REVIEW 2 major objections 4 minor 19 references
On sets of $n$ points in general position that determine lines that can be pierced by $n$ points
T0 review · 2 major / 4 minor · reviewed 2026-08-14 · deepseek-v4-flash
Pith's one-line read Under the outside-segment piercing condition, any configuration P∪R of 2n points must lie on a single cubic curve.
desk verdict A genuinely new structural result with an elegant two-step proof, but two real small-case gaps (k=4 hulls and n=5,6) need patching before the theorem as stated is proved. read the letter →
The pith
A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.
The reading
What carries the argument
The machinery is two-step. First, the outside-segment hypothesis becomes a rigid combinatorial structure: the points of $P$ are in convex position, the points of $R$ lie outside their convex hull, and collinearities are exactly the modular rule $i+j+k\equiv 0\pmod n$. Second, this rule feeds a Chasles-theorem grid: whenever three lines $\ell_1,\ell_2,\ell_3$ meet three lines $m_1,m_2,m_3$ in nine distinct points, a cubic through any eight of those points passes through the ninth. The paper applies this repeatedly to the nine-point grids formed by the cyclic labels, so that one cubic containing seven $P$-points and two $R$-points must contain every point of $P\cup R$.
What would settle it
A direct way to test the theorem is to search oriented matroids for small $n$, especially hull size $k=4$: if any configuration satisfying the outside-segment condition has $x_i,x_j,r_k$ collinear with $i+j+k\not\equiv 0\pmod n$, or has all $2n$ points not lying on a cubic, the claim is false.
Extended reading notes
Core claim
The central claim, Theorem 1.3, is that the outside-segment piercing condition is strong enough to determine the entire combinatorial geometry of $P\cup R$ and then force it onto an algebraic curve of degree at most three. More precisely, for every pair $x,y\in P$ the required piercing point $r\in R$ lies outside the interval $xy$; the paper shows $P$ is in convex position, $R$ lies outside $\operatorname{conv}(P)$, and after labelling $P$ cyclically as $x_0,\dots,x_{n-1}$ and $R$ as $r_0,\dots,r_{n-1}$, the collinearity condition is $x_i,x_j,r_k$ collinear iff $i+j+k\equiv 0\pmod n$. It then proves, using Chasles' theorem in the form that a cubic through eight of the nine intersections of two triples of lines passes through the ninth, that any cubic through seven consecutive $P$-points and two $R$-points must contain all of $P\cup R$. Hence $P\cup R$ is contained in a cubic curve.
Load-bearing premise
The load-bearing premise is that for every pair $x,y\in P$ the required piercing point in $R$ lies strictly outside the open segment between $x$ and $y$; without this, the convex-position and cyclic-collinearity rigidity that feeds the Chasles argument has no basis.
Editorial extensions
If this is right
- Under the outside-segment hypothesis the extremal piercing configuration is completely rigid: $P$ is in convex position, each point of $R$ lies outside the convex hull, and the only collinear triples are those with $i+j+k\equiv 0\pmod n$.
- Consequently every such configuration is algebraic of degree at most three: there is a single cubic curve containing all $2n$ points, so the combinatorial extremum cannot be realized by generic point sets.
- The theorem recovers the known characterization of general-position sets with exactly $n$ directions, because the required $R$ can be placed on the line at infinity, outside every finite segment.
- The same Chasles propagation works in the bipartite setting: if blue and green points alternate on the convex hull and every blue-green line is pierced by a red point outside the segment, then the union lies on a cubic.
Reading between the lines
- A natural next test is whether the outside-segment condition can be weakened to allow at most one piercing point inside the segment per line; if the cubic conclusion persists, the full Conjecture 1.2 would be much closer.
- The cyclic rule $i+j+k\equiv 0\pmod n$ suggests that equality configurations are essentially residues on a cubic curve, so a plausible classification is that they are affine images of regular polygons on a cubic.
- A small-case search should target hull sizes $k=3$ and $k=4$, since the counting step invokes a lower bound stated only for even $n>4$; if a counterexample exists, it is most likely there.
Signed reviews
Editorial analysis
A structured set of objections, weighed in public.
Referee Report
Summary. The paper studies the following problem: given a set P of n points in general position in the plane and a disjoint set R of n points such that every line through two points of P contains a point of R outside the segment determined by those two points, show that P∪R lies on a cubic curve. The main theorem (Theorem 1.3) is proved in two steps. The first step is combinatorial: the outside-segment condition forces P to be in convex position, all points of R to lie outside the convex hull of P, and the collinear triples to follow the explicit rule i+j+k=0 modulo n after suitable labeling. The second step is algebraic: using Chasles's theorem in the style of Green and Tao, the authors construct a cubic curve through seven consecutive points of P and two points of R and then extend containment to the whole set P∪R via two claims (Claims 2.2 and 2.3). A bipartite analogue (Theorem 3.1) is also stated and proved along similar lines.
Significance. If correct, this is a strong and elegant result. It proves a special case of Conjecture 1.2, the conjecture attributed to Erdos and Purdy that equal-size piercing sets force a cubic curve, and it generalizes Jamison's characterization of extremal slope configurations. The proof is largely explicit: the combinatorial classification is clear and the Chasles applications are checkable, following the Green-Tao ordinary-lines framework. The outside-segment hypothesis is the key new input and it is used in a principled way to derive a rigid collinearity structure. No circularity or free parameters appear; external theorems are used as black boxes. The main caveat is that the proof as written has genuine small-case gaps that are not acknowledged in the manuscript.
major comments (2)
- [Section 2, paragraph after the definition of P′] The claim that |R′| ≥ k is justified by an application of Theorem 1.1 to the extreme subset P′. However, Theorem 1.1 is stated only for even cardinalities larger than 4. Thus the argument does not cover the case where P′ has exactly four extreme vertices. For odd k the trivial counting bound |R′| ≥ k is indeed available, but this is not stated, and for k = 4 the trivial bound gives only |R′| ≥ 3, which is insufficient. The equality |R′| = k and all subsequent steps depend on this lower bound, so the case of a convex hull with exactly four extreme vertices is not handled. A separate argument for k = 4 (or a modification of the counting argument) is needed for the theorem to hold for all n.
- [Section 2.1, construction of the cubic Γ and Claims 2.2–2.3] The algebraic step requires a cubic Γ passing through x_{n−1}, …, x_{n−7} and then repeatedly applies Claim 2.2, whose statement presupposes seven distinct consecutive points of P for the initial configuration and six distinct consecutive points plus three distinct points of R for the Chasles grid. For n = 5 and n = 6 these indices are not distinct modulo n (for example, with n = 6 the list x_5, ..., x_{−1} repeats x_5), so the claims are not applicable. The paper does not provide a separate treatment of n = 5 and n = 6, and the same omission affects the even-n case in Claim 2.3 for n = 6. Thus the main theorem is not proved for n = 5 and n = 6. The same issue affects Theorem 3.1 when 2n is small.
minor comments (4)
- [Theorem 1.1 statement] The statement contains a grammatical typo: “Let P is a set” should read “Let P be a set”.
- [Throughout the proof] A few typographical errors are present, including “colllinear” in the proof of the collinearity structure and “the the line” in the proof of Claim 2.3. These are harmless but should be corrected.
- [Section 2, notational clarity] The term “relevant” is introduced informally. A short formal definition (e.g., “r is relevant for x if there exists y ∈ P \ {x} such that r was chosen for the pair (x,y)”) would improve readability, though the intended meaning is clear from context.
- [Figure 4] In the illustration for Claim 2.3, the label “m3” appears to be repeated in a way that may confuse the reader; the intended three m-lines should be labeled distinctly.
Circularity Check
No significant circularity: Theorem 1.3 is derived from the outside-segment hypothesis using external theorems, with no fitted inputs or self-citation chain forcing the conclusion.
full rationale
The paper's central derivation does not reduce to its inputs. The outside-segment condition is used to prove that P is in convex position and that collinearities obey i+j+k=0 modulo n; this structural step is a genuine consequence of the hypothesis, not an assumption dressed as a result. The algebraic step then uses Chasles' theorem (an external geometric fact) to propagate containment in a cubic curve from nine initial points to all of P∪R. No parameter is fitted to data, no quantity called a prediction is defined in terms of the target, and no normalization or ansatz is imported through citation. The cited Theorem 1.1 is used as a black-box lower bound and has independent peer-reviewed proofs ([ABK+08], [Mil18]); the manuscripts [Pin18] and [PP19] merely provide alternative proofs and are not load-bearing for the present argument. The skeptical concerns about n=5,6 and k=4 are proof gaps concerning edge cases, not circularity: they do not show that the conclusion is equivalent to the hypothesis. Therefore the appropriate circularity score is 0.
Assumptions & free parameters
assumptions (4)
- standard math Theorem 1.1: for n>4 even, any R piercing all lines determined by n general-position points has |R| at least n.
- standard math Chasles theorem: if two triples of lines form nine distinct intersections, any cubic through eight of them passes through the ninth.
- standard math Any nine points in the plane lie on some cubic curve.
- domain assumption General position: no three points of P or of B union G are collinear.
Cite this review
Pith. "Pith review of On sets of $n$ points in general position that determine lines that can be pierced by $n$ points." pith.science (2026). https://pith.science/paper/77FVBG2H
@misc{pith2026190806390,
author = {Pith},
title = {Pith review of: On sets of $n$ points in general position that determine lines that can be pierced by $n$ points},
year = {2026},
howpublished = {\url{https://pith.science/paper/77FVBG2H}},
note = {Machine review of arXiv:1908.06390}
}
abstract
Let $P$ be a set of $n$ points in general position in the plane. Let $R$ be a set of $n$ points disjoint from $P$ such that for every $x,y \in P$ the line through $x$ and $y$ contains a point in $R$ outside of the segment delimited by $x$ and $y$. We show that $P \cup R$ must be contained in a cubic curve.
Figures
Reference graph
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Reviewed August 14, 2026 · model on record in the stance chip above.
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