Finding a Mediocre Player

TL;DR

This paper introduces improved deterministic and randomized algorithms for finding a mediocre element in a totally ordered set, outperforming Yao's classic approach in comparison bounds for a wide range of parameters.

Contribution

The authors present new algorithms that surpass the comparison bounds of Yao's algorithms for finding mediocre elements, providing definitive answers to Yao's open question.

Findings

Deterministic algorithm beats Yao's comparison bound for large i

Randomized algorithm has lower average comparisons for symmetric cases

Provides a definitive answer to Yao's question on algorithm optimality

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.)

Equations

$$\mathrm{Prob}\left(|X-\mathrm{E}[X]|\ge a\right)\le \frac{\mathrm{Var}[X]}{a^2}.$$

$$c_{\mathrm{Dor}-\mathrm{Zwick}}=c_{\mathrm{Dor}-\mathrm{Zwick}}(\alpha ,l)=1+(l+2)\left(\alpha +\frac{1-\alpha }{2^l}\right),$$

$$c_{\mathrm{Dor}-\mathrm{Zwick}}(\alpha )$$

$$=1+\alpha \log \frac{1}{\alpha }+\alpha \log \log \frac{1}{\alpha }+O(\alpha ).$$

$$g(\alpha ,l)=\left(1+(l+2)\left(\alpha +\frac{1-\alpha }{2^l}\right)\right).$$

$$l$$

$$f(\alpha )$$

$$c_{A1}$$

$$c_{\mathrm{Yao}}$$

$$f(\alpha )=\min \left(g(\alpha ,l),g(\alpha ,l+1),3\right),$$

$$f(\alpha )=\min \left(g(\alpha ,l),g(\alpha ,l+1),2.95\right).$$

$$c_{\mathrm{A}}$$

$$c_{\mathrm{Yao}}$$

$$2(i+j{)}^{3/4}\ge 2{\left(\frac{m}{2}\right)}^{3/4}\ge m^{3/4}.$$

$$j=m-i-2(i+j{)}^{3/4}\le m-i-m^{3/4}.$$

$$m^{1/2}/2\le k\le (m-2(i+j{)}^{3/4})m^{-1/4}+m^{1/2}/2\le m^{3/4}-m^{1/2}/2.$$

$$X_r$$

$$Y_r$$

$$p:=\mathrm{Prob}(X_r=1)=\frac{j}{m}\text{ and }q:=\mathrm{Prob}(Y_r=1)=\frac{i}{m}.$$

$$\mathrm{E}[X]$$

$$\mathrm{E}[Y]$$

$$|R|-k$$

$$=i{m}^{-1/4}+m^{1/2}/2$$

$$\mathrm{Prob}(E_1),\mathrm{Prob}(E_2)\le m^{-1/4}.$$

$$\mathrm{Var}(X)$$

$$\mathrm{Var}(Y)$$

$$\mathrm{Prob}(E_1)$$

$$\le \frac{\mathrm{Var}(X)}{m/4}\le \frac{m^{3/4}/4}{m/4}=m^{-1/4}.$$

$$\mathrm{Prob}(E_2)$$

$$\le \frac{\mathrm{Var}(Y)}{m/4}\le \frac{m^{3/4}/4}{m/4}=m^{-1/4}.$$

$$\mathrm{Prob}(E_1\cup E_2)\le \mathrm{Prob}(E_1)+\mathrm{Prob}(E_2)\le 2m^{-1/4}.$$

$$\frac{1}{1-2m^{-1/4}}\le 1+O\left(m^{-1/4}\right).$$

$$\left(1+O\left(m^{-1/4}\right)\right)\left(m+O\left(m^{3/4}\right)\right)=m+O(m^{3/4}).$$

$$e(P)=i!j!(n-i-j-1)!(\genfrac{}{}{0ex}{}{n}{i+j+1}).$$

$$C(P)\geq\bigg{\lceil}\log(n!)-\left(\log(n!)+\log{\frac{i!j!}{(i+j+1)!}}\right)\bigg{\rceil}=\bigg{\lceil}\log{\frac{(i+j+1)!}{i!j!}}\bigg{\rceil}.$$

Full text

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.)

Keywords: comparison algorithm, randomized algorithm, approximate selection, $i$-th order statistic, mediocre element, Yao’s hypothesis, tournaments, quantiles.

1 Introduction

Consider a totally ordered set $S$ of $n$ elements. Following Yao222Throughout this paper we frequently use the name Yao to refer to Foong Frances Yao; we also use the name A. C.-C. Yao to refer to Andrew Chi-Chih Yao., an element is said to be $(i,j)$-mediocre if it is neither among the top (i.e., largest) $i$ nor among the bottom (i.e., smallest) $j$ elements of $S$. The notion of mediocre element introduced by Yao is essentially an approximate solution to the selection problem, which can be stated as follows: Given a sequence $A$ of $n$ elements and an integer (selection) parameter $1\leq i\leq n$, the selection problem is that of finding the $i$-th smallest element in $A$. If the $n$ elements are distinct, the $i$-th smallest is larger than $i-1$ elements of $A$ and smaller than the other $n-i$ elements of $A$. By symmetry, the problems of determining the $i$-th smallest and the $i$-th largest are equivalent.

Together with sorting, the selection problem is one of the most fundamental problems in computer science. Sorting trivially solves the selection problem; however, a higher level of sophistication is required in order to obtain a deterministic linear time algorithm. This was accomplished in the early 1970s, when Blum et al. [6] gave a $O(n)$-time algorithm for the problem. Their algorithm performs at most $5.43n$ comparisons and its running time is linear irrespective of the selection parameter $i$. Their approach was to use an element in $A$ as a pivot to partition $A$ into two smaller subsequences and recurse on one of them with a (possibly different) selection parameter $i$. The pivot was set as the (recursively computed) median of medians of small disjoint groups of the input array (of constant size at least $5$). More recently, several variants of Select with groups of $3$ and $4$, also running in $O(n)$ time, have been obtained by Chen and Dumitrescu and independently by Zwick [7]. On the other hand, a randomized linear time algorithm for selection is relatively easy to obtain, simply by using a random pivot at each partitioning stage.

The selection problem and computing the median in particular are in close relation with the problem of finding the quantiles of a set. The $k$-th quantiles of an $n$-element set are the $k-1$ order statistics that divide the sorted set in $k$ equal-sized groups (to within $1$); see, e.g., [8, p. 223]. The $k$-th quantiles of a set can be computed by a recursive algorithm running in $O(n\log{k})$ time. For $k=100$ the quantiles are called percentiles.

In an attempt to drastically reduce the number of comparisons done for selection (down from $5.43n$), Schönhage et al. [32] designed a non-recursive algorithm based on different principles, most notably the technique of mass production. Their algorithm finds the median (the $\lceil n/2\rceil$-th largest element) using at most $3n+o(n)$ comparisons; as noted by Dor and Zwick [12], it can be adjusted to find the $i$-th largest, for any $i$, within the same comparison count. In a subsequent work, Dor and Zwick [13] managed to reduce the $3n+o(n)$ comparison bound to about $2.95n$; this however required new ideas and took a great deal of effort.

Mediocre elements.

Following Yao, an element is said to be $(i,j)$-mediocre if it is neither among the top (i.e., largest) $i$ nor among the bottom (i.e., smallest) $j$ of a totally ordered set $S$ of $n$ elements. Yao remarked that historically finding a mediocre element is closely related to finding the median, with a common motivation being selecting an element that is not too close to either extreme. Observe also that $(i,j)$-mediocre elements where $i=\lfloor\frac{n-1}{2}\rfloor$, $j=\lfloor\frac{n}{2}\rfloor$ (and symmetrically exchanged) are medians of $S$. Intuitively, the larger the difference $n-(i+j)$ is, the more candidates for an $(i,j)$-mediocre exist and so the easier the task of finding one should become.

In her PhD thesis [34], Yao suggested a very simple scheme (method) for finding an $(i,j)$-mediocre element: Pick $i+j+1$ elements arbitrarily and select the $(i+1)$-th largest among them. It is easy to check that this element satisfies the required condition. It should be noted that this scheme becomes an algorithm when the selection algorithm is specified. By slightly abusing notation, we call this scheme an algorithm throughout this paper.

Yao asked whether this algorithm is optimal. No improvements over this algorithm were known. An interesting feature of this algorithm is that its complexity does not depend on $n$ (unless $i$ or $j$ do). Yao proved her algorithm is optimal for $i=1$. For $i+j+1\leq n$, let $S(i,j,n)$ denote the minimum number of comparisons needed in the worst case to find an $(i,j)$-mediocre element. Yao [34, Sec. 4.3] proved that $S(1,j,n)=V_{2}(j+2)=j+\lceil\log(j+2)\rceil$, and so $S(1,j,n)$ is independent of $n$. Here $V_{2}(j+2)$ denotes the minimum number of comparisons needed in the worst case to find the second largest out of $j+2$ elements.

The question of whether this algorithm is optimal for all values of $i$ and $j$ has remained open ever since. Another question is whether $S(i,j,n)$ is independent of $n$ for other values of $i$ and $j$ (as is the case for $i=1$). Here we provide two alternative schemes for finding a mediocre element, one deterministic and one randomized, and thereby propose alternatives that can be compared and contrasted with Yao’s algorithm. It should be pointed out that Yao’s general question leads to several more specific questions:

Assuming that an optimal deterministic algorithm for exact selection is available, is Yao’s algorithm that uses this subroutine optimal? 2. 2.

Assuming that the current best deterministic algorithm for exact selection is used, is Yao’s algorithm that uses this subroutine optimal? 3. 3.

How do the answers change if randomized algorithms are considered?

Since recent progress on deterministic exact selection has been lagging and it is quite possible that an optimal deterministic algorithm may be out of reach, here we concentrate on the last two questions. However, it is perfectly possible that the answer to the first question is within reach even if an optimal deterministic algorithm for selection has not been identified.

Background and related problems.

Determining the comparison complexity for computing various order statistics including the median has lead to many exciting questions, some of which are still unanswered today. In this respect, Yao’s hypothesis on selection [32], [34, Sec. 4] has stimulated the development of such algorithms [12, 29, 32]. That includes the seminal algorithm of Schönhage et al. [32], which introduced principles of mass production for deriving an efficient comparison-based algorithm.

Due to its primary importance, the selection problem has been studied extensively; see for instance [5, 9, 11, 12, 13, 17, 18, 19, 20, 22, 23, 29, 35]. A comprehensive review of early developments in selection is provided by Knuth [24]. The reader is also referred to dedicated book chapters on selection in [1, 4, 8, 10] and the more recent articles [7, 15], including experimental work [3].

In many applications (e.g., sorting), it is not important to find an exact median or any other precise order statistic for that matter, and an approximate median suffices [16]. For instance, quick-sort type algorithms aim at finding a balanced partition without much effort; see e.g., [20]. Various strategies that attempt to optimize the sample size towards this goal have been studied in [25, 26].

Our results.

Our main results are summarized in the two theorems stated below. The comparison count of our deterministic algorithm for approximate selection can be as low as about $89\%$ of the corresponding count for Yao’s algorithm that employs the current best deterministic algorithm for exact selection (for certain $i,j$ combinations). Similarly, the comparison count of our randomized algorithm for approximate selection can be as low as about $66\%$ of the corresponding count for the current best randomized algorithm for exact selection (for certain $i,j$ combinations).

Theorem 1**.**

Given $n$ elements, an $(i,j)$-mediocre element where $i=\alpha n$, $j=(1-2\alpha)n-1$, and $0<\alpha<1/3$, can be found by a deterministic algorithm A1 using $c_{\rm A1}\cdot n+o(n)$ comparisons in the worst case. If the number of comparisons done by Yao’s algorithm is $c_{\rm Yao}\cdot n+o(n)$, we have $c_{\rm A1}<c_{\rm Yao}$ for each of the percentiles $1$ through $33$ (i.e., $\alpha_{s}=s/100$, $s=1,\ldots,33$).

The constants $c_{\rm A1}=c_{\rm A1}(\alpha)$ and $c_{\rm Yao}=c_{\rm Yao}(\alpha)$ for the percentiles $1$ through $33$ are given in Fig. 2. See also Fig 3.

Theorem 2**.**

Given $n$ elements and $i,j\geq 1$ where $i+j+2(i+j)^{3/4}\leq n$ and $i+j=\omega(1)$, an $(i,j)$-mediocre element can be found by a randomized algorithm using $(i+j)+O((i+j)^{3/4})$ comparisons on average.

For example: (i) an $(i,j)$-mediocre element, where $i=j=n/2-n^{3/4}$, can be found using $n+O(n^{3/4})$ comparisons on average; (ii) if $\alpha,\beta>0$ are fixed constants with $\alpha+\beta<1$, an $(\alpha n,\beta n)$-mediocre element can be found using $(\alpha+\beta)n+O(n^{3/4})$ comparisons on average.

Note that finding an element near the median requires about $3n/2$ comparisons for any previous algorithm (including Yao’s), and finding the precise median requires $3n/2+o(n)$ comparisons on average, while the main term in this expression cannot be improved  [9]. In contrast, our randomized algorithm finds an element near the median in about $n$ comparisons on average, thereby achieving a substantial savings of about $n/2$ comparisons.

Remarks.

Whereas the deterministic algorithm in Theorem 1 calls an algorithm for exact selection as a subroutine, the randomized algorithm in Theorem 2 uses random sampling but does not use exact selection as a subroutine. It is worth recalling that presently no optimal algorithm for exact selection is known with respect to the number of comparisons; e.g., the best known bounds for median finding are as follows: this task can be accomplished with $(2.95+o(1))n$ comparisons by the algorithm of Dor and Zwick [13] and requires at least $(2+2^{-80})n$ comparisons by the result of the same authors [14]. On the other hand an optimal randomized algorithm for exact selection is known: the $k$-th largest element out of $n$ given can be found using at most $n+\min(k,n-k)+o(n)$ comparisons on average by the algorithm of Floyd and Rivest [17] and requires $n+\min(k,n-k)+o(n)$ comparisons on average by the result of Cunto and Munro [9].

Preliminaries and notation.

Without affecting the results, the following two standard simplifying assumptions are convenient: (i) the input $A$ contains $n$ distinct elements; and (ii) the floor and ceiling functions are omitted in the descriptions of the algorithms and their analyses. For example if $\alpha\in(0,1)$, for simplicity we write the $\alpha n$-th element instead of the more precise $\lfloor\alpha n\rfloor$-th (or $\lceil\alpha n\rceil$-th) element. In the same spirit, for convenience we treat $n^{1/4}$, $j\,n^{-1/4}$ (for $j\in\mathbb{N}$) and other algebraic expressions that appear in the description of the algorithms as integers. Unless specified otherwise, all logarithms are in base $2$.

Let ${\rm E}[X]$ and ${\rm Var}[X]$ denote the expectation and respectively, the variance, of a random variable $X$. If $E$ is an event in a probability space, ${\rm Prob}(E)$ denotes its probability. Chebyshev’s inequality is the following: For any $a>0$,

[TABLE]

See for instance [27, p. 49].

2 Deterministic approximate selection

Consider the problem of finding an $(i,j)$-mediocre element. Without loss of generality (by considering the complementary order), it can be assumed that $i\leq j$; and consequently $i<n/2$. In addition, our algorithm is designed to work for a specific range $i\leq j\leq n-2i-1$ (hence $i<n/3$); outside this range our algorithm simply proceeds as in Yao’s algorithm. With anticipation, we note that our test values for purpose of comparison are contained in the specified range.

Yao’s algorithm is very simple:

Algorithm Yao.

• Step 1: Choose an arbitrary subset of $i+j+1$ elements from the given $n$.
• Step 2: Select and return the $(i+1)$-th largest element from the chosen subset.

As mentioned earlier, it is easy to check that the element output by Yao’s algorithm is $(i,j)$-mediocre. Our algorithm (for the specified range) is also simple:

Algorithm A1.

• Step 1: Choose an arbitrary subset of $2i+j+1$ elements from the given $n$ and group them into $m=i+\lfloor\frac{j+1}{2}\rfloor$ pairs and at most one leftover element (if $j$ is even).
• Step 2: Perform the $m$ comparisons and include the largest from each pair in a pool of $m$ elements. Add the leftover element, if any, to the pool.
• Step 3: Select and return the $(i+1)$-th largest from this pool.

Let us briefly argue about its correctness and refer to Fig. 1. Denote the selected element by $x$. Assume first that $n$ is even. On one hand, $x$ is smaller than $i$ (upper) elements in disjoint pairs; on the other hand, $x$ is larger than $n-2i-1\geq j$ (upper and lower) elements in disjoint pairs by the range assumption. The argument is similar for odd $n$; it turns out that the final place in the poset diagram where the singleton element ends up is irrelevant. It follows that the algorithm returns an $(i,j)$-mediocre element, as required.

It should be noted that both algorithms (ours as well as Yao’s) make calls to exact selection, however with different input parameters. As such, we use the current best deterministic algorithm and corresponding worst-case bound for (exact) selection available. In particular, selecting the median can be accomplished with at most $2.95n$ comparisons, by using the algorithm of Dor and Zwick [13].

The algorithm of Dor and Zwick.

Consider the problem of selecting the $\alpha n$-th largest element out of given $n$. By symmetry one may assume that $0<\alpha\leq 1/2$. If $l\geq 0$ is any fixed integer, this task can be accomplished with at most $c_{\rm Dor-Zwick}\cdot n+o(n)$ comparisons, where

[TABLE]

by using an algorithm due to the same authors [12]. By letting $l=\lfloor\log{\frac{1}{\alpha}}+\log{\log{\frac{1}{\alpha}}}\rfloor$ in Equation (2), the authors obtain the following upper bound:

[TABLE]

Note that Equations (2) and (3) only lead to upper bounds in asymptotic terms.

Next we show that algorithm A1 outperforms Yao’s algorithm for finding an $(\alpha n,\beta n)$-mediocre element for large $n$ and for a broad range of values of $\alpha$ and suitable $\beta=\beta(\alpha)$, when using a slightly modified version of the algorithm of Dor and Zwick. A key difference between our algorithm and Yao’s lies in the amount of effort put into processing the input. Whereas Yao’s algorithm chooses an arbitrary subset of elements of a certain size and ignores the remaining elements, our algorithm looks at more (possibly all) input elements and gathers initial information based on grouping the elements into disjoint pairs and performing the respective comparisons.

Fine-tuning the algorithm of Dor and Zwick.

For $0<\alpha\leq 1/2$, let $f(\alpha)$ denote the multiplicative constant in the current best upper bound on the number of comparisons in the algorithm of Dor and Zwick for selection of the $\alpha n$-th largest element out of $n$ elements, according to (2), with one improvement. Instead of considering only one value for $l$, namely $l=\lfloor\log{\frac{1}{\alpha}}+\log{\log{\frac{1}{\alpha}}}\rfloor$, we also consider the value $l+1$, and let the algorithm choose the best (i.e., the smallest of the two resulting values in (2) for the number of comparisons in terms of $\alpha$). This simple change yields an advantage of Algorithm A1 over Yao’s algorithm for an extended range of inputs. (Without this change Yao’s algorithm wins over A1 for some small $\alpha$.)

We first note that the algorithm of Dor and Zwick [12], which is a refinement of the algorithm of Schönhage et al. [32], is non-recursive. As such, the selection target remains the same during its execution, and so choosing the best value for $l$ can be done at the beginning of the algorithm. (Recall that the seminal algorithm of Schönhage et al. [32] is non-recursive as well.)

To be precise, let

[TABLE]

For a given $0<\alpha\leq 1/2$, let

[TABLE]

Problem instances and analysis of the number of comparisons.

Consider the instance $(\alpha n,(1-2\alpha)n-1)$ of the problem of selecting a mediocre element, where $\alpha$ is a constant $0<\alpha<1/3$. The comparison counts for Algorithm A1 and Algorithm Yao on this instance are bounded from above by $c_{\rm A1}\cdot n+o(n)$ and $c_{\rm Yao}\cdot n+o(n)$, respectively, where

[TABLE]

Indeed, Algorithm A1 performs $\alpha n+(n/2-\alpha n)=n/2$ initial comparisons followed by a selection problem with a fraction $\alpha^{\prime}=2\alpha$ from the $n/2$ available. The element returned by Yao’s algorithm corresponds to a selection problem with a fraction $\alpha^{\prime}=\frac{\alpha}{1-\alpha}$ from the $(1-\alpha)n$ available.

Since Dor and Zwick [13] managed to reduce the $3n+o(n)$ comparison bound to about $2.95n$, the expression of $f(\alpha)$ in (6) can be replaced by

[TABLE]

or even by

[TABLE]

We next show that Algorithm A1 outperforms Algorithm Yao with respect to the (worst-case) number of comparisons in selecting a mediocre element for $n$ large enough and for all instances $(\alpha_{s}n,(1-2\alpha_{s})n-1)$, where $\alpha_{s}=s/100$, and $s=1,\ldots,33$; that is, for all percentiles $s=1,\ldots,33$ and suitable values of the second parameter. This is proven by the data in the two tables in Fig. 2, where the entries are computed using Equations (7) and (8), respectively. Moreover, the results remain the same, regardless of whether one uses the expression of $f(\alpha)$ in (9) or (10); to avoid the clutter, we only included the results obtained by using the expression of $f(\alpha)$ in (9).

Note that the computation in (7) may need the value of $f$ for arguments $x>1/2$; and such values are computed from $f(x)=f(1-x)$ by the symmetry assumption (e.g., $f(0.6)=f(0.4)=3$, $f(0.8)=f(0.2)=2.5$). The functions $c_{\rm A1}(\alpha)$ and $c_{\rm Yao}(\alpha)$ for $\alpha\in(0,1/3)$, as given by (7) and (8), are plotted in Fig 3; however, a proof that $c_{\rm A1}(\alpha)<c_{\rm Yao}(\alpha)$ on this interval is missing.

Extension to hyperpairs.

It turns out that Algorithm A1 is not a singular case but a member of larger family. Instead of working with pairs, a more general scheme that works with hyperpairs (algorithm A below) can be designed. (See [7, 12, 32] for other uses of hyperpairs in selection.) The algorithm picks a group size that is a power of $2$ and divides the elements into groups of that size. It then computes the largest in each group by a tournament method and then calls a suitable selection procedure on these set of largest group elements. We omit a formal algorithm description but provide a figure instead (Fig. 4).

Consider the instance $(\alpha n,(1-4\alpha)n-1)$ of the problem of selecting a mediocre element, where $\alpha$ is a constant $0<\alpha\leq 1/5$. Suppose that Algorithm A works with groups of $4$ elements. Analogous to (7) and (8), the comparison counts for Algorithm A and Algorithm Yao on this instance are bounded from above by $c_{\rm A}\cdot n+o(n)$ and $c_{\rm Yao}\cdot n+o(n)$, respectively, where

[TABLE]

We have $c_{\rm A}<c_{\rm Yao}$ for each of the percentiles $9$ through $16$ (i.e., $\alpha_{s}=s/100$, $s=9,\ldots,16$). For example: if $\alpha=0.1$, a $(0.1n,0.6n-1)$-mediocre element is desired; then $c_{\rm A}=1.5$ and $c_{\rm Yao}=1.525$. If $\alpha=0.13$, a $(0.13n,0.48n-1)$-mediocre element is desired; then $c_{\rm A}=1.5$ and $c_{\rm Yao}=1.56$. These results are obtained by using the expression of $f(\alpha)$ in (9).

3 Randomized approximate selection

Consider the problem of selecting an $(i,j)$-mediocre element where $i+j+2(i+j)^{3/4}\leq n$ and $i+j=\omega(1)$. In particular, assume that $i+j\geq 16$.

Algorithms.

We next specify our algorithm and then compare it with Yao’s algorithm running on the same input. Our algorithm relies on random sampling and is similar to the Floyd-Rivest randomized algorithm for selection [17]. For example, if $i=j=n/2-n^{3/4}$, an element in the near vicinity of the median is sought. In this case, the algorithm also resembles the strategies used by quicksort algorithms that attempt to optimize the sample size for obtaining balanced partitions [25, 26].

Algorithm A2.

Input: A set $S$ of $n$ elements over a totally ordered universe and a pair $i,j$ where $i+j+2(i+j)^{3/4}\leq n$.

Output: An $(i,j)$-mediocre element.

• Step 0: Choose an arbitrary subset $S^{\prime}$ of $m:=i+j+2(i+j)^{3/4}$ elements from the given $n$. (Note that $i+j\geq m/2$ by the assumption.)
• Step 1: Pick a (multi)-set $R$ of $m^{3/4}$ elements in $S^{\prime}$, chosen uniformly and independently at random with replacement.
• Step 2: Let $k=jm^{-1/4}+m^{1/2}/2$. Let $x$ be the $k$-th smallest element of $R$, computed by a linear-time deterministic selection algorithm.
• Step 3: Compare each of the remaining elements of $S^{\prime}\setminus R$ to $x$.
• Step 4: If there are at least $i$ elements of $S^{\prime}$ larger than $x$ and at least $j$ elements of $S^{\prime}$ smaller than $x$ return $x$, otherwise FAIL.

Since $i+j\geq 16$ we have $2(i+j)^{3/4}\leq i+j$ or $i+j\geq m/2$ and thus

[TABLE]

In particular $j$ is bounded from above as follows

[TABLE]

Note that $k$ in Step 2 satisfies

[TABLE]

Observe that (i) Algorithm A2 performs at most $m+O(m^{3/4})$ comparisons; and (ii) it either correctly outputs an $(i,j)$-mediocre element or FAIL.

Analysis of the number of comparisons.

Our analysis is an adaptation of that of the classic randomized algorithm for selection; see [17], but also [28, Sec. 3.3] and [27, Sec. 3.4]. In particular, the randomized selection algorithm and Algorithm A2 both fail for similar reasons.

For $r=1,\ldots,m^{3/4}$, define random variables $X_{r}$ and $Y_{r}$ by

[TABLE]

The variables $X_{r}$ and $Y_{r}$ are independent, since the sampling is done with replacement. It is easily seen that

[TABLE]

Let $X=\sum_{r=1}^{m^{3/4}}X_{r}$ and $Y=\sum_{r=1}^{m^{3/4}}Y_{r}$ be the random variables counting the number of samples in $R$ of rank at most $j$ and at least $m-i+1$, respectively. By the linearity of expectation, we have

[TABLE]

Observe that the randomized algorithm A2 fails if and only if the rank of $x$ in $S^{\prime}$ is outside the interval $[j+1,m-i]$, i.e., the rank of $x$ is at most $j$ or at least $m-i+1$. Note that if algorithm A2 fails then at least $jm^{-1/4}+m^{1/2}/2$ elements of $R$ have rank at most $j$ or at least

[TABLE]

elements of $R$ have rank at least $m-i+1$. Denote these two bad events by $E_{1}$ and $E_{2}$, respectively. We next bound from above their probability. (Sharper bounds on the failure probability can be obtained by using Chernoff bounds [27, Ch. 4]; however, they do not affect the asymptotics of our algorithm.)

Lemma 1**.**

[TABLE]

Proof.

Since $X_{r}$ is a Bernoulli trial, $X$ is a binomial random variable with parameters $m^{3/4}$ and $p$. Similarly $Y$ is a binomial random variable with parameters $m^{3/4}$ and $q$.

Observing that $x(1-x)\leq 1/4$ for every $x\in[0,1]$, it follows (see for instance [27, Sec. 3.2.1]) that

[TABLE]

Applying Chebyshev’s inequality (1) to $X$ yields

[TABLE]

Similarly, applying Chebyshev’s inequality (1) to $Y$ yields

[TABLE]

The two inequalities have been proved. ∎

By the union bound, the probability that one execution of Algorithm A2 fails is bounded from above by

[TABLE]

As in [27, Sec 3.4], Algorithm A2 can be converted (from a Monte Carlo algorithm) to a Las Vegas algorithm by running it repeatedly until it succeeds. By Lemma 1, the FAIL probability is significantly small, and so the expected number of comparisons of the resulting algorithm is still $m+o(m)$. Indeed, the expected number of repetitions until the algorithm succeeds is at most

[TABLE]

Since the number of comparisons in each execution of the algorithm is $m+O\left(m^{3/4}\right)$, the expected number of comparisons until success is at most

[TABLE]

We now analyze the average number of comparisons done by Yao’s algorithm. On one hand, the $k$-th largest element out of $n$ given can be found using at most $n+\min(k,n-k)+o(n)$ comparisons on average [17]. On the other hand, this task requires $n+\min(k,n-k)+o(n)$ comparisons on average [9]. Consequently, Yao’s algorithm performs $i+j+\min(i,j)+o(i+j)$ comparisons on average.

Comparison.

Consider the problem of selecting an $(i,j)$-mediocre element where $i+j+$ $2(i+j)^{3/4}\leq n$ and $i+j\geq\left(\frac{2}{3}+\varepsilon\right)n$ for some constant $\varepsilon>0$. Algorithm A2 performs $n+o(n)$ comparisons on average. If $i\approx j$, Yao’s algorithm performs $\frac{3}{2}\left(\frac{2}{3}+\varepsilon\right)n+o(n)=n+\frac{3}{2}\varepsilon n+o(n)$ comparisons on average, strictly more than Algorithm A2 for large $n$.

For example, let $i=j=n/2-n^{3/4}$. Whereas Algorithm A2 performs $n+o(n)$ comparisons on average, Yao’s algorithm performs $3n/2+o(n)$ comparisons on average. Indeed, the median of $i+j+1=n-2n^{3/4}+1$ elements can be found in at most $3n/2+o(n)$ comparisons on average; and the main term in this expression cannot be improved.

Let $\alpha,\beta>0$ be two constants, where $\alpha+\beta<1$. Algorithm A2 can find an $(\alpha n,\beta n)$-mediocre element out of $n$, for large $n$, in $n+o(n)$ comparisons on average, whereas Yao’s algorithm performs $(\alpha+\beta)n+\min(\alpha,\beta)n+o(n)$ comparisons on average. If $\alpha\approx\beta$ and $\alpha+\beta$ is close to $1$ a significant $33\%$ savings results.

4 Lower bounds

We compute lower bounds by leveraging the work of Schönhage on a related problem, namely partial order production. In the partial order production problem, we are given a poset $P$ partially ordered by $\preceq_{1}$, and another set $S$ of $n$ elements with an underlying, unknown, total order $\preceq_{2}$; with $|P|\leq|S|$. The goal is to find a monotone injection from $P$ to $S$ by querying the total order $\preceq_{2}$ and minimizing the number of such queries. Alternatively, the partial order production problem can be (equivalently) formulated with $|P|=|S|$, by padding $P$ with $|S|-|P|$ singleton elements.

This problem was first studied by Schönhage [31], who showed by an information-theoretic argument that $C(P)\geq\lceil\log(n!/e(P))\rceil$, where $C(P)$ is the minimax comparison complexity of $P$ and $e(P)$ is the number of linear extensions (i.e., total orders) of $P$. Further results on poset production were obtained by Aigner [2]. A. C.-C. Yao [33] proved that Schönhage’s lower bound can be achieved asymptotically in the sense that $C(P)=O(\log(n!/e(P))+n)$, confirming a conjecture of Saks [30].

Finding an $(i,j)$-mediocre element amounts to a special case of the partial order production problem, where $P$ consists of a center element, $i$ elements above it, and $j$ elements below it. For applying Schönhage’s lower bound we have

[TABLE]

This yields

[TABLE]

Interestingly enough, the resulting lower bound does not depend on $n$; observe here the connection with Yao’s hypothesis, namely the question on the independence of $S(i,j,n)$ on $n$ mentioned in Section 1. Moreover, since

[TABLE]

the above lower bound is rather weak, namely $i+j-o(i+j)$. We remark that this is not unusual for selection problems and note that a lower bound of $(i+j+1)-1=i+j<n$ for selecting an $(i,j)$-mediocre element is immediate by a connectivity argument applied to $P$; see also [17, Eq. (1)] and [33, Lemma 2]. On the other hand, observe that the coefficients of the linear terms in the upper bounds in the right table in Fig. 2 are all strictly greater than $1$.

The situation is similar for randomized algorithms but only in part. Schönhage’s lower bound on the minimax comparison complexity of $P$ in the problem of partial order production was extended to minimean comparison complexity by A. C.-C. Yao [33]. Denoting this complexity by $\overline{C}(P)$, he showed that $\overline{C}(P)\geq\lceil\log(n!/e(P))\rceil$. As such, the lower bound in (13) holds for randomized algorithms as well. On the other hand, the trivial lower bound $i+j$ mentioned previously also holds. Consequently, the upper bound in Theorem 2 is optimal up to lower order terms.

5 Conclusion

In Sections 2 and 3 we presented two alternative algorithms—one deterministic and one randomized—for finding a mediocre element, i.e., for approximate selection.

The deterministic algorithm outperforms Yao’s algorithm for large $n$ with respect to the worst-case number of comparisons for about one third of the percentiles (as the first parameter), and suitable values of the second parameter, using the best known complexity bounds for exact selection due to Dor and Zwick [12]. Moreover, we suspect that this extends to the entire range of $\alpha\in(0,1/3)$ and suitable $\beta=\beta(\alpha)$ in the problem of selecting an $(\alpha n,\beta n)$-mediocre element for large $n$. Whether Yao’s algorithm can be beaten by a deterministic algorithm in the symmetric case $i=j$ remains an interesting question.

The randomized algorithm outperforms Yao’s algorithm for large $n$ with respect to the expected number of comparisons for the entire range of $\alpha\in(0,1/2)$ in the problem of finding an $(\alpha n,\alpha n)$-mediocre element for large $n$. As shown in Section 3, these ideas can be also used to generate asymmetric instances (i.e., with $i\neq j$) with a gap.

Acknowledgments.

The author thanks Jean Cardinal for stimulating discussions on the topic. In particular, the idea of examining the existent lower bounds for the partial order production problem is due to him. The author is also grateful to an anonymous reviewer for constructive comments and exquisite attention to detail. Finally, thanks go to another anonymous reviewer for suggesting that our deterministic algorithm is not a singular example; the extension of the algorithm described at the end of Section 2 was inspired by this suggestion.