Kristina Ago | Searcharxiv

arXiv2026

Closing the gap and settling the problem of queens on an $n\times n$ board, each attacking at most one other

Let $q(n)$ denote the largest number of queens that can be placed on an $n\times n$ chessboard so that no queen attacks more than one other queen. We prove that $q(n)=\lfloor4n/3\rfloor$ for every $n\geqslant6$, and that $q(n)=n$ for $n\leqslant5$, which settles a previously conjectural value. As a corollary, we also settle that, in the version of the problem where each queen attacks \emph{exactly} one other queen, the answer is $2\lfloor2n/3\rfloor$, again as previously conjectured.

math.CO↗

arXiv2026

Intersecting families with bounded intersections

Let $\mathcal F\subset 2^{[n]}$ be an $s$-uniform family such that every two distinct sets have a nonempty intersection but intersect in at most $k$ elements. By the well-known Ray-Chaudhuri--Wilson theorem, since the intersections can take at most $k$ different values, we have $|\mathcal F|\leq \binom{n}{k}$. We give a stronger upper bound under our assumptions above, when $n$ is large enough compared to $s$ (and $k+1<s$): $|\mathcal F|\leq \frac{\binom{n-1}{k}}{\binom{s-1}{k}}$. This is a special case of an old theorem of Deza, Erd\H os and Frankl, but our proof is simpler and gives a better threshold for $n$. Furthermore, we prove a generalization of the Erd\H os--Ko--Rado theorem for non-uniform families. Let $\mathcal F\subset \binom{[n]}{k}\cup\binom{[n]}{k+1}\cup\dots\cup\binom{[n]}{s}$, $3\leq k\leq s$, be a family such that for every two distinct sets the size of the intersection is between 1 and $k-1$ and $n$ is large enough then $|\mathcal F|\leq {n-1 \choose k-1}$. \emph{Mathematics Subject Classification (2020):} 05D05 \emph{Keywords: intersecting families, uniform families, Ray-Chaudhuri--Wilson theorem, Erd\H os--Ko--Rado theorem}

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