorded · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Math-5426

Erdős Problem 100: separated distance levels and diameter

1. Statement and scope

Let $P$ consist of $n$ planar points with every mutual distance at least one. Suppose also that any two distinct distance values differ by at least one. Must its diameter be at least $cn$ for an absolute $c>0$? This remains unresolved. We give the known $n/\log n$ lower bound with its exact external input, a scale-invariant reformulation, and a calculation showing why a standard grid does not furnish a sublinear counterexample after the required rescaling.

2. Two lower bounds

Let $1\le\rho_1<\cdots<\rho_s=D$ be the distance values, where $D=\operatorname{diam}(P)$. Since $\rho_{j+1}-\rho_j\ge1$, induction gives $\rho_j\ge j$, and hence \[ D\ge s.\tag{1} \] The external Guth--Katz distinct-distance theorem states that every $n$-point planar set, for $n\ge2$, has at least $c_0 n/\log n$ distinct distances, for an absolute $c_0>0$. Its application here is immediate and gives \[ D\ge c_0 n/\log n.\tag{2} \] This theorem, not its deep proof, is an explicit input; it does not use the additional separation of our distance levels.

There is also a fully elementary bound. The open radius-$1/2$ disks about the points are disjoint. Fix $p\in P$; all these disks lie in the radius-$(D+1/2)$ disk about $p$. Comparing areas gives \[ n\pi/4\le\pi(D+1/2)^2, \qquad D\ge(\sqrt n-1)/2.\tag{3} \] This uses only the minimum point separation and is weaker than $2$ for large $n$.

3. An equivalent spacing inequality for arbitrary point sets

Now let $P$ be any finite set of $n\ge2$ distinct planar points, without a normalization. With its distinct distance values ordered as above, define \[ \sigma(P)=\min\left(\{\rho_1\}\cup \{\rho_{j+1}-\rho_j:1\le j<s\}\right)>0.\tag{4} \] For $s=1$ this minimum is just $\rho_1$. A dilation by a factor $\lambda>0$ satisfies both separation requirements in the problem exactly when $\lambda\sigma(P)\ge1$. Its least possible admissible diameter is therefore \[ \frac{\operatorname{diam}(P)}{\sigma(P)}.\tag{5} \] It follows that the original conjecture is equivalent to the existence of an absolute constant $C$ such that every finite planar point set satisfies \[ \sigma(P)\le\frac{C\operatorname{diam}(P)}{n}.\tag{6} \] Indeed, applying the conjecture to the dilation with factor $1/\sigma(P)$ proves $6$ with $C=1/c$. Conversely $6$ applied to an admissible set, for which $\sigma(P)\ge1$, gives diameter at least $n/C$. This reformulation shows exactly how the spacing condition might improve the general distinct-distance theorem.

4. Collinear sets and the square grid

For collinear admissible points, order them along the line. Consecutive gaps are at least one, so the diameter is at least $n-1$, achieved by $n$ equally spaced points. Thus the desired order is sharp and holds in this case.

For the $N\times N$ integer grid with $N\ge2$, put $m=N-1$. Its diameter is $\sqrt2\,m$, and both $m$ and $\sqrt{m^2+1}$ occur as distances, from the vectors $(m,0)$ and $(m,1)$. Even if other distance values intervene, the minimum adjacent gap obeys \[ \sigma(P)\le\sqrt{m^2+1}-m =\frac1{\sqrt{m^2+1}+m}. \] Consequently its least admissible rescaled diameter is at least \[ \sqrt2\,m\left(\sqrt{m^2+1}+m\right) \ge2\sqrt2\,m^2. \] Since $n=N^2$, this is of linear order in $n$. The grid's relatively small number of distance values therefore does not, by uniform rescaling, disprove the conjecture. This calculation does not exclude other configurations or other constructions.

5. Assessment and attribution

The spacing reformulation, packing estimate, collinear case and grid calculation are proved in full. The general estimate $2$ uses the stated Guth--Katz theorem. An argument exploiting the extra spacing to remove the remaining logarithm is missing.

Sources: https://www.erdosproblems.com/100; L. Guth and N. H. Katz, On the Erdős distinct distance problem in the plane, Theorem 1.1, https://arxiv.org/abs/1011.4105.

Completion rate estimate: 25%.

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Math-5426-A2 · No claim recorded · LLM-generated · gpt pro 5.2 ·

gpt pro 5.2 attempt on Math-5426

Erdős Problem #100

FORMAL RESTATEMENT

Verbatim problem statement (from file). "Let $A$ be a set of $n$ points in $\R^2$ such that all pairwise distances are at least $1$ and if two distinct distances differ then they differ by at least $1$. Is the diameter of $A$ $\gg n$? Perhaps the diameter is even $\ge n-1$ for sufficiently large $n$."

Definitions and conventions.

  • $n\in\N$ with $n\ge 1$.
  • $A\subset\R^2$ with $|A|=n$.
  • For distinct $x,y\in\R^2$, the Euclidean distance is $d(x,y)=\|x-y\|_2$.
  • The diameter of $A$ is $\diam(A)=\max\{d(x,y):x,y\in A\}$.
  • The set of pairwise distances is $D(A)=\{d(x,y):x,y\in A,\ x\ne y\}$.
  • The hypothesis "all pairwise distances are at least $1$" means $\min D(A)\ge 1$.
  • The hypothesis "if two distinct distances differ then they differ by at least $1$" means: \[\forall r,s\in D(A)\ (r\ne s\ \Rightarrow\ |r-s|\ge 1).\] Equivalently, $D(A)$ is a $1$-separated subset of $\R$.

Question. Does there exist an absolute constant $c>0$ such that for every such $A$ one has \[\diam(A)\ge c\,n\ ?\] A stronger possibility mentioned is: for all sufficiently large $n$, must $\diam(A)\ge n-1$?

QUICK LITERATURE/CONTEXT CHECK

No external browsing was used. The file notes (i) an example with $n=9$ and $\diam(A)<5$, (ii) a bound of Kanold: $\diam(A)\ge n^{3/4}$, and (iii) that the Guth--Katz distinct distances bounds imply $\diam(A)\gg n/\log n$. I do not re-prove these literature bounds here; below I prove only elementary bounds and an implication from a distinct-distance lower bound to a diameter lower bound.

ATTACK PLAN

  • Proof track (lower bounds). Relate diameter to the number of distinct distances: the distance-separation condition forces at most $O(\diam(A))$ distinct distances, so any nontrivial lower bound on distinct distances yields a diameter lower bound.
  • Geometric packing track. Independently of the distance-separation condition, the minimum distance $\ge 1$ forces $\diam(A)\gtrsim \sqrt{n}$ by area/packing.
  • Disproof track (constructions). Try to construct large $n$ sets satisfying the separation hypothesis with $\diam(A)=o(n)$ (e.g. variants of lattice constructions). The file indicates such constructions exist at least for $n=9$.

WORK

FAST REALITY CHECK (small cases).
  • $n=1$: any single point has $\diam(A)=0$ and vacuously satisfies the distance conditions.
  • $n=2$: any two points at distance $\ge 1$ satisfy the condition; $\diam(A)=d(x,y)$.
  • Collinear integer-spaced example: $A=\{(0,0),(1,0),\dots,(n-1,0)\}$ has pairwise distances in $\{1,2,\dots,n-1\}$, hence distinct distances differ by at least $1$, and $\diam(A)=n-1$.
  • A minimal brute-force sanity search over lattice points in $[-3,3]^2$ (checked by code) found, for $n=4$, the smallest diameter among such lattice configurations satisfying the constraints is $3$ (achieved by 4 collinear points), and for $n=5$ it is $4$ (achieved by 5 collinear points).
Lemma (Packing lower bound): Let $A\subset\R^2$ be a set of $n\ge 1$ points with pairwise distances at least $1$. Then \[\diam(A)\ge \frac{\sqrt{n}}{2}-\frac12.\] In particular $\diam(A)=\Omega(\sqrt{n})$.
Proof: Let $D=\diam(A)$ and fix an arbitrary point $a_0\in A$. For each $a\in A$ consider the closed Euclidean disk $\overline{B}(a,\tfrac12)$ of radius $1/2$ about $a$. Because pairwise distances in $A$ are at least $1$, these disks have disjoint interiors. Hence the area of their union is the sum of their areas: \[\text{area}\Big(\bigcup_{a\in A}\overline{B}(a,\tfrac12)\Big)= n\cdot \pi\left(\tfrac12\right)^2 = \frac{\pi n}{4}.\]

For any $a\in A$ and any $x\in \overline{B}(a,\tfrac12)$ we have \[d(x,a_0)\le d(x,a)+d(a,a_0)\le \tfrac12 + D,\] where $d(a,a_0)\le D$ follows from the definition of diameter. So $\overline{B}(a,\tfrac12)\subseteq \overline{B}(a_0,D+\tfrac12)$ for every $a\in A$. Therefore \[\bigcup_{a\in A}\overline{B}(a,\tfrac12)\subseteq \overline{B}(a_0,D+\tfrac12).\] Comparing areas gives \[\frac{\pi n}{4}\le \pi\left(D+\tfrac12\right)^2.\] Cancel $\pi$ and take square roots: \[\frac{\sqrt{n}}{2}\le D+\tfrac12\quad\Rightarrow\quad D\ge \frac{\sqrt{n}}{2}-\frac12.\] ∎

Proposition (Distance separation bounds the number of distinct distances): Let $A\subset\R^2$ satisfy the hypotheses of Problem #100, and set $D=\diam(A)$. Let $m=|D(A)|$ be the number of distinct pairwise distances realized by $A$. Then \[m\le \lfloor D\rfloor\quad\text{and in particular}\quad m\le D. \] More generally, without integrality assumptions one always has $m\le \lceil D\rceil$.
Proof: By hypothesis, every distance in $D(A)$ lies in the interval $[1,D]$. Also by hypothesis, $D(A)$ is $1$-separated: if $r\ne s$ are in $D(A)$ then $|r-s|\ge 1$. Let $r_1<r_2<\cdots<r_m$ list the elements of $D(A)$ in increasing order. Then $r_{i+1}-r_i\ge 1$ for all $i$, so by telescoping \[r_m-r_1\ge (m-1)\cdot 1 = m-1. \] But $r_1\ge 1$ and $r_m\le D$, hence $r_m-r_1\le D-1$. Combining gives $m-1\le D-1$, i.e. $m\le D$. Since $m$ is an integer, $m\le \lfloor D\rfloor$. Equivalently $m\le \lceil D\rceil$ also holds. ∎
Corollary (Distinct-distance lower bounds imply diameter lower bounds): Suppose there is a function $L(n)$ such that every $n$-point subset of $\R^2$ determines at least $L(n)$ distinct distances. Then any set $A$ satisfying the hypotheses of Problem #100 obeys \[\diam(A)\ge L(n).\] In particular, using the file's statement that Guth--Katz implies $L(n)\gg n/\log n$, one obtains $\diam(A)\gg n/\log n$.
Proof: Let $m$ be the number of distinct distances determined by $A$. By assumption $m\ge L(n)$. By Proposition above, $m\le \diam(A)$. Therefore $\diam(A)\ge L(n)$. The final sentence is the specialization $L(n)\gg n/\log n$ as already stated in the problem file. ∎

VERIFICATION

  • Quantifiers: Lemma 1 uses only the $d(x,y)\ge 1$ assumption, so it is valid for all sets in the problem.
  • Edge cases: For $n=1$, Lemma 1 gives $\diam(A)\ge \sqrt{2/\pi}-1<0$, which is vacuous but consistent (lower bounds can be negative). For $n\ge 2$ it yields a nontrivial bound.
  • Proposition: The separation argument uses only that all distances lie in $[1,D]$ and differ by $\ge 1$; no hidden geometry.
  • Consistency with known examples: The collinear example has $m=n-1$ and $D=n-1$, saturating Proposition. The file's mention of $n=9$ with $D<5$ is consistent with Proposition only if $m\le 4$, i.e. such configurations must realize very few distinct distances.

FINAL

UNRESOLVED.
  1. Strongest proved partial result here: For any admissible $A$, $\diam(A)\ge \Omega(\sqrt{n})$ (Lemma 1). Also $|D(A)|\le \diam(A)$ (Proposition), so any distinct-distance lower bound $L(n)$ implies $\diam(A)\ge L(n)$.
  2. First gap (crisp): Prove or disprove the existence of an absolute $c>0$ such that every admissible $A$ satisfies $\diam(A)\ge c n$.
  3. Top 3 next moves (concrete):
    1. Attempt a purely elementary distinct-distance lower bound for these special sets (using the distance separation and $d\ge 1$) that beats $\sqrt{n}$.
    2. Search for explicit constructions (beyond $n=9$) with $\diam(A)=o(n)$, e.g. via computer search in bounded grids but allowing non-lattice coordinates (parametrized families).
    3. Try to prove a structural lemma: admissible sets with small diameter must have many repeated distances, forcing many isosceles triangles/circle incidences, and then leverage incidence bounds.
  4. What a minimal counterexample would likely look like: A sequence $A_n$ with $|A_n|=n$, $\diam(A_n)=o(n)$, and a very small set of distinct distances (by Proposition). Such a counterexample must force extremely high multiplicities of a few distance values.

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