On the maximum number of k-holes in point sets with no (k + 1)-hole
Pith reviewed 2026-06-28 01:06 UTC · model grok-4.3
The pith
For each fixed k >= 6, the maximum number of k-holes in an n-point set with no (k+1)-hole satisfies polynomial bounds with exponents between floor(k/3) and ceil(k/2).
A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.
Core claim
The authors prove that for each fixed k >= 6 there exist absolute constants c1, c2 > 0 such that (c1/k)^{floor(k/3)} n^{floor(k/3)} <= h_k(n) <= (c2/k)^{ceil(k/2)} n^{ceil(k/2)}, where h_k(n) denotes the maximum number of k-holes in any set of at most n points in the plane in general position with no (k+1)-hole.
What carries the argument
The extremal function h_k(n) that records the maximum number of empty convex k-gons over all n-point sets avoiding empty convex (k+1)-gons.
If this is right
- For k=6 the number of 6-holes lies between order n squared and order n cubed in any set without a 7-hole.
- For each larger fixed k the gap between the lower exponent floor(k/3) and upper exponent ceil(k/2) remains open.
- Point sets that avoid all (k+1)-holes can still realize polynomially many k-holes whose count grows with n.
- The upper bound limits the density of k-holes once the next size is forbidden, independent of the minimum-number results in the Erdős problem.
Where Pith is reading between the lines
- The lower-bound constructions may be obtained by taking suitable subsets or perturbations of Horton sets that already avoid 7-holes.
- Closing the exponent gap for small fixed k would require either stronger constructions or tighter counting arguments that track convex layers more precisely.
- The same style of bound might apply to the maximum number of empty convex polygons of size k when one forbids all empty polygons larger than some fixed m > k+1.
Load-bearing premise
The point sets under study are required to be in general position with no three points collinear.
What would settle it
A family of point sets of size n with no (k+1)-hole but strictly more than (c2/k)^{ceil(k/2)} n^{ceil(k/2)} many k-holes for arbitrarily large n, or a matching upper bound below the stated lower bound that holds for infinitely many n.
Figures
read the original abstract
The classical problem of Erd\H{o}s asks for the minimum number of empty convex $k$-gons determined by an $n$-element point set in the plane. The celebrated empty hexagon theorem, proved independently by Gerken and Nicol\'as, shows that every sufficiently large planar point set contains a $6$-hole, while Horton's famous construction shows the existence of arbitrarily large point sets with no $7$-hole. In this paper, we initiate the study of the maximum number of $k$-holes in planar point sets with no $(k+1)$-hole. More precisely, for each fixed $k\geq 6$, let $h_k(n)$ be the maximum number of $k$-holes determined by a planar point set in general position, of size at most $n$, and with no $(k+1)$-hole. We prove that there are absolute constants $c_1,c_2>0$ such that $(c_1/k)^{\lfloor k/3\rfloor} n^{\lfloor k/3\rfloor} \leq h_k(n)\leq (c_2/k)^{\lceil k/2\rceil} n^{\lceil k/2\rceil}$.
Editorial analysis
A structured set of objections, weighed in public.
Referee Report
Summary. The paper defines h_k(n) as the maximum number of k-holes in an n-point set in general position with no (k+1)-hole. For each fixed k≥6 it claims the existence of absolute constants c1,c2>0 such that (c1/k)^⌊k/3⌋ n^⌊k/3⌋ ≤ h_k(n) ≤ (c2/k)^⌈k/2⌉ n^⌈k/2⌉.
Significance. If the claimed bounds hold, the work would initiate a new quantitative direction in extremal combinatorial geometry by bounding the number of k-holes under the no-(k+1)-hole constraint, consistent with the empty-hexagon theorem and Horton sets. The exponents match known qualitative results in the area.
major comments (1)
- Abstract: the central claim asserts the existence of the stated bounds with the given exponents and absolute constants, but the provided text supplies no proof details, key constructions, or verification steps for either the lower or upper bound.
Simulated Author's Rebuttal
We thank the referee for their review. The abstract is a concise summary of the main result, as is conventional; the full proofs, constructions, and arguments appear in the body of the manuscript.
read point-by-point responses
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Referee: Abstract: the central claim asserts the existence of the stated bounds with the given exponents and absolute constants, but the provided text supplies no proof details, key constructions, or verification steps for either the lower or upper bound.
Authors: The abstract states the theorem without proof details, which is standard practice. The lower bound construction (a suitable Horton set variant with controlled k-holes) is given in Section 3, while the upper bound (via charging arguments and the no-(k+1)-hole assumption) is proved in Section 4. We are happy to insert forward references to these sections in a revised abstract. revision: partial
Circularity Check
No significant circularity identified
full rationale
The paper defines h_k(n) explicitly as the maximum number of k-holes in an n-element point set in general position with no (k+1)-hole. It then states and proves asymptotic bounds on this quantity for fixed k≥6, with the lower bound using a construction and the upper bound using combinatorial counting arguments that reference external results (empty hexagon theorem, Horton sets). No step reduces the claimed bounds to a fitted parameter, self-definition, or self-citation chain; the exponents floor(k/3) and ceil(k/2) arise from known extremal geometry rather than from re-expressing the input definition. The derivation is therefore self-contained against external benchmarks.
Axiom & Free-Parameter Ledger
free parameters (2)
- c1
- c2
axioms (1)
- domain assumption Point sets lie in general position in the Euclidean plane
Forward citations
Cited by 1 Pith paper
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Many holes but no large one: maximizing $k$-holes while forbidding $(k+1)$-holes
For n = k + a with a ≤ k/2 - 1, the maximum number of k-holes in a point set with no (k+1)-hole is exactly 2^a.
Reference graph
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discussion (0)
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