Disproportionate division

TL;DR

This paper improves the understanding of dividing a cake among agents with different demands, showing a linear bound on cuts needed and proposing a method to achieve it, with a conjecture for further optimality.

Contribution

It proves that disproportionate division can always be achieved with 3n-4 cuts and provides a constructive procedure, improving upon the previous O(n log n) bound.

Findings

Disproportionate division can be achieved with 3n-4 cuts.

A combinatorial procedure for constructing such divisions is provided.

A topological conjecture suggests 2n-2 cuts may suffice, which would be optimal.

Abstract

We study the disproportionate version of the classical cake-cutting problem: how efficiently can we divide a cake, here $[0,1]$, among $n$ agents with different demands $\alpha_1, \alpha_2, \dots, \alpha_n$ summing to $1$? When all the agents have equal demands of $\alpha_1 = \alpha_2 = \dots = \alpha_n = 1/n$, it is well-known that there exists a fair division with $n-1$ cuts, and this is optimal. For arbitrary demands on the other hand, folklore arguments from algebraic topology show that $O(n\log n)$ cuts suffice, and this has been the state of the art for decades. Here, we improve the state of affairs in two ways: we prove that disproportionate division may always be achieved with $3n-4$ cuts, and give an effective combinatorial procedure to construct such a division. We also offer a topological conjecture that implies that $2n-2$ cuts suffice in general, which would be optimal.