  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.
