Finding a Mediocre Player

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 Yao [29], an element (player) is said to be (i, j)-mediocre if it is neither among the top i nor among the bottom j elements of S. More than 40 years ago, Yao suggested a stunningly simple 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 such an 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 $j=j(i)$). 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 $j=j(i)$); 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.)