Envy-free cake division without assuming the players prefer nonempty pieces

Consider $n$ players having preferences over the connected pieces of a cake, identified with the interval $[0,1]$. A classical theorem, found independently by Stromquist and by Woodall in 1980, ensures that, under mild conditions, it is possible to divide the cake into $n$ connected pieces and assign these pieces to the players in an envy-free manner, i.e, such that no player strictly prefers a piece that has not been assigned to her. One of these conditions, considered as crucial, is that no player is happy with an empty piece. We prove that, even if this condition is not satisfied, it is still possible to get such a division when $n$ is a prime number or is equal to $4$. When $n$ is at most $3$, this has been previously proved by Erel Segal-Halevi, who conjectured that the result holds for any $n$. The main step in our proof is a new combinatorial lemma in topology, close to a conjecture by Segal-Halevi and which is reminiscent of the celebrated Sperner lemma: instead of restricting the labels that can appear on each face of the simplex, the lemma considers labelings that enjoy a certain symmetry on the boundary

Piercing axis-parallel boxes

Let \F be a finite family of axis-parallel boxes in $\R^d$ such that \F contains no $k+1$ pairwise disjoint boxes. We prove that if \F contains a subfamily \M of $k$ pairwise disjoint boxes with the property that for every F\in \F and M\in \M with $F \cap M \neq \emptyset$, either $F$ contains a corner of $M$ or $M$ contains $2^{d-1}$ corners of $F$, then \F can be pierced by $O(k)$ points. One consequence of this result is that if $d=2$ and the ratio between any of the side lengths of any box is bounded by a constant, then \F can be pierced by $O(k)$ points. We further show that if for each two intersecting boxes in \F a corner of one is contained in the other, then \F can be pierced by at most $O(k\log\log(k))$ points, and in the special case where \F contains only cubes this bound improves to $O(k)$