No-three-in-line problem
The no-three-in-line problem is a classic question in discrete geometry that seeks the maximum number of points that can be placed on an n × n integer lattice grid such that no three points are collinear.[1] Posed by the English mathematician and puzzle designer Henry Ernest Dudeney in 1917, the problem requires avoiding collinear triples not only in horizontal, vertical, or diagonal directions but along any straight line in the plane.[2][3] A straightforward upper bound on the maximum number t(n) of such points is 2n, as no row or column can contain more than two points without forming a collinear triple.[4] Constructions achieving exactly 2n points exist for all n up to 58, including computational solutions using branch-and-bound algorithms and constraint satisfaction methods.[2] However, it remains an open question whether t(n) = 2n for every positive integer n, with the challenge intensifying for larger n due to the increasing complexity of avoiding all possible lines.[1][5] Asymptotic lower bounds guarantee at least (3/2 - ε)n points for any ε > 0 and sufficiently large n, building on early results by Paul Erdős and later improvements.[3] The problem has inspired extensive research, including generalizations to higher dimensions—where up to Θ(n2) points are possible in 3D without three collinear—and connections to combinatorial geometry, such as Roth's theorem on arithmetic progressions.[3] Despite over a century of study, no counterexample to the 2n conjecture has been found, making it one of the oldest unsolved problems in combinatorial geometry.[2]Problem statement
Definition
The no-three-in-line problem asks for the maximum number t(n) of points that can be placed on the n \times n integer grid \{1, 2, \dots, n\} \times \{1, 2, \dots, n\} such that no three points are collinear.[6] This grid consists of n^2 lattice points with integer coordinates in the first quadrant, bounded by the lines x = 1, x = n, y = 1, and y = n.[7] Three points (x_i, y_i), (x_j, y_j), and (x_k, y_k) on this grid are collinear if they lie on the same straight line in the Euclidean plane, meaning any slope is possible—not just horizontal, vertical, or grid-aligned directions. Mathematically, this occurs precisely when the following equation holds: x_i (y_j - y_k) + x_j (y_k - y_i) + x_k (y_i - y_j) = 0. This condition arises from the vanishing of the signed area of the triangle formed by the three points, equivalent to the determinant criterion for collinearity in the plane. The problem requires selecting a subset of the grid points that avoids any such triple, maximizing the subset size while preserving this general position property. Trivially, t(n) \leq n^2, as this is the total number of grid points, but a tighter bound follows from the pigeonhole principle: since any three points in the same row or column are collinear, at most two points can be chosen per row (or per column), yielding t(n) \leq 2n.[6] The no-three-in-line problem is a classic question in discrete geometry concerning the maximum size of a set in general position within a finite lattice.[7]Conjecture
The central conjecture of the no-three-in-line problem, attributed to Henry Dudeney, states that the maximum number of points t(n) that can be placed on an n \times n grid with no three collinear is exactly $2nfor every positive [integer](/page/Integer)n. This would imply both that t(n) \geq 2n is achievable via suitable constructions and that $2n is the tight upper bound. The upper bound t(n) \leq 2n follows immediately from the pigeonhole principle applied to the rows (or columns) of the grid: with n rows available, placing more than $2n points would force at least three points into some single row by the [pigeonhole principle](/page/Pigeonhole_principle), and points in the same row are collinear. Achieving exactly $2n points thus requires distributing precisely two points per row (and per column) while ensuring no three points are collinear in any other direction. Supporting evidence for the lower bound t(n) \geq 2n includes explicit constructions that place $2npoints with no three collinear for everynup to 58 as of 2025,[2] with ongoing computational verifications.[22] For primen = p, an algebraic construction places ppoints at coordinates(i, 2^i \mod p)fori = 0, 1, \dots, p-1, ensuring no three are collinear due to properties of exponential functions modulo p; later generalizations of such algebraic methods, often combining multiple curves or permutations, extend this to achieve $2n points for general n.[8] If the conjecture holds, it provides the exact solution to the problem and highlights deep connections to finite geometry, where sets of $2npoints with no three collinear correspond to maximum [cap set](/page/Cap_set)s in the affine plane over\mathbb{Z}/n\mathbb{Z}$, analogous to cap set problems in higher-dimensional vector spaces over finite fields.Small instances
Configurations for n ≤ 10
For small values of n \leq 10, the maximum number t(n) of points that can be placed on an n \times n grid with no three collinear is t(1) = 1 and t(n) = 2n for $2 \leq n \leq 10. This value is achieved by constructions that place precisely two points per row (and typically per column as well), while ensuring no three points align on any line of other slopes. Moreover, $2n is maximal for these n (and 1 for n=1), since placing more than $2n points would force at least one row to contain three points by the pigeonhole principle, and any three points in the same row are collinear.[1] The following table summarizes these exact values:| n | t(n) |
|---|---|
| 1 | 1 |
| 2 | 4 |
| 3 | 6 |
| 4 | 8 |
| 5 | 10 |
| 6 | 12 |
| 7 | 14 |
| 8 | 16 |
| 9 | 18 |
| 10 | 20 |