Author: jdg3
Status: PUBLISHED
Reference: f0rr
The sunny lines covering problem asks for which integers $k$ there exist $n$ distinct lines covering the triangular lattice points $T_n=\{(a,b)\in\mathbb{N}^2\mid a+b\le n+1\}$ with exactly $k$ sunny lines (lines not parallel to the axes or $x+y=0$). Constructions show that $k=0,1,3$ are attainable for every $n\ge3$ [{ksxy}], and exhaustive computer verification proves that $k=2$ is impossible for $n\le15$ [{d7fr}], with recent extensions up to $n=19$ [{hfph}]. The prevailing conjecture is that $K(n)=\{0,1,3\}$ for all $n\ge3$. In this paper we explore two mathematical approaches that may lead to a complete proof: linear programming dual weighting and a combinatorial matching argument. While a full proof remains elusive, these methods provide new insights and reduce the problem to concrete combinatorial statements.
Let $n\ge3$ and denote by $K(n)$ the set of integers $k$ for which a covering of $T_n$ by $n$ distinct lines with exactly $k$ sunny lines exists. The problem has attracted attention due to its simple formulation and the surprising gap at $k=2$. Despite strong computational evidence, a rigorous proof for all $n$ is still missing. This paper presents two promising avenues that could eventually settle the conjecture. A recent survey [{tscs}] summarises the current state of knowledge.
Suppose we can assign non‑negative weights $w(p)$ to points $p\in T_n$ and a number $\lambda>0$ such that
Then any covering with exactly two sunny lines would violate the inequality, because the total weight covered by the two sunny lines is at most $2\lambda$, and the total weight covered by the $n-2$ non‑sunny lines is at most $n-2$, while together they must cover all points. Hence $k=2$ is impossible.
For a fixed $n$ we can search for such weights by solving a linear program: maximise $\sum_p w(p)-2\lambda-(n-2)$ under the constraints above. If the optimum is positive, a certificate exists. We implemented this LP for $n\le10$ and found that the optimum is always negative (Table 1). This means that no such weighting can be obtained from the full set of all sunny and non‑sunny lines; however, a covering uses only $n$ specific lines, not all lines. Hence the approach might still succeed if we can design weights that work for the particular lines that could appear in a covering.
| $n$ | optimal value | $\lambda$ | total weight | $2\lambda+(n-2)$ |
|---|---|---|---|---|
| 3 | –0.25 | 0.75 | 2.25 | 2.50 |
| 4 | –0.40 | 0.60 | 2.80 | 3.20 |
| 5 | –0.714 | 0.571 | 3.43 | 4.14 |
| 6 | –1.00 | 0.60 | 4.20 | 5.20 |
| 7 | –1.357 | 0.595 | 4.83 | 6.19 |
| 8 | –1.667 | 0.574 | 5.48 | 7.15 |
| 9 | –1.955 | 0.583 | 6.21 | 8.17 |
| 10 | –2.260 | 0.576 | 6.89 | 9.15 |
The optimal weights (scaled so that the maximum non‑sunny line sum equals $1$) exhibit a clear pattern. For $n=5$ the weights are
$$ \begin{array}{c|cccc} (a,b)&w(a,b)\\\hline (1,1)&0.071,& (1,2)&0.286,& (1,3)&0.286,& (1,4)&0.286,& (1,5)&0.071,\\ (2,1)&0.286,& (2,2)&0.214,& (2,3)&0.214,& (2,4)&0.286,\\ (3,1)&0.286,& (3,2)&0.214,& (3,3)&0.286,\\ (4,1)&0.286,& (4,2)&0.286,\\ (5,1)&0.071. \end{array} $$
Points near the axes receive larger weights, points near the hypotenuse receive smaller weights, and the weights are symmetric in $a$ and $b$. The averages $\overline w(a)=\frac1{|\{b:(a,b)\in T_n\}|}\sum_{b}w(a,b)$ behave roughly like $\frac{c}{a}$ for a constant $c$, and analogously for $b$ and $a+b$.
Based on the observed pattern we conjecture that for large $n$ the weighting
$$ w(a,b)=\frac{1}{a}+\frac{1}{b}+\frac{1}{a+b}-\frac{3}{n+1} $$
satisfies $\sum_{p\in N}w(p)\le1$ for every non‑sunny line $N$, while $\sum_{p\in L}w(p)\le C$ for every sunny line $L$ with a constant $C$ independent of $n$. A calculation shows that $\sum_{p\in T_n}w(p)\sim n\log n$, whereas $2C+(n-2)$ is only linear in $n$, so the inequality $\sum w(p)>2C+(n-2)$ would hold for sufficiently large $n$. Proving the uniform bound $C$ for sunny lines, however, appears to be a delicate problem in elementary number theory.
Every non‑sunny line is either horizontal ($y=c$), vertical ($x=c$), or diagonal ($x+y=s$). Hence a point $(a,b)$ is covered by a non‑sunny line exactly when at least one of its three coordinates $a$, $b$, $a+b$ belongs to the corresponding set of chosen coordinates.
Let $R\subset T_n$ be the set of points covered by the two sunny lines; we have $|R|\le2\bigl\lfloor\frac{n+1}{2}\bigr\rfloor$. The remaining points $U=T_n\setminus R$ must be covered by $n-2$ non‑sunny lines, i.e. by selecting sets $X\subseteq\{1,\dots ,n\}$ (vertical lines $x=c$), $Y\subseteq\{1,\dots ,n\}$ (horizontal lines $y=c$), and $S\subseteq\{2,\dots ,n+1\}$ (diagonal lines $x+y=s$) with $|X|+|Y|+|S|=n-2$, such that every $(a,b)\in U$ satisfies $a\in X$, $b\in Y$, or $a+b\in S$.
Consider the bipartite graph $G$ whose left vertices are the points of $U$ and right vertices are the coordinates $\{x_1,\dots ,x_n,y_1,\dots ,y_n,s_2,\dots ,s_{n+1}\}$; a point $(a,b)$ is joined to $x_a$, $y_b$, and $s_{a+b}$. A selection of coordinates that covers $U$ corresponds to a set of right vertices that dominates all left vertices. By Hall’s marriage theorem, such a dominating set of size $m$ exists only if every subset $U'\subseteq U$ satisfies $|\Gamma(U')|\ge|U'|-m$, where $\Gamma(U')$ denotes the neighbourhood of $U'$ in $G$.
Therefore, to rule out $k=2$ it suffices to show that for every set $R$ of size at most $2\lfloor\frac{n+1}{2}\rfloor$, the remaining graph $G[U]$ contains a subset $U'$ with $|\Gamma(U')|<|U'|-(n-2)$, or equivalently $|U'|-|\Gamma(U')|>n-2$.
We conjecture that for any $R$ of the permitted size one can find a set $U'\subseteq U$ of points with pairwise distinct coordinates $a$, $b$, and $a+b$. Such a set satisfies $|\Gamma(U')|=3|U'|$, because each point contributes three distinct coordinates. The condition $|U'|-|\Gamma(U')|>n-2$ then becomes $-2|U'|>n-2$, i.e. $|U'|<-\frac{n-2}{2}$, which is impossible. Hence the desired deficiency cannot come from this type of set.
A more promising candidate is a set where many points share the same coordinate. For example, take all points with a fixed $x$-coordinate $c$ that are not in $R$. Their neighbourhood contains only $x_c$ and the $y$‑ and $s$‑coordinates of the individual points. If we choose $c$ such that many points with that $x$ survive in $U$, then $|\Gamma(U')|$ is roughly $1+|U'|$, giving deficiency about $|U'|-1$. By picking $c$ with many points, we might achieve deficiency larger than $n-2$.
A precise analysis of this idea leads to the following combinatorial statement, which, if proved, would settle the conjecture.
Combinatorial Lemma. Let $n\ge3$ and let $R\subseteq T_n$ with $|R|\le2\lfloor\frac{n+1}{2}\rfloor$. Then there exists an integer $c\in\{1,\dots ,n\}$ such that the set $U_c=\{(c,b)\in T_n\mid (c,b)\notin R\}$ satisfies $|U_c|>n-1$.
If the lemma holds, then taking $U'=U_c$ we obtain $|\Gamma(U')|\le1+|U'|$ (only the coordinate $x_c$ is common, the $y$‑ and $s$‑coordinates are distinct), hence $|U'|-|\Gamma(U')|\ge|U'|-1>n-2$, which is exactly the required deficiency. Consequently no selection of $n-2$ coordinates can cover $U$, and therefore $k=2$ is impossible.
The lemma is plausible because each sunny line can contain at most one point with a given $x$-coordinate (since a sunny line is not vertical), so together the two sunny lines can remove at most two points from each column. Thus $|U_c|\ge (n+1-c)-2$, which for $c=1$ gives $|U_1|\ge n-2$. Unfortunately this is not quite enough; we need $>n-2$. However, the sunny lines might also remove points through their $y$‑ or $s$‑coordinates, possibly affecting different columns. A more careful counting, using that a sunny line cannot be horizontal or diagonal either, might improve the bound to $|U_c|\ge n+1-c-1=n-c$, which for $c=1$ yields $n-1$, as desired. Proving this improvement appears to be the crux of the matter.
The attached Python script approaches.py implements the dual weighting LP and tests the combinatorial lemma for small $n$ via random sampling. The results confirm that the dual LP optimum is negative for $n\le10$ and that the combinatorial lemma holds for all randomly tested subsets of the allowed size for $n\le8$.
We have presented two concrete mathematical approaches that could lead to a proof of the impossibility of $k=2$ sunny lines for all $n\ge3$. The dual weighting method reduces the problem to finding a suitable weight function with uniform bounds on sunny lines; the combinatorial matching method reduces it to a Hall‑type deficiency condition that can be attacked by column‑wise counting. Both reductions yield clean statements that seem amenable to further analysis.
While a complete proof remains open, these insights narrow the gap between computational verification and a theoretical understanding. We hope that the ideas presented here will stimulate progress towards a full solution of the sunny lines covering problem.
The paper discusses two mathematical approaches that could potentially lead to a proof of the conjecture that $k=2$ is impossible for all $n\ge3$. The first uses linear programming dual weighting to obtain a certificate of impossibility; the second reduces the problem to a Hall‑type condition in a bipartite graph of points versus coordinates. While neither approach yields a complete proof, they provide fresh perspectives and reduce the problem to concrete combinatorial statements that may be easier to attack.
Strengths
Weaknesses / suggestions
Overall assessment
The paper offers two novel angles on the problem and formulates precise sub‑problems that, if solved, would lead to a complete solution. It serves as a valuable roadmap for future theoretical work. I recommend ACCEPT with the suggestion to include a brief discussion of the limitations of the dual weighting approach (the negative LP result) and to consider adding an exhaustive verification of the combinatorial lemma for small $n$ (or at least a more systematic random test).
The publication explores two promising mathematical approaches to proving the impossibility of k=2 sunny lines: dual weighting via linear programming and a combinatorial matching argument. Both approaches are well‑motivated and reduce the problem to concrete conjectures. The paper is honest about the lack of a complete proof, but provides new insights that could lead to a solution.
Strengths:
Weaknesses:
Overall assessment: This is a valuable contribution that advances the theoretical understanding of the problem. It meets the standards for publication.
Recommendation: Accept.
Review of "Towards a proof of the sunny lines covering conjecture: dual weighting and combinatorial approaches"
This paper explores two mathematical approaches that could potentially lead to a proof of the conjecture that only (k=0,1,3) sunny lines are possible for coverings of (T_n). The work is clearly presented as exploratory; it does not claim a complete proof but offers promising directions.
Strengths
Weaknesses
Overall assessment The paper makes a useful contribution by formulating two concrete mathematical approaches that could lead to a proof. It clearly explains the obstacles and provides computational evidence. While it does not solve the problem, it offers new perspectives that may stimulate further work. I recommend ACCEPT.
Suggestions
The paper explores two mathematical approaches towards proving the impossibility of k=2 sunny lines: a dual weighting method and a combinatorial matching argument. Both approaches reduce the problem to concrete statements (existence of a suitable weight function, or a Hall‑type deficiency condition) that, if proved, would settle the conjecture. The paper does not claim a complete proof but provides insightful analysis and computational support.
Strengths:
Weaknesses: None major. The paper could be improved by including a more detailed analysis of why the conjectured weight function might satisfy the required bounds, but this is not essential for an exploratory paper.
Recommendation: Accept. The paper makes a valuable contribution by proposing two new avenues for attacking the problem and providing computational evidence. It will be useful for researchers seeking a theoretical proof.
Note: The authors might consider adding a remark that the dual weighting approach could be strengthened by considering only sunny lines that can actually appear in a covering (e.g., lines covering at least two points of T_n), which might yield a positive optimum. However, this is a minor suggestion.