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Constraint Satisfaction Programming for the No-three-in-line Problem

Thomas Prellberg

TL;DR

This work solves the classical no-three-in-line problem on $n\times n$ grids for all $n\le 60$ by formulating it as a constraint-satisfaction problem and employing a symmetry-reduced CP-SAT model. The authors exploit $90^\circ$ rotational symmetry to reduce the search space, enabling construction of $2n$-point configurations and proving $D(n)=2n$ for all $2\le n\le60$, thereby shifting the unknown boundary to $n=61$. They analyze solve-time behavior under a run-until-first-success parallel protocol and introduce a shifted-exponential model to describe early-time performance, providing practical extrapolations for planning large-scale runs. The work also documents data availability and suggests directions for further speedups and deeper structural insights to guide exact search in combinatorial geometry.

Abstract

Using a constraint satisfaction approach, we exhibit configurations of $2n$ points on the $n\times n$ grid for all $n\le60$ with no three collinear. Consequently, the smallest $n$ for which it is unknown whether $D(n)=2n$ increases from $47$ to $61$.

Constraint Satisfaction Programming for the No-three-in-line Problem

TL;DR

This work solves the classical no-three-in-line problem on grids for all by formulating it as a constraint-satisfaction problem and employing a symmetry-reduced CP-SAT model. The authors exploit rotational symmetry to reduce the search space, enabling construction of -point configurations and proving for all , thereby shifting the unknown boundary to . They analyze solve-time behavior under a run-until-first-success parallel protocol and introduce a shifted-exponential model to describe early-time performance, providing practical extrapolations for planning large-scale runs. The work also documents data availability and suggests directions for further speedups and deeper structural insights to guide exact search in combinatorial geometry.

Abstract

Using a constraint satisfaction approach, we exhibit configurations of points on the grid for all with no three collinear. Consequently, the smallest for which it is unknown whether increases from to .
Paper Structure (18 sections, 1 theorem, 23 equations, 6 figures, 3 tables)

This paper contains 18 sections, 1 theorem, 23 equations, 6 figures, 3 tables.

Key Result

Theorem 1

For $2\le n\le60$ there exists a configuration in $\mathcal{F}_n$. In particular, $D(n)=2n$ for $2\le n\le60$.

Figures (6)

  • Figure 1: $2n$ lattice points with no three in line for $n=60$. The $120$ occupied sites (red) are shown together with the $\binom{120}{2}=7140$ straight-line segments joining each pair of occupied sites (blue), illustrating the $90^\circ$ rotational symmetry.
  • Figure 2: Wall-clock time (log scale) to the first solution in a single experiment consisting of $384$ independent CP-SAT runs executed in parallel on an AMD EPYC 9965 CPU. Red squares: direct model \ref{['CPSAT']} (no symmetry restriction). Blue circles: symmetry-reduced model \ref{['CPSATred']} enforcing rotational symmetry.
  • Figure 3: Empirical cumulative distribution function $F(t)$ of the time to success for a single CP-SAT run (blue), obtained from $10{,}436$ independent runs of \ref{['CPSATred']} at $n=34$. Also shown is the derived CDF $F_{384}(t)$ for the first success among $M=384$ parallel runs (orange), computed via \ref{['FM']}, and a shifted-exponential fit \ref{['expmodel']} to $F_{384}(t)$ near the origin (green). The inset magnifies the small-time region relevant for $F_{384}$.
  • Figure 4: For $n=44$: derived CDF $F_{384}(t)$ for the first success among $384$ parallel runs (orange) together with a shifted-exponential fit \ref{['expmodel']} (green). On this time window the empirical single-run CDF $F(t)$ is extremely small and is therefore not visually prominent.
  • Figure 5: Even sizes: semi-log plot of solve-time statistics versus grid size $n$. Blue circles: observed wall-clock times to first solution in single runs. Green triangles and cyan inverted triangles: estimated $50\%$ and $98\%$ completion times for $M=384$ parallel runs, derived from the shifted-exponential fit \ref{['expmodel']} via \ref{['meanquant']}. Solid lines: least-squares fits of $\log t$ versus $n$. The horizontal line indicates a ten-day wall-clock budget. Reported times over 10 days are aggregated from multiple job runs.
  • ...and 1 more figures

Theorems & Definitions (2)

  • Theorem 1
  • proof