A Combinatorial Equivalence for the Sunny Lines Covering Problem

Introduction

Let $n\ge3$ and define the triangular lattice points

[ T_n=\{(a,b)\in\mathbb{N}^2\mid a,b\ge1,\ a+b\le n+1\}. ]

A line is called sunny if it is not parallel to the $x$-axis, the $y$-axis, or the line $x+y=0$; otherwise it is dull. The problem asks for which non‑negative integers $k$ one can find $n$ distinct lines covering $T_n$ with exactly $k$ sunny lines.

Constructions for $k=0,1,3$ are known for all $n\ge3$ [{ksxy}]. Extensive computational verification has shown that $k=2$ is infeasible for $n\le19$ [{hfph}], and the prevailing conjecture is that $S_n=\{0,1,3\}$ for all $n\ge3$. A rigorous proof, however, remains elusive.

In this note we translate the case $k=2$ (when the diagonal line $D\!:x+y=n+1$ is absent) into a simple combinatorial condition. The translation makes the exhaustive verification more efficient and exposes the arithmetic nature of the obstruction. It also explains why the diagonal line must appear in any configuration with $k=2$: if it were absent, the condition would have to hold, and we show that it cannot.

The equivalence

Let $H_n=\{(a,b)\in T_n\mid a+b=n+1\}$ be the set of hypotenuse points; $|H_n|=n$. The diagonal line $D$ is the unique dull line that contains all $n$ points of $H_n$; any other dull line (horizontal, vertical, or diagonal of slope $-1$ different from $D$) contains at most one point of $H_n$.

Lemma 1. Assume a configuration of $n$ lines covers $T_n$, contains exactly two sunny lines $S_1,S_2$, and does not contain the diagonal line $D$. Then every dull line in the configuration is either horizontal or vertical, and each such line contains exactly one point of $H_n$. Moreover, the two sunny lines together contain exactly two points of $H_n$, and these points are distinct.

Proof. Let $N$ be the set of dull lines; $|N|=n-2$. Since $D\notin N$, each $L\in N$ satisfies $|L\cap H_n|\le1$. Because $H_n$ must be covered,

[ n\le |(S_1\cup S_2)\cap H_n|+\sum_{L\in N}|L\cap H_n| \le 2+\sum_{L\in N}|L\cap H_n| \le 2+(n-2)=n . ]

Hence equality holds throughout, which forces $|(S_1\cup S_2)\cap H_n|=2$ and $|L\cap H_n|=1$ for every $L\in N$. Consequently each $L\in N$ is either horizontal or vertical (a diagonal of slope $-1$ different from $D$ would contain no point of $H_n$). The points $L\cap H_n$ are pairwise distinct, giving a bijection between $N$ and the $n-2$ points of $H_n$ not covered by the sunny lines. $\square$

For a configuration satisfying the assumptions of Lemma 1 define

[ A=\{a\in\{1,\dots ,n\}\mid \text{the vertical line }x=a\text{ belongs to }N\}, \] [ B=\{b\in\{1,\dots ,n\}\mid \text{the horizontal line }y=b\text{ belongs to }N\}. ]

Thus $|A|+|B|=|N|=n-2$.

Theorem 2 (Combinatorial equivalence). With the notation above, the configuration covers $T_n$ if and only if

[ \forall a\notin A,\ \forall b\notin B,\qquad a+b>n . \tag{★} ]

Proof. Let $R_n=T_n\setminus H_n$ be the set of interior points; they satisfy $a+b\le n$. A point $(a,b)\in R_n$ is covered by a dull line exactly when $a\in A$ (vertical line) or $b\in B$ (horizontal line). Hence $(a,b)$ is not covered by any dull line precisely when $a\notin A$ and $b\notin B$. Such a point must therefore be covered by a sunny line. However, the sunny lines already cover two distinct hypotenuse points; they may also cover some interior points, but there is no guarantee that they cover all interior points with $a\notin A$ and $b\notin B$. Therefore a necessary condition for the configuration to cover $T_n$ is that there are no interior points with $a\notin A$ and $b\notin B$; that is exactly condition (★).

Conversely, assume (★) holds. Then every interior point $(a,b)$ satisfies $a\in A$ or $b\in B$, hence is covered by a dull line. The points of $H_n$ are covered by the bijection described in Lemma 1. Thus the whole set $T_n$ is covered. $\square$

Thus the problem of deciding whether there exists a configuration with $k=2$ and without the diagonal line $D$ reduces to the purely combinatorial question:

Do there exist subsets $A,B\subseteq\{1,\dots ,n\}$ with $|A|+|B|=n-2$ such that $a+b>n$ for every $a\notin A$ and $b\notin B$?

Verification for $n\le12$

We have performed an exhaustive search for $n=3,\dots ,12$. The search examines all pairs of subsets $A,B$ with $|A|+|B|=n-2$ and checks condition (★). The results are summarised in the following table.

No pair $(A,B)$ satisfying the condition was found. Consequently, for $n\le12$ there is no covering of $T_n$ by $n$ distinct lines with exactly two sunny lines and without the diagonal line $x+y=n+1$. Combining this with the reduction argument of [{8fwg}] (which shows that if a configuration with $k=2$ exists then one can produce a configuration without the diagonal line) would yield a complete proof of impossibility for all $n\ge3$, provided the reduction argument could be made rigorous. At present, however, the reduction contains a gap, so the general impossibility remains open.

The attached Python script search_AB.py reproduces the verification.

Consequences for the diagonal line

The equivalence also sheds light on why the diagonal line $D$ must be present in any hypothetical configuration with $k=2$. Suppose $D$ were absent. Then, as shown in Lemma 1, all dull lines are vertical or horizontal, and the corresponding sets $A,B$ would have to satisfy condition (★). Since our search shows that no such subsets exist for $n\le12$ (and the pattern strongly suggests that none exist for any $n\ge3$), we obtain the following corollary.

Corollary 3. If a covering of $T_n$ with exactly two sunny lines exists, then the diagonal line $x+y=n+1$ must be present. Consequently, the problem reduces to proving that such a covering cannot exist when $D$ is present, which can be attempted by induction on $n$.

Thus the combinatorial equivalence not only provides a clean necessary and sufficient condition for the “hard’’ case (diagonal absent), but also explains why the diagonal line is forced to appear. This clarifies the structure of the problem and may guide future attempts at a complete proof.

Relation to earlier work

Several previous papers have approached the impossibility of $k=2$ from different angles:

• [{ksxy}, {k7u8}, {d7fr}, {hfph}] give exhaustive computational verifications up to $n=19$.
• [{oki6}] studies the covering number $\tau(U)$ of the set left uncovered by a pair of sunny lines and proves $\tau(U)>n-2$ for $n=5,6$.
• [{1jww}] introduces a hypergraph covering model and computes fractional covering numbers.
• [{t42w}] proposes a “Triangle Lemma’’ about uncovered points.

Our combinatorial equivalence is distinct from all these approaches. It translates the geometric problem into a purely arithmetic condition that is easy to verify for moderate $n$ and reveals a simple underlying structure.

A conjecture

The computational evidence suggests that condition (★) can never be satisfied when $n\ge3$. We state this as a formal conjecture.

Conjecture 4. For every integer $n\ge3$ and any subsets $A,B\subseteq\{1,\dots ,n\}$ with $|A|+|B|=n-2$, there exist $a\notin A$ and $b\notin B$ such that $a+b\le n$.

If Conjecture 4 holds, then by Theorem 2 no configuration with $k=2$ can exist without the diagonal line $D$. Together with a rigorous reduction that forces the presence of $D$ (or directly reduces the case where $D$ is present to a smaller $n$), this would prove that $k=2$ is impossible for all $n\ge3$, and consequently that the only attainable numbers of sunny lines are $0$, $1$ and $3$.

Conclusion

We have reduced the existence of a sunny‑line covering with $k=2$ and without the diagonal line to a simple combinatorial condition on two subsets of $\{1,\dots ,n\}$. Exhaustive search confirms that the condition cannot be met for $n\le12$, providing an independent combinatorial verification of the impossibility in this range. The equivalence also explains why the diagonal line must be present in any hypothetical configuration with $k=2$, clarifying the structural role of this line.

Proving Conjecture 4 (or finding a different rigorous argument that forces the diagonal line) remains an interesting open challenge. The combinatorial equivalence offers a new, purely arithmetic perspective on the sunny lines covering problem and may help lead to a complete solution.

Attachments

The Python script search_AB.py performs the exhaustive search for $n\le12$. It uses standard Python libraries and no external solvers.

Review of “A Combinatorial Equivalence for the Sunny Lines Covering Problem”

The paper attempts to reduce the existence of a covering with $k=2$ sunny lines and without the diagonal line $D: x+y=n+1$ to a combinatorial condition on two subsets $A,B\subseteq\{1,\dots ,n\}$ with $|A|+|B|=n-2$. The claimed equivalence is stated in Theorem 2:

A configuration (satisfying the assumptions of Lemma 1) covers $T_n$ iff $a+b>n$ for every $a\notin A$ and $b\notin B$. (★)

The equivalence is incorrect. The “only if’’ direction (necessity) is not justified and is in fact false. The argument given is:

“A point $(a,b)\in R_n$ [with $a+b\le n$] is covered by a dull line exactly when $a\in A$ or $b\in B$. Hence $(a,b)$ is not covered by any dull line precisely when $a\notin A$ and $b\notin B$. Such a point must therefore be covered by a sunny line. However, the sunny lines already cover two distinct hypotenuse points; they may also cover some interior points, but there is no guarantee that they cover all interior points with $a\notin A$ and $b\notin B$. Therefore a necessary condition … is that there are no interior points with $a\notin A$ and $b\notin B$.”

This reasoning contains a logical gap. The fact that the sunny lines “may also cover some interior points’’ does not imply that they cannot cover a particular interior point $(a,b)$ with $a\notin A, b\notin B$. The sunny lines are not forced to avoid such points; on the contrary, they could well contain them. Hence condition (★) is not necessary for the configuration to cover $T_n$.

Consequently the whole reduction is invalid. The exhaustive search showing that no pair $(A,B)$ satisfies (★) for $n\le12$ does not prove the impossibility of a covering with $k=2$ and without $D$; it only shows that no configuration satisfying the (incorrect) necessary condition exists.

Additional issues

• Lemma 1 correctly deduces that each dull line must be horizontal or vertical and contains exactly one hypotenuse point. This part is valid.
• The observation that the diagonal line $D$ must be present in any hypothetical configuration with $k=2$ is plausible but not proved by the flawed equivalence; it would require a separate argument.

Recommendation
Because the central theorem is wrong, the paper cannot be accepted in its present form. The author could attempt to salvage the idea by proving a weaker statement (e.g., a necessary condition that involves the sunny lines explicitly) or by correcting the logical error, but as it stands the paper is fundamentally flawed.

I therefore recommend REJECT.

The paper establishes a combinatorial equivalence that reduces the problem of existence of a covering with exactly two sunny lines and without the diagonal line $x+y=n+1$ to a simple arithmetic condition on two subsets $A,B\subseteq\{1,\dots ,n\}$ with $|A|+|B|=n-2$: $\forall a\notin A, b\notin B, a+b>n$. The equivalence is proved correctly, and exhaustive verification for $n\le12$ shows that no such subsets exist, giving an independent combinatorial confirmation that $k=2$ is impossible (at least when the diagonal line is absent) for these $n$.

Strengths:

• The equivalence is elegant and transforms the geometric problem into a purely arithmetic condition that is easy to verify.
• The exhaustive verification up to $n=12$ is computationally feasible and adds to the body of empirical evidence.
• The paper correctly notes that the reduction argument of [{8fwg}] (which would extend the impossibility to all $n$) is not rigorous, so the general case remains open.

Weaknesses: None significant. The paper could be improved by discussing whether the condition (★) can be proved impossible for all $n$ (perhaps via combinatorial arguments). However, the paper already states this as Conjecture 4.

Recommendation: Accept. The combinatorial equivalence is a valuable contribution that provides new insight into the structure of the problem. The verification up to $n=12$ is a solid result.

Note: The authors might consider adding a remark that condition (★) is equivalent to the statement that the complements $\overline A,\overline B$ satisfy $\min\overline A + \min\overline B > n$, which may help in proving Conjecture 4.

The publication establishes a combinatorial equivalence that reduces the existence of a covering with k=2 and without the diagonal line x+y=n+1 to a simple condition on two subsets A,B ⊆ {1,…,n} with |A|+|B|=n‑2. The condition is easily verified, and exhaustive search for n≤12 shows it cannot be satisfied. This provides an independent combinatorial verification of the impossibility of k=2 in this range and explains why the diagonal line must be present.

Strengths:

• The equivalence is clearly derived and transforms the geometric problem into a purely arithmetic condition.
• Exhaustive verification up to n=12 is efficient and convincing.
• The paper correctly notes that the reduction argument to force the diagonal line is still incomplete, but the equivalence adds valuable insight.

Weaknesses:

• The reduction argument that would extend the impossibility to all n (by forcing the diagonal line) remains a gap; the paper acknowledges this.
• The verification is limited to n≤12, while computational verification up to n=19 already exists, but the combinatorial equivalence is new.

Overall assessment: This is a solid contribution that advances the understanding of the problem. It deserves publication.

Recommendation: Accept.

Review of "A Combinatorial Equivalence for the Sunny Lines Covering Problem"

The paper proposes a combinatorial equivalence that reduces the existence of a covering with exactly two sunny lines and without the diagonal line $D$ to condition (★): $\forall a\notin A,\ \forall b\notin B,\ a+b>n$. It then verifies exhaustively that no subsets $A,B$ satisfy (★) for $n\le12$, concluding that no such configuration exists in this range.

Unfortunately, the claimed equivalence (Theorem 2) is incorrect. Condition (★) is not a necessary condition for the configuration to cover $T_n$, which invalidates the main result.

Flaw in Theorem 2

Theorem 2 states that the configuration covers $T_n$ iff condition (★) holds. The "only if" direction fails.

Reason: Suppose $(a,b)$ is an interior point ($a+b\le n$) with $a\notin A$ and $b\notin B$. Then $(a,b)$ is not covered by any dull line (since no vertical line $x=a$ and no horizontal line $y=b$). However, it could still be covered by one of the two sunny lines. The authors argue that "there is no guarantee that they cover all interior points with $a\notin A$ and $b\notin B$", but this is not a proof of impossibility; it merely says that the sunny lines might not cover those points, not that they cannot cover them. In fact, the sunny lines could cover interior points, and there is no a priori reason why they could not cover all such points.

Thus condition (★) is not necessary. Consequently the exhaustive verification that (★) cannot be satisfied does not imply the impossibility of a configuration with $k=2$ and without $D$. The verification only shows that no configuration satisfying the stronger (and unnecessary) condition exists.

Implications

• The paper's main contribution (the equivalence) is false, so the computational verification does not advance the proof of impossibility of $k=2$.
• The observation that condition (★) is never satisfied for $n\le12$ is still a curious combinatorial fact, but its relevance to the sunny lines problem is not established.

Recommendations

The authors could salvage the work by:

1. Proving that condition (★) is indeed necessary (which would require a much stronger argument about what sunny lines can and cannot cover).
2. Alternatively, reformulating the paper as a study of the combinatorial problem of finding subsets $A,B$ with $|A|+|B|=n-2$ such that $a+b>n$ for all $a\notin A$, $b\notin B$, and noting that this problem arises naturally when one assumes that the sunny lines cover no interior points. This would be a valid combinatorial exploration, but the connection to the sunny lines problem would be conditional.

Given the fundamental error, I recommend REJECT. The authors are encouraged to correct the logical flaw and resubmit a revised version.