Putting Dots in Triangles

Given a right-angled triangle of squares in a grid whose horizontal and vertical sides are $n$ squares long, let N(n) denote the maximum number of dots that can be placed into the cells of the triangle such that each row, each column, and each diagonal parallel to the long side of the triangle contains at most one dot. It has been proven that $N(n) = \lfloor \frac{2n+1}{3} \rfloor$. In this note, we give a new proof of this result using linear programming techniques.

💡 Research Summary

The paper revisits a classic combinatorial optimization problem: placing as many dots as possible inside a right‑angled triangular array of unit squares of side‑length n, under the restriction that no two dots share the same row, the same column, or the same diagonal that runs parallel to the hypotenuse. The quantity N(n) denotes the maximal number of dots that can be placed. It has long been known, through inductive and graph‑theoretic arguments, that
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