Exact Cut Complexity of Equal-Length Proportional Cake Cutting

Review failed. It will be retried automatically.

We study proportional cake cutting under an equal-length constraint, where each of $n$ agents must receive a bundle of length exactly $1/n$ that is worth at least $1/n$ to that agent. Our main result determines the exact worst-case number of cuts needed to guarantee such an allocation, namely $2n-2$ for every $n$. This improves on the previously known quadratic upper bound. The proof combines the constrained necklace-splitting theorem with a recursive assignment argument based on Hall's theorem. The same bound extends to the stronger requirement of envy-freeness when $n$ is a prime power, and an $O(n^{1.525})$ upper bound holds for arbitrary $n$. We then ask whether such allocations can be computed. We first show that no deterministic protocol in the Robertson--Webb model terminates after finitely many queries on every instance and always returns an exact equal-length proportional allocation, even for two agents with identical valuations. For every $\varepsilon>0$, however, $O(n^2/\varepsilon)$ queries suffice to produce an allocation in which every agent receives a bundle of length exactly $1/n$ and value at least $1/n-\varepsilon$, again using at most $2n-2$ cuts. Finally, we augment the Robertson--Webb model with two-knife queries. We show that recursive balanced halving attains the optimal bound of $2n-2$ cuts when $n$ is a power of two. For arbitrary $n$, we give a procedure using $O(n\log n)$ cuts.

Pick a starting point or write your own. Challenges run in the background, so you can keep reading while the AI investigates.

No challenges yet. Disagree with the review? Ask the AI to revisit a specific claim.