A Ratio Inequality For Binary Trees And The Best Secretary


Let T be the complete binary tree of height n considered as the Hasse diagram of a poset with its root 1n as the maximum element. Define A(n;T ) = |{S ⊆ Tn : 1n ∈ S, S ∼= T }|, and B(n;T ) = |{S ⊆ Tn : 1n / ∈ S, S ∼= T }|. In this note we prove that A(n;T1) B(n;T1) 6 A(n;T2) B(n;T2) for any fixed n and rooted binary trees T1, T2 such that T2 contains a subposet isomorphic to T1. We conjecture that the ratio A/B also increases with T for arbitrary trees. These inequalities imply natural behaviour of the optimal stopping time in a poset extension of the secretary problem.

DOI: 10.1017/S0963548301004977

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@article{Kubicki2002ARI, title={A Ratio Inequality For Binary Trees And The Best Secretary}, author={Grzegorz Kubicki and Jen{\"{o} Lehel and Michal Morayne}, journal={Combinatorics, Probability & Computing}, year={2002}, volume={11}, pages={149-161} }