Fixed-point cycles and EFX allocations

January 21, 2022 Β· Declared Dead Β· πŸ› arXiv.org

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Authors Benjamin Aram Berendsohn, Simona Boyadzhiyska, LΓ‘szlΓ³ Kozma arXiv ID 2201.08753 Category cs.DS: Data Structures & Algorithms Cross-listed math.CO Citations 7 Venue arXiv.org Last Checked 4 months ago
Abstract
We study edge-labelings of the complete bidirected graph $\overset{\tiny\leftrightarrow}{K}_n$ with functions from the set $[d] = \{1, \dots, d\}$ to itself. We call a cycle in $\overset{\tiny\leftrightarrow}{K}_n$ a fixed-point cycle if composing the labels of its edges results in a map that has a fixed point, and we say that a labeling is fixed-point-free if no fixed-point cycle exists. For a given $d$, we ask for the largest value of $n$, denoted $R_f(d)$, for which there exists a fixed-point-free labeling of $\overset{\tiny\leftrightarrow}{K}_n$. Determining $R_f(d)$ for all $d >0$ is a natural Ramsey-type question, generalizing some well-studied zero-sum problems in extremal combinatorics. The problem was recently introduced by Chaudhury, Garg, Mehlhorn, Mehta, and Misra, who proved that $d \leq R_f(d) \leq d^4+d$ and showed that the problem has close connections to EFX allocations, a central problem of fair allocation in social choice theory. In this paper we show the improved bound $R_f(d) \leq d^{2 + o(1)}$, yielding an efficient ${(1-\varepsilon)}$-EFX allocation with $n$ agents and $O(n^{0.67})$ unallocated goods for any constant $\varepsilon \in (0,1/2]$; this improves the bound of $O(n^{0.8})$ of Chaudhury, Garg, Mehlhorn, Mehta, and Misra. Additionally, we prove the stronger upper bound $2d-2$, in the case where all edge-labels are permulations. A very special case of this problem, that of finding zero-sum cycles in digraphs whose edges are labeled with elements of $\mathbb{Z}_d$, was recently considered by Alon and Krivelevich and by MΓ©szΓ‘ros and Steiner. Our result improves the bounds obtained by these authors and extends them to labelings from an arbitrary (not necessarily commutative) group, while also simplifying the proof.
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