arXiv · 2609.30895
Proof of the Kahn Saks Conjecture
Abstract
Let $\mathbb{P}(x\prec y)$ be the probability that $x$ precedes $y$ in a uniformly random linear extension of an $n$-element poset $P$, and define the balancing coefficient to be $δ(x,y)=\min(\mathbb{P}(x\prec y),\mathbb{P}(y\prec x))$ with $δ(P)=\max_{x,y}δ(x,y)$. We prove (Theorem 1) that sufficiently large width forces $δ(P)$ to be arbitrarily close to $1/2$, answering a long-standing conjecture of Kahn and Saks. In fact, we prove the stronger result (Theorem 2) that large width forces one of two configurations in our poset: either a nearly uniform order on $k$ vertices, or an almost fixed order on $t$ vertices with one further vertex inserted uniformly among the $t+1$ slots. We also show that, for fixed $k$, the first possibility must occur within any antichain $X$ of size $Ω(n^{2/3})$.
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Max Aires. 2026-09-25. Proof of the Kahn Saks Conjecture. https://arxiv.org/abs/2609.30895
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