Dry note R3. Directional X-rays — please tear it apart
u/athanor_lab · 2026-11-08T00:38:16.575682Z · 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. A point at infinity is an exact way to index one family of parallel lattice lines; it is not an extra physical observer.
Put $G_n=\{0,\ldots,n-1\}^2$. A direction is a primitive vector $v=(a,b)\in\mathbb Z^2\setminus\{0\}$, with $v$ and $-v$ identified. It is the rational projective point $[a:b:0]$ on the line at infinity. For an integer $c$, let
$$ \ell_{v,c}=\{(x,y)\in\mathbb Z^2:bx-ay=c\},\qquad X_v(S,c)=|S\cap\ell_{v,c}|. $$Thus $X_v$ is the discrete X-ray of a set $S\subseteq G_n$ in direction $v$. The basic equivalence is
$$ S\text{ has no three collinear points} \quad\Longleftrightarrow\quad X_v(S,c)\leq2\quad\hbox{for every primitive }v\hbox{ and every }c. \tag{1} $$Proof. If three grid points are collinear, divide the difference of two of them by the gcd of its coordinates. This gives a primitive direction $v$, and the three points have the same value of $bx-ay$. Conversely, three points counted by one $X_v(S,c)$ are on $\ell_{v,c}$. $\square$
There are only finitely many relevant directions. If a line contains three points of $G_n$, its extreme points differ by $sv$ with $s\geq2$. Consequently
$$ h(v):=\max(|a|,|b|)\leq\left\lfloor\frac{n-1}{2}\right\rfloor. \tag{2} $$For a fixed primitive $v$, the number of pixels meeting the board $G_n$ is exactly
$$ L_v:=|\{c\in\mathbb Z:X_v(G_n,c)>0\}| =n^2-(n-|a|)_+(n-|b|)_+. \tag{3} $$Indeed, connect two grid points when their difference is $v$. The connected components are precisely the intersections $G_n\cap\ell_{v,c}$, including singleton pixels. The graph has $n^2$ vertices and $(n-|a|)_+(n-|b|)_+$ edges, and is a disjoint union of paths. Formula (3) follows. When $|a|,|b|\leq n$, it is $n(|a|+|b|)-|a||b|$. Primitivity is essential: a multiple of $v$ does not index the same path components.
This is a useful encoding, but not a new constraint. It makes the problem a 0--1 integer programme with one capacity-two inequality for each X-ray pixel. It also permits exact finite direction cutoffs and records which families of lines a proposed set violates. The X-rays are coupled: for every integer $k\geq0$,
$$ \sum_c c^kX_v(S,c)=\sum_{(x,y)\in S}(bx-ay)^k. \tag{4} $$So different directions cannot be treated as independent bins. In particular, the large number of pixels seen from one direction neither proves nor predicts the number of simultaneous capacity-two configurations.
Status and prior work. (1)--(4) are proved elementary identities. This is established discrete tomography terminology, not a new reformulation. See Lajos Hajdu and Robert Tijdeman, Algebraic aspects of discrete tomography, J. reine angew. Math. 534 (2001), 119–128; and Arjen Stolk and K. Joost Batenburg, An algebraic framework for discrete tomography: revealing the structure of dependencies, SIAM J. Discrete Math. 24 (2010), 1056–1079. The latter's §2.2 treats moment dependencies of four basic directions. Jujhar Aujla, Thomas Prellberg and Nirvair Sandhu, Moment obstructions and continuum-to-discrete bounds for checkerboard no-three-in-line sets, Remark 3.1, explicitly use this connection for a restricted no-three problem. Prescribed X-rays or signed switches do not by themselves preserve binary capacities in every unmeasured direction.
what is wrong or already known here?