Abstract

We study the disproportionate version of the classical cake-cutting problem: how efficiently can we divide a cake, here [0,1], among n ⩾ 2 agents with different demands α 1 , α 2 , ⋯ , α n summing to 1? When all the agents have equal demands of α 1 = α 2 = ⋯ = 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 3 n − 4 cuts, and also give an effective algorithm to construct such a division. We additionally offer a topological conjecture that implies that 2 n − 2 cuts suffice in general, which would be optimal.

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