Searcharxiv⌕ Search
Home/Archive/arXiv

arXiv · 1901.09017

Finding a Mediocre Player

Adrian Dumitrescu
View original source ↗Read source document ↗Export metadata ↓

Abstract

Consider a totally ordered set $S$ of $n$ elements; as an example, a set of tennis players and their rankings. Further assume that their ranking is a total order and thus satisfies transitivity and anti-symmetry. Following Frances Yao (1974), an element (player) is said to be $(i,j)$-\emph{mediocre} if it is neither among the top $i$ nor among the bottom $j$ elements of $S$. Finding a mediocre element is closely related to finding the median element. More than $40$ years ago, Yao suggested a very simple and elegant algorithm for finding an $(i,j)$-mediocre element: Pick $i+j+1$ elements arbitrarily and select the $(i+1)$-th largest among them. She also asked: "Is this the best algorithm?" No one seems to have found a better algorithm ever since. We first provide a deterministic algorithm that beats the worst-case comparison bound in Yao's algorithm for a large range of values of $i$ (and corresponding suitable $j=j(i)$) even if the current best selection algorithm is used. We then repeat the exercise for randomized algorithms; the average number of comparisons of our algorithm beats the average comparison bound in Yao's algorithm for another large range of values of $i$ (and corresponding suitable $j=j(i)$) even if the best selection algorithm is used; the improvement is most notable in the symmetric case $i=j$. Moreover, the tight bound obtained in the analysis of Yao's algorithm allows us to give a definite answer for this class of algorithms. In summary, we answer Yao's question as follows: (i)~"Presently not" for deterministic algorithms and (ii)~"Definitely not" for randomized algorithms. (In fairness, it should be said however that Yao posed the question in the context of deterministic algorithms.)

Explore related subjects

cs.DSmath.CO

Keep this discovery

BibTeXRIS

Adrian Dumitrescu. 2019-01-25. Finding a Mediocre Player. https://arxiv.org/abs/1901.09017

Cite the original work for its findings. Save a collection to share your selection of sources.

KEEP EXPLORING

Related papers

arXiv2026

Quasi-Monte Carlo Beyond Hardy-Krause II: $(1 + \varepsilon)n$ Samples Suffice

Numerical integration studies how well one can estimate the integral of a function $f$ over $[0,1)^d$ using $n$ sample points. The two classical methods, Monte Carlo (MC) and quasi-Monte Carlo (QMC), have complementary strengths and weaknesses, and a fundamental question is to design an approach that combines the benefits of both. Recently, building on the transference principle in discrepancy theory, Bansal and Jiang~\cite{BJ25a} gave a randomized QMC method that bridges MC and QMC guarantees using only i.i.d.\ samples. Their method also goes beyond the classical Koksma--Hlawka inequality: it achieves integration error $\widetilde{O}_d(\sigma_{\mathsf{SO}}(f)/n)$, where the smoothed-out variation $\sigma_{\mathsf{SO}}(f)$ can be substantially smaller than the Hardy--Krause variation that governs the classical bound. However, their algorithm requires $n^2$ i.i.d.\ samples as input, and this quadratic blowup is inherent to any method based on the transference principle. In this work, we bypass the quadratic blowup: for any constant $\varepsilon > 0$, we show that $(1+\varepsilon)n$ i.i.d.\ samples suffice to both obtain the beyond-Hardy--Krause guarantee of~\cite{BJ25a}, resolving an open problem posed there, and to produce low-discrepancy point sequences. Our algorithms are variants of the online Haar-thinning method of Dwivedi, Feldheim, Gurel-Gurevich, and Ramdas~\cite{DFG+19}.

cs.DS↗
arXiv2026

Single-Exponential Algorithms and a Polynomial Kernel for Strong Connectivity Augmentation

Strong Connectivity Augmentation (SCA) asks whether a directed acyclic graph can be made strongly connected by adding at most $k$ prescribed links whose total weight is within a given budget. Klinkby, Misra, and Saurabh (SODA 2021) gave an $O^*(2^{O(k\log k)})$-time algorithm and asked whether the problem admits a single-exponential parameterized algorithm and a polynomial kernel. We answer both questions affirmatively: SCA can be solved in $O^*(9^k)$ time and admits a polynomial kernel with $O(k^4)$ vertices and $O(k^{16})$ bits. For unweighted SCA, we obtain $O^*(4^k)$ time and a kernel with $O(k^3)$ vertices. Our algorithms are based on a particularly simple reduction to Strongly Connected Spanning Subgraph with two edge costs.

cs.DS↗