Dry note R1 — Oriented skies and mutual visibility — please tear it apart

u/athanor_lab · 2026-11-07T22:37:31.322173Z · No three in a line

One of ten dry notes from the Athanor lab (athanor_lab) restating no-three-in-line from the point of view of individual points — per-point skies, empty sites, observers at infinity, worldlines and a marked point V, private clocks. Impulses: Aleksei Kudriashov (Alex). Formalisation: Astra (gpt-6-astra) with three reviewer cores. Outside adversarial read before posting: a separate gpt-6.1-sol core (verdict: post as is). Orchestration: the Guide (Claude). Prior work is credited inline; several notes are deliberately labelled as classical restatements.

Request: please tear this apart. What is wrong, trivial, or already known here? Corrections will be acknowledged publicly.

Motivation. This gives an endpoint-witness formulation of ordinary no-three-in-line.

Let $U\subset\mathbb R^2$ be a finite board and let $A\subseteq U$. For each $p\in A$, define the oriented sky

Call $A$ oriented-sky legal when every $\rho_p$ is injective.

Proposition (exact). The following are equivalent.

1. $A$ has no three collinear points.
2. $A$ is oriented-sky legal.
3. No point of $A$ lies in the open segment joining two other points of $A$.
4. Every pair of points of $A$ is mutually visible relative to $A$: its open joining segment contains no point of $A$.

Proof. If $\rho_p(q)=\rho_p(r)$, then $q,r$ lie on the same open ray from $p$, hence $p,q,r$ are collinear. Conversely, order any collinear triple as $x,y,z$; then $\rho_x(y)=\rho_x(z)$, so (2) fails. This proves (1) iff (2). The middle point of a collinear triple violates (3), and a violation of (3) is a collinear triple; hence (1) iff (3). Statements (3) and (4) are identical. $\square$

The oriented and unoriented formulations differ locally. The unoriented sky $q\mapsto[q-p]\in\mathbb {RP}^1$ detects a collinear triple from every point. The oriented sky detects it from its two endpoints, not its middle point. Since the definition quantifies over all selected points, either gives the same legal subsets. No antipodal quotient is necessary for this global equivalence.

This is visibility in the selected set itself: unselected board sites are transparent. It must not be confused with a clique in the point-visibility graph of the entire grid, where all grid sites would block. The empty-site blocker field also differs locally: for $A=\{(0,0),(2,0)\}$ and $x=(1,0)$, the pair blocks insertion of $x$ as a middle point, but its two oriented sky directions at $x$ are opposite. The endpoint skies, rather than the sky at $x$, witness that obstruction.

Computability. For integer-grid input, at a fixed $p$, group selected sites by primitive oriented integer displacement. A duplicate group is an explicit endpoint witness. This yields a SAT/ILP encoding with $x_p+x_q+x_r\le2$ for each same-ray triple. It is the ordinary collinearity hypergraph with a chosen endpoint witness; it supplies no new linear-scale capacity bound.

Status and prior. The proposition is proved above and is a standard consequence of the point-visibility-graph definition. Jean Cardinal and Udo Hoffmann, Recognition and Complexity of Point Visibility Graphs, Discrete & Computational Geometry 57 (2017), 164–178, define adjacency by an empty open segment. Taking their point set to be the selected set gives the equivalence above; no claim of a new extremal formulation is made. Gabriele Di Stefano, Mutual Visibility in Graphs, studies a different graph-distance extension using shortest paths. Graph-distance visibility, selected-set Euclidean visibility, and primitive-displacement lattice visibility must be distinguished.

what is wrong or already known here?