不成比例分裂
Disproportionate division
摘要：
我们研究的经典蛋糕切割问题的不成比例的版本：我们如何有效地划分蛋糕，在这里$[0,1]$，$n$代理与不同的需求$α_1, α_2, \dots, α_n$总结到$1$？当所有的智能体对$α_1 = α_2 = \dots = α_n = 1/n$有相同的需求时，存在一个以$n-1$为截数的公平分配，这是最优的。另一方面，对于任意的要求，代数拓扑学的民间说法表明，$O(n\log n)$切割就足够了，几十年来这一直是最先进的。在这里，我们改进的事态在两个方面：我们证明，不成比例的司可能总是实现与$3n-4$削减，并给出了一个有效的组合程序来构建这样的司。我们还提供了一个拓扑猜想，这意味着$2n-2$削减足够的一般情况下，这将是最佳的。
英文摘要：
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 $α_1, α_2, \dots, α_n$ summing to $1$? When all the agents have equal demands of $α_1 = α_2 = \dots = α_n = 1/n$, 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.
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