% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %
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% The Project Gutenberg EBook of Topics on Tournaments, by John W. Moon   %
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% almost no restrictions whatsoever.  You may copy it, give it away or    %
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% Title: Topics on Tournaments                                            %
%                                                                         %
% Author: John W. Moon                                                    %
%                                                                         %
% Release Date: June 5, 2013 [EBook #42833]                               %
% Most recently updated: June 11, 2021                         %
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% Language: English                                                       %
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% *** START OF THIS PROJECT GUTENBERG EBOOK TOPICS ON TOURNAMENTS ***     %
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%%%%%%%%%%%%%%%%%%%%%%%% START OF DOCUMENT %%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}
%% PG BOILERPLATE %%
\PGBoilerPlate
\begin{center}
\begin{minipage}{\textwidth}
\small
\begin{PGtext}
The Project Gutenberg EBook of Topics on Tournaments, by John W. Moon

This eBook is for the use of anyone anywhere at no cost and with
almost no restrictions whatsoever.  You may copy it, give it away or
re-use it under the terms of the Project Gutenberg License included
with this eBook or online at www.gutenberg.org

** This is a COPYRIGHTED Project Gutenberg eBook, Details Below **
**     Please follow the copyright guidelines in this file.     **

Title: Topics on Tournaments

Author: John W. Moon

Release Date: June 5, 2013 [EBook #42833]
Most recently updated: June 11, 2021

Language: English

Character set encoding: UTF-8     

*** START OF THIS PROJECT GUTENBERG EBOOK TOPICS ON TOURNAMENTS ***
\end{PGtext}
\end{minipage}
\end{center}
\newpage
%% Credits and transcriber's note %%
\begin{center}
\begin{minipage}{\textwidth}
\begin{PGtext}
Produced by Sean Muller, Andrew D. Hwang, John W. Moon,
and Brenda Lewis.
\end{PGtext}
\end{minipage}
\vfill
\end{center}

\begin{minipage}{0.85\textwidth}
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\BookMark{0}{Copyright Notice.}
\subsection*{\centering\normalfont\scshape%
\normalsize\MakeLowercase{\TransNote}}%

\raggedright
\TransNoteText
\end{minipage}
%%%%%%%%%%%%%%%%%%%%%%%%%%% FRONT MATTER %%%%%%%%%%%%%%%%%%%%%%%%%%
\PageSep{v}%
\FrontMatter
\begin{flushright}
\Large John W. Moon \\
\normalsize \textit{University of Alberta}
\vfill
\begin{flushleft}
\LARGE \textit{Topics on Tournaments}
\end{flushleft}
\vfill\vfill\vfill

{\large HOLT, RINEHART AND WINSTON} \\
New York\ .\ Chicago\ .\ San Francisco\ .\ Atlanta \\
Dallas\ .\ Montreal\ .\ Toronto\ .\ London
\end{flushright}
\clearpage
\PageSep{vi}%
\null\vfill
\begin{flushright}
Copyright © 1968 by Holt, Rinehart and Winston, Inc. \\
All Rights Reserved
\medskip

Library of Congress Catalog Card Number: 68-13611 \\
2676302
\medskip

Printed in the United States of America \\
1\quad 2\quad 3\quad 4\quad 5\quad 6\quad 7\quad 8\quad 9
\end{flushright}
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\small
\subsection*{\centering\normalfont\scshape%
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\Contents

\ToCSect{1.}{Introduction}{1}

\ToCSect{2.}{Irreducible Tournaments}{2}

\ToCSect{3.}{Strong Tournaments}{4}

\ToCSect{4.}{Cycles in a Tournament}{5}

\ToCSect{5.}{Strong Subtournaments of a Tournament}{8}

\ToCSect{6.}{The Distribution of $3$-cycles in a Tournament}{11}

\ToCSect{7.}{Transitive Tournaments}{14}

\ToCSect{8.}{Sets of Consistent Arcs in a Tournament}{19}

\ToCSect{9.}{The Parity of the Number of Spanning Paths of a
Tournament}{21}

\ToCSect{10.}{The Maximum Number of Spanning Paths of a
Tournament}{26}

\ToCSect{11.}{An Extremal Problem}{28}

\ToCSect{12.}{The Diameter of a Tournament}{32}

\ToCSect{13.}{The Powers of Tournament Matrices}{34}

\ToCSect{14.}{Scheduling a Tournament}{39}

\ToCSect{15.}{Ranking the Participants in a Tournament}{42}
\PageSep{viii}%

\ToCSect{16.}{The Minimum Number of Comparisons Necessary to
Determine a Transitive Tournament}{47}

\ToCSect{17.}{Universal Tournaments}{49}

\ToCSect{18.}{Expressing Oriented Graphs as the Union of Bilevel
Graphs}{51}

\ToCSect{19.}{Oriented Graphs Induced by Voting Patterns}{56}

\ToCSect{20.}{Oriented Graphs Induced by Team Comparisons}{57}

\ToCSect{21.}{Criteria for a Score Vector}{61}

\ToCSect{22.}{Score Vectors of Generalizations of Tournaments}{63}

\ToCSect{23.}{The Number of Score Vectors}{65}

\ToCSect{24.}{The Largest Score in a Tournament}{71}

\ToCSect{25.}{A Reversal Theorem}{73}

\ToCSect{26.}{Tournaments with a Given Automorphism Group}{74}

\ToCSect{27.}{The Group of the Composition of Two Tournaments}{78}

\ToCSect{28.}{The Maximum Order of the Group of a Tournament}{81}

\ToCSect{29.}{The Number of Nonisomorphic Tournaments}{84}

\ToCApp{Appendix}{91}

\ToCApp{References}{96}

\ToCApp[author]{Author Index}{102}

\ToCApp[subject]{Subject Index}{104}
\PageSep{1}%
\MainMatter

\Section{1.}{Introduction}

A (round-robin) \emph{tournament}~$T_{n}$ consists of $n$~nodes $p_{1}, p_{2}, \dots, p_{n}$ such that
\index{Tournament!round-robin}%
each pair of distinct nodes $p_{i}$ and~$p_{j}$ is joined by one and only one of the
oriented arcs $\Arc{p_{i}p_{j}}$ or~$\Arc{p_{j}p_{i}}$. If the arc~$\Arc{p_{i}p_{j}}$ is in~$T_{n}$, then we say that $p_{i}$ \emph{dominates}~$p_{j}$
(symbolically, $p_{i} \Dom p_{j}$). The relation of dominance thus defined is a
\index{Dominance relation}%
complete, irreflexive, antisymmetric, binary relation. The \emph{score} of~$p_{i}$ is the
\index{Score of a node}%
number~$s_{i}$ of nodes that $p_{i}$~dominates. The \emph{score vector} of $T_{n}$ is the ordered
\index{Score vector}%
$n$-tuple $(s_{1}, s_{2}, \dots, s_n)$. We usually assume that the nodes are labeled
in such a way that $s_1 \leq s_2 \leq \dots \leq s_n$.

Tournaments provide a model of the statistical technique called the
\emph{method of paired comparisons}. This method is applied when there are a
\index{Paired comparisons, the method of}%
number of objects to be judged on the basis of some criterion and it is
impracticable to consider them all simultaneously. The objects are compared
two at a time and one member of each pair is chosen. This method and
related topics are discussed in David (1963) and Kendall (1962). Tournaments
\index[xauthor]{David, H. A.}%
\index[xauthor]{Kendall, M. G.}%
have also been studied in connection with sociometric relations in
small groups. A survey of some of these investigations is given by Coleman
\index[xauthor]{Coleman, J. S.}%
(1960). Our main object here is to derive various structural and combinatorial
properties of tournaments.


\Exercises

\Ex{1.} {\Loosen Two tournaments are \emph{isomorphic} if there exists a one-to-one dominance-preserving
\index{Isomorphic tournaments}%
correspondence between their nodes. The non\-isomorphic
tournaments with three and four nodes are illustrated in \Figure{1}. Determine}
\begin{figure}[hbt!]
\centering
\Graphic[\textwidth]{fig1}
\FCaption{Figure 1}
\end{figure}
the number of ways of assigning the labels to the nodes of these
tournaments and verify that there are a total of $2^{\ebinom{n}{2}}$ labeled tournaments~$T_{n}$
when $n = 3$,~$4$.

\Ex{2.} The \emph{complement} of a tournament is obtained by reversing the orientation
\index{Complement of a tournament}%
of all its arcs. A tournament is \emph{self-complementary} if it is isomorphic to its
\index{Self-complementary tournament}%
\PageSep{2}%
complement. Show that self-complementary\Erratum{}{ cyclic} tournaments~$T_{n}$ exist if and
only if $n$~is odd.\Erratum{}{ A tournament is \emph{cyclic}, in the present context, if it is isomorphic to itself under some cyclic permutation of the labels of its node.} [Sachs (1965).]
\index[xauthor]{Sachs, H.}%


\Section{2.}{Irreducible Tournaments}
\index{Irreducible tournament}%

A tournament~$T_{n}$ is \emph{reducible} if it is possible to partition its nodes into
\index{Reducible tournament}%
two nonempty sets $B$~and~$A$ in such a way that all the nodes in~$B$ dominate
all the nodes in~$A$; the tournament is \emph{irreducible} if this is not possible. It is
very easy to determine whether a tournament~$T_{n}$ is reducible; if $(s_1, s_2, \dots, s_n)$
is the score vector of~$T_n$ and $s_1 \leq s_2 \leq \dots \leq s_n$, then $T_{n}$~is reducible if and
\index{Score vector}%
only if the equation
\[
\sum_{i=1}^k s_i = \binom{k}{2}
\]
holds for some value of~$k$ less than~$n$.

The (dominance) \emph{matrix of the tournament~$T_{n}$} is the $n$~by~$n$ matrix $M(T_{n}) = [a_{ij}]$
\index{Matrix!of a tournament}%
in which $a_{ij}$~is~$1$ if $p_{i} \Dom p_{j}$ and $0$~otherwise. All the diagonal entries are~$0$.
A tournament matrix satisfies the equation
\[
M + M^T = J - I,
\]
where $J$~is the matrix of~$1$'s and $I$~is the identity matrix. If the tournament~$T_{n}$
is reducible and the scores $s_i = \sum_{j=1}^n a_{ij}$ are in nondecreasing order, then its
matrix has the structure
\[
M(T_n) = \left|
  \begin{array}{c|c}
    M_1 & 0 \\
    \hline
    \Strut
    1 & M_2
  \end{array}
\right|,
\]
where $M_1$~and~$M_2$ are the matrices of the tournaments defined by the sets
$A$~and~$B$ of the preceding paragraph.

There are $2^{\ebinom{n}{2}}$~labeled tournaments~$T_n$. We now derive an approximation
for~$P(n)$, the probability that a random tournament~$T_{n}$ is irreducible.

Every reducible tournament~$T_n$ has a unique decomposition into irreducible
subtournaments $T^{(1)}$, $T^{(2)}$,~\dots, $T^{(l)}$ such that every node in~$T^{(j)}$ dominates
every node in~$T^{(i)}$ if $1 \leq i \leq j \leq l$. The probability that $T^{(1)}$~has $t$~nodes
is
\[
\frac{\dbinom{n}{t} P(t) 2^{\ebinom{t}{2}} 2^{\ebinom{n-t}{2}}}{2^{\ebinom{n}{2}}}
  = \binom{n}{t} P(t) \left(\frac{1}{2}\right)^{t(n - t)}.
\]
For each of the $\dbinom{n}{t}$~subsets of $t$~nodes, there are $P(t) 2^{\ebinom{t}{2}}$ possible choices
\PageSep{3}%
for~$T^{(1)}$; the $\dbinom{n - t}{2}$~arcs joining the remaining $n - t$ nodes may be
oriented arbitrarily.

It is possible for $t$ to be any positive integer less than~$n$. It follows that
\[
P(n) = 1 - \sum_{t=1}^{n-1} \binom{n}{t} P(t) \left(\frac{1}{2}\right)^{t(n-t)},
\Tag{(1)}
\]
since these cases are mutually exclusive. R.~A. MacLeod used this formula
\index[xauthor]{MacLeod, R. A.}%
to calculate the first few values of~$P(n)$ given in \Table{1}.
\begin{table}
\TCaption{Table 1. $P(n)$, the probability that a tournament~$T_{n}$ is irreducible.}
\begin{gather*}
\begin{array}{*{10}{c}}
n  & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline
\Strut
P(n) & 1 & 0 & .25 & .375 & .53125 & .681152 & .799889 & .881115 & .931702
\end{array} \\
\setlength{\arraycolsep}{4.5pt}
\begin{array}{@{}*{8}{c}@{}}
n  & 10 & 11 & 12 & 13 & 14 & 15 & 16 \\
\hline
\Strut
P(n) & .961589 & .978720 & .988343 & .993671 & .996587 & .998171 & .999024
\end{array}
\end{gather*}
\end{table}
The following bound is a stronger form of a result due to Moon and Moser
\index[xauthor]{Moon, J. W.}%
\index[xauthor]{Moser, L.}%
(1962b).

\begin{Theorem}{1.}
If $Q(n)$ denotes the probability that a random tournament~$T_{n}$
is reducible, then
\[
\left|Q(n) - \frac{n}{2^{n-2}}\right|
  < \frac{1}{2} \left(\frac{n}{2^{n-2}}\right)^{2}
  \quad\text{if $n \geq 2$.}
\]
\end{Theorem}

\Proof. It follows from~\Eq{(1)} that
\[
P(n) > 1 - 2n \left(\frac{1}{2}\right)^{n-1}
  - \binom{n}{2} \left(\frac{1}{2}\right)^{2(n-2)}
  - \sum_{t=3}^{n-3} \binom{n}{t} \left(\frac{1}{2}\right)^{t(n-t)}.
\]
The terms in the sum are largest for the extreme values of~$t$. Consequently,
\[
P(n) > 1 - n(\nf{1}{2})^{n-2} - n^{2}(\nf{1}{2})^{2n-3}
  + \left(n(\nf{1}{2})^{2n-3} - (n - 5) \binom{n}{3} (\nf{1}{2})^{3n-9}\right).
\]
If $n \geq 14$, then
\[
P(n) > 1 - n(\nf{1}{2})^{n-2} - n^{2}(\nf{1}{2})^{2n-3},
\Tag{(2)}
\]
since the expression within the parenthesis is positive.

To obtain an upper bound for~$P(n)$, we retain only the three largest
terms in the sum in~\Eq{(1)}. Therefore,
\[
P(n) < 1 - n(\nf{1}{2})^{n-1} - nP(n - 1) (\nf{1}{2})^{n-1} - \binom{n}{2} P(n - 2) (\nf{1}{2})^{2(n-2)}.
\]
\PageSep{4}%
If $n - 2 \geq 14$, we can use inequality~\Eq{(2)} to bound $P(n - 1)$ and $P(n - 2)$
from below; the resulting expression may be simplified to yield the inequality
\[
P(n) < 1 - n(\nf{1}{2})^{n-2} + n^{2} (\nf{1}{2})^{2n-3}.
\Tag{(3)}
\]

\ThmRef{1} now follows from \Eq{(2)},~\Eq{(3)}, and the data in \Table{1}.

\Exercises

\Ex{1.} Verify that $P(3) = \nf{1}{4}$ and $P(4) = \nf{3}{8}$ by examining the tournaments in
\Figure{1}.

\Ex{2.} Deduce inequality~\Eq{(3)} from the two preceding inequalities.

\Ex{3.} Prove that $T_{n}$~is irreducible if the difference between
\index{Irreducible tournament}%
every two scores in~$T_{n}$
is less than $\nf{1}{2} n$. [L.~Moser.]
\index[xauthor]{Moser, L.}%

\Ex{4.} Let the score vector of~$T_{n}$ be $(s_{1}, s_{2}, \dots, s_{n})$, in nondecreasing order.
Show that $p_{i}$ and~$p_{j}$ are in the same irreducible subtournament of~$T_{n}$ if
$0 \leq s_{j} - s_{i} < (j - i + 1)/2$. [L.~W. Beineke.]
\index[xauthor]{Beineke, L. W.}%

\Ex{5.} If $T(x) = \sum_{n=1}^{\infty} 2^{\ebinom{n}{2}} x^{n}/n!$ and $t(x) = \sum_{n=1}^{\infty} P(n) 2^{\ebinom{n}{2}} x^{n}/n!$, then show
that $t(x) = T(x)/[1 + T(x)]$. (These are only formal generating functions,
so questions of convergence may be ignored.)

\Ex{6.} Let $T_{n} - p_{i}$ denote the tournament obtained from~$T_{n}$ by removing the
node~$p_{i}$ (and all arcs incident with~$p_{i}$). If $T_{n}$~and $H_{n}$ are two tournaments with
$n$ ($n \geq 5$) nodes $p_{i}$ and~$q_{i}$ respectively, such that $T_{n} - p_{i}$ is isomorphic to
\index{Isomorphic tournaments}%
$H_{n} - q_{i}$ for all~$i$, then is $T_{n}$ necessarily isomorphic to~$H_{n}$? (Consider first
the case in which $T_{n}$~and~$H_{n}$ are reducible.) Consider the analogous problem
when arcs instead of nodes are removed from $T_{n}$ and~$H_{n}$. [See Harary and
\index[xauthor]{Harary, F.}%
Palmer (1967).]
\index[xauthor]{Palmer, E. M.}%


\Section{3.}{Strong Tournaments}
\index{Strong tournament}%

For any subset~$X$ of the nodes of a tournament~$T_{n}$, let
\[
\Gamma(X) = \{q : p \to q \text{ for some $p \in X$}\},
\]
and, more generally, let
\[
\Gamma^{m}(X) = \Gamma\bigl(\Gamma^{m-1}(\Typo{x}{X})\bigr),
\quad \text{for $m = 2, 3, \dots$.}
\]
Notice that $T_{n}$~is reducible if and only if $\Gamma(X) \subseteq X$ for some nonempty proper
subset~$X$ of the nodes. A tournament~$T_{n}$ is \emph{strongly connected} or \emph{strong} if
and only if for every node~$p$ of~$T_{n}$ the set
\[
\{p\} \cup \Gamma(p) \cup \Gamma^{2}(p) \cup \dots \cup \Gamma^{n-1}(p)
\]
\PageSep{5}
contains every node of~$T_{n}$. The following theorem apparently appeared first
in a paper by Rado (1943); it was also found by Roy (1958) and others.
\index[xauthor]{Rado, R.}%
\index[xauthor]{Roy, B.}%

\begin{Theorem}{2.}
A tournament~$T_{n}$ is strong if and only if it is irreducible.
\end{Theorem}

\Proof. If $T_{n}$~is reducible, then it is obviously not strong. If $T_{n}$~is not strong,
then for some node~$p$ the set
\[
A = \{p\} \cup \Gamma(p) \cup \Gamma^{2}(p) \cup \dots \cup \Gamma^{n-1}(p)
\]
does not include all the nodes of~$T_{n}$. But then all the nodes not in~$A$ must
dominate all the nodes in~$A$; consequently $T_{n}$~is reducible.

In view of the equivalence between strong connectedness and irreducibility,
it is a simple matter to determine whether or not a given tournament
is strongly connected by using the observation in the introduction to \SecRef{2}.
For types of graphs other than tournaments in which, for example, not
every pair of nodes is joined by an arc, it is sometimes convenient to use
the properties of matrix multiplication to deal with problems of connectedness.
\index{Matrix!of a tournament}%


\Exercises

\Ex{1.} Show that the tournament~$T_{n}$ is strong if and only if all the entries in the
matrix
\[
M(T_{n}) \dotplus M^{2}(T_{n}) \dotplus \dots \dotplus M^{n-1}(T_{n})
\]
are~$1$s. (Boolean addition is to be used here, that is, $1 \dotplus 1 = 1 \dotplus 0 = 1$
and $0 \dotplus 0 = 0$.) [Roy (1959) and others.]

\Ex{2.} Prove that, in any tournament~$T_{n}$, there exists at least one node~$p$ such
that the set
\[
\{p\} \cup \Gamma(p) \cup \Gamma^{2}(p) \cup \dots \cup \Gamma^{n-1}(p)
\]
contains every node of~$T_{n}$. [See Bednarek and Wallace (1966) for a relation-theoretic
\index[xauthor]{Bednarek, A. R.}%
\index[xauthor]{Wallace, A. D.}%
extension of this result.]


\Section{4.}{Cycles in a Tournament}
\index{Cycles in a tournament}%

A sequence of arcs of the type $\Arc{ab}, \Arc{bc}, \dots, \Arc{pq}$ determines a \emph{path} $P(a, q)$ from
\index{Paths in a tournament}%
$a$ to~$q$. We assume that the nodes $a, b, c, \dots, q$ are all different. If the arc~$\Arc{qa}$ is
in the tournament, then the arcs in~$P(a, q)$ plus the arc~$\Arc{qa}$ determine a
\emph{cycle}. The \emph{length} of a path or a cycle is the number of arcs it contains.
\index{Length of a path or cycle}%
A \emph{spanning} path or cycle is one that passes through every node in a tournament.
\index{Spanning path or cycle}%
A tournament is strong if and only if each pair of nodes is contained
in some cycle.
\PageSep{6}

Moser and Harary (1966) proved that an irreducible tournament~$T_{n}$
\index{Irreducible tournament}%
\index[xauthor]{Harary, F.}%
\index[xauthor]{Moser, L.}%
contains a $k$-cycle (a cycle of length~$k$) for $k = 3, 4, \dots, n$. [Their argument
was a refinement of the argument Camion (1959) used to prove that a
\index[xauthor]{Camion, P.}%
tournament~$T_{n}$ ($n \geq 3$) contains a spanning cycle if and only if it is irreducible.]
The following slightly stronger result is proved in essentially the
same way.

\begin{Theorem}{3.}
Each node of an irreducible tournament~$T_{n}$ is contained in
some $k$-cycle, for $k = 3, 4, \dots, n$.
\end{Theorem}

\Proof. Let $a$~be an arbitrary node of the irreducible tournament~$T_{n}$.
There must be some arc~$\Arc{pq}$ in~$T_{n}$ where $q \to a$ and $a \to p$; otherwise $T_{n}$
would be reducible. Consequently, node~$a$ is contained in some $3$-cycle.

Let
\[
C = \{\Arc{ab}, \Arc{bc}, \dots, \Arc{lm}, \Arc{ma}\Typo{,}{}\}
\]
be a $k$-cycle containing the node~$a$, where $3 \leq k < n$. We shall show that
there also exists a $(k + 1)$-cycle containing~$a$.

We first suppose there exists some node~$p$ not in the cycle such that $p$
both dominates and is dominated by nodes that are in the cycle. Then there
must be two consecutive nodes of the cycle, $e$~and~$f$ say, such that $e \to p$ and
$p \to f$. We can construct a $(k + 1)$-cycle containing node~$a$ simply by
replacing the arc~$\Arc{ef}$ of~$C$ by the arcs $\Arc{ep}$ and~$\Arc{pf}$. (This case is illustrated in
\Figure{2}(a).)
\begin{figure}[hbt!]
\centering
\Graphic{fig2}
\FCaption{Figure 2}
\end{figure}

Now let $A$ and $B$ denote, respectively, the sets of nodes of~$T_{n}$ not in
cycle~$C$ that dominate, or are dominated by, every node of~$C$. We may
assume that every node of~$T_{n}$ not in~$C$ belongs either to $A$ or to~$B$. Since
$T_{n}$~is irreducible, both $A$ and~$B$ must be nonempty and some node~$u$ of~$B$
\PageSep{7}
must dominate some node~$v$ of~$A$. But then we can construct a $(k + 1)$-cycle
containing node~$a$ by replacing the arcs $\Arc{ab}$ and $\Arc{be}$ of~$C$ by the arcs
$\Arc{au}$,~$\Arc{uv}$, and~$\Arc{vc}$. (This case is illustrated in \Figure{2}(b).) This completes the
proof of the theorem by induction.


\Exercises

\Ex{1.} Examine the argument Foulkes (1960) gave to show that an irreducible
\index{Irreducible tournament}%
\index[xauthor]{Foulkes, J. D.}%
tournament has a spanning cycle. [See also Fernández de~Trocóniz (1966).]
\index[xauthor]{Fernández de Troconiz, A.}%

\Ex{2.} Let us say that a tournament has property~$P_{k}$ if every subset of $k$~nodes
determines at least one $k$-cycle. Show that $T_{n}$~has a spanning cycle if it
has property~$P_{k}$ for some~$k$ such that $3 \leq k < n$.

\Ex{3.} What is the maximum number of arcs~$\Arc{pq}$ that an irreducible tournament~$T_{n}$
can have such that, if the arc~$\Arc{pq}$ is replaced by the arc~$\Arc{qp}$, then the resulting
tournament is reducible?
\index{Reducible tournament}%

\Ex{4.} What is the least integer $r = r(n)$ such that any irreducible tournament~$T_{n}$
can be transformed into a reducible tournament by reversing the orientation
of at most $r$ arcs?

\Ex{5.} A tournament~$T_{r}$ is a \emph{subtournament} of a tournament~$T_{n}$ if there exists a
\index{Subtournament}%
one-to-one mapping~$f$ between the nodes of~$T_{r}$ and a subset of the nodes of~$T_{n}$
such that, if $p \to q$ in~$T_{r}$, then $f(p) \to f(q)$ in~$T_{n}$. Let $T_{r}$ ($r > 1$) denote an
irreducible subtournament of an irreducible tournament~$T_{n}$. Prove that
there exist $k$-cycles~$C$ such that every node of~$T_{r}$ belongs to~$C$ for $k = r,
r + 1, \dots, n$ with the possible exception of $k = r + 1$. Characterize the
exceptional cases.

\Ex{6.} Let $T_{r}$ ($r > 1$) denote a reducible subtournament of an irreducible
tournament~$T_{n}$. Determine bounds for~$h(T_{n})$, the least integer~$h$ for which
there exists an $h$-cycle~$C$ that contains every node of~$T_{r}$.

\Ex{7.} A \emph{regular} tournament is one in which the scores of the nodes are all as
\index{Regular tournament}%
\index{Score of a node}%
nearly equal as possible. Let $T_{r}$ ($r > 1$) denote a subtournament of a
regular tournament~$T_{n}$. Prove that there exist $k$-cycles~$C$ such that every
node of~$T_{r}$ belongs to~$C$ for $k = r, r + 1, \dots, n$ with the possible exception
of $k = r$ or $k = r + 1$. [See Kotzig (1966).]
\index[xauthor]{Kotzig, A.}%

\Ex{8.} Prove that every arc of a regular tournament~$T_{n}$ with an odd number of
nodes is contained in a $k$-cycle, for $k = 3, 4, \dots, n$. [Alspach (1967).]
\index[xauthor]{Alspach, B.}%

\Ex{9.} P.~Kelly has asked the following question: Is it true that the arcs of every
\index[xauthor]{Kelly, P. J.}%
regular tournament~$T_{n}$ with an odd number of nodes can be partitioned into
$\nf{1}{2}(n - 1)$ arc-disjoint spanning cycles?
\PageSep{8}


\Section{5.}{Strong Subtournaments of a Tournament}
\index{Strong tournament}%

Let $S(n, k)$ denote the maximum number of strong subtournaments~$T_{k}$
that can be contained in a tournament~$T_{n}$ ($3 \leq k \leq n$). The following
result is due to Beineke and Harary (1965).
\index[xauthor]{Beineke, L. W.}%
\index[xauthor]{Harary, F.}%

\begin{Theorem}{4.}
If $[x]$ denotes the greatest integer not exceeding~$x$, then
\[
S(n, k) = \binom{n}{k}
  - \left[\frac{1}{2}(n + 1)\right] \binom{[\nf{1}{2} n]}{k - 1}
  - \left[\frac{1}{2}n\right] \binom{[\nf{1}{2}(n - 1)]}{k - 1}.
\]
\end{Theorem}

\Proof. Let $(s_{1}, s_{2}, \dots, s_{n})$ be the score vector of a tournament~$T_{n}$. The
\index{Score vector}%
number of strong subtournaments~$T_{k}$ in~$T_{n}$ certainly cannot exceed
\[
\binom{n}{k} - \sum_{i=1}^{n} \binom{s_{i}}{k - 1}.
\]
This is because the terms subtracted from~$\dbinom{n}{k}$, the total number of subtournaments~$T_{k}$
in~$T_{n}$, enumerate those subtournaments~$T_{k}$ in which one
node dominates all the remaining $k - 1$ nodes; such tournaments are
certainly not strong. It is a simple exercise to show that the sum attains its
minimum value when the~$s_{i}$'s are as nearly equal as possible. Consequently,
\[
S(n, k) \leq \binom{n}{k}
  - \left[\frac{1}{2}(n + 1)\right] \binom{[\nf{1}{2} n]}{k - 1}
  - \left[\frac{1}{2}n\right] \binom{[\nf{1}{2}(n - 1)]}{k - 1}.
\]

To show that equality actually holds, it suffices to exhibit a regular
\index{Regular tournament}%
tournament~$R_{n}$ with the following property.
% [** TN: List indentation added with author's approval]
\begin{itemize}
\item[(a)] Every subtournament~$T_{k}$ of~$R_{n}$ is either strong or has one node that
\index{Subtournament}%
dominates all the remaining $k - 1$ nodes.
\end{itemize}

If $n$~is odd, let $R_{n}$~denote the regular tournament in which $p_{i} \to p_{j}$ if
and only if $0 < j - i \leq (n - 1)/2$ (the subtraction is modulo~$n$). We shall
show that $R_{n}$~has the following property.
\begin{itemize}
\item[(b)] Every subtournament~$T_{k}$ of~$R_{n}$ either is strong or has no cycles.
\index{Cycles in a tournament}%
\end{itemize}

Property~(b) holds for any tournament when $k = 3$. We next show that it
holds for~$R_{n}$ when $k = 4$. If it did not, then $R_{n}$~would contain a subtournament~$T_{4}$
consisting of a $3$-cycle and an additional node that either dominates
every node of the cycle or is dominated by every node of the cycle. We may
suppose that the former is the case without loss of generality. Let $p_{1}$,~$p_{i}$ and
$p_{j}$ be the nodes of the cycle, proceeding according to its orientation, and let
$p_{k}$~be the fourth node of~$T_{4}$. Then $i \leq 1 + \nf{1}{2}(n - 1)$, $j \leq i + \nf{1}{2}(n - 1)$,
$j > 1 + \nf{1}{2}(n - 1)$, and $n \geq k > i + \nf{1}{2}(n - 1)$, from the definition of~$R_{n}$.
\PageSep{9}
Hence, $n \geq k > j > \nf{1}{2}(n + 1)$, and $p_{j}$~must dominate~$p_{k}$ in $R_{n}$ and~$T_{4}$.
This contradiction shows that $R_{n}$~satisfies~(b) when $k = 4$.

If $n$~is even, let $R_{n}$~denote the regular tournament in which $p_{i} \to p_{j}$ if and
only if $0 < j - i \leq \nf{1}{2} n$ (the subtraction is modulo~$n + 1$). This tournament
satisfies~(b) when $k = 3$ or~$4$, since it is a subtournament of the tournament~$R_{n+1}$
defined earlier with an odd number of nodes.

Let $T_{k}$~be any subtournament with $k$~nodes of~$R_{n}$ ($k > 4$). If $T_{k}$~has any
cycles at all, then it must have a $3$-cycle~$C$. If $T_{k}$~is not strong, then there
must be a node~$p$ that either dominates every node of~$C$ or is dominated
by every node of~$C$. In either case, this would imply the existence of a
subtournament~$T_{4}$ in~$R_{n}$ that is not strong but which has a cycle. This
contradicts the result just established. Hence, every subtournament~$T_{k}$ of~$R_{n}$
is either strong or has no cycles. We have shown that the regular tournament~$R_{n}$
has Property~(b). It is easy to show that Property~(b) implies Property~(a),
so the theorem is now proved.

There are only two different types of tournaments~$T_{3}$ (see \Figure{1}); it
follows that equality holds in the second statement in the proof of \ThmRef{4}
when $k = 3$. This implies the following result, found by Kendall and
\index[xauthor]{Kendall, M. G.}%
Babington Smith (1940), Szele (1943), Clark [see Gale (1964)], and others.
\index[xauthor]{Clark, A. H.}%
\index[xauthor]{Gale, D.}%
\index[xauthor]{Smith, B. B.}%
\index[xauthor]{Szele, T.}%

\begin{Corollary}
Let $c_{3}(T_{n})$ denote the number of $3$-cycles in the tournament~$T_{n}$.
If $(s_{1}, s_{2}, \dots, s_{n})$ is the score vector of~$T_{n}$, then
\[
c_{3}(T_{n}) = \binom{n}{3} - \sum_{i=1}^{n} \binom{s_{i}}{2}
  \leq \begin{cases}
    \nf{1}{24}(n^{3} - n)  & \text{if $n$~is odd,} \\
    \nf{1}{24}(n^{3} - 4n) & \text{if $n$~is even.}
  \end{cases}
\]
Equality holds throughout only for regular tournaments.
\end{Corollary}

The next corollary follows from the observation that every strong
tournament~$T_{4}$ has exactly one $4$-cycle.

\begin{Corollary}
The maximum number of $4$-cycles possible in a tournament~$T_{n}$
is
\[
S(n, 4) = \begin{cases}
    \nf{1}{48}n(n + 1)(n - 1)(n - 3) & \text{if $n$~is odd,} \\
    \nf{1}{48}n(n + 2)(n - 2)(n - 3) & \text{if $n$~is even.}
  \end{cases}
\]
\end{Corollary}

Colombo (1964) proved this result first by a different argument. It seems
\index[xauthor]{Colombo, U.}%
to be difficult to obtain corresponding results for cycles of length greater
than four. (See Exercise~3 at the end of \SecRef{10}.)

The following result is proved by summing the equation in \ThmRef{4}
over~$k$.

\begin{Corollary}
The maximum number of strong subtournaments with
at least three nodes in any tournament~$T_{n}$ is
\[
\begin{cases}
  2^{n} -  n2^{(1/2)(n - 1)} - 1 & \text{if $n$~is odd,} \\
  2^{n} - 3n2^{(1/2)(n - 4)} - 1 & \text{if $n$~is even.}
\end{cases}
\]
\end{Corollary}
\PageSep{10}

Let $s(n, k)$ denote the minimum number of strong subtournaments~$T_{k}$
that a strong tournament~$T_{n}$ can have. (If $T_{n}$~is not strong, then it need not
have any nontrivial strong subtournaments.) Moon (1966) discovered the
\index[xauthor]{Moon, J. W.}%
following result.

\begin{Theorem}{5.}
If $3 \leq k \leq n$, then $s(n, k) = n - k + 1$.
\end{Theorem}

\Proof. We first show that $s(n, k) \geq n - k + 1$. This inequality certainly
holds when $n = k$ for any fixed value of~$k$. It follows from \ThmRef{3} that
any strong tournament~$T_{n}$ has a strong subtournament~$T_{n-1}$. Now $T_{n-1}$~has
at least $s(n - 1, k)$ strong subtournaments~$T_{k}$ by definition and the node
not in~$T_{n-1}$ is in at least one $k$-cycle. This $k$-cycle determines a strong
subtournament~$T_{k}$ not contained in~$T_{n-1}$. Consequently,
\[
s(n, k) \geq s(n - 1, k) + 1.
\]
The earlier inequality now follows by induction on~$n$.

To show that $s(n, k) < n - k + 1$, consider the tournament~$T_{n}$ in which
$p_{i} \to p_{j}$ when $i = j - 1$ or $i \geq j + 2$. (This tournament is illustrated in
\Figure{3}.) It is easy to see that this tournament has precisely $n - k + 1$
strong subtournaments~$T_{k}$ if $3 \leq k \leq n$. This completes the proof of the
theorem.
\begin{figure}[hbt!]
\centering
\Graphic[0.5in]{fig3}
\FCaption{Figure 3}
\end{figure}

Notice that \ThmRef{5} remains true if the phrase ``strong subtournaments~$T_{k}$''
is replaced by ``$k$-cycles'' in the definition of~$s(n, k)$. The case $k = 3$
of this result was given by Harary, Norman, and Cartwright (1965, p.~306).
\index[xauthor]{Cartwright, D.}%
\index[xauthor]{Harary, F.}%
\index[xauthor]{Norman, R. Z.}%

\begin{Corollary}
The minimum number of cycles a strong tournament~$T_{n}$
can have is~$\dbinom{n - 1}{2}$.
\end{Corollary}

This is proved by summing the expression for~$s(n, k)$ from $k = 3$ to $k = n$.


\Exercises

\Ex{1.} Show that a sum of the type $\sum_{i=1}^{n} \dbinom{s_{i}}{t}$, where $t$~is a fixed integer and
$(s_{1}, s_{2}, \dots, s_{n})$ is the score vector of a tournament~$T_{n}$, attains its minimum
when the scores are as nearly equal as possible.
\PageSep{11}

\Ex{2.} Prove that Property~(b) implies Property~(a).

\Ex{3.} Prove that, if a tournament has any cycles, then it has some $3$-cycles.
\index{Cycles in a tournament}%

\Ex{4.} Suppose the tournament~$T_{n}$ has at least one cycle; if every subtournament
of~$T_{n}$ either is strong or has no cycles, then is $T_{n}$ necessarily regular?
\index{Regular tournament}%

\Ex{5.} Show that
\[
c_{3}(T_{n})
  = \frac{n(n^{2} - 1)}{24}
  - \frac{1}{2} \sum_{i=1}^{n} \left(s_{i} - \frac{n - 1}{2}\right)^{2}.
\]

\Ex{6.} Prove that $T_{n}$~must be strong if it has more $3$-cycles than a tournament~$T_{n-1}$
can have. [L.~W. Beineke.]
\index[xauthor]{Beineke, L. W.}%

\Ex{7.} Use \ThmRef{4} to determine for what values of $n$~and~$k$ there exist
tournaments~$T_{n}$ with property~$P_{k}$. (See Exercise~4.2.) %**xref

\Ex{8.} An $m$~by~$n$ \emph{bipartite tournament} consists of two sets of nodes $P$ and~$Q$,
\index{Tournament!bipartite}%
where $|P| = m$ and $|Q| = n$, such that each node of~$P$ is joined with each
node of~$Q$ by an arc; there are no arcs joining nodes in the same node-set.
The score vectors of a bipartite tournament are defined in the same way as
the score vector of an ordinary tournament. Give an example of two
bipartite tournaments with the same score vectors that do not have the
same number of $4$-cycles.

\Ex{9.} Show that the maximum number of $4$-cycles an $m$~by~$n$ bipartite tournament
can have is $[m^{2}/4] · [n^{2}/4]$. [Moon and Moser (1962a).]
\index[xauthor]{Moon, J. W.}%
\index[xauthor]{Moser, L.}%


\Section{6.}{The Distribution of \texorpdfstring{$3$}{3}-cycles in a Tournament}

In applying the method of paired \Typo{comparisions}{comparisons}, it is usually assumed that
\index{Paired comparisons, the method of}%
the objects can be ordered on a linear scale. Hence, the presence of cycles in
the tournament representing the outcomes of the comparisons indicates
inconsistency on the part of the judge or that the underlying assumption is
inappropriate. Kendall and Babington Smith (1940) proposed the number
\index[xauthor]{Kendall, M. G.}%
\index[xauthor]{Smith, B. B.}%
of $3$-cycles (\Typo{suitable}{suitably} normalized, so as to equal~$1$ when there are no $3$-cycles
and $0$~when there are as many $3$-cycles as possible) as a measure of the
consistency of the comparisons. The number of $3$-cycles was chosen,
because it is a simple function of the score vector of the tournament and
hence is easy to calculate. To define significance tests for the departure
from consistency, it is necessary to know the distribution of the number of
$3$-cycles in a random tournament. Tables of the distribution of the number
of $3$-cycles in a random tournament~$T_{n}$ have been given by Kendall and
Babington Smith when $3 \leq n \leq 7$ and by Alway (1962b) when $8 \leq n \leq 10$.
\index[xauthor]{Alway, G. G.}%
Kendall and Babington Smith conjectured and Moran (1947) proved that
\index[xauthor]{Moran, P. A. P.}%
the distribution of the number of $3$-cycles is asymptotically normal.
\PageSep{12}

\begin{Theorem}{6.}
Let $c_{n}$ denote the number of $3$-cycles in a random tournament~$T_{n}$.
Then the distribution of $(c_{n} - \mu')/\sigma$ tends to the normal distribution
with zero mean and unit variance, where
\[
\mu' = \frac{1}{4} \binom{n}{3}\quad\text{and}\quad
\sigma^{2} = \frac{3}{16} \binom{n}{3}.
\]
\end{Theorem}

\Proof. In a tournament~$T_{n}$, let the variable $t(ijk)$ be $1$~or~$0$ according as
the distinct nodes $p_{i}$,~$p_{j}$, and~$p_{k}$ do or do not determine a $3$-cycle. Since only
two of the eight equally likely ways of orienting the arcs joining these nodes
yield a $3$-cycle, it follows that $E[t(ijk)] = \nf{1}{4}$. Therefore,
\[
\mu' = E(c_{n}) = \frac{1}{4} \binom{n}{3}.
\]

If $r(ijk) = t(ijk) - \nf{1}{4}$, then
\[
\sigma^{2} = E\bigl[(c_{n} - \mu')^{2}\bigr]
  = E\bigl[\bigl(\sum r(ijk)\bigr)^{2}\bigr],
\]
where the sum is over the triples $i$,~$j$, and~$k$. The products in this expansion
are of the following types:
\[
r(ijk) · r(ijk),\quad
r(ijk) · r(ijw),\quad
r(ijk) · r(ivw),\quad\text{and}\quad
r(ijk) · r(uvw).
\]
The variables appearing in each of the last two products are independent;
hence, the expectation of their product equals the product of their individual
expectations, zero. The variables in the second product are also independent,
since the probability that three nodes form a $3$-cycle is still~$\nf{1}{4}$, even when the
orientation of one of the arcs joining them is specified. Thus the only
nonzero contributions to~$\sigma^{2}$ come from products of the first type. Therefore,
\[
\sigma^{2} = \binom{n}{3} E\bigl[r(ijk)^{2}\bigr]
  = \frac{3}{16} \binom{n}{3}.
\]

We shall now show that for each fixed positive integer~$h$
\[
\Inum{(i)} \frac{\mu_{2h}}{\sigma^{2h}} \to \frac{(2h)!}{2^{h} h!}
\quad\text{and}\quad
\Inum{(ii)} \frac{\mu_{2h+1}}{\sigma^{2h+1}} \to 0
\]
as $n$~tends to infinity ($\mu_{k}$~denotes the $k$th central moment of~$c_{n}$). It will then
follow from the second limit theorem of probability theory [see Kendall
\index[xauthor]{Kendall, M. G.}%
and Stuart (1958, p.~115)] that the distribution of $(c_{n} - \mu')/\sigma$ tends to the
\index[xauthor]{Stuart, A.}%
normal distribution with zero mean and unit variance.

If we combine all terms in the expansion of $\mu_{2h} = E\bigl[\bigl(\sum r(ijk)\bigr)^{2h}\bigr]$ in which
similar combinations of values of $i$,~$j$, and~$k$ occur, as we did earlier for the
second moment, the number of times terms of a given type appear will be a
polynomial in~$n$ whose degree equals the number of different values of
$i$,~$j$, and~$k$ that occur in the terms. Therefore, $\mu_{2h}$~is a polynomial in~$n$ whose
coefficients depend only on~$h$. To calculate its degree and leading coefficient,
we must identify the terms with a nonzero expectation in the expansion that
involve the largest number of different values of the indices $i$,~$j$, and~$k$.
\PageSep{13}

We may suppose that each term in the expansion has been split into
classes of products such that different classes of the same term have no
indices in common; we may also suppose that any factor in a class of more
than one factor has at least one index in common with at least one other
factor in the same class. If a term contains a factor~$r$ involving one or more
indices that occur nowhere else in the term, then the term has expectation
zero, since $r$~is independent of the remaining factors in the term. Hence,
if a term is to have a nonzero expectation, each index in it must occur
at least twice.

We need only consider terms in which each index occurs exactly twice,
because, as we shall show, the leading term in~$\mu_{2h}$ is of order~$3h$. Suppose
such a term contains fewer than $h$ classes. Some class of this term must
contain at least three different factors. For any factor~$r(ijk)$ of this class,
either the indices $i$,~$j$, and~$k$ occur in three other factors or one index, $i$~say,
occurs in another factor and $j$~and~$k$ occur together in another. In either
case $r(ijk)$~is independent of the remaining factors and the expectation of the
term is zero. Consequently, the terms with a nonzero expectation that
involve the largest number of different values of $i$,~$j$, and~$k$ are of the type
$\prod_{\nu=1}^{h} r(i_{v}j_{v}k_{v})^{2}$. There are
\[
\frac{(2h)!}{2^{h} h!} \binom{n}{3} \binom{n - 3}{3} \dots \binom{n - 3(h - 1)}{3}
\]
% [** TN: Slant fractions in the original]
terms of this type and the expected value of each is~$(3/16)^{h}$. Therefore,
\[
\mu_{2h} = \frac{(2h)!}{2^{6h} h!} n^{3h} + O(n^{3h-1}).
\]
Since
\[
\sigma^{2h} = \frac{n^{3h}}{2^{5h}} + O(n^{3h-1}),
\]
it follows that (i)~holds for each fixed positive integer~$h$ as $n$~tends to infinity.

Similarly, if a term in the expansion of~$\mu_{2h+1}$ is to have a nonzero expectation,
each index in it must occur at least twice. Hence, $\mu_{2h+1}$~is a polynomial
in~$n$ of degree at most $\bigl[3(2h + 1)/2\bigr] = 3h + 1$. Since $\sigma^{2h+1}$~is of degree
$3h + 3/2$ in~$n$, it follows that (ii)~also holds for each fixed positive integer~$h$
as $n$~tends to infinity. This completes the proof of the theorem.

Notice that the expected value of~$c_{n}$ is asymptotically equal to the
maximum value of~$c_{n}$ (see the first corollary to \ThmRef{4}). This \Typo{phenomenom}{phenomenon}
occurs frequently in problems of this type.

The following theorem may be proved in the same way.

\begin{Theorem}{7.}
Let $q_{n}$ denote the number of $4$-cycles in a random tournament~$T_{n}$.
Then the distribution of $(q_{n} - \mu')/\sigma$ tends to the normal distribution
with zero mean and unit variance, where
\[
\mu' = \frac{3}{8} \binom{n}{4}\quad\text{and}\quad
\sigma^{2} = \frac{3}{64} \binom{n}{4} (4n - 11).
\]
\end{Theorem}
\PageSep{14}

Moon and Moser (1962a) proved the following analogous result for
\index[xauthor]{Moon, J. W.}%
\index[xauthor]{Moser, L.}%
bipartite tournaments.

\begin{Theorem}{8.}
Let $c(m, n)$ denote the number of $4$-cycles in a random $m$~by~$n$
\index{Cycles in a tournament}%
bipartite tournament. Then, subject to certain mild conditions on the
\index{Tournament!bipartite}%
relative rates of growth of $m$ and~$n$, the distribution of $\bigl(c(m, n) - \mu'\bigr)/\sigma$
tends to the normal distribution with zero mean and unit variance, where
\[
\mu' = \frac{1}{8} \binom{m}{2} \binom{n}{8}\quad\text{and}\quad
\sigma^{2} = \frac{1}{64} \binom{m}{2} \binom{n}{2} (2m + 2n - 1).
\]
\end{Theorem}


\Exercises

\Ex{1.} The quantity
\[
h = \frac{12}{n^{3} - n}
  · \sum_{i=1}^{n} \left(s_{i} - \frac{1}{2}(n - 1)\right)^{2}
\]
is called the \emph{hierarchy index} of a tournament. Show that the mean and
\index{Hierarchy index}%
variance of~$h$ satisfy the equations $\mu' = 3/(n + 1)$ and $\sigma^{2} = 18(n - 2)/(n + 1)^{2} n(n - 1)$. [Landau (1951~a,~b).]
\index[xauthor]{Landau, H. G.}%

\Ex{2.} Prove \ThmRef{7}.

\Ex{3.} Prove that the third and fourth moments of the distribution of~$c_{n}$ are
given by the formulas:
\[
\mu_{3} = -\frac{3}{32} \binom{n}{3} (n - 4)
\]
and
\[
\mu_{4} = \frac{3}{256} \binom{n}{3} \left\{
  9\binom{n - 3}{3} + 39\binom{n - 3}{2} + 9\binom{n - 3}{1} + 7
\right\}.
\]
[Moran (1947).]
\index[xauthor]{Moran, P. A. P.}%

\Ex{4.} Let $q = q(n, k)$ denote the number of $k$-cycles in a random tournament~$T_{n}$
($k \geq 3$). Show that the mean and variance of~$q$ satisfy the equations
\[
\mu' = \binom{n}{k} \frac{(k - 1)!}{2^{k}}\quad\text{and}\quad
\sigma^{2} = \Typo{0}{O}(n^{2k-3})
\]
for each fixed value of~$k$ as $n$~tends to infinity. What can be deduced about
the number of $k$-cycles in most tournaments~$T_{n}$?

\Ex{5.} Obtain a similar result for paths of length~$k$.


\Section{7.}{Transitive Tournaments}
\index{Transitive tournament}%

A tournament is \emph{transitive} if, whenever $p \to q$ and $q \to r$, then $p \to r$ also.
Transitive tournaments have a very simple structure. The following theorem
\PageSep{15}
gives some properties of a transitive tournament~$T_{n}$ whose scores
$(s_{1}, s_{2}, \dots, s_{n})$ are in nondecreasing order.

\begin{Theorem}{9.}
The following statements are equivalent.

(1) $T_{n}$ is transitive.

(2) Node~$p_{j}$ dominates node~$p_{i}$ if and only if $j > i$.

(3) $T_{n}$~has score vector $(0, 1, \dots, n - 1)$.
\index{Score of a node}%
\index{Score vector}%

(4) The score vector of~$T_{n}$ satisfies the equation
\[
\sum_{i=1}^{n} s_{i}^{2} = \frac{n(n - 1)(2n - 1)}{6}.
\]

(5) $T_{n}$~contains no cycles.

(6) $T_{n}$~contains exactly $\dbinom{n}{k + 1}$ paths of length~$k$, if $1 \leq k \leq n - 1$.

(7) {\Loosen $T_{n}$~contains exactly $\dbinom{n}{k}$ transitive subtournaments~$T_{k}$, if
$1 \leq k \leq n$.}

(8) Each principal submatrix of the dominance matrix~$M(T_{n})$ contains
\index{Matrix!of a tournament}%
a row and column of zeros.
\end{Theorem}

Every tournament~$T_{n}$ ($n \geq 4$) contains at least one transitive subtournament~$T_{3}$,
\index{Subtournament}%
but not every tournament~$T_{n}$ is itself transitive. The following
question arises: What is the largest integer $v = v(n)$ such that every tournament~$T_{n}$
contains a transitive subtournament~$T_{v}$? Erdös and Moser (\Erratum{1964}{1964a})
\index[xauthor]{Erdös, P.}%
\index[xauthor]{Moser, L.}%
gave the following bounds for~$v(n)$. [The lower bound was first found by
Stearns (unpublished).]
\index[xauthor]{Stearns, R.}%

\begin{Theorem}{10.}
$[\log_{2} n] + 1 \leq v(n) \leq [2\log_{2} n] + 1$.
\end{Theorem}

\Proof. Consider a tournament~$T_{n}$ in which the node~$p_{n}$ has the largest
score~$s_{n}$. It must be that $s_{n} \geq [\nf{1}{2} n]$, so there certainly exists a subtournament
$T_{[(1/2)n]}$ in~$T_{n}$ each node of which is dominated by~$p_{n}$. We may suppose that
this subtournament contains a transitive subtournament with at least
$\bigl[\log_{2} [\nf{1}{2} n]\bigr] + 1$ nodes. These nodes together with~$p_{n}$ determine a transitive
subtournament of~$T_{n}$ with at least
\[
\bigl[\log_{2} [\nf{1}{2} n]\bigr] + 2 = [\log_{2} n] + 1
\]
nodes. The lower bound now follows by induction.

There are $2^{\ebinom{n}{2} - \ebinom{v}{2}}$ tournaments~$T_{n}$, containing a given transitive subtournament~$T_{v}$,
and there are $\dbinom{n}{v} v!$ such subtournaments~$T_{v}$ possible. Therefore,
\[
\binom{n}{v} v! 2^{\ebinom{n}{2} - \ebinom{v}{2}} \geq 2^{\ebinom{n}{2}},
\]
{\Loosen since every tournament~$T_{n}$ contains at least one transitive subtournament~$T_{v}$.
\PageSep{16}
This inequality implies that $n^{v} \geq 2^{\ebinom{v}{2}}$. Consequently, $v \leq [2\log_{2} n] + 1$,
and theorem is proved.}

The exact value of~$v(n)$ is known only for some small values of~$n$. For
example, $v(7) \geq 3$ by \ThmRef{10}. The tournament~$T_{7}$ in which $p_{i} \to p_{j}$ if
and only if $j - i$ is a quadratic residue modulo~$7$ contains no transitive
subtournament~$T_{4}$. It follows that $v(7) = 3$. Bent (1964) examined other
\index[xauthor]{Bent, D. H.}%
similarly constructed tournaments and deduced the information about~$v(n)$
given in \Table{2}. Erdös and Moser conjecture that $v(n) = [\log_{2} n] + 1$
\index[xauthor]{Erdös, P.}%
\index[xauthor]{Moser, L.}%
for all~$n$.
\begin{table}[hbt!]
\TCaption{Table 2. $v(n)$, the largest integer~$v$ such that every tournament~$T_{n}$
contains a transitive subtournament~$T_{v}$.}
\begin{align*}
% [** TN: No punctuation in the original]
v(2) &=  v(3) = 2 \\
v(4) &=  \dots = v(7) = 3 \\
v(8) &=  \dots = v(11) = 4 \\
4 \leq v(12) &\leq \dots \leq v(15) \leq 5 \\
v(16) &=  \dots = v(23) = 5 \\
5 \leq v(24) &\leq \dots \leq v(31) \leq 7 \\
6 \leq v(32) &\leq \dots \leq v(43) \leq 7
\end{align*}
\end{table}

Moon (1966) found the next two theorems.
\index[xauthor]{Moon, J. W.}%

Let $u(n, k)$ denote the maximum number of transitive subtournaments~$T_{k}$
that a strong tournament~$T_{n}$ can have. (The problem is trivial if $T_{n}$~is
\index{Strong tournament}%
not strong.)

\begin{Theorem}{11.}
If $3 \leq k \leq n$, then $u(n, k) = \dbinom{n}{k} - \dbinom{n - 2}{k - 2}$.
\end{Theorem}

\Proof. When $k = 3$ the theorem follows from \ThmRef{5}, since every
subtournament~$T_{3}$ is either strong or transitive. We now show that
$u(n, k) \leq \dbinom{n}{k} - \dbinom{n - 2}{k - 2}$ for any larger fixed value of~$k$. This inequality
certainly holds when $n = k$. If $n > k \geq 4$, then any strong tournament~$T_{n}$
contains a strong subtournament~$T_{n-1}$ by \ThmRef{3}. Let $p$~be the node not
in~$T_{n-1}$; there are at most $u(n - 1, k - 1)$ transitive subtournaments~$T_{k}$
of~$T_{n}$ that contain the node~$p$ and at most $u(n - 1, k)$ that do not. We
may suppose
\[
u(n - 1, k - 1) \leq \binom{n - 1}{k - 1} - \binom{n - 3}{k - 3}
\]
and
\[
u(n - 1, k) \leq \binom{n - 1}{k} - \binom{n - 3}{k - 2}.
\]
\PageSep{17}
Therefore,
\begin{align*}
u(n, k) &\leq u(n - 1, k - 1) + u(n - 1, k)\displaybreak[0] \\
  &\leq \binom{n - 1}{k - 1} + \binom{n - 1}{k}
      - \binom{n - 3}{k - 3} - \binom{n - 3}{k - 2}\displaybreak[0] \\
  &= \binom{n}{k} - \binom{n - 2}{k - 2}.
\end{align*}
The inequality now follows by induction.

To show that $u(n, k) \geq \dbinom{n}{k} - \dbinom{n - 2}{k - 2}$ consider the strong tournament~$T_{n}$
in which $p_{1} \to p_{n}$ but otherwise $p_{j} \to p_{i}$ if $j > i$. (This tournament is
illustrated in \Figure{4}.) This tournament has exactly $\dbinom{n}{k} - \dbinom{n - 2}{k - 2}$
transitive subtournaments~$T_{k}$ if $3 \leq k \leq n$, because every subtournament~$T_{k}$
is transitive except those containing both $p_{1}$ and~$p_{n}$. This completes the
proof of the theorem.
\begin{figure}[hbt!]
\centering
\Graphic[0.75in]{fig4}
\FCaption{Figure 4}
\end{figure}

\begin{Corollary}
The maximum number of transitive subtournaments a
strong tournament~$T_{n}$ ($n \geq 3$) can contain, including the trivial tournaments
$T_{1}$ and~$T_{2}$, is~$3 · 2^{n-2}$.
\end{Corollary}

Let $r(n, k)$ denote the minimum number of transitive subtournaments~$T_{k}$
a tournament~$T_{n}$ can have. It follows from \ThmRef{10} that $r(n, k) = 0$
if $k > [2\log_{2} n] + 1$ and that $r(n, k) > 0$ if $k \leq [\log_{2} n] + 1$.

\begin{Theorem}{12.}
Let
\[
\tau(n, k) = \begin{cases}
  n · \dfrac{(n - 1)}{2} · \dfrac{(n - 3)}{4} \dots \dfrac{(n - 2^{k-1} + 1)}{2^{k-1}} & \text{if $n > 2^{k-1} - 1$}, \\
  0 & \text{if $n \leq 2^{k-1} - 1$}.
\end{cases}
\]
Then
\[
r(n, k) \geq \tau(n, k).
\]
\end{Theorem}
\PageSep{18}

\Proof. When $k = 1$, the result is certainly true if we count the trivial
tournament~$T_{1}$ as transitive. If $k \geq 2$, then clearly
\[
r(n, k) \geq \sum_{i=1}^{n} r(s_{i}, k - 1),
\]
where $(s_{1}, s_{2}, \dots, s_{n})$ denotes the score vector of the tournament~$T_{n}$. We may
suppose that $r(s_{i}, k - 1) \geq \tau(s_{i}, k - 1)$; since $\tau(n, k)$~is a convex function
of~$n$ for fixed values of~$k$, we may apply Jensen's inequality and conclude that
\[
r(n, k) \geq \sum_{i=1}^{n} \tau(s_{i}, k - 1)
  \geq n\tau\bigl(\nf{1}{2} (n - 1), k - 1\bigr)
  = \tau(n, k).
\]
The theorem now follows by induction on~$k$.

Notice that the lower bound in \ThmRef{10} follows from \ThmRef{12}.


\Exercises

\Ex{1.} Prove \ThmRef{9}. [The equivalence between (4) and~(5) is due to Moser;
\index[xauthor]{Moser, L.}%
see Bush (1961). In connection with~(6), see \Typo{(}{}Szele (1943) and Camion
\index[xauthor]{Bush, L. E.}%
\index[xauthor]{Camion, P.}%
\index[xauthor]{Szele, T.}%
(1959); in connection with~(8), see Marimont (1959) and Harary (1960).]
\index[xauthor]{Harary, F.}%
\index[xauthor]{Marimont, R. B.}%

\Ex{2.} Construct the tournament~$T_{7}$ defined after the proof of \ThmRef{10} and
verify that it contains no transitive subtournament~$T_{4}$.

\Ex{3.} Prove the corollary to \ThmRef{11}.

\Ex{4.} Use \ThmRef{10} to prove that
\[
r(n, k) \geq \binom{n}{2^{k-1}}\bigg/ \binom{n - k}{2^{k-1} - k}
  \geq \left(\frac{n}{2^{k-1}}\right)^{k}
\quad\text{if $n \geq 2^{k-1}$.}
\]
How does this bound compare with the one in \ThmRef{12}?

\Ex{5.} Prove that $r(n, 3) = \tau(n, 3)$ when $n$~is odd.

\Ex{6.} What is the maximum number of arcs~$\Arc{pq}$ a tournament~$T_{n}$ can have such
that $\Arc{pq}$~is not contained in any transitive subtournament~$T_{3}$?.

\Ex{7.} A transitive subtournament of~$T_{n}$ is said to be \emph{maximal} if it is not
\index{Maximal transitive subtournament}%
contained in any larger transitive subtournament of~$T_{n}$. Let $m(n)$ denote the
maximum number of maximal transitive subtournaments that a tournament~$T_{n}$
can have. Show that $m(n) \geq (7)^{n/5} > (1.47)^{n}$ if $n \equiv 0 \pmod{5}$. Can this
lower bound be improved? Try to obtain a nontrivial upper bound for~$m(n)$.

\Ex{8.} If $k$~is any fixed integer greater than two, let $t = t(n, k)$ denote the number
of transitive subtournaments~$T_{k}$ contained in a random tournament~$T_{n}$.
\PageSep{19}
Show that the distribution of $(t - \mu')/\sigma$ tends to the normal distribution
with zero mean and unit variance as $n$~tends to infinity, where
\[
\mu' = (n)_{k} 2^{-\ebinom{k}{2}}
\]
and
\[
\sigma^{2} = \left(k! 2^{-\ebinom{k}{2}}\right)^{2}
  \sum_{r=3}^{k} \binom{n}{k} \binom{k}{r} \binom{n - k}{k - r}
    \biggl(\frac{2^{\ebinom{r}{2}}}{r!} - 1\biggr).
\]


\Section{8.}{Sets of Consistent Arcs in a Tournament}
\index{Consistent arcs}%

In the last section, we considered subsets of nodes of a tournament such
that the subtournament determined by these nodes was transitive. We may
also consider subsets of arcs of a tournament~$T_{n}$ such that these arcs, by
themselves, define no intransitivities. More specifically, we shall call the
arcs in a set~$S$ \emph{consistent} if it is possible to relabel the nodes of~$T_{n}$ in such a
way that, if the arc~$\Arc{p_{j}p_{i}}$ is in~$S$, then $j > i$. (An equivalent definition is that
$T_{n}$~contains no cycles all of whose arcs belong to~$S$.) Erdös and Moon (1965)
\index[xauthor]{Erdös, P.}%
\index[xauthor]{Moon, J. W.}%
gave the following result.

\begin{Theorem}{13.}
Let $w(n)$ denote the largest integer~$w$ such that every tournament~$T_{n}$
contains a set of $w$ consistent arcs. Then
\[
w(n) \geq \left[\frac{n}{2}\right] · \left[\frac{n + 1}{2}\right]
\quad\text{for all $n$ and }
w(n) \leq \frac{(1 + \eps)}{2} \binom{n}{2}
\]
for any positive~$\eps$ and all sufficiently large~$n$.
\end{Theorem}

\Proof. When $n = 1$, the lower bound is trivially true. In any tournament~$T_{n}$,
there exists at least one node, say~$p_{n}$, whose score is at least~$\left[\dfrac{n}{2}\right]$. We
\index{Score of a node}%
may suppose that the tournament defined by the remaining $n - 1$ nodes
contains a set~$S$ of at least $\left[\dfrac{n - 1}{2}\right] · \left[\dfrac{n}{2}\right]$ consistent arcs. The arcs in~$S$ and
the arcs oriented away from~$\Typo{P_{n}}{p_{n}}$ are clearly consistent. Therefore, $T_{n}$~contains
a set of at least
\[
\left[\frac{n}{2}\right]
  + \left[\frac{n - 1}{2}\right] · \left[\frac{n}{2}\right]
  = \left[\frac{n}{2}\right] · \left[\frac{n + 1}{2}\right]
\]
consistent arcs. The lower bound follows by induction.

We now prove the upper bound. Let $\eps$ be chosen satisfying the inequality
$0 < \eps < 1$. The tournament~$T_{n}$ has $N = \dbinom{n}{2}$ pairs of distinct nodes and
\PageSep{20}
the nodes can be labeled in $n!$~ways. Hence, there are at most $n! · \dbinom{N}{k}$
tournaments~$T_{n}$, whose largest set of consistent arcs contains $k$~arcs. So, an
upper bound for the number of tournaments~$T_{n}$ that contain a set of more
than $\dfrac{(1 + \eps)}{2} N$ consistent arcs is given by
%[** TN: Added line break at second inequality]
\begin{align*}
n! · \sum_{k > (1 + \eps) N/2} \binom{N}{k}
  &< n! N \binom{N}{\bigl[(1 + \eps) N/2\bigr]} \\
  &\leq n! N 2^{N} \binom{N}{\bigl[(1 + \eps) N/2\bigr]}
   · \binom{N}{\left[\dfrac{1}{2} N\right]}^{-1}\displaybreak[0] \\
  &= n! N 2^{N}
    \frac{\left(N - \left[\dfrac{1}{2}N\right]\right)_{\bigl[(1 + \eps) N/2\bigr] - \left[\dfrac{1}{2}N\right]}}
         {\bigl[(1 + \eps) N/2\bigr]_{\bigl[(1 + \eps) N/2\bigr] - \left[\dfrac{1}{2}N\right]}}\displaybreak[0] \\
%
  &\leq n! N 2^{N} \left(\frac{N - \left[\dfrac{1}{2} N\right]}
                              {\bigl[(1 + \eps) N/2\bigr]}\right)^{[\eps N/2]}\displaybreak[0] \\
  &\leq n! N 2^{N} \left(1 - \frac{1}{2} \eps\right)^{[\eps N/2]},
\end{align*}
provided that $n$~is sufficiently large. Since $1 - x < e^{-x}$ when $0 < x < 1$,
it follows that this last expression is less than
\[
n! N 2^{N} e^{-(\eps/2)[\eps N/2]};
\]
but this in turn is less than~$2^{N}$ for all sufficiently large values of~$n$. Hence,
there must be at least one tournament~$T_{n}$ that does not contain any set of
more than $\dfrac{(1 + \eps)}{2} N$ consistent arcs. This completes the proof of the
theorem.

A more careful analysis of the above inequalities shows that the proportion
of tournaments~$T_{n}$ that contain more than
\[
\frac{1}{2} \binom{n}{2}
   + (1 + \eps) \left(\frac{1}{2} n^{3} \log n\right)^{1/2}
\]
consistent arcs, for any fixed positive~$\eps$, tends to zero as $n$~tends to infinity.


\Exercises

\Ex{1.} Determine the exact value of~$w(n)$ for $3 \leq n \leq 5$.

\Ex{2.} Let $z = z(n)$ denote the least integer such that any two tournaments with
$n$~nodes can be made isomorphic by reversing at most $z$ arcs of one of the
tournaments. Prove that $z(n) = \left(\dfrac{1}{2} + o(1)\right) \dbinom{n}{2}$.
\PageSep{21}

\Ex{3.} Prove the assertion in the last paragraph of this section.

\Ex{4.} An \emph{oriented graph}, or incomplete tournament, $T(n, m)$ consists of $n$~nodes,
\index{Oriented graph}%
$m$~pairs of which are joined by a single arc. Let $w = w(n, m)$ denote
the largest integer such that every oriented graph $T(n, m)$ contains a set of $w$
consistent arcs. Prove that $\lim_{n \to \infty} w(n, m)/m = 1/2$, under suitable assumptions
on the relative rates of growth of $m$~and~$n$. [Erdös and Moon (1965).]
\index[xauthor]{Erdös, P.}%
\index[xauthor]{Moon, J. W.}%

\Ex{5.} Let $u(T_{n})$ denote the least number of arcs possible in a subset~$U$ of the
arcs of~$T_{n}$ if at least one arc of every cycle of~$T_{n}$ belongs to~$U$. Let $r(T_{n})$
denote the smallest integer~$r$ such that $T_{n}$~can be transformed into a transitive
\index{Transitive tournament}%
tournament by reversing the orientation of $r$~arcs. Finally, let $w(T_{n})$
denote the maximum number of consistent arcs in~$T_{n}$. Prove that $u(T_{n}) = r(T_{n}) = \dbinom{n}{2} - w(T_{n})$
for any tournament~$T_{n}$. [Remage and Thompson
\index[xauthor]{Remage, R., Jr.}%
\index[xauthor]{Thompson, W. A., Jr.}%
(1966) gave an algorithm for determining~$r(T_{n})$.]

\Ex{6.} Let $f(n, i)$ denote the number of tournaments~$T_{n}$ such that $r(T_{n}) = i$.
Prove the following equations [Slater (1961)].
\index[xauthor]{Slater, P.}%
\begin{align*}
&f(n, 0) = n!, \\
&f(n, 1) = \frac{n! (3n^{2} - 13n + 14)}{6},\displaybreak[0] \\
&f(n, 2) = \frac{n! (9n^{4} - 78n^{3} + 235n^{2} - 438n + 680)}{72}
\quad\text{if $n \geq 4$},\displaybreak[0] \\
&f(n, 3) = \\
&\left\{
\begin{aligned}
  &24\quad\text{if $n = 5$}, \\
  &\Squeeze[0.925]{\dfrac{n! (135n^{6} - 1755n^{5} + 8685n^{4} - 27,185n^{3} + 77,820n^{2} - 157,204n + 210,336)}{6480}} \\
%[** TN: Width-dependent spacing]
&  \rule{4in}{0pt}\text{if $n \geq 6$}.
\end{aligned}
\right.
\end{align*}


\Section[The Parity of the Number of Spanning Paths]
{9.}{The Parity of the Number of Spanning Paths
of a Tournament}
\index{Paths in a tournament}%
\index{Spanning path or cycle}%

Every tournament has a spanning path. This is an immediate consequence
of \ThmRef{3}, if the tournament~$T_{n}$ is strong. If $T_{n}$~is not strong, then it has
\index{Strong tournament}%
\index{Subtournament}%
a unique decomposition into strong subtournaments $T^{(1)}, T^{(2)}, \dots, T^{(l)}$ such
that every node of~$T^{(j)}$ dominates every node of~$T^{(i)}$ if $1 \leq i < j \leq l$.
Each of these strong subtournaments has a spanning path and these
spanning paths can be combined to yield a path spanning~$T_{n}$.

The following theorem is a special case of a result proved by Rédei (1934)
\index[xauthor]{Redei@Rédei, L.}%
and generalized by Szele (1943).
\index[xauthor]{Szele, T.}%

\begin{Theorem}{14.}
Every tournament has an odd number of spanning paths.
\end{Theorem}
\PageSep{22}

\Proof. Let $A = [a_{ij}]$ denote the dominance matrix of an arbitrary
\index{Matrix!of a tournament}%
tournament~$T_{n}$. Recall that $a_{ij} = 1$ if $p_{i} \to p_{j}$ and $a_{ij} = 0$ otherwise; in
particular, $a_{ii} = 0$. The number~$h$ of spanning paths in~$T_{n}$ is equal to the sum
of all products of the type
\[
a_{i_{1}\pi(i_{1})} a_{i_{2}\pi(i_{2})} \dots a_{i_{n-1}\pi(i_{n-1})},
\]
where $1 \leq i_{1} < i_{2} \Chg{}{<} \dots < i_{n-1} \leq n$ and $\pi$~is a permutation of the set $N = \{1, 2, \dots, n\}$
such that $\pi^{k}(i) \neq i$ for $1 \leq k \leq n - 1$.

We introduce more notation before continuing the proof. The symbols
$e, e_{1}, \dots$ denote subsets of the set~$N$; $e'$~is the complement of~$e$ with respect to~$N$;
$e(M)$~is the determinant of the submatrix of the $n$~by~$n$ matrix~$M$ whose
row and column indices belong to~$e$; if $e$~is the empty set, then $e(M) = 1$.
The determinant of~$M$ will be denoted by~$|M|$.

Let $A_{k}$ denote the matrix obtained from~$A$ by replacing the $k$th~column
by a column of~$1$'s. Set
\[
S_{k} = \sum_{1 \in e} e(A_{k}) · e'(A_{k}),
\]
where the sum is over all subsets~$e$ of~$N$ that contain~$1$. (The restriction that
$1 \in e$ merely serves to distinguish between $e$ and~$e'$.) We now show that
\[
h \equiv \sum_{k=1}^{m} S_{k} \pmod{2}.
\Tag{(1)}
\]

It is clear that any term~$g$ in the expansion of~$S_{k}$ corresponds to a term in
the expansion of the determinant of~$A_{k}$. (Since the summation is modulo~$2$,
we may disregard the signs of the terms.) Suppose the permutation of~$N$
associated with this term can be expressed as the product of the disjoint
cycles $c_{1}, c_{2}, \dots, c_{r}$. Those nonzero terms associated with permutations for
which $r = 1$, arising only when $e = N$, are precisely those terms described
earlier whose sum is~$h$. To prove assertion~\Eq{(1)}, it will suffice to show that
the sum of the other terms is even.

Consider any term~$g$ for which $r > 1$. There is no loss of generality if we
assume that $1 \in c_{1}$. Now, all the elements in~$c_{i}$ belong to~$e$ or they all belong
to~$e'$, where $e$~is a subset of~$N$ giving rise to~$g$ in the expansion of~$S_{k}$. Since
there are two choices for each cycle $c_{2}, c_{3}, \dots, c_{r}$ it follows that there are
$2^{r-1}$~subsets~$e$ giving rise to~$g$. Consequently, the term~$g$ appears $2^{r-1}$~times
in the sum. This completes the proof of~\Eq{(1)}, since $2^{r-1}$~is even if $r > 1$.

Let $\bar{A} = [\bar{a}_{ij}]$ denote the matrix whose elements satisfy the equation
$\bar{a}_{ij} = a_{ij} + 1$ for all $i$~and~$j$. This implies that
\[
|\bar{A}| = |A| + \sum_{k=1}^{n} |A_{k}|.
\Tag{(2)}
\]
We observe that $\bar{a}_{ij} = a_{ji} \pmod{2}$ if $i \neq j$ and that $\bar{a}_{ii} = 1$. Hence, the
transpose of~$\bar{A}$, which has the same determinant as~$\bar{A}$, has its entries all
\PageSep{23}
congruent to the corresponding entries of~$A$ with the exception of the
diagonal entries. Consequently,
\[
\sum_{k=1}^{n} |A_{k}| = |\bar{A}^{T}| - |A|
  \equiv \sum_{e \neq N} e(A) \pmod{2},
\Tag{(3)}
\]
since the last sum effectively involves those terms in the expansion of~$|\bar{A}^{T}|$
that contain diagonal entries. (Notice that the diagonal entries in a
term in the expansion of~$|A^{T}|$ corresponding to some term in the expansion
of~$e(A)$ are those~$\bar{a}_{ii}$ for which $i \in e'$; this is why the sum is over the proper
subsets~$e$ of~$N$.)

Let $\sum_{k}^{*} e(A_{k})$ denote the sum of~$e(A_{k})$ over all~$k$ such that $k \in e$ with the
convention that an empty sum equals zero. Then,
\begin{align*}
h &\equiv \sum_{k=1}^{n} \left\{\sum_{1 \in e} e(A_{k}) · e'(A_{k})\right\} \\
  &\equiv \sum_{1 \in e} \left\{\sum_{k \in e} e(A_{k}) \Chg{}{·} e'(A_{k})
    + \sum_{k \not\in e} e(A_{k}) · e'(A_{k})\right\} \\
  &\equiv \sum_{1 \in e} \left\{e'(A) {\sum_{k}}^{*} e(A_{k})
    + e(A) {\sum_{k}}^{*} e'(A_{k})\right\}
\Tag{(4)} \\
  &\equiv \sum_{e} e'(A) {\sum_{k}}^{*} e(A_{k}) \pmod{2}.
\end{align*}

Let us apply \Eq{(3)} to~$e(A)$ instead of to~$|A|$; then
\[
{\sum_{k}}^{*} e(A_{k}) \equiv \sum_{e_{1} \subset e} e_{1}(A) \pmod{2},
\Tag{(5)}
\]
where the second summation is over all proper subsets $e_{1}$ of~$e$. (If $e$~is the
empty set, then both sides equal zero.) If we substitute \Eq{(5)} into~\Eq{(4)}, we find
that
\[
h \equiv \sum_{e} \sum_{e_{1} \subset e_{1}} e'(A) · e_{1}(A) \pmod{2}.
\Tag{(6)}
\]

If $e - e_{1} \neq N$, then the term arising from $e$ and~$e_{1}$, where $e_{1} \subset e$ and
$e_{1} \neq e$, is equal to the term arising from $e_{1}'$ and~$e'$. Hence, the terms in the
sum come in pairs except for the terms arising when $e - e_{1} = N$, that is,
when $e = N$ and $e_{1}$~is the empty set. Since $e'(A) = e_{1}(A) = 1$ in this case,
it follows that
\[
h \equiv 1 \pmod{2}.
\]

This completes the proof of the theorem.

\begin{Corollary}
A tournament~$T_{n}$ has at least $\dbinom{n}{k}$ paths of length~$k - 1$ if
$2 \leq k \leq n$; furthermore, the number of paths of length~$k - 1$ in~$T_{n}$ is
congruent to $\dbinom{n}{k}$ modulo~$2$.
\end{Corollary}
\PageSep{24}

This is proved by applying \ThmRef{14} to each subtournament~$T_{k}$ of~$T_{n}$
and then summing over all such subtournaments.

Let $G$ denote a graph that differs from a tournament~$T_{n}$ in that some of
\index{Oriented graph}%
the arcs $\Arc{p_{i}p_{j}}$ of~$T_{n}$ may have been either removed entirely or replaced by
unoriented edges~$p_{i}p_{j}$. A path in such a graph~$G$ may pass through any
unoriented edge~$p_{i}p_{j}$ in either direction. Let $(G)_{k}$ denote the number of ways
of labeling the nodes of~$G$ as $q_{1}, q_{2}, \dots, q_{n}$ in such a way that there are $k$~pairs
of consecutive nodes $q_{i}$~and~$q_{i+1}$ for which the arc~$\Arc{q_{i+1}q_{i}}$ is in~$G$ (notice that
$(G)_{0} + (G)_{1} + \dots + (G)_{n-1} = n!$), and let $(G)$ denote the number of spanning
paths of~$G$. Finally, let $[G]_{k}$ denote the number of sets of $n - k$
disjoint paths of~$G$ with the property that every node of~$G$ belongs to
one path in the set. (Notice that there are $k$~arcs or edges involved in each
such set and that $[G]_{n-1} = (G)_{0} = (G)$ if $G$~is a tournament.)

Szele (1943) generalized \ThmRef{14} by showing that if $G$~is a tournament~$T$
\index[xauthor]{Szele, T.}%
with $n$~nodes, then
\[
(T)_{k} \equiv \binom{n - 1}{k} \pmod{2},
\Tag{(7)}
\]
and
\[
[T]_{k} \equiv \binom{\left[\dfrac{n + k - 1}{2}\right]}{k} \pmod{2}.
\Tag{(8)}
\]


\Exercises

\Ex{1.} Give a direct proof of the result that every (finite) tournament has a
spanning path.

\Ex{2.} Give an example of a tournament with an infinite number of nodes that
does not have a spanning path. [See Korvin (1966).]
\index[xauthor]{Korvin, G.}%

\Ex{3.} Prove Equation~\Eq{(2)}.

\Ex{4.} Let $G$ denote the graph obtained from the tournament~$T_{n}$ by replacing an
arbitrary arc~$\Arc{p_{i}p_{j}}$ by an unoriented edge $e = p_{i}p_{j}$ and let $H$ denote the graph
obtained from~$T_{n}$ simply by removing the arc~$\Arc{p_{i}p_{j}}$. Prove the following
statements and use them to give another proof of \ThmRef{14}.
\begin{itemize}
\item[(a)] $(G) = \sum\limits_{i=0}^{n-1} (-1)^{i} (n - i)! [H]_{i}$.

\item[(b)] $(G) \equiv (H) \pmod{2}$.

\item[(c)] The number of spanning paths of~$G$ that contain the edge~$e$ is even.

\item[(d)] The parity of the number of spanning paths of~$T_{n}$ is not changed
if the arc~$\Arc{p_{i}p_{j}}$ is replaced by the arc~$\Arc{p_{j}p_{i}}$. [Szele (1943).]
\end{itemize}
\PageSep{25}

\Ex{5.} Prove Equation~\Eq{(7)} by first showing that
\[
(T)_{k} = \sum_{i=0}^{n-1-k} (-1)^{i} \binom{k + i}{k} (n - k - i)! [T]_{k+i}.
\]

\Ex{6.} If $T^{*}$ denotes a transitive tournament~$T_{n}$, then show that $[T^{*}]_{k}$~is equal to
\index{Transitive tournament}%
the number of ways of partitioning $n$~different objects into $n - k$~indistinguishable
nonempty subsets. Deduce from this that
\[
[T^{*}]_{k} = \frac{1}{(n - k)!}
  \sum_{i=0}^{n-k-1} (-1)^{i} \binom{n - k}{i} (n - k - i)^{n}
  = S(n, n - k),
\]
where $S(m, t)$ denotes a Stirling number of the second kind. (The Stirling
numbers $S(m, t)$ may be defined by the identity
\[
x^{m} = \sum_{t=1}^{m} S(m, t)(x)_{t},
\]
where $(x)_{t} = x(x - 1) \dots (x - t + 1)$; they satisfy the recurrence relation
\[
S(m, t) = S(m - 1, t - 1) + tS(m - 1, t).)
\]

\Ex{7.} Prove that
\[
(T^{*})_{k} = \sum_{t=0}^{n-k-1} (-1)^{t} \binom{n + 1}{t} (n - k - t)^{n}.
\]

\Ex{8.} Use \ThmRef{14} to prove that, if $T$~is any tournament with $n$~nodes, then
\[
\Typo{]}{[}T]_{k} \equiv [T^{*}]_{k} \pmod{2}.
\]

\Ex{9.} Show that
\[
[T^{*}]_{k} = S(n, n - k) =  \sum j_{1} j_{2} \dots j_{k},
\]
where the sum is over all sets of $k$~integers $j_{i}$ such that $1 \leq j_{1} \leq j_{2} \leq \dots \leq j_{k} \leq n - k$.

\Ex{10.} Use the results in Exercises 8 and~9 to prove Equation~\Eq{(8)}. [Hint: Which
terms in the sum in Exercise~9 are odd? Recall that the number of ways of
choosing $t$~objects from $m$~objects when repetitions are allowed is
$\dbinom{m + t - 1}{t}$.]
% [** TN: Large closing brace in the original]

\Ex{11.} Let $k$~denote the maximum number of nodes that can be chosen from
the oriented graph~$T$ such that no two of the chosen nodes are joined by an
arc. Prove that there exists a set of $t$~disjoint paths in~$T$, where $t \leq k$, such
that every node of~$T$ belongs to exactly one path in the set. [Gallai and
\index[xauthor]{Gallai, T.}%
Milgram (1960).]
\index[xauthor]{Milgram, A. N.}%
\PageSep{26}


\Section[The Maximum Number of Spanning Paths]
{10.}{The Maximum Number of Spanning Paths
of a Tournament}
\index{Paths in a tournament}%
\index{Spanning path or cycle}%

Let $t(n)$ denote the maximum number of spanning paths a tournament~$T_{n}$
can contain. The following result is due to Szele (1943).
\index[xauthor]{Szele, T.}%

\begin{Theorem}{15.} If
\[
\alpha = \lim_{n \to \infty} \left(\frac{t(n)}{n!}\right)^{1/n},
\]
then $\alpha$~exists and satisfies the inequality $.5 = 2^{-1} \leq \alpha \leq 2^{-3/4} \leq .6$.
\end{Theorem}

\Proof. Let us assume for the time being that the limit does exist.

The expected number of spanning paths in a random tournament~$T_{n}$
is~$\dfrac{n!}{2^{n-1}}$. Consequently,
\[
t(n) \geq \frac{n!}{2^{n-1}}\quad\text{or}\quad
\left(\frac{t(n)}{n!}\right)^{1/n} \geq \frac{1}{2^{1 - 1/n}}
\Tag{(1)}
\]
for all positive integers~$n$. This implies the first inequality of the theorem.

Let $A$, $B$, $C$, and~$D$ denote the number of subtournaments~$T_{4}$ a given
tournament~$T_{n}$ has of the four types depicted in \Figure{1}. Subtournaments
of these types have $1$,~$3$,~$3$, and~$5$ spanning paths, respectively. Hence, the
total number of $3$-paths (paths of length three) in the tournament~$T_{n}$ is
equal to
\[
A + 3(B + C) + 5D = \binom{n}{4} + 2(n - 3) c_{3}(T_{n}),
\]
where $c_{3}(T_{n})$ is the number of $3$-cycles in~$T_{n}$. (The equality is established by
checking that each subtournament~$T_{4}$ is counted the appropriate number
\index{Subtournament}%
of times by the right-hand expression.)

But the corollary to \ThmRef{4} gives an upper bound for~$c_{3}(T_{n})$. We find,
therefore, that an upper bound for the number of $3$-paths in any tournament~$T_{n}$
is given by
\[
\begin{cases}
\dfrac{n^{2}(n - 1)(n - 3)}{8} & \text{if $n$~is odd,} \\
\dfrac{(n + 1)n(n - 2)(n - 3)}{8} & \text{if $n$~is even.}
\end{cases}
\]
It follows that
\[
t(n) \leq \left\{
\begin{aligned}
&\dfrac{n^{2}(n - 1)(n - 3)}{8} · \dfrac{(n - 4)^{2}(n - 5)(n - 7)}{8}\dots
\quad\text{if $n$~is odd,} \\
&\dfrac{(n + 1)n(n - 2)(n - 3)}{8} · \dfrac{(n - 3)(n - 4)(n - 6)(n - 7)}{8}\dots \\
% [** TN: Width-dependent spacing]
&\rule{3in}{0pt}\text{if $n$~is even.}
\end{aligned}
\right.
\]
\PageSep{27}
The last factors in these products will depend on the residue class modulo~$4$
to which $n$~belongs. We find in all cases, however, that
\[
t(n) \leq \frac{(n + 1)!}{8^{[n/4]}}
  \leq \frac{(n + 1)!}{2^{(3/4)n - 3}}.
\]
Therefore,
\[
\left(\frac{t(n)}{n!}\right)^{1/n}
  \leq \frac{(n + 1)^{1/n}}{2^{(3/4) - (3/n)}}
\Tag{(2)}
\]
for all positive integers~$n$. This implies the last inequality of the theorem.

We shall use the following result on subadditive functions [see Fekete
\index[xauthor]{Fekete, M.}%
(1923)] to show that the limit actually exists.

\begin{Lemma}
{\Loosen Let $\{g(n) : n = 1, 2, \dots\}$ be a sequence such that $g(a + b) \leq g(a) + g(b)$
for all $a, b = 1, 2, \dots$. If $\phi = \inf_{n} g(n)/n$ is finite, then
$\lim_{n \to \infty} g(n)/n$ exists and equals~$\phi$.}
\end{Lemma}

\Proof. Choose the integer~$r$ so that
\[
\frac{g(r)}{r} \leq \phi + \eps,
\]
for an arbitrary positive value of~$\eps$. For any given value of~$n$, let $n = rs + t$,
where $0 < t \leq r$. Then
\begin{align*}
\frac{g(n)}{n}
  &= \frac{g(rs + t)}{n}
   \leq \frac{g(rs)}{n} + \frac{g(t)}{n} \\
  &\leq \frac{sg(r)}{n} + \frac{g(t)}{n}
   \leq \frac{rs}{n}(\phi + \eps) + \frac{g(t)}{n} \\
  &\leq \phi + \eps + \frac{1}{n}\max\{g(1), \dots, g(r)\}.
\end{align*}
Therefore,
\[
\lim_{n \to \infty} \sup \frac{g(n)}{n} \leq \phi + \eps
\]
for every positive~$\eps$. The lemma follows immediately.

In forming a spanning path of a tournament~$T_{n}$, we may choose the first
$k$ nodes of this path, form a path spanning these $k$~nodes, and then join
this path to a path spanning the remaining $n - k$ nodes. Consequently,
\[
t(n) \leq \binom{n}{k} t(k) · t(n - k)\quad\text{or}\quad
\frac{t(n)}{n!} \leq \frac{t(k)}{k!} · \frac{t(n - k)}{(n - k)!}
\]
for all positive integers $k$ and $n$ with $k < n$. If we let $h(n) = \dfrac{t(n)}{n}$, then
\[
\log h(a + b) \leq \log h(a) + \log h(b)
\]
\PageSep{28}
for all positive integers $a$ and~$b$. The existence of $\lim\limits_{n \to \infty} \left(\dfrac{t(n)}{n!}\right)^{1/n}$ may now be
demonstrated by applying the lemma to the function $\log h(n)$.

Szele conjectured that the limit actually is~$\nf{1}{2}$ and gave the following
\index[xauthor]{Szele, T.}%
table of values of~$t(n)$.
\begin{table}[hbt!]
\TCaption{Table 3. $t(n)$, the maximum number of spanning paths
\index{Spanning path or cycle}%
in a tournament~$T_{n}$.}
\[
\begin{array}{*{6}{c}}
n & 3 & 4 & 5 & 6 & 7 \\
\hline
\Strut
t(n) & 3 & 5 & 15 & 45 & 189
\end{array}
\]
\end{table}


\Exercises

\Ex{1.} Supply the details omitted in the proof of inequality~\Eq{(2)}.

\Ex{2.} The inequality $t(n) \leq n!\bigl(t(k)/k!\bigr)^{[n/k]}$ holds for any fixed value of~$k$. How
does this upper bound compare with~\Eq{(2)} for small values of~$k$?

\Ex{3.} Let $c(n, k)$ denote the maximum number of $k$-cycles possible in a
\index{Cycles in a tournament}%
tournament~$T_{n}$. Prove that
\[
\binom{n}{k} \frac{(k - 1)!}{2^{k}}
  \leq c(n, k)
  \leq \binom{n}{k} \frac{(k + 1)(k - 1)!}{2^{(3/4)k-3}}.
\]
[G.~Korvin and others.]
\index[xauthor]{Korvin, G.}%

\Ex{4.} The expected number of spanning cycles in a random tournament~$T_{n}$ is
approximately $\dfrac{2\pi}{n} \left(\dfrac{n}{2e}\right)^{n}$. It seems difficult to give explicit examples of
tournaments having this many spanning cycles in general. Construct a
tournament having at least $\left(\dfrac{n}{3e}\right)^{n}$ spanning cycles for large values of~$n$.
[L.~Moser.]
\index[xauthor]{Moser, L.}%


\Section{11.}{An Extremal Problem}

A tournament $T_{n}$ has \emph{property $S(k, m)$}, where $k \leq n$, if for every subset~$A$
\index{Property $S(k, m)$}%
of $k$~nodes there exist at least $m$~nodes~$p$ such that $p$~dominates every node
of~$A$. In 1962, K.~Schütte raised the question of determining the least
\index[xauthor]{Schutte@Schütte, K.}%
integer~$n$ such that there exist tournaments~$T_{n}$ with property~$S(k, 1)$, if
such tournaments exist at all. Erdös (1962) showed that such tournaments
\index[xauthor]{Erdös, P.}%
do indeed exist and that if the tournament~$T_{n}$ has property~$S(k, 1)$ then
\[
n \geq 2^{k+1} - 1,\quad\text{for $k = 1, 2, \dots$.}
\Tag{(1)}
\]
E.~and G. Szekeres (1965) deduced a stronger inequality as a consequence
\index[xauthor]{Szekeres, E. and G.}%
of the following result.
\PageSep{29}

\begin{Theorem}{16.}
If the tournament~$T_{n}$ has property $S(k, m)$, then
\[
n \geq 2^{k}(m + 1) - 1.
\Tag{(2)}
\]
\end{Theorem}

\Proof. If $p$~is any node of the tournament~$T_{n}$, let $T_{d}$~denote the subtournament
determined by the $d$~nodes~$q$ that dominate~$p$. We show that $T_{d}$~has
property $S(k - 1, m)$.

Suppose that $d \geq k - 1$. Let $A$ be any set of $k - 1$ nodes of~$T_{d}$; since $T_{n}$~has
property $S(k, m)$, there must exist a set~$B$ of at least $m$ nodes~$q$ such that
$q$~dominates every node in the set $A \cup \{p\}$. Since all these nodes dominate~$p$,
it follows that they all are in the tournament~$T_{d}$. Therefore $T_{d}$~has property
$S(k - 1, m)$.

The other alternative is that $d < k - 1$. In this case, adjoin to~$T_{d}$ any
$k - 1 - d$ other nodes of~$T_{n}$ (but not~$p$) to form a subtournament~$T_{k-1}$. The
same argument as used before shows that $T_{k-1}$~has property $S(k - 1, m)$.
But this is impossible, since it would imply that the nodes of~$T_{k-1}$ dominate
themselves.

We now prove inequality~\Eq{(2)} by induction; it certainly holds when $k = 0$
and $m \geq 1$, since we may say that a tournament~$T_{n}$ has property $S(0, m)$
if and only if $n \geq m$. Suppose therefore that $k > 0$ and that \Eq{(2)}~has been
proved for all tournaments with property $S(k - 1, m)$. It follows from
the remarks in the preceding paragraphs and the induction hypothesis that
at least $2^{k-1} (m + 1) - 1$ arcs are oriented toward each node of~$T_{n}$.
Therefore,
\[
n\bigl(2^{k-1} (m + 1) - 1\bigr) < \binom{n}{2},
\]
or
\[
n \geq 2^{k} (m + 1) - 1.
\]

\begin{Corollary}
If the tournament~$T_{n}$ has property $S(k, 1)$, then
\[
n \geq 2^{k-1} (k + 2) - 1,\quad\text{for $k = 2, 3, \dots$.}
\]
\end{Corollary}

\Proof. If $T_{n}$~has property $S(k, 1)$ where $k \geq 2$, then $T_{n}$~has property
$S(k - 1, k + 1)$. For, let $A$~be any set of $k - 1$ nodes of~$T_{n}$ and suppose that
the set~$B$ of nodes that dominate every node of~$A$ contains only $h$ nodes,
where $h \leq k$. Then there exists a node~$x$ that dominates every node in~$B$.
Furthermore, there exists a node~$y$ that dominates every node in $A \cup \{x\}$.
Then $y$~must be in~$B$ and $x$~and~$y$ must dominate each other, an impossibility.
The corollary now follows from \ThmRef{16}.

This corollary is best possible when $k = 2, 3$ (see Exercises 1~and~2) but
it is not known if it is best possible when $k > 3$.

Erdös (1962) showed that there exist tournaments~$T_{n}$ with property $S(k, 1)$
\index[xauthor]{Erdös, P.}%
whenever
\[
n > 2^{k} k^{2} \log(2 + \eps)
\Tag{(3)}
\]
\PageSep{30}
for any positive~$\eps$, provided that $k$~is sufficiently large. The ideas used in
establishing this and an analogous result for tournaments with property
$S(k, m)$ are similar to those used in proving the following result due to Erdös
\index[xauthor]{Erdös, P.}%
and Moser (1964b).
\index[xauthor]{Moser, L.}%

If $A$ and $B$ are disjoint subsets of the tournament~$T_{n}$, let $t = t(A, B)$ denote
the number of nodes~$p$ that dominate every node of~$A$ and are dominated by
every node of~$B$. If $\eps$~is any positive number, let $F(n, k, \eps)$ denote the number
of tournaments~$T_{n}$ such that
\[
\left|t - \frac{n - k}{2^{k}}\right| < \frac{\eps(n - k)}{2^{k}}
\Tag{(4)}
\]
for every pair of subsets $A$ and $B$ for which $|A| + |B| = k$.

\begin{Theorem}{17.}
For every positive~$\eps$ there exist constants $c$ and~$K$, which
depend on~$\eps$, such that if $k > K$ and $n > ck^{2}2^{k}$, then
\[
\lim_{n \to \infty} \frac{F(n, k, \eps)}{2^{\ebinom{n}{2}}} = 1.
\Tag{(5)}
\]
\end{Theorem}

\Proof. Suppose $0 < \eps < 1$. For a particular choice of $A$~and~$B$, there are
\[
\binom{n - k}{t} (2^{k} - 1)^{n - k - t} · 2^{\ebinom{n}{2}-k(n-k)}
  = 2^{\ebinom{n}{2}} \binom{n - k}{t}
    \left(\frac{1}{2^{k}}\right)^{t}
    \left(1 - \frac{1}{2^{k}}\right)^{n-k-t}
\]
tournaments~$T_{n}$ for which $t(A, B) = t$. For, there are $\dbinom{n - k}{t}$ choices for the
$t$~nodes that dominate the nodes of~$A$ and are dominated by the nodes
of~$B$; the other $n - k - t$ nodes may each be joined to the $k$~nodes of
$A \cup B$ in one of $2^{k} - 1$ ways; the remaining $\dbinom{n}{2} - k(n - k)$ arcs may be
oriented arbitrarily. There are $\Erratum{\dbinom{n}{2}}{\dbinom{n}{k}} 2^{k}$ ways of selecting the sets $A$~and~$B$.
Consequently, if $S$~denotes the number of tournaments~$T_{n}$ that do not
satisfy~\Eq{(4)} for every choice of $A$~and~$B$, then
\[
S \leq \binom{n}{k} 2^{\ebinom{n}{2}+k}
  {\sum}' \binom{n - k}{t}
        \left(\frac{1}{2^{k}}\right)^{t}
        \left(1 - \frac{1}{2^{k}}\right)^{n-k-t},
\]
where the sum is over all~$t$ such that $\left|t - \dfrac{n - k}{2^{k}}\right| \geq \dfrac{\eps(n - k)}{2^{k}}$.

The largest term in the sum occurs when $t = T = \left[(1 - \eps) \dfrac{(n - k)}{2^{k}}\right]$,
$\dbinom{n}{k} 2^{k} \leq n^{k}$ if $k \geq 4$, and $1 - x < e^{-x}$ if $0 < x < 1$. Therefore,
\[
S \leq 2^{\ebinom{n}{2}} n^{k+1} \binom{n - k}{T} \frac{1}{2^{kT}} e^{-(n-k-T)/2^{k}}
\]
\PageSep{31}
when $k \geq 4$. But
\[
\binom{n - k}{T} \frac{1}{2^{kT}}
  < \left(\frac{(n - k)e}{T2^{k}}\right)^{T}
  \leq c_{1} \left(\frac{e}{1 - \eps}\right)^{T},
\]
where $c_{1}$~denotes some constant. Consequently, to prove~\Eq{(5)}, we need only
show that
\[
n^{k+1} \left(\frac{e}{1 - \eps}\right)^{(n/2^{k})(1-\eps)} e^{-(n/2^{k})+(n/2^{2k})(1-\eps)}
\]
tends to zero as $n$~tends to infinity, since $\dfrac{k}{2^{k}}$~is bounded.

If $L$ denotes the logarithm of this expression, then
\[
L = (k + 1) \log n - \frac{n}{2^{k}} \left[(1 - \eps) \log (1 - \eps) + \eps - \frac{(1 - \eps)}{2^{k}}\right].
\]
Since
\begin{align*}
\log (1 - \eps)
  &= -\eps - \nf{1}{2} \eps^{2} - \nf{1}{3} \eps^{3} - \dots
  > -\eps - \nf{1}{2} \eps^{2}(1 + \eps + \dots) \\
  &= \frac{-\eps + \nf{1}{2} \eps^{2}}{1 - \eps},
\end{align*}
when $0 < \eps < 1$, it follows that, for sufficiently large values of~$k$, the
quantity inside the brackets will exceed some positive constant $c_{2} = c_{2}(\eps)$.
Therefore,
\[
L \leq (k + 1) \log n - \frac{c_{2}n}{2^{k}}
  \leq n(k + 1) \left(\frac{\log n}{n} - \frac{c_{3}}{k2^{k}}\right),
\]
where $c_{3}$~is another constant equal, say, to~$\nf{1}{2}c_{2}$. Now let us suppose that
\[
n > ck^{2} 2^{k},
\]
where the constant $c = c(\eps)$ will be specified presently. Then it follows that
\[
L \leq n(k + 1) \left[\frac{\log(ck^{2} 2^{k})}{ck^{2} 2^{k}} - \frac{c_{3}}{k2^{k}}\right]
  = \frac{n(k + 1)}{ck2^{k}} \left[\frac{\log(ck^{2})}{k} + \log 2 - cc_{3}\right],
\]
since $\log n/n$ is a decreasing function for $n \geq 3$. It is clear that $c$~can be
defined as a function of~$\eps$ in such a way that, if $k$~is sufficiently large, then
the expression within the brackets is negative. Consequently, $L$~will tend to~$-\infty$
as $n$~tends to~$+\infty$. This completes the proof of the theorem.

When this type of argument is applied to the simpler problem of establishing~\Eq{(3)},
it turns out that the constant~$c$ may be any quantity greater than~$\log 2$.


\Exercises

\Ex{1.} Let $T_{7}$ denote the tournament in which $p_{i} \to p_{j}$ if and only if $j - i$ is a
quadratic residue modulo~$7$. Prove that $T_{7}$~has property $S(2, 1)$. [Erdös
\index[xauthor]{Erdös, P.}%
(1962).]
\PageSep{32}

\Ex{2.} Let $T_{19}$~denote the tournament in which $p_{i} \to p_{j}$ if and only if $j - i$ is a
quadratic residue modulo~$19$. Prove that $T_{19}$~has property $S(3, 1)$. [E.~and
G.~Szekeres (1965).]
\index[xauthor]{Szekeres, E. and G.}%

\Ex{3.} Prove that if a tournament~$T_{n}$ with property $S(k, 1)$ exists, then a tournament~$T_{r}$
with property $S(k, 1)$ exists whenever $r > n$. [See the remark at the
end of the second paragraph of Erdös and Moser (1964b).]
\index[xauthor]{Erdös, P.}%
\index[xauthor]{Moser, L.}%

\Ex{4.} Let $A$ and $B$ denote two disjoint subsets of nodes of a tournament~$T_{n}$
such that $|A| = a$ and $|B| = b$. If $t(A, B) \geq m$ for every choice of $A$~and~$B$,
then we will say that $T_{n}$~has property $T(a, b; m)$. (Notice that property
$T(a, 0; m)$ is the same as property $S(a, m)$.) Generalize \ThmRef{16} by
obtaining a lower bound for the number of nodes a tournament must have
if it has property $T(a, b; m)$.

\Ex{5.} Let $h = h(n)$ denote the smallest integer such that in any tournament~$T_{n}$
there exists some set~$E$ of $h$~nodes such that any node not in~$E$ is dominated
by at least one node in~$E$. Show that $[\log_{2} n - 2\log_{2}\log_{2} n] \leq h(n) \leq \bigl[\log_{2} (n + 1)\bigr]$
if $n \geq 2$. [L.~Moser.]


\Section{12.}{The Diameter of a Tournament}
\index{Diameter of a tournament}%

The \emph{diameter} of a strong tournament~$T_{n}$ ($n > 3$) is the least integer~$d$ such
\index{Strong tournament}%
that, for every ordered pair of nodes $p$ and~$q$ of~$T_{n}$, there exists a nontrivial
path $P(p, q)$ of length at most~$d$. (The diameter of a reducible tournament
is not defined.) No node is connected to itself by a nontrivial path of length
less than three so the diameter of every strong tournament is at least three.

We shall prove a result that implies that almost all tournaments~$T_{n}$ (that
is, all but a fraction that tends to zero as $n$~tends to infinity) have the following
property: the number of paths of length two from $p_{i}$ to $p_{j}$ lies between
\index{Paths in a tournament}%
$(\nf{1}{4} - \eps)(n - 2)$ and $(\nf{1}{4} + \eps)(n - 2)$ for every ordered pair of distinct
nodes $p_{i}$ and~$p_{j}$ and for any positive~$\eps$. Consequently, almost all tournaments
have diameter three, since there is a path of length three connecting every
node to itself in every strong tournament (and almost all tournaments are
strong by \ThmRef{1}).

Let $r_{ij}$ denote the number of paths of length two from $p_{i}$ to~$p_{j}$ ($p_{i} \neq p_{j}$)
in a tournament~$T_{n}$ and let $E(n, \lambda)$ denote the expected number of ordered
pairs of distinct nodes $p_{i}$ and~$p_{j}$ in a random tournament~$T_{n}$ for which
$|r_{ij} - \nf{1}{4}(n - 2)| > \lambda$. The following theorem is very similar to a theorem
proved by Moon and Moser (1966).
\index[xauthor]{Moon, J. W.}%

\begin{Theorem}{18.}
If $\lambda = \bigl\{\nf{3}{4}(n - 2) \bigl[\log(n - 2) + w(n)\bigr]\bigr\}^{1/2}$, where $\log(n - 2) + w(n) \to \infty$
and $w(n) n^{-1/3} \to 0$, then
\[
E(n, \lambda) \sim \bigl\{2\pi \bigl[\log(n - 2) + w(n)\bigr]\bigr\}^{-1/2} e^{-2w(n)}
\]
as $n \to \infty$.
\end{Theorem}
\PageSep{33}

\Proof. There are
\[
\binom{n - 2}{r} 3^{n-2-r}\, 2^{\ebinom{n}{2}-2(n-2)}
  = 2^{\ebinom{n}{2}} \binom{n - 2}{r}
    \left(\frac{1}{4}\right)^{r} \left(\frac{3}{4}\right)^{n-2-r}
\]
tournaments~$T_{n}$ such that $r_{ij} = r$ for any admissible choice of $i$,~$j$, and~$r$.
For, there are $\dbinom{n - 2}{r}$ ways of selecting the $r$~nodes that are to be on a
path of length two from $p_{i}$ to~$p_{j}$ and there are three ways of orienting the
arcs joining $p_{i}$ and~$p_{j}$ to any node that is not on such a path; the remaining
$\dbinom{n}{2} - 2(n - 2)$ arcs may be oriented arbitrarily. Since there are $n(n - 1)$
ordered pairs of distinct nodes $p_{i}$ and~$p_{j}$, it follows that
\[
E(n, \lambda) = n(n - 1) \sum \binom{n - 2}{r}
   \left(\frac{1}{4}\right)^{r} \left(\frac{3}{4}\right)^{n-2-r}
\]
where the sum is over all~$r$ such that $|r - \nf{1}{4}(n - 2)| > \lambda$. We may apply
the De~Moivre-Laplace Theorem to the sum [see Feller (1957; p.~172)]
\index[xauthor]{Feller, W.}%
and conclude that
\[
E(n, \lambda) \sim 2n(n - 1)\bigl(1 - \Phi(X)\bigr),
\]
where $X = 2(2\lambda + 1)\bigl[3(n - 2)\bigr]^{-1/2}$ and $\Phi$~is the normal distribution function.
(We need the hypothesis that $w(n)n^{-1/3} \to 0$ at this step.)

Since $X \to \infty$ as $n \to \infty$, we may use the relation
\[
1 - \Phi(X) \sim \frac{1}{(2\pi)^{1/2} X} e^{-(1/2)X^{2}}
\]
[see Feller (1957; p.~166)]. After further simplification, we find that
\begin{multline*}
E(n, \lambda) \sim \left(\frac{3}{8\pi}\right)^{1/2}
  \frac{(n - 2)^{5/2}}{\lambda} e^{-(8/3)[\lambda^{2}/(n-2)]} \\
  = \bigl\{2\pi \bigl[\log(n - 2) + w(n)\bigr]\bigr\}^{-1/2} e^{-2w(n)}.
\end{multline*}
This completes the proof of the theorem.

\begin{Corollary}[1.]
If $\lambda = \bigl((\nf{3}{4} + \eps) (n - 2) \log(n - 2)\bigr)^{1/2}$, where $\eps$~is any
positive constant, then almost all tournaments~$T_{n}$ have the property that
\[
|r_{ij} - \nf{1}{4}(n - 2)| < \lambda
\]
for every pair of distinct nodes $p_{i}$ and~$p_{j}$.
\end{Corollary}

\Proof. Let $N(k)$ denote the number of tournaments~$T_{n}$ such that the
inequality $|r_{ij} - \nf{1}{4}(n - 2)| > \lambda$ holds for exactly $k$ ordered pairs of distinct
nodes $p_{i}$ and~$p_{j}$. Then
\begin{align*}
0 &\leq \frac{2^{\ebinom{n}{2}} - N(0)}{2^{\ebinom{n}{2}}}
  = \frac{N(1) + N(2) + \dots + N\bigl(n(n - 1)\bigr)}{2^{\ebinom{n}{2}}} \\
  &\leq \frac{0 · N(0) + 1 · N(1) + \dots + n(n - 1) · N\bigl(n(n - 1)\bigr)}{2^{\ebinom{n}{2}}}
  = E(n, \lambda).
\end{align*}
\PageSep{34}

If $w(n) = \nf{4}{3} \eps \log(n - 2)$ in \ThmRef{18}, then
\[
E(n, \lambda) \sim \bigl(2\pi(1 + \nf{4}{3}\eps) \log(n - 2)\bigr)^{-1/2} (n - 2)^{-(8/3)\eps},
\]
and
\[
\lim_{n \to \infty} E(n, \lambda) = 0.
\]
Therefore,
\[
\lim_{n \to \infty} \left(1 - N(0) · 2^{-\ebinom{n}{2}}\right) = 0,
\]
and the corollary is proved.

\begin{Corollary}{2.}
Almost all tournaments have diameter three.
\end{Corollary}

Notice that, if the definition of the diameter of a tournament involved
ordered pairs of distinct nodes only, then we could assert that almost all
tournaments have diameter two.


\Exercises

\Ex{1.} Supply the details omitted in the proof of \ThmRef{18}.

\Ex{2.} Prove that $d \leq n - 1$ for any strong tournament~$T_{n}$ with at least four
nodes.

\Ex{3.} Prove that $d \leq \max \{3, s_{n} - s_{1} + 2\}$, where $s_{n}$~and~$s_{1}$ denote the largest
and smallest score in a strong tournament~$T_{n}$.
\index{Score of a node}%


\Section{13.}{The Powers of Tournament Matrices}

Let $M$ denote a square matrix with nonnegative elements. If there exists
\index{Matrix!of a tournament}%
\index{Matrix!of zeros and ones}%
an integer~$t$ such that $M^{t}$~has only positive elements, then $M$~is \emph{primitive}
\index{Primitive!matrix}%
and the least such integer~$t$ is called the \emph{exponent} of~$M$. We may assume that
\index{Exponent!of a matrix}%
the elements of~$M$ are zeros and ones, and we shall use Boolean arithmetic
in calculating the powers of~$M$. A well-known result due to Wielandt (1950)
\index[xauthor]{Wielandt, H.}%
states that, if the primitive $n$~by~$n$ matrix~$M_{n}$ has exponent~$e$, then
$e < (n - 1)^{2} + 1$.

The \emph{directed graph}~$D_{n}$ defined by a matrix $M_{n} = [a_{ij}]$ of zeros and ones
\index{Directed graph}%
consists of $n$~nodes $p_{1}, p_{2}, \dots, p_{n}$ such that an arc~$\Arc{p_{i}p_{j}}$ goes from $p_{i}$ to~$p_{j}$
if and only if $a_{ij} = 1$. The $(i, j)$-entry in~$M_{n}^{t}$ is one if and only if there exists
a path $P(p_{i}, p_{j})$ of length~$t$ in the graph~$D_{n}$. (We cannot assume, however,
that all the elements in~$P(p_{i}, p_{j})$ are distinct.)

Several authors have used properties of directed graphs to obtain results
on primitive matrices and their exponents. [See, for example, the expository
\PageSep{35}
article by Dulmage and Mendelsohn (1965).] Most of these results are
\index[xauthor]{Dulmage, A. L.}%
\index[xauthor]{Mendelsohn, N. S.}%
unnecessarily weak or complicated when applied to tournament matrices.
Moon and Pullman (1967) showed that some fairly sharp results for
\index[xauthor]{Moon, J. W.}%
\index[xauthor]{Pullman, N. J.}%
primitive tournament matrices could be deduced as simple consequences of
\index{Primitive!tournament}%
\ThmRef{3}. If a tournament matrix~$M_{n}$ has a certain property (for example,
if it is primitive), we shall say that the corresponding tournament~$T_{n}$ has the
same property.

\begin{Theorem}{19.}
If the tournament~$T_{n}$ is irreducible and $n \geq 5$, then $T_{n}$~is
\index{Irreducible tournament}%
primitive; if $d$~and~$e$ denote the diameter and exponent of~$T_{n}$, then
\index{Diameter of a tournament}%
\index{Exponent!of a tournament}%
$d \leq e \leq d + 3$.
\end{Theorem}

\Proof. It is obvious that $d \leq e$ if $T_{n}$~is primitive; therefore, we need
\index{Paths in a tournament}%
only show that there exists a path $P(p, q)$ of length $d + 3$ for any ordered
pair of nodes $p$~and~$q$ of~$T_{n}$. There exists at least one nontrivial path $P(p, q)$,
since $T_{n}$~is irreducible and $n \neq 1$. Let $P_{1}(p, q)$ be the shortest such path and
let $l$ denote its length. Then $3 \leq d - l + 3 \leq n + 2$, since $0 \leq l \leq d \leq n - 1$.
If $3 \leq d - l + 3 \leq n$, then there exists a cycle $P_{2}(p, p)$ of length
$d - l + 3$ by \ThmRef{3}. Then the path $P(p, q) = P_{2}(p, p) + P_{1}(p, q)$ has
length $d + 3$. If $d - l + 3 = n + h$, where $h = 1$ or~$2$, then $3 \leq n + h - 3 \leq n - 1$
since $n \geq 5$. Hence, there exist cycles $P_{3}(p, p)$ and $P_{4}(p, p)$ of
\index{Cycles in a tournament}%
lengths $3$ and $n + h - 3$. The path
\[
P(p, q) = P_{3}(p, p) + P_{4}(p, p) + P_{1}(p, q)
\]
has length $3 + (n + h - 3) + l = d + 3$. This completes the proof of the
theorem.

\begin{Corollary}
A tournament~$T_{n}$ is primitive if and only if $n \geq 4$ and $T_{n}$~is
irreducible.
\end{Corollary}

This result, apparently first stated by Thompson (1958), follows from
\index[xauthor]{Thompson, G. L.}%
\ThmRef{19} and the obvious fact that a primitive tournament must be
irreducible. (It is easily verified that the only primitive tournament with
fewer than five nodes is the strong tournament~$T_{4}$; it has exponent nine.)

\begin{Corollary}
If $T_{n}$ ($n \geq 5$) is a primitive tournament with exponent~$e$,
then $3 \leq e \leq n + 2$.
\end{Corollary}

There are six irreducible tournaments~$T_{5}$ (see the Appendix) and they
realize the exponents four, six, and seven. However, there are no gaps in
the exponent set of larger primitive tournaments.

\begin{Theorem}{20.}
If $3 \leq e \leq n + 2$, where $n \geq 6$, then there exists a primitive
tournament~$T_{n}$ with exponent~$e$.
\end{Theorem}

\Proof. Let $n$ and $k$ be integers such that $3 \leq k \leq n - 1$ and $n \geq 6$.
Consider the tournament~$T_{n}$ defined as follows: The arcs $\Arc{p_{1}p_{n}}$, $\Arc{p_{n}p_{n-1}}$,~\dots,
\PageSep{36}
$\Arc{p_{3}p_{2}}$, $\Arc{p_{2}p_{1}}$ and all arcs $\Arc{p_{j}p_{i}}$ where $k \leq i < j \leq n$ are in~$T_{n}$; the remaining
arcs are all oriented toward the node with the larger subscript. (The structure
of~$T_{n}$ is illustrated in \Figure{5}.) This tournament contains a spanning
cycle so it is irreducible and hence primitive.
\begin{figure}[hbt!]
\centering
\Graphic{fig5}
\FCaption{Figure 5}
\end{figure}

If $k \geq 3$, then it is easy to see that the diameter of~$T_{n}$ is~$k$. The exponent of~$T_{n}$
is not $k$ or~$k + 1$, however, since there are no paths $P(p_{k}, p_{1})$ of these
lengths; neither is it~$k + 2$, since there is no path $P(p_{k+1}, p_{1})$ of this length.
Therefore, the exponent of~$T_{n}$ is~$k + 3$, by \ThmRef{19}. This shows that
there exists a primitive tournament~$T_{n}$ with exponent~$e$ if $6 \leq e \leq n + 2$
and $n \geq 6$.

If $k = 2$ or~$3$ let $T_{n}'$ differ from~$T_{n}$ in that the arc joining $p_{k+1}$ and~$p_{n}$
is oriented toward~$p_{n}$. It is not difficult to verify that $T_{n}'$~has exponent
$k + 2$. Therefore, to complete the proof of the theorem, we need only show
that there exist primitive tournaments~$T_{n}$ with exponent three when $n \geq 6$.
If $n = 6$, let $T_{n}$ be the tournament whose matrix is
\[
\left|
\begin{array}{*{6}{c}}
0 & 1 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 & 1 & 0 \\
1 & 0 & 0 & 1 & 0 & 0 \\
1 & 1 & 0 & 0 & 1 & 0 \\
1 & 0 & 1 & 0 & 0 & 1 \\
0 & 1 & 1 & 1 & 0 & 0
\end{array}
\right|;
\]
if $n > 6$, let $T_{n}$~be the regular tournament~$R_{n}$ that was defined in \SecRef{5}.
A simple and direct argument shows that these tournaments have exponent
three. This completes the proof of the theorem.

The \emph{index} (\emph{of maximum density}) of a square matrix~$M$ is the least integer~$k$
\index{Index!of maximum density of a matrix}%
such that the number of nonzero entries in~$M^{k}$ is maximized. If $M$~is
primitive, then its exponent and index are equal.

Before stating the next result, we observe that every tournament~$T_{n}$ has a
unique decomposition into subtournaments~$T^{i}$ ($i = 1, 2, \dots, l$) such that
% [** TN: List indentation added with author's approval]
\begin{itemize}
\item[(a)] every node in~$T^{(j)}$ dominates every node in~$T^{(i)}$ if $1 \leq i < j \leq l$;

\item[(b)] every subtournament~$T^{(i)}$ is either irreducible or transitive;
\index{Subtournament}%

\item[(c)] no two consecutive subtournaments $T^{(i)}$ and $T^{(i+1)}$ are both transitive.
\end{itemize}
When $T_{n}$~is itself irreducible or transitive, then $l = 1$ and $T_{n} = T^{(1)}$.
\PageSep{37}

\begin{Theorem}{21.}
Let $k = \text{index}(M)$ and $k_{i} = \text{index}(M_{i})$, where $M$~and~$M_{i}$
denote the matrices of the tournament  $T_{n}$ and~$T^{(i)}$. Then
\[
k \leq \max (k_{i} : i = 1, 2, \dots, l).
\]
\end{Theorem}

\Proof. We may suppose that the matrix~$M$ has the form
\[
M = \left|
\begin{array}{*{4}{c}}
M_{1} &       &        & 0 \\
      & M_{2} &        & \\
      &       & \ddots & \\
1     &       &        & M_{l}
\end{array}
\right|,
\]
where the entries above and below the diagonal blocks are zeros and ones,
respectively. Then
\[
M^{t} = \left|
\begin{array}{*{4}{c}}
M_{1}^{t} &           &        & 0 \\
          & M_{2}^{t} &        & \\
          &           & \ddots & \\
1         &           &        & M_{l}^{t}
\end{array}
\right|,
\Tag{(1)}
\]
for $t = 1, 2, \dots$.

Let $|B|$ denote the number of nonzero entries in the matrix~$B$, and let
$J = \{j : T^{(j)} \text{ is transitive}\}$. It follows from~\Eq{(1)} that
\[
\Typo{|M^{T}|}{|M^{t}|}
  = \sum_{j \in J} |M_{j}^{t}| + \sum_{i \not\in J} |M_{i}^{t}| + K,
\]
where the constant~$K$ denotes the number of ones below the diagonal blocks
of~$M$.

If $m = \max (k_{i} : i = 1, 2, \dots, l)$, then $|M_{i}^{t}| \leq |M_{i}^{m}|$ for all $i \not\in J$ if
$t \geq 1$, since every subtournament~$T^{(i)}$ ($i \not\in J$) is either primitive or a $3$-cycle.
Furthermore, $|M_{j}^{t}| \leq |M_{j}^{m}|$ for all $j \in J$ if $t \geq m$, since $|M_{j}^{t+1}| \leq |M_{j}^{t}|$ for
all $j \in J$ if $t \geq 1$. Therefore, $|M^{t}| \leq |M^{m}|$ if $t \geq m$, and the theorem is
proved. Strict equality holds in \ThmRef{21} if no transitive subtournament~$T^{(j)}$
has two or more nodes.

\begin{Corollary}
If $\eta$ denotes the maximum number of nodes in any of the
irreducible subtournaments~$T^{(i)}$, then
\[
k \leq \left\{
\begin{aligned}
&1 \\
&9 \\
&\max(\eta + 2, 9) \\
&\Typo{n + 2}{\eta + 2}
\end{aligned}
\quad\text{if}\quad
\begin{aligned}
&\eta \leq 3, \\
&\eta = 4, \\
&\eta > 4, \\
&\eta \geq 7.
\end{aligned}
\right.
\]
\end{Corollary}

This follows from the second corollary to \ThmRef{19} and the fact that
the irreducible tournaments $T_{1}$, $T_{3}$, and $T_{4}$ have index one, one, and nine.
\PageSep{38}

Since we are using Boolean arithmetic, there are only finitely many
distinct matrices among the powers of a given matrix~$M$ of zeros and ones.
The \emph{index of convergence} and the \emph{period} of~$M$ are the smallest positive
\index{Index!of convergence of a matrix}%
\index{Period of a matrix}%
integers $\gamma = \gamma(M)$ and $\rho = \rho(M)$ such that $M^{\rho+\gamma} = M^{\gamma}$. If $M$~is primitive,
then $\gamma$~equals the exponent and the index of maximum density of~$M$, and
$\rho = 1$.

Let $\alpha$ denote the maximum number of nodes in any of the transitive
subtournaments~$T^{(j)}$ in the decomposition of~$T_{n}$ and let $\beta = \max (k_{i} : i \not\in J)$.
(We adopt the convention that the maximum of an empty set is zero.)

\begin{Theorem}{22.}
If $M$~is the matrix of the tournament~$T_{n}$, then
\[
\gamma(M) = \max(\alpha, \beta) \leq n + 2
\]
and
\[
\rho(M) = \begin{cases}
  3 & \text{if some irreducible subtournament~$T^{(i)}$ has three nodes,} \\
  1 & \text{otherwise.}
\end{cases}
\]
\end{Theorem}

\Proof. It follows from~\Eq{(1)} that
\[
\gamma(M) = \max \bigl(\gamma(M_{i}) : i = 1, 2, \dots, l\bigr).
\]
However,
\[
\alpha = \max \bigl(\gamma(M_{j}) : j \in J\bigr)
\]
and
\[
\beta = \max (k_{i} : i \not\in J) = \max \bigl(\gamma(M_{i}) : i \not\in J\bigr).
\]
This suffices to establish the first part of the theorem; the second part is
obvious.

\begin{Theorem}{23.}
If $n \geq 16$ and $1 \leq k \leq n + 2$, then there exists a tournament~$T_{n}$
with index~$k$.
\end{Theorem}

\Proof. The transitive tournament~$T_{n}$ has index one for all~$n$. In view of
\ThmRef{20}, it remains only to exhibit a tournament~$T_{n}$ with index two
for $n \geq 16$.

If $n \geq 16$, let $n = 3h + r$, where $r = 16$,~$17$, or~$18$, and $h$~is an integer.
Let $T_{n}$~be the tournament that can be decomposed in the manner described
earlier into the subtournaments~$T^{(i)}$ ($i = 1, 2, \dots, h + 2$), where $T^{(i)}$~is the
transitive tournament~$T_{r-7}$, $T^{(2)}$~is the tournament~$T_{7}$ in which $p_{i} \to p_{j}$ if and
only if $j - i$ is a quadratic residue modulo seven, and the tournament~$T^{(i)}$
is a $3$-cycle, for $i = 3, 4, \dots, h + 2$. It is a simple exercise to verify that $T_{n}$~has
index two. (Notice that the $3$-cycles $T^{(i)}$ ($i \geq 3$) have no \Typo{affect}{effect} on the
\PageSep{39}
index of~$T_{n}$; they merely extend the basic construction to tournaments with
an arbitrary number~$n$ of nodes, where $n \geq 16$.)


\Exercises

\Ex{1.} If $s_{n}$ and $s_{1}$ denote the largest and smallest scores in a primitive tournament~$T_{n}$
\index{Score of a node}%
($n \geq 5$), then show that the exponent satisfies the inequality
$e \leq 5 + s_{n} - s_{1}$.

\Ex{2.} Determine the exponents of the primitive tournaments with at most
five nodes.

\Ex{3.} Supply the details omitted in the proofs of Theorems 20~and~23.

\Ex{4.} Verify Equation~\Eq{(1)} in the proof of \ThmRef{21}.

\Ex{5.} Determine whether there exist tournament matrices~$M_{n}$ with index two
when $n < 16$.

\Ex{6.} Prove that almost all tournament matrices~$M_{n}$ are primitive with
exponent three.


\Section{14.}{Scheduling a Tournament}

If $n$~players are to participate in a round-robin tournament~$T_{n}$, then the
\index{Tournament!round-robin}%
problem arises of scheduling the matches between the players. When $n$~is
even, the $\nf{1}{2} n(n - 1)$ matches can be split into $n - 1$ rounds of $\nf{1}{2} n$~matches
each in such a way that every pair of players meet exactly once; when $n$~is
odd, an extra round is needed. We may suppose that $n$~is even, for if $n$~is
odd we may introduce an imaginary player~$P$ and let each player sit out
the round in which he is matched with~$P$. We now outline a simple scheme
Reisz (1859) gave for scheduling the matches of a round-robin tournament
\index[xauthor]{Reisz, M.}%
between an even number~$n$ of players.

First write the pairs $(1, 2)$, $(1, 3)$,~\dots, $(1, n)$ in the first column of an
$n - 1$~by $n - 1$ table. Then, for $i = 2, 3, \dots, n - 1$, write the pairs
$(i, i + 1)$, $(i, i + 2)$,~\dots, $(i, n)$ in the $i$th~column, beginning in the first row
below the row containing $(i - 1, i)$ that does not already contain an~$i$ or
an~$i + 1$, and returning to the top of the column when the bottom is
reached. Whenever this rule would place an entry $(i, j)$ in a position in
a row that already contains an~$i$ or a~$j$, leave the position vacant and
place $(i, j)$ in the next admissible position. When all the pairs $(i, j)$ with
$1 \leq i <j \leq n$ have been entered, then the pairs in the $r$th~row indicate
the matches of the $r$th~round.
\PageSep{40}

The following table illustrates Reisz's construction for the case $n = 8$.
\index[xauthor]{Reisz, M.}%
\[
\begin{array}{*{7}{c}}
(1, 2) &        & (3, 7) & (4, 6) & (5, 8) \\
(1, 3) & (2, 8) &        & (4, 7) & (5, 6) \\
(1, 4) & (2, 3) &        &        & (5, 7) & (6, 8) \\
(1, 5) & (2, 4) & (3, 8) &        &        & (6, 7) \\
(1, 6) & (2, 5) & (3, 4) &        &        &        & (7, 8) \\
(1, 7) & (2, 6) & (3, 5) & (4, 8) \\
(1, 8) & (2, 7) & (3, 6) & (4, 5) \\
\end{array}
\]
The table can be put in the following, more compact, form.
\[
\begin{array}{*{4}{c}}
(1, 2) & (3, 7) & (4, 6) & (5, 8) \\
(1, 3) & (2, 8) & (4, 7) & (5, 6) \\
(1, 4) & (2, 3) & (5, 7) & (6, 8) \\
(1, 5) & (2, 4) & (3, 8) & (6, 7) \\
(1, 6) & (2, 5) & (3, 4) & (7, 8) \\
(1, 7) & (2, 6) & (3, 5) & (4, 8) \\
(1, 8) & (2, 7) & (3, 6) & (4, 5)
\end{array}
\]

This construction matches players $i$~and~$j$, where $1 \leq i, j \leq n - 1$, in
round~$r$ if
\[
i + j \equiv r + 2 \pmod{n - 1},
\]
and it matches players $k$~and~$n$ in round~$r$ if
\[
2k \equiv r + 2 \pmod{n - 1},
\]
for $r = 1, 2, \dots, n - 1$. If the $r + 2$ in these congruences is replaced by~$r$,
the only effect on the schedule is that the rounds are numbered differently.
König (1936, p.~157) gave this construction [see also Freund (1959) and
\index[xauthor]{Freund, J.}%
\index[xauthor]{Konig@König, D.}%
Lockwood (1962)]. Slightly different schemes for generating similar schedules
\index[xauthor]{Lockwood, E. H.}%
have also been given by Lockwood (1936), Kraitchik (1950, p.~230),
\index[xauthor]{Kraitchik, M.}%
and Ore (1963, p.~50). This problem is only a special case of the more general
\index[xauthor]{Ore, O.}%
problem of finding factors of a graph that satisfy certain conditions.

Some more complicated problems on the design of other types of tournaments
where more than two players meet at a time have been treated by
Kraitchik, Watson (1954), Scheid (1960), Gilbert (1961), and Yalavigi
\index[xauthor]{Gilbert, E. N.}%
\index[xauthor]{Scheid, F.}%
\index[xauthor]{Watson, G. L.}%
\index[xauthor]{Yalavigi, C. C.}%
(1963). There are also a number of design problems connected with various
modifications of the method of paired comparisons where not every pair of
\index{Paired comparisons, the method of}%
objects is compared or where there is more than one judge. Certain conditions
on balance and symmetry are usually imposed. Material on these
problems and other references may be found in Kendall (1955), Bose
\index[xauthor]{Bose, R. C.}%
\index[xauthor]{Kendall, M. G.}%
(1956), Tietze (1957), and David (1963).
\index[xauthor]{David, H. A.}%
\index[xauthor]{Tietze, H.}%
\PageSep{41}


\Exercises

\Ex{1.} The following array illustrates for the case $n = 7$ a construction Kraitchik
\index[xauthor]{Kraitchik, M.}%
gives for scheduling a tournament between an odd number of players.
(Players $i$~and~$j$ meet in round~$r$ if the pair~$ij$ occurs in row~$r$ and player~$k$
sits out round~$r$ if $k$~occurs in the $r$th~row of column~$1$.)
\[
\begin{array}{*{4}{c}}
1 & 72 & 63 & 54 \\
5 & 46 & 37 & 21 \\
2 & 13 & 74 & 65 \\
6 & 57 & 41 & 32 \\
3 & 24 & 15 & 76 \\
7 & 61 & 52 & 43 \\
4 & 35 & 26 & 17
\end{array}
\]
Extend this construction to the general case. Notice that a schedule for a
tournament on $8$~players can be obtained from this by writing an~$8$ next to
the numbers in the first column. Is the resulting schedule different from the
schedule obtained by Reisz's construction?
\index[xauthor]{Reisz, M.}%

\Ex{2.} Describe how the diagram in \Figure{6} can be used to schedule a tournament
between seven players. How can the construction be generalized?
\begin{figure}[hbt!]
\centering
\Graphic[1.75in]{fig6}
\FCaption{Figure 6}
\end{figure}

\Ex{3.} A tennis match is played between two teams. Each player plays one
\index{Teams}%
or more members of the other team. Any two members of the same team
have exactly one opponent in common and no two members of the same
team play all the members of the other team between them. Prove that, if
two players on different teams do not play each other, then they have the
same number of opponents. Deduce from this that all the players have the
same number of opponents and that, if this number is~$n$, then there are
$n^{2} - n + 1$ players on each team. [Boyd (1961).]
\index[xauthor]{Boyd, A. V.}%

\Ex{4.} A tennis match is played between two teams $A$~and~$B$. Each member of~$A$
plays at least one member of~$B$, and no member of~$B$ plays every member
\PageSep{42}
\index{Ranking problems}%
of~$A$. Prove that there exist players $a_{1}$,~$a_{2}$, $b_{1}$~and $b_{2}$ such that $a_{i}$~and~$b_{j}$ have
played each other if and only if $i = j$ for $i, j = 1, 2$. [See McKay (1966).]
\index[xauthor]{MacKay@McKay, J. H.}%

\Ex{5.} The members of a bridge club participated over a period of several days
in a tournament that satisfied the following conditions:
\begin{itemize}
\item[(a)] Each pair of members appeared together at exactly one day's play.

\item[(b)] For any two days' play there was exactly one member who played on
both days.

\item[(c)] At least four players were scheduled to play on each day.

\item[(d)] The president, vice president, secretary, and treasurer were the only
members scheduled to play on the first day.
\end{itemize}
%[** TN: Block indentation continues in the original]
How many members did the club have, how many days did the
play last, and how could the tournament have been scheduled?
[Mendelsohn (1953).]
\index[xauthor]{Mendelsohn, N. S.}%

\Ex{6.} The nine members of a whist club want to arrange a tournament among
themselves. There are nine chairs in their club room, four at each of two
tables and an extra one for the person who sits out each round. Is it possible
to arrange the tournament so that all three of the following conditions are
satisfied?
\begin{itemize}
\item[(a)] Every two members play together as partners once.

\item[(b)] Every two members play together as opponents twice.

\item[(c)] Each member spends one round in each chair.
\end{itemize}
Consider the analogous problem for clubs with five and thirteen members.
[See Watson (1954) and Scheid (1960).]
\index[xauthor]{Scheid, F.}%
\index[xauthor]{Watson, G. L.}%

\Ex{7.} When is it possible to schedule a round robin tournament between $n$~chess
players in such a way that all players alternate in playing the white
and black pieces? [Tietze (1957).]
\index[xauthor]{Tietze, H.}%


\Section{15.}{Ranking the Participants in a Tournament}

The simplest way of ranking the participants in a tournament is according
to the number of games they have won. This, however, will lead to ties,
except when the tournament is transitive. One could consider the subtournament
determined by all players who have the same score and then try to
rank these players on the basis of their performance within this subtournament,
but this will not necessarily remove all the ties. In fact, there is no
reason to expect that all ties can be removed. For example, if the result of a
tournament between three players is a $3$-cycle, the most plausible conclusion
is that they are of equivalent strength.

Various schemes for ranking and comparing the participants in a tournament
have been proposed, none of which is entirely satisfactory. Zermelo\index[xauthor]{Zermelo, E.}
\PageSep{43}
(1929) developed a method based on the maximum likelihood principle.
\index{Maximum likelihood principle}%
[The same method was rediscovered later by Bradley and Terry (1952) and
\index[xauthor]{Bradley, R. A.}%
\index[xauthor]{Terry, M. E.}%
Ford (1957).] We shall describe his procedure only for the case of a round-robin
\index[xauthor]{Ford, L. R., Jr.}%
tournament, although there is no difficulty in extending it to more
general situations where the number of encounters between various players
is arbitrary.

We assume that each player~$p_{i}$ has a positive strength~$w_{i}$ such that in the
\index{Strength of a player}%
encounter between $p_{i}$~and~$p_{j}$ the probability that $p_{i}$~will win is given by
$w_{i}/(w_{i} + w_{j})$. (We assume the strengths are normalized so that their sum
is one.)

If we assume that the outcomes of the various matches are independent,
then the probability that a particular tournament~$T_{n}$ is obtained is given by
the formula
\begin{align*}
Pr(T_{n}: w_{1}, w_{2}, \dots, w_{n})
  &= \prod_{i<j} \left(\frac{w_{i}}{w_{i} + w_{j}}\right)^{a_{ij}} \left(\frac{w_{j}}{w_{i} + w_{j}}\right)^{a_{ji}} \\
  &= \frac{\prod_{i} w_{i}^{s_{i}}}{\prod_{i<j} (w_{i} + w_{j})},
\end{align*}
where $a_{ij} = 1$ or~$0$ according as $p_{i}$~does or does not beat~$p_{j}$ and $s_{i}$~is the
number of matches won by~$p_{i}$. The problem now is to determine the value
of the~$w_{i}$'s that maximize the probability of obtaining the tournament~$T_{n}$.

If the players can be split into two nonempty classes $A$~and~$B$ such that
every player in~$A$ beats every player in~$B$, then it is obvious that every
player in~$A$ should be ranked ahead of every player in~$B$, but there exist no
nonzero maximizing values of the~$w_{i}$'s for the players in~$B$ (see Exercise~1).
Therefore, we assume that $T_{n}$~is irreducible.
\index{Irreducible tournament}%

If $T_{n}$~is irreducible, then it can be shown that there exists a unique set of
positive strengths $(w_{i} : i = 1, 2, \dots, n)$ that maximize $Pr(T_{n} : w_{1}, w_{2}, \dots, w_{n})$.
Upon taking partial derivatives of the logarithm of the likelihood
function, we see that the maximizing strengths satisfy the equations
\[
\frac{s_{i}}{w_{i}} = {\sum_{j}}' \frac{1}{w_{i} + w_{j}}
\Tag{(1)}
\]
and
\[
\sum_{i=1}^{n} w_{i} = 1,
\Tag{(2)}
\]
where for each~$i$ the first sum is over all $j$ not equal to~$i$.

There is no direct way of solving these equations in general, but an
iterative scheme can be used. For any trial solution, say $w_{i}^{(0)} = 1/n$, let
\[
w_{i}^{(1)} = \frac{s_{i}}{{\sum_{j}}' (w_{i}^{(0)} + w_{j}^{(0)})^{-1}},
\]
\PageSep{44}
for $i = 1, 2, \dots, n$. If this procedure is repeated, the trial solutions will
converge (slowly) to a solution of~\Eq{(1)} if $T_{n}$~is irreducible [see David (1963)].
\index[xauthor]{David, H. A.}%
This solution may not satisfy~\Eq{(2)}, but since the equations are homogeneous
this can be remedied by multiplying through by the appropriate constant.

We now illustrate Zermelo's method on the irreducible tournament~$T_{6}$ in
\index[xauthor]{Zermelo, E.}%
\Figure{7}. We would expect players $3$,~$4$, and~$5$ to have equal strengths,
but it is not obvious what the relative strengths of the other players should
be.

{\Loosen We find that $w_{1} = w_{2} = .38735$, $w_{3} = w_{4} = w_{5} = .06580$, and $w_{6} = .02689$
is an approximate solution to Equations \Eq{(1)}~and~\Eq{(2)}. Notice that this
gives exactly the same ranking as would be obtained simply by considering
the score of each player. This is always the case for ordinary round-robin
\index{Tournament!round-robin}%
tournaments, as pointed out by Zermelo and Ford (see Exercise~2). Hence,
\index[xauthor]{Ford, L. R., Jr.}%
all we really achieve in this case are the maximum likelihood estimates for
the relative strengths of the players.}

The preceding method of ranking the players in a tournament and assigning
their relative strengths takes into account only the number of matches
won by each player. A method due to Wei (1952) and Kendall (1955) also
\index[xauthor]{Kendall, M. G.}%
\index[xauthor]{Wei, T. H.}%
takes into account the quality of the opponents in the matches won by each
player. For example, it seems plausible that Player~$1$ is stronger than Player~$2$
in the tournament in \Figure{7}, even if they do have the same number of
\begin{figure}[hbt!]
\centering
\Graphic[1.75in]{fig7}
\FCaption{Figure 7}
\end{figure}
wins, because Player~$1$ beats Player~$2$. Also, the relative position of the sixth
player is not clear, because although he wins only one match he wins
against one of the strongest players. We now illustrate the Kendall-Wei
method, or rather a slight simplification of it, for this tournament.

First assign to each player the number of matches he has won. This
gives the initial strength vector
\[
w_{1} = (4, 4, 2, 2, 2, 1).
\]
\PageSep{45}

Next assign to each player the sum of the initial strengths of the players
he has beaten. This gives the second vector
\[
w_{2} = (10, 7, 3, 3, 3, 4).
\]

The next four strength vectors, defined in a similar way, are
\begin{align*}
w_{3} &= (16, 13, 7, 7, 7, 10), \\
w_{4} &= (34, 31, 17, 17, 17, 16), \\
w_{5} &= (82, 67, 33, 33, 33, 34), \\
w_{6} &= (166, 133, 67, 67, 67, 82).
\end{align*}
Notice that the first player has the largest strength now and that the sixth
player has the third largest strength. It seems unlikely that further repetitions
of this process will alter the relative positions of the players.

In the general case, the $i$th~strength vector is given by
\[
w_{i} = M^{i}e,
\Tag{(3)}
\]
where $M$~is the matrix of the tournament being considered and $e$~is a column
\index{Matrix!of a tournament}%
vector of~$1$'s. It follows from a theorem of Frobenius (1912) that, if the
\index[xauthor]{Frobenius, G.}%
matrix~$M$ is primitive, then
\index{Primitive!matrix}%
\[
\lim_{i \to \infty} \left(\frac{M}{\lambda}\right)^{i} e= y,
\Tag{(4)}
\]
where $\lambda$~is the unique positive characteristic root of~$M$ with the largest
absolute value and $y$~is a positive characteristic vector of~$M$ corresponding to~$\lambda$.
Therefore, in view of the corollary to \ThmRef{19}, the normalized vector~$y$
can be taken as the vector of relative strengths of the players in~$T_{n}$ if $T_{n}$~is
irreducible and $n \geq 4$.

The characteristic polynomial of the matrix of the tournament in \Figure{7}
is $\lambda^{6} - 5\lambda^{3} - 6\lambda^{2} - 6\lambda - 2$. The dominant root of this is approximately
$\lambda = 2.1106295$. Upon calculating the corresponding characteristic vector
and normalizing, we find that the relative strengths assigned to the players
by this method are approximately $w_{1} = .27899$, $w_{2} = .23179$, $w_{3} = w_{4} = w_{5} = .11901$,
and $w_{6} = .13218$. It would seem that these strengths are
somewhat more realistic than those obtained by the maximum likelihood
method.

Kendall's original version also applies to more general situations. He
\index[xauthor]{Kendall, M. G.}%
first assigns to each player the number of matches he has won, plus~$\nf{1}{2}$ for
% [** TN: [sic] "tieing"]
tieing with himself. At each subsequent stage of adjusting the strengths, he
assigns to each player $\nf{1}{2}$~his old strength plus the sum of the strengths of the
players he beats. Thompson (1958) showed that the $\nf{1}{2}$ is arbitrary and
\index[xauthor]{Thompson, G. L.}%
that, if the $\nf{1}{2}$ is replaced by~$r$, then the final ranking is independent of~$r$
if $r > 0$. If this method is used, then the matrix~$M$ in Equations \Eq{(3)}~and~\Eq{(4)}
is replaced by $M + rI$. This is still somewhat arbitrary and it seems that the
only reason for the~$rI$ is to ensure that the matrix used in \Eq{(3)}~and~\Eq{(4)} is
\PageSep{46}
primitive. But the matrix of an irreducible tournament~$T_{n}$ is itself primitive
\index{Irreducible tournament}%
if $n \geq 4$ so that there is no need for the~$rI$ in this case, especially since it has
no effect upon the final relative strength vector obtained.

Katz (1953) and Thompson (1958) proposed another method for comparing
\index[xauthor]{Katz, L.}%
\index[xauthor]{Thompson, G. L.}%
\index[xauthor]{Thompson, W. A., Jr.}%
\index{Ranking problems}%
the participants in a tournament. If $M$~is the matrix of the tournament,
they let the vector of relative strengths be proportional to
\[
(M + rM^{2} + r^{2} M^{3} + \dots) e = M(I - rM)^{-1} e,
\]
{\Loosen where $r$~is some positive constant for which the series converges (that is to
say, $r < \lambda^{-1}$ where $\lambda$~is the dominant characteristic root of~$M$). Thompson
shows that the normalized relative strengths given by the vector
$M(I - \lambda^{-1}M)^{-1} e$ are the same as those given by the Kendall-Wei method.}
\index[xauthor]{Kendall, M. G.}%
\index[xauthor]{Wei, T. H.}%

Additional material on the problem of ranking a collection of objects
on the basis of binary comparisons between them may be found, for
example, in Brunk (1960), Slater (1961), Hasse (1961), \Typo{Buhlman}{Buhlmann} and
\index[xauthor]{Brunk, H. D.}%
\index[xauthor]{Buhlmann@\Typo{Buhlman}{Buhlmann}, N. H.}%
\index[xauthor]{Hasse, M.}%
\index[xauthor]{Slater, P.}%
Huber (1963), Huber (1963b), Gridgeman (1963), David (1963), Thompson
\index[xauthor]{David, H. A.}%
\index[xauthor]{Gridgeman, N. T.}%
\index[xauthor]{Huber, P.}%
and Remage (1964), and Kadane (1966). Good (1955) has treated a related
\index[xauthor]{Good, I. J.}%
\index[xauthor]{Kadane, J. B.}%
\index[xauthor]{Remage, R., Jr.}%
problem concerning the grading of chess players.


\Exercises

\Ex{1.} Show that there is no positive solution of the equation $\sum_{i=1}^{n} w_{i} = 1$ that
maximizes $Pr(T_{n} : w_{1}, w_{2}, \dots, w_{n})$ if the tournament~$T_{n}$ is reducible.
\index{Reducible tournament}%

\Ex{2.} Suppose $(w_{i} : i = 1, 2, \dots, n)$ satisfies Equations \Eq{(1)}~and~\Eq{(2)}. Show that
if $s_{i} < s_{j}$, then
\[
0 < {\sum_{k}}' \frac{w_{j}}{w_{j} + w_{k}}
  - {\sum_{k}}' \frac{w_{i}}{w_{i} + w_{k}}
  = (w_{j} - w_{i}) \sum_{k} \frac{w_{k}}{(w_{i} + w_{k})(w_{j} + w_{k})},
\]
so $w_{i} < w_{j}$ also.

\Ex{3.} Thompson raises the following questions about his method of ranking
the participants in a tournament:
\begin{itemize}
\item[(a)]  Does the final ranking depend on the choice of~$r$?

\item[(b)]  Can the method lead to a tie between an even number of players?
\end{itemize}

\Ex{4.} Slater (1961) suggests that the participants in a tournament~$T_{n}$ be ranked
in such a way as to minimize the number of \emph{upsets}, that is, the number of
\index{Upsets in a tournament}%
matches in which the losing player is ranked ahead of the winning player.
If the players are ranked in this way, does this imply that the players are
ranked according to the number of matches they have won?

\Ex{5.} The relative weakness vector of the participants in a tournament~$T_{n}$ may
be defined as the vector obtained by applying the Kendall-Wei method to
the transpose of the matrix of~$T_{n}$. Ramanujacharyula (1964) suggests that
\index[xauthor]{Ramanujacharyula, C.}%
\PageSep{47}
the participants be ranked according to the quotients of their relative
strengths and weaknesses. Apply this method to the tournament~$T_{6}$ considered
in this section.


\Section[The Minimum Number of Comparisons Necessary]
{16.}{The Minimum Number of Comparisons
Necessary to Determine a Transitive Tournament}
\index{Transitive tournament}%

If one knows in advance that a dominance relation defined on a set of~$n$
\index{Dominance relation}%
\index{Ranking problems}%
($n \geq 4$) objects is transitive, then it is not necessary to compare all the
$\dbinom{n}{2}$ pairs of the objects in order to determine the transitive tournament~$T_{n}$
that this relation defines. (For example, one might use a balance scale
to rank according to weight $n$~objects, no two of which had the same weight.)
The number of comparisons necessary to rank the $n$~objects will depend on
the order in which they are compared, in general. We now give bounds
for~$M(n)$, the least integer~$M$ such that at most $M$ comparisons are necessary
to rank $n$~objects according to some transitive relation.

\begin{Theorem}{24.}
If $n = 2^{t-1} + r$, where $0 \leq r < 2^{i-1}$ and $n > 2$, then
\[
1 + \bigl[\log_{2} (n!)\bigr] \leq M(n)
  \leq 1 + nt - 2^{t}
  \leq 1 + n[\log_{2} n].
\]
\end{Theorem}

\Proof. The lower bound follows immediately from the observation
[see Ford and Johnson (1959)] that, if $M$~comparisons will always suffice
\index[xauthor]{Ford, L. R., Jr.}%
\index[xauthor]{Johnson, S. M.}%
to rank $n$~objects, then it must be that $2^{M} \geq n!$, since after $M$~comparisons
we can distinguish at most $2^{M}$~alternatives and we must be able to distinguish
between the $n!$~possible rankings.

The upper bound follows from a construction due to Steinhaus (1950).
\index[xauthor]{Steinhaus, H.}%
Compare any two of the objects at the outset. If $k$~objects have been
ranked relative to each other, compare any $(k + 1)$st object~$A$ with one
of the objects, say~$B$, ranked in the middle of these $k$ objects. Next, compare~$A$
with one of the objects ranked in the middle of those objects already
ranked above or below~$B$, according as $A$~dominates~$B$ or $B$~dominates~$A$.
If this procedure is repeated, the position of~$A$ relative to the $k$~objects
already ranked will be determined after $1 + [\log_{2} k]$ comparisons.

Thus, if $n = 2^{t-1} + r$, where $0 \leq r < 2^{i-1}$, the total number of matches
necessary to rank the $n$~objects is at most
\begin{multline*}
(n - 1) + [\log_{2} 2] + [\log_{2} 3] + \dots + \bigl[\log_{2} (n - 1)\bigr] \\
\begin{aligned}
&= (n - 1) + 2(1) + 2^{2}(2) + 2^{3}(3) + \dots + 2^{t-2} (t - 2) + r(t - 1) \\
&= (n - 1) + 2\bigl[1 + (t - 3) 2^{t-2}\bigr] + (n - 2^{t-1})(t - 1) \\
&= 1 + nt - 2^{t}
  \leq 1 + n[\log_{2} n].
\end{aligned}
\end{multline*}
\PageSep{48}

Ford and Johnson (1959) have given a somewhat sharper upper bound by
\index[xauthor]{Ford, L. R., Jr.}%
\index[xauthor]{Johnson, S. M.}%
means of a more efficient and complicated procedure. The top two rows of
\Table{4} give the number of comparisons sufficient to rank $n$~objects by the
procedures of Steinhaus, and Ford and Johnson, for $n \leq 13$; the bottom
\index[xauthor]{Steinhaus, H.}%
row gives the lower bound for the number of comparisons necessary.
\begin{table}[hbt!]
\TCaption{Table 4. $M(n)$, the minimum number of comparisons
necessary to determine a transitive tournament~$T_{n}$.}
\[
\begin{array}{*{13}{c}}
n = 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 \\
\hline
\Strut
0 & 1 & 3 & 5 & 8 & 11 & 14 & 17 & 21 & 25 & 29 & 33 & 37 \\
0 & 1 & 3 & 5 & 7 & 10 & 13 & 16 & 19 & 22 & 26 & 30 & 34 \\
0 & 1 & 3 & 5 & 7 & 10 & 13 & 16 & 19 & 22 & 26 & 29 & 33
\end{array}
\]
\end{table}

%[** TN: No paragraph indentation in the original]
Ford and Johnson conjecture that $M(n)$~equals the number of comparisons
required by their procedure. Steinhaus (1963) conjectures that $M(n) = 1 + \bigl[\log_{2} (n!)\bigr]$
if $n > 2$ and discusses recent progress on this conjecture. The
first unsettled case is $n = 12$.

Kislicyn (1963) has given an asymptotic bound for the least mean number
\index[xauthor]{Kislicyn, S. S.}%
of comparisons necessary to rank $n$~objects.

Let $M_{k}(n)$ denote the least integer~$M$ such that at most $M$ comparisons
are necessary to determine the $k$th~highest ranking object in a set of $n$~objects.
It is not difficult to see that $M_{1}(n) = n - 1$. More generally,
Kislicyn (1964) has shown that
\[
M_{k}(n) \leq (n - 1) + \sum_{i=1}^{k-1} \bigl[\log_{2} (n - i)\bigr]
\Tag{(1)}
\]
if $k \leq \bigl[\nf{1}{2}(n + 1)\bigr]$. The special cases $k = 1$ and~$2$ of this result were obtained
earlier by Schreier (1932) and Slupecki (1951).
\index[xauthor]{Schreier, J.}%
\index[xauthor]{Slupecki, J.}%


\Exercises

\Ex{1.} Prove inequality~\Eq{(1)} for $k = 1, 2$.

\Ex{2.} A \emph{knock-out tournament} between $n$~players may be conducted as follows:
\index{Tournament!knock-out}%
If $n = 2^{t} + r$, where $1 \leq r \leq 2^{t}$, then $2r$~players are matched off in the first
round. The $2^{t}$~players not yet defeated are then matched off in the second
round. In the $i$th~round, the $2^{t+2-i}$~players not yet defeated are matched off.
The undefeated winner emerges after $t + 1$~rounds. What is the probability
that two given players will be matched against each other in the course of a
random knock-out tournament? [Hartigan (1966) has dealt with some
\index[xauthor]{Hartigan, J. A.}%
problems of estimating the relative ranks of participants in a knock-out
tournament on the basis of the known outcomes of the matches that are
played. David (1963) gives more material on this type of tournament.]
\index[xauthor]{David, H. A.}%
\PageSep{49}

\Ex{3.} A certain tournament consisted of five rounds, a semifinal, and a final.
In the five rounds, there were $8$,~$6$, $0$, $1$, $0$~byes, respectively. If there were $100$~entrants,
then how many matches were played? [Chisholm (1948).]
\index[xauthor]{Chisholm, J. S. R.}%

\Ex{4.} Prove the identity
\[
\left[\frac{n}{2}\right]
  + \left[\frac{n + 1}{4}\right]
  + \left[\frac{n + 3}{8}\right]
  + \left[\frac{n + 7}{16}\right]
  + \dots = n - 1.
\]
Hint: Consider a knock-out tournament on $n$~players in which as many
players as possible are matched off in each round and the losers of these
matches are eliminated from further play. [Mendelsohn (1949).]
\index[xauthor]{Mendelsohn, N. S.}%


\Section{17.}{Universal Tournaments}
\index{Universal tournament}%

A tournament~$T_{N}$ is said to be \emph{$n$-universal} ($n \leq N$) if every tournament~$T_{n}$
is isomorphic to some subtournament of~$T_{N}$. For every positive integer~$n$, let
\index{Isomorphic tournaments}%
\index{Subtournament}%
$\lambda(n)$ denote the least integer~$N$ for which there exists an $n$-universal tournament~$T_{N}$.
(It is clear that $\lambda(n)$~is finite, since any tournament that contains
disjoint copies of all the different tournaments~$T_{n}$ is $n$-universal.)

%[** TN: Added period after "25" for cross-referencing]
\begin{Theorem}{25.}
\[
2^{(1/2)(n-1)} \leq \lambda(n)
  \leq \begin{cases}
    n · 2^{(1/2)(n-1)} & \text{if $n$~is odd,} \\
    \dfrac{3}{2\sqrt{2}} n · 2^{(1/2)(n-1)} & \text{if $n$~is even.}
  \end{cases}
\]
\end{Theorem}

\Proof. There are at least $2^{\ebinom{n}{2}}/n!$ different tournaments~$T_{n}$ (the labelings
assigned to the nodes are immaterial to the problem). Hence, if $T_{N}$~is
$n$-universal, it must be that
\[
2^{\ebinom{n}{2}}/n! \leq \binom{N}{n} \leq \frac{N^{n}}{n!},
\]
since different tournaments~$T_{n}$ must be isomorphic to different subtournaments
with $n$~nodes of~$T_{N}$. The lower bound for~$\lambda(n)$ is an immediate
consequence of this inequality.

To obtain an upper bound for~$\lambda(n)$, we proceed as follows. Let $R_{n}$~be any
tournament with nodes $y_{1}, y_{2}, \dots, y_{n}$ and let $Y_{i} = \{y_{k} : y_{k} \to y_{i}\}$ for each~$i$.
Construct a tournament~$H$ whose nodes~$q_{i,A}$ are in one-to-one correspondence
with the ordered pairs~$(i, A)$ where $A \subset Y_{i}$. If $A \subset Y_{i}$, $B \subset Y_{j}$, and
$y_{i} \to y_{j}$, then $q_{1,A} \to q_{j,B}$ in~$H$ if $y_{i} \in B$, and $q_{j,B} \to q_{i, A}$ if $y_{i} \not\in B$. The arcs
joining nodes of the type $q_{i,A_{1}}$ and $q_{i,A_{2}}$ may be oriented arbitrarily. (The
tournament obtained when $R_{n}$~is a $3$-cycle is illustrated in \Figure{8}. The
notation used should be obvious.)
\PageSep{50}
\begin{figure}[hbt!]
\centering
\Graphic{fig8}
\FCaption{Figure 8}
\end{figure}

We now show that $H$~is $n$-universal. If the nodes of a tournament~$T_{n}$ are
$p_{1}, p_{2}, \dots, p_{n}$, we set $f(p_{i}) = q_{i,A(i)}$, where $A(i) = \{y_{k} : y_{k} \to y_{i} \text{ in } R_{n} \text{ and }
p_{k} \to p_{i} \text{ in } T_{n}\}$. The subtournament of~$H$ determined by the nodes~$f(p_{i})$ is
isomorphic to~$T_{n}$, since, if $y_{i} \to y_{j}$ in~$R_{n}$, then
\[
(p_{i} \to p_{j}) \implies \bigl(p_{i} \in  A(j)\bigr)
  \implies (q_{i,A(i)} \to q_{j,A(j)})
  \implies \bigl(f(p_{i}) \to f(p_{j})\bigr)
\]
and
\[
(p_{j} \to p_{i}) \implies \bigl(p_{i} \not\in  A(j)\bigr)
  \implies (q_{j,A(j)} \to q_{i,A(i)})
  \implies \bigl(f(p_{j}) \to f(p_{i})\bigr).
\]

Therefore,
\[
\lambda(n) \leq (\text{number of nodes of~$H$})
  = 2^{|Y_{1}|} + 2^{|Y_{2}|} + \dots + 2^{|Y_{n}|}.
\]
To minimize this sum, let $R_{n}$ be the regular tournament defined in \SecRef{5}
\index{Regular tournament}%
for which
\begin{align*}
|Y_{1}| &= \dots = |Y_{n}| = \nf{1}{2}(n - 1),\quad\text{if $n$~is odd,}
\intertext{and}
|Y_{1}| &= \dots = |Y_{(1/2)n}| = \nf{1}{2}n \\
|Y_{(1/2)n+1}| &= \dots = |Y_{n}| = \nf{1}{2}(n - 2),\quad\text{if $n$~is even.}
\end{align*}
Hence,
\[
\lambda(n) \leq n · 2^{\Chg{1/2}{(1/2)}(n-1)}\quad\text{if $n$ is odd,}
\]
and
\[
\lambda(n) \leq \nf{1}{2} n · 2^{(1/2)n} + \nf{1}{2} n · 2^{(1/2)(n-2)}
  = \frac{3}{2\sqrt{2}} n · 2^{(1/2)(n-1)},\quad\text{if $n$~is even.}
\]

This completes the proof of the theorem.
\PageSep{51}
\index[xauthor]{Debruijn@de Bruijn, N. G.}%

Rado (1964) and de~Bruijn were the first to study universal graphs;
\index[xauthor]{Rado, R.}%
they restricted their attention to infinite graphs. \ThmRef{25} is closely
related to a result Moon (1965) obtained for ordinary finite graphs.
\index[xauthor]{Moon, J. W.}%


\Exercises

\Ex{1.} Verify that the graph~$H$ in \Figure{8} is $3$-universal.

\Ex{2.} Determine the exact values of~$\lambda(n)$ for $n \leq 4$.

\Ex{3.} Obtain a result for directed graphs that is analogous to \ThmRef{25}.


%[** TN: "Expressing Oriented Graphs as Union of Bilevel Graphs" in the original]
\Section[Oriented Graphs as the Union of Bilevel Graphs]
{18.}{Expressing Oriented Graphs as the Union
of Bilevel Graphs}
\index{Oriented graph}%
\index{Directed graph}%

The rather complicated results in this section will be used to prove other
results in the next two sections. In order to prove these results for tournaments,
it is necessary to prove them for oriented graphs in general. (Recall
that an oriented graph differs from a tournament in that not all pairs of
\index{Tournament!bipartite}%
nodes need be joined by an arc.)

Recall that an $a$~by~$b$ bipartite tournament consists of two disjoint sets
$A$~and~$B$ containing $a$~and $b$ nodes, respectively, such that each node in~$A$
is joined by an arc to each node in~$B$. We shall consider special $a$~by~$b$
bipartite graphs $H(a, b)$ in which all the arcs are similarly oriented, say,
from the nodes in~$A$ to the nodes in~$B$. A \emph{bilevel graph}~$H$ is any oriented graph
\index{Bilevel graph}%
that can be expressed as the union of disjoint special graphs $H(a_{i}, b_{i})$; the
graphs $H(a_{i}, b_{i})$ are called the \emph{components} of~$H$. (We admit the possibility
\index{Components of a bilevel graph}%
that one of the node sets of one of the components of~$H$ is empty.) The
structure of a typical bilevel graph with four components is illustrated in
\Figure{9}.
\begin{figure}[hbt!]
\centering
\Graphic[3in]{fig9}
\FCaption{Figure 9}
\end{figure}

We are going to prove that any oriented graph with $n$~nodes can be
expressed as the union of $cn/\log n$~or fewer arc-disjoint bilevel graphs, where
$c$~is a certain constant. First, however, we prove several lemmas.

The (\emph{total}) \emph{degree} of a node in an oriented graph is the total number of
\index{Degree of a node}%
arcs incident with it. The following lemma and some subsequent statements
\PageSep{52}
are valid only for large values of~$n$; when this is clear from the context, we
shall not mention this qualification explicitly.

\begin{Lemma}[1.]
Let $G$~be an oriented graph with $n$~nodes and $e$~arcs such that
\[
\frac{n^{2}}{2^{2r+4}} < e \leq \frac{n^{2}}{2^{2r+1}}
\]
for some nonnegative integer~$r$ where $\dfrac{\log n}{3(r + 3)} \geq 1$. Then $G$~contains a
special bipartite tournament $H(a, b)$ with $a = [\sqrt{n}]$ and $b = \left[\dfrac{\log n}{3(r + 3)}\right]$
where the degrees of the nodes of~$A$ do not exceed~$\dfrac{16n}{2^{r}}$ in~$G$.
\end{Lemma}

\Proof. There can be at most $n/2^{r+4}$ nodes whose degrees exceed $16n/2^{r}$ in~$G$,
for otherwise there would be too many arcs in~$G$. Hence, there are fewer
than $n^{2}/2^{2r+9}$ arcs joining two such nodes. If we disregard such arcs, there
will remain more than $n^{2}/2^{2r+5}$ arcs with the property that at least one of the
nodes they join has degree at most~$16n/2^{r}$. Let $y_{1}, y_{2}, \dots, y_{t}$ denote the
nodes whose degrees in~$G$ were at most $16n/2^{r}$ originally. Then it must be
that
\[
n \left(1 - \frac{1}{2^{r+4}}\right) \leq t \leq n.
\]

If $s_{i}$ denotes the number of arcs oriented away from~$y_{i}$, then we may
assume that
\[
\sum_{i=1}^{t} s_{i} \geq \frac{n^{2}}{2^{2r+6}}.
\]
(If this inequality is not satisfied, then in the special graph $H(a, b)$ all the
arcs will be oriented from the nodes of~$B$ to the nodes of~$A$, but this will not
affect the result.)

The number~$N$ of sets~$B$ of $b$~nodes for which there exists some node~$y_{i}$
($1 \leq i \leq t$) such that $y_{i}$~dominates every node of~$B$ is given by the formula
\[
N = \sum_{i=1}^{t} \binom{s_{i}}{b}.
\]
{\Loosen (If there are $h$~nodes~$y_{i}$ that dominate every node of~$B$, then $B$~is counted $h$~times
in this sum.) This sum will be minimized when $t = n$ and $s_{i} = \bigl[n/(2^{\Erratum{2r+5}{2r+6}})\bigr]$.
If we can show that}
\[
N \geq n \binom{\left[\dfrac{n}{2^{2r+6}}\right]}{b}
  > (a - 1) \binom{n}{b},
\Tag{(1)}
\]
then it will follow from the box principle that at least one set~$B$ is counted
at least $a$~times; this means that there exists a set~$A$ of $a$~nodes (chosen from
\PageSep{53}
the nodes whose degrees in~$G$ were at most $16n/2^{r}$ originally) and a set~$B$
of $b$~nodes such that every node in~$A$ dominates every node in~$B$.

It is not difficult to show (see Exercise~1) that inequality~\Eq{(1)} does in fact
hold when
\[
a = [\sqrt{n}]\quad\text{and}\quad
b = \left[\frac{\log n}{3(r + 3)}\right].
\]
This completes the proof of Lemma~1.

\begin{Lemma}[2.]
Let $G$ be an oriented graph with $n$~nodes and $e$~arcs such that
\[
\frac{n^{2}}{2^{2r+3}} < e \leq \frac{n^{2}}{2^{2r+1}}
\]
for some nonnegative integer~$r$ such that $r \leq 4\log\log n$. Then $G$~contains a
bilevel graph with at least $(n \log n)/(r + 3) 2^{r+11}$ arcs.
\end{Lemma}

\Proof. We first disregard all arcs that join two nodes whose degrees
exceed $16n/2^{r}$; as before, there are fewer than $n^{2}/2^{2r+9}$ such arcs so there
certainly remain more than $n^{2}/2^{2r+13/4}$ arcs. Let $H_{1}$~be the special subgraph
of~$G$ described in Lemma~1. Since the degrees of the nodes in $A_{1}$ and~$B_{1}$
are at most $16n/2^{r}$ and $n - 1$, respectively, it follows that the number of
arcs incident with nodes of $A_{1} \cup B_{1}$ is less than
\[
n \left(\frac{16\sqrt{n}}{2^{r}} + \log n\right)
  < \frac{17 n^{3/2}}{2^{r}}.
\]

If we disregard these arcs, there still remain more than
\[
\frac{n^{2}}{2^{2r+13/4}} - \frac{17 n^{3/2}}{2^{r}}
  > \frac{n^{2}}{2^{2r+4}}
\]
arcs. We now apply Lemma~1 again to obtain a special subgraph~$H_{2}$.
(The arcs in~$H_{i}$ need not all be oriented from nodes in~$A_{i}$ to nodes in~$B_{i}$; it
may be that they are all oriented from nodes in~$B_{i}$ to nodes in~$A_{i}$.) We now
disregard the arcs incident with the nodes of~$H_{2}$ and apply Lemma~1 again.
When we have repeated this procedure $\bigl[\sqrt{n}/(17 · 2^{r+5})\bigr]$ times, we are left with
a graph that still has more than
\[
\frac{n^{2}}{2^{2r+13/4}} - \frac{17 n^{3/2}}{2^{r}} \left[\frac{\sqrt{n}}{17 · 2^{r+5}}\right]
  > \frac{n^{2}}{2^{2r+4}}
\]
arcs. If we apply Lemma~1 once more, then the bilevel graph with components~$H_{i}$, $i = 1, 2, \dots, \bigl[\sqrt{n}/(17·2^{r+5})\bigr] + 1$, has at least
\[
\left(\left[\frac{\sqrt{n}}{17 · 2^{r+5}}\right] + 1\right)
  · [\sqrt{n}] · \left[\frac{\log n}{3(r + 3)}\right]
  > \frac{n \log n}{(r + 3) · 2^{r+11}}
\]
arcs. This proves the lemma.
\PageSep{54}

\begin{Lemma}[3.]
If $G$~is an oriented graph with $e$~arcs, then $G$~contains a bilevel
subgraph with at least $\nf{1}{2}\sqrt{e}$ arcs.
\end{Lemma}

\Proof. This is trivially true when $e = 0$; the proof of the general case is by
induction. Let $p$ be one of the nodes of largest degree~$d$ in~$G$. We may
suppose that at least $\nf{1}{2}d$~arcs are oriented away from~$p$; use these arcs to
form one component of a bilevel subgraph of~$G$. If $\alpha$~denotes the number of
arcs that are incident with any nodes that are joined to~$p$, then $0 < \alpha \leq \min(d^{2}, e)$.
We may now apply the induction hypothesis to the other $e - \alpha$ arcs
and assert that $G$~has a bilevel subgraph with at least
\[
\nf{1}{2}(d + \sqrt{e - \alpha}) \geq \nf{1}{2} \sqrt{e}
\Tag{(2)}
\]
arcs.

\begin{Lemma}[4.]
If $G$~is an oriented graph with $e$~arcs, then $G$~can be expressed as
the union of fewer that $4\sqrt{e}$ arc-disjoint bilevel graphs.
\end{Lemma}

\Proof. Suppose that $G$~is the union of $t$~arc-disjoint bilevel graphs with
$e_{1}, e_{2}, \dots, e_{t}$ arcs, respectively, where $e_{1} \leq e_{2} \leq \dots \leq e_{t}$. In view of
Lemma~3, we may assume that
\[
e_{i} \geq \nf{1}{2}\sqrt{e_{1} + e_{2} + \dots + e_{i}}
\]
or
\[
4e_{i}^{2} - e_{i} \geq e_{1} + e_{2} + \dots + e_{i-1}
\]
for each~$i$. From this inequality it follows by induction that $e_{i} \geq \nf{1}{8}i$ for
each~$i$. Therefore
\[
e = e_{1} + e_{2} + \dots + e_{i}
  \geq \nf{1}{8} \binom{t + 1}{2}
  > (\nf{1}{4}t)^{2},
\]
or $t < 4\sqrt{e}$.

We can now prove the following theorem due to Erdös and Moser (1964a).
\index[xauthor]{Erdös, P.}%
\index[xauthor]{Moser, L.}%

\begin{Theorem}{26.}
There exists a constant~$c$ such that any oriented graph~$G$ with
$n$~nodes can be expressed as the union of $l$~arc-disjoint bilevel graphs, all of
which have the same $n$~nodes, where
\[
l \leq \frac{cn}{\log n}.
\]
\end{Theorem}

\Proof. We define oriented graphs $G_{i}$ and~$G^{(i)}$ inductively for $i = 1,
2, \dots, \bigl[2^{13}(n/\log n)\bigr]$.
The graph~$G^{(i)}$ is obtained from~$G$ by removing the
arcs of $G_{1}, G_{2}, \dots, G_{i}$. Let $G_{1}$~be a bilevel subgraph of~$G$ with a maximal
number of arcs and let $G_{i+1}$~be similarly defined with respect to~$G^{(i)}$. (We may
suppose that all the graphs~$G_{i}$ have $n$~nodes.)
\PageSep{55}

If there are $e_{i}$~arcs in~$G^{(i)}$ and if $i_{r}$~is the smallest integer such that
\[
e_{i_{r}} \leq \frac{n^{2}}{2^{2r+1}},
  \quad\text{for $r = 0, 1, \dots, [4\log\log n]$,}
\]
then we shall show that
\[
i_{r+1} - i_{r}
  \leq 2^{11} · \frac{r + 3}{2^{r+1}} · \frac{n}{\log n}.
\Tag{(3)}
\]
We may suppose that
\[
e_{i_{r}} > \frac{n^{2}}{2^{2r+3}},\quad\text{for $i_{r+1} - i_{r} = 0$}
\]
otherwise. If $i_{r} \leq j < i_{r+1}$, then
\[
\frac{n^{2}}{2^{2r+3}} < e_{j} \leq \frac{n^{2}}{2^{2r+1}}.
\]
Hence, $G^{(j)}$~contains a bilevel subgraph with at least
\[
\frac{n \log n}{(r + 3) 2^{r+11}}
\]
arcs, by Lemma~2. This implies that
\[
e_{j} - e_{j+1} \geq \frac{n \log n}{(r + 3) 2^{r+11}}.
\]
If we sum this inequality over all~$j$ such that $i_{r} \leq j < i_{r+1}$, we find that
\[
\frac{n^{2}}{2^{2r+1}} \geq e_{i_{r}} - e_{i_{r+1}}
  = \sum (e_{j} - e_{j+1})
  \geq (i_{r+1} - i_{r}) · \frac{n \log n}{(r + 3) 2^{r+11}}.
\]
This implies inequality~\Eq{(3)}.

Therefore, upon removing at most
\[
\sum_{0 \leq r \leq [4\log\log n]} (i_{r+1} - i_{r})
  < \frac{2^{11} n}{\log n} \sum_{r=0}^{\infty} \frac{r + 3}{2^{r+1}}
  = \frac{2^{13} n}{\log n}
\]
arc-disjoint bilevel graphs~$G_{i}$, we are left with a graph~$G'$ that has at most
\[
\frac{n^{2}}{2^{2([4\log\log n] + 1) + 1}}
  < \frac{n^{2}}{(\log n)^{5}}
\]
arcs. But $G'$~is the union of fewer than $4n/(\log n)^{5/2}$ arc-disjoint bilevel
graphs by Lemma~4. This completes the proof of \ThmRef{25}.


\Exercises

\Ex{1.}  Prove that inequality~\Eq{(1)} holds when $a = [\sqrt{n}]$ and $b = \left[\dfrac{\log n}{3(r + 3)}\right]$.

\Ex{2.} Verify inequality~\Eq{(2)}.
\PageSep{56}


\Section{19.}{Oriented Graphs Induced by Voting Patterns}
\index{Oriented graph}%

Suppose that $m$~voters each rank $n$~objects $1, 2, \dots, n$ in order of preference.
\index{Voters}%
Their preferences may be represented by an $m$~by~$n$ matrix~$M$, each
row of which is a permutation of the objects $1, 2, \dots, n$. The collective
preferences of these voters induce an oriented graph~$H_{n}$ with $n$~nodes
$p_{1}, p_{2}, \dots, p_{n}$ in which the arc~$\Arc{p_{i}p_{j}}$ goes from node~$p_{i}$ to node~$p_{j}$ if and only
if $i$~precedes~$j$ in a majority of the $m$~rows of the matrix~$M$. This graph will be
a tournament, unless $m$~is even and some ties occur; then certain nodes will
not be joined by an arc.

Even though each voter's preferences are transitive, the collective
\index{Preference matrix}%
preferences determined by the majority rule need not be transitive. For
example, three voters who rank three objects according to the preference
matrix
\[
M = \left|\begin{array}{*{3}{c}}
1 & 2 & 3 \\
2 & 3 & 1 \\
3 & 1 & 2
\end{array}\right|
\]
induce a $3$-cycle that is certainly not transitive.

McGarvey (1953) has shown that at most $n(n - 1)$ voters are necessary to
\index[xauthor]{MacGarvey@McGarvey, D. S.}%
induce any oriented graph~$H_{n}$. He associates a pair of voters with each arc
in the graph. The two voters associated with the arc~$\Arc{p_{i}p_{j}}$ rank the $n$~objects
in the orders $(i, j, 1, 2, \dots, n)$ and $(n, n - 1, \dots, 1, i, j)$. Each pair of
voters thus determines one arc~$\Arc{p_{i}p_{j}}$, since the preferences of the remaining
voters cancel out with respect to $i$~and~$j$.

Stearns (1959) showed, by a more complicated construction, that at most
\index[xauthor]{Stearns, R.}%
$n + 2$ voters are necessary to induce any oriented graph~$H_{n}$; he also showed
that at least $\nf{1}{2} \log 3(n/\log n)$ voters are necessary in some cases. This lower
bound is combined with an upper bound due to Erdös and Moser (1964a)
\index[xauthor]{Erdös, P.}%
\index[xauthor]{Moser, L.}%
in the following result.

\begin{Theorem}{27.}
If $m(n)$ denotes the least integer~$m$ such that at most $m$
voters are necessary to induce any oriented graph~$H_{n}$ ($n > 1$), then there
exist constants $c_{1}$ and~$c_{2}$ such that
\[
\frac{c_{1} n}{\log n} < m(n) < \frac{c_{2} n}{\log n}.
\]
\end{Theorem}

\Proof. If no more than $m$ voters are necessary to induce each of the
$3^{\ebinom{n}{2}}$ oriented graphs~$H_{n}$, then it must be that
\[
3^{\ebinom{n}{2}} \leq (n! + 1)^{m},
\]
\PageSep{57}
since each of the $m$ possible voters can vote in one of $n!$ ways or not vote
at all. The lower bound is obtained by solving for~$m$.

We assert that any bilevel graph~$H$ can be induced by two voters. For,
\index{Bilevel graph}%
suppose that $H$~has three components and that $A_{1}, A_{2}, A_{3}$ and $B_{1}, B_{2}, B_{3}$ are
\index{Components of a bilevel graph}%
the node sets of the components. If
\[
A_{i} = \{a_{ij} : j = 1, 2, \dots, j_{i}\}\quad\text{and}\quad
B_{i} = \{b_{ik} : k = 1, 2, \dots, k_{i}\},
\]
then $H$~can be induced by the preference matrix
\[
\setlength{\arraycolsep}{0pt}
\Squeeze{\left|\begin{array}{*{13}{c}}
a_{11}a_{12} & \dots &
a_{\Typo{i}{1}j_{1}}b_{11}b_{12} & \dots &
b_{1k_{1}}a_{21}a_{22} & \dots &
a_{2j_{2}}b_{21}b_{22} & \dots &
b_{2k_{2}}a_{31}a_{32} & \dots &
a_{3j_{3}}b_{31}b_{32} & \dots &
b_{3k_{3}} \\
a_{3j_{3}} & \dots &
a_{32}a_{31}b_{3k_{3}} & \dots &
b_{32}b_{31}a_{2j_{2}} & \dots &
a_{22}a_{21}b_{2k_{2}} & \dots &
b_{22}b_{21}a_{1j_{1}} & \dots &
a_{12}a_{11}b_{1k_{1}} & \dots &
b_{12}b_{11}
\end{array}\right|.}
\]
This construction can easily be extended to the general case.

Therefore, if an oriented graph~$H_{n}$ can be expressed as the union of $l$~arc-disjoint
bilevel graphs with $n$~nodes, then $H_{n}$~can be induced by $2l$~voters.
The upper bound for~$m(n)$ now follows from \ThmRef{26}.


\Exercises

\Ex{1.} Prove the following results about the least integer $g = g(n)$ such that at
most $g$ voters are necessary to induce any tournament~$T_{n}$, where $n \geq 2$.
\begin{itemize}
\item[(a)] $g(n)$ is always odd.

\item[(b)] $g(3) = g(4) = g(5) = 3$.

\item[(c)] $g(n + 1) \leq g(n) + 2$.

\item[(d)] $m(n) < 2g(n)$. [Stearns (unpublished).]
\index[xauthor]{Stearns, R.}%
\end{itemize}

\Ex{2.} H.~Robbins raised the following question: If an odd number~$m$ of
\index[xauthor]{Robbins, H.}%
voters rank $n$~objects randomly and independently, then what is the probability
$p(m, n)$ that the tournament~$T_{n}$ they induce is transitive? Is it true that
\[
\lim_{m \to \infty} p(m, n) = \frac{n!}{2^{\ebinom{n}{2}}}
\]
for each fixed value of~$n$? [See DeMeyer and Plott (1967).]
\index[xauthor]{DeMeyer, F. R.}%
\index[xauthor]{Plott, C. R.}%

\Ex{3.} Characterize those oriented graphs that can be induced by $m$~voters, for
\index{Oriented graph}%
$m = 1, 2, 3$.


\Section{20.}{Oriented Graphs Induced by Team Comparisons}

Two teams of players can be compared by matching each player on one
\index{Teams}%
team against each player on the other team. If the players on Team~$A$
collectively win more games than they lose against players on Team~$B$,
then we say that Team~$A$ is stronger than Team~$B$ (symbolically, $A > B$).
\PageSep{58}
We admit the possibility of draws, both between individual players and
between teams. We denote both the names and the strengths of players
by a single number and we assume that the stronger player wins in any
match between two players. [Steinhaus and Trybula (1959) have mentioned a
\index[xauthor]{Steinhaus, H.}%
\index[xauthor]{Trybula, S.}%
possible industrial application of this method of comparing two samples
of objects.]

Let $N$~players $p_{1}, p_{2}, \dots, p_{N}$ be split into~$n$ ($n > 1$) nonempty teams
$T_{1}, T_{2}, \dots, T_{n}$ and suppose that every team is compared with every other
team. This induces an oriented graph~$H_{n}$ with $n$~nodes $t_{1}, t_{2}, \dots, t_{n}$ in which
the arc~$\Arc{t_{i}t_{j}}$ goes from node~$t_{i}$ to node~$t_{j}$ if and only if $T_{i} > T_{j}$. For example,
the teams $T_{1} = \{6, 7, 2\}$, $T_{2} = \{1, 5, 9\}$, and $T_{3} = \{8, 3, 4\}$ induce the
$3$-cycle in \Figure{10}.
\begin{figure}[hbt!]
\centering
\Graphic[1.25in]{fig10}
\FCaption{Figure 10}
\end{figure}

We now show that any oriented graph can be induced by comparing
appropriate teams.

If the teams $T_{1}, T_{2}, \dots, T_{n}$ generate the oriented graph~$H_{n}$, let $\alpha(i, j)$
denote the number of games won minus the number of games lost by
players of~$T_{i}$ against players of~$T_{j}$. (For example, $\alpha(1, 2) = \alpha(2, 3) = \alpha(3, 1) = 1$
in the illustration in \Figure{10}.) We call $\alpha(i, j)$ the \emph{net score} of~$T_{i}$ against~$T_{j}$;
\index{Net score}%
notice that $\alpha(j, i) = -\alpha(i, j)$.

If $w$ and $s$ denote the strengths of the weakest and strongest players on
the $n$~teams, let $w_{1}, w_{2}, s_{1}$, and~$s_{2}$ be any numbers such that $w_{1} = w_{2} < w$
and $s_{1} > s_{2} > s$. Add two players of strength $w_{1}$~and $s_{1}$ to~$T_{i}$ and two players
\index{Strength of a player}%
$w_{2}$~and $s_{2}$ to~$T_{j}$. This will \Erratum{insrease}{increase} $\alpha(i, j)$ by one, but the net scores between
all other pairs of teams will not be affected (see Exercise~1).

We may assume that initially there are $n$~players of equal strength, one on
each team; the net scores are all zero and the induced graph has no arcs.
The process described in the preceding paragraph can now be repeated as
often as necessary. We conclude, therefore, that if the (nonnegative) net
scores between the teams are prescribed in advance and their sum is~$\beta$, then
no more than $n + 4\beta$ players are necessary to realize these scores. In
particular, any oriented graph~$H_{n}$ (for which the corresponding net scores
would all be zero or one) may be induced by $n + 4 \dbinom{n}{2} = 2n^{2} - n$ or
fewer players.

The preceding argument is due to Moser; Moon and Moser (1967)
\index[xauthor]{Moon, J. W.}%
\index[xauthor]{Moser, L.}%
gave the following sharper result that, in a sense, is the  best possible.
\PageSep{59}

\begin{Theorem}{28.}
If $N(n)$ denotes the least integer~$N$ such that at most $N$
players are necessary to induce any oriented graph~$H_{n}$ ($n > 1$), then there
exist constants $c_{1}$~and $c_{2}$ such that
\[
\frac{c_{1} n^{2}}{\log n} < N(n) < \frac{c_{2} n^{2}}{\log n}.
\]
\end{Theorem}

\Proof. If $m$~players can induce the graph~$H_{n}$, let $h_{i}$~denote the number of
players that have strength~$i$; we may suppose there exists an integer~$t$ not
exceeding~$m$ such that $h_{i} > 0$ for $i = 1, 2, \dots, t$ and
\[
h_{1} + h_{2} + \dots + h_{t} = m.
\]
The number of solutions in positive integers to this equation is~$2^{m-1}$ and
the $m$~players corresponding to each solution can be split into $n$~teams in at
most $n^{m}$~ways. Consequently, if $N$~or fewer players suffice to induce every
oriented graph~$H_{n}$, it must be that
\[
3^{\ebinom{n}{2}} \leq \nf{1}{2} \sum_{m=1}^{N} (2n)^{m} < (2n)^{N},
\]
or
\[
N > \frac{\log 3}{2}\, \frac{n(n - 1)}{\log (2n)},
\]
since each allocation of players determines at most one graph. This implies
the lower bound of the theorem.

We assert that any bilevel graph~$H_{n}$ can be induced by $2n$~players, two
\index{Bilevel graph}%
on each team. For, suppose that $H_{n}$~has three components and that $A_{1}, A_{2}, A_{3}$
\index{Components of a bilevel graph}%
and $B_{1}, B_{2}, B_{3}$ are the node sets of the components; then $H_{n}$~is induced if
each node in the various node sets is associated with the team indicated in
the following list.
\begin{alignat*}{3}
&A_{1}: (1, 10)\quad &&A_{2}: (2, 8)\quad &&A_{3}: (3, 6) \\
&B_{1}: (1, 9)       &&B_{2}: (2, 7)      &&B_{3}: (3, 5)
\end{alignat*}
It is easy to extend this construction to the general case.

Suppose the oriented graph~$H_{n}$ can be expressed as the union of $l$~arc-disjoint
bilevel graphs~$B^{(k)}$ ($k = 1, 2, \dots, l$), all of which have the same $n$
nodes. We know that there exist teams~$R_{ki}$ of two players each such that
the teams~$R_{ki}$ ($i = 1, 2, \dots, n$) induce the bilevel graph~$B^{(k)}$ ($k = 1, 2, \dots, l$).
We may assume that every player on any team~$R_{kj}$ is stronger than every
player on any team~$R_{hi}$ ($1 \leq h < k \leq l$). [This property can be ensured by
adding a suitable constant~$c_{k}$ to the strength of every player on the team~$R_{ki}$
($k = 1, 2, \dots, l$), if necessary.] The teams
\[
T_{i} = \bigcup_{k=1}^{l} R_{ki},\quad i = 1, 2, \dots, n,
\]
\PageSep{60}
have $2l$~players each and it is not difficult to see that they generate the
oriented graph~$H_{n}$.

Therefore, if an oriented graph~$H_{n}$ can be expressed as the union of $l$~arc-disjoint
bilevel graphs with $n$~nodes, then $H_{n}$~can be induced by $2ln$~players.
This implies the upper bound for~$N(n)$ since, by \ThmRef{26}, we
may suppose that $l \leq cn/\log n$.

There are certain rather curious aspects of this mode of comparing
teams. In the example illustrated in \Figure{10}, the teams $T_{1}, T_{2}$, and~$T_{3}$
were such that $T_{1} > T_{2}$ and $T_{2} > T_{3}$. One might expect that $T_{1} \cup T_{2} > T_{2} \cup T_{3}$,
and this is indeed the case. However, since $T_{1} < T_{3}$, one might equally
well expect that $T_{1} \cup T_{2} < T_{2} \cup T_{3}$, and this is false.

The following example is perhaps more striking. If $A = \{2, 3, 10\}$ and
$B = \{1, 8, 9\}$ then $A > B$ by $5$~wins to~$4$. If $A_{1} = A \cup \{5\}$ and $B_{1} = B \cup \{4\}$,
then the teams $A_{1}$~and~$B_{1}$ are tied with $8$~wins each. If $A_{2} = A_{1} \cup \{7\}$
and $B_{2} = B_{1} \cup \{6\}$, then $B_{2} > A_{2}$ by $13$~wins to~$12$. Notice that at each
stage we added the stronger player to the team that was stronger originally,
yet the net effect was to reverse the relative strengths of the two teams.
This process can be continued. If $A_{3} = A_{2} \cup \{12\}$ and $B_{3} = B_{2} \cup \{11\}$,
then $A_{3}$~and~$B_{3}$ are tied with $18$~wins each. Finally, if $A_{4} = A_{3} \cup \{14\}$
and $\Typo{B_{4} \cup \{13\}}{B_{4} = B_{3} \cup \{13\}}$, then $A_{4} > B_{4}$ by $25$~wins to~$24$.


\Exercises

\Ex{1.} Verify the unproved assertion about the effect upon the net scores of
adding players $w_{1}, s_{1}, w_{2}$, and~$s_{2}$ to $T_{i}$ and~$T_{j}$.

\Ex{2.} Show that $N(3) = 7$.

\Ex{3.} The first argument in this section shows that $N(n) \leq 2n^{2} - n$. Refine this
argument and show by induction that $N(n) \leq n^{2} + 3n - 11$ if $n \geq 3$.
[Notice that this upper bound for~$N(n)$ is superior to the upper bound in
\ThmRef{28}, unless $n$~is very large.]

\Ex{4.} Let $w(i, j)$ and $l(i, j)$ denote the number of games won and lost by players
of~$T_{i}$ against players of~$T_{j}$. We have shown that the net scores $\alpha(i, j) = w(i, j) - l(i, j)$
for the matches between $n$~teams can be prescribed arbitrarily.
Can the win-loss ratios $w(i, j)/\Typo{w}{l}(i, j)$ be prescribed arbitrarily?
[See Steinhaus and Trybula (1959), Trybula (1961), and Usiskin (1964).]
\index[xauthor]{Steinhaus, H.}%
\index[xauthor]{Trybula, S.}%
\index[xauthor]{Usiskin, Z.}%

\Ex{5.} Modify the construction used in the proof of \ThmRef{28} and show
that any bilevel graph~$H_{n}$ can be induced by $2n$~players, two on each team,
such that different players do not have the same strength.

\Ex{6.} Characterize those oriented graphs that can be induced by comparing
teams with $k$~players each, for $k = 1, 2, 3$.
\PageSep{61}

\Ex{7.} Suppose two teams $A$~and~$B$ are compared as follows. The $i$th~strongest
\index{Teams}%
player of~$A$ is matched against the $i$th~strongest player of~$B$, for $i = 1, 2, \dots, h$,
where $h$~denotes the number of players on the smaller team. The stronger
team is the one whose players win the most matches. Try to obtain a result
analogous to \ThmRef{28} when this method of comparing teams is used.


\Section{21.}{Criteria for a Score Vector}
\index{Score vector}%

The following result was first proved by Landau (1953); the proof we give
\index[xauthor]{Landau, H. G.}%
here is due to Ryser (1964). [See also Alway (1962a) and Fulkerson (1966).]
\index[xauthor]{Alway, G. G.}%
\index[xauthor]{Fulkerson, D. R.}%
\index[xauthor]{Ryser, H. J.}%

\begin{Theorem}{29.}
A set of integers $(s_{1}, s_{2}, \dots, s_{n})$, where $s_{1} \leq s_{2} \leq \dots \leq s_{n}$,
is the score vector of some tournament~$T_{n}$ if and only if
\[
\sum_{i=1}^{k} s_{i} \geq \binom{k}{2},
\Tag{(1)}
\]
for $k = 1, 2, \dots, n$ with equality holding when $k = n$.
\end{Theorem}

\Proof. Any $k$ nodes of a tournament are joined by $\dbinom{k}{2}$~arcs, by definition.
Consequently, the sum of the scores of any $k$ nodes of a tournament must be
at least~$\dbinom{k}{2}$. This shows the necessity of~\Eq{(1)}.

The sufficiency of~\Eq{(1)} when $n = 1$ is obvious. The proof for the general
case will be by induction. Let $j$~and $k$ be the smallest and largest indices less
than~$n$ such that $s_{j} = s_{s_{n}} = s_{k}$. Consider the set of integers $(s_{1}', s_{2}', \dots, s_{n-1}')$
defined as follows:
\begin{align*}
s_{i}' &= s_{i}\quad\text{if}\quad
  \begin{aligned}[t]
    i &= 1, 2, \dots, j - 1\quad\text{or} \\
    i &= k - (s_{n} - j), \dots, k - 1, k;
  \end{aligned} \\
s_{i}' &= s_{i} - 1\quad\text{if}\quad
  \begin{aligned}[t]
    i &= j, j + 1, \dots, k - (s_{n} - j) - 1\quad\text{or} \\
    i &= k + 1, k + 2, \dots, n - 1.
  \end{aligned}
\end{align*}
From this definition, it follows that
\[
s_{1}' \leq s_{2}' \leq \dots \leq s_{n-1}',
\]
that $s_{1}' = s_{i}$ for $s_{n}$~values of~$i$, and that $s_{i}' = s_{i} - 1$ for $(n - 1) - s_{n}$ values
of~$i$. Consequently,
\[
\sum_{i=1}^{n-1} s_{i}' = \sum_{i=1}^{n} s_{i} - (n - 1) = \binom{n - 1}{2}.
\]
\PageSep{62}

{\Loosen If there exists a tournament $T_{n-1}$ with score vector $(s_{1}', s_{2}', \dots, s_{n-1}')$,
then there certainly exists a tournament~$T_{n}$ with score vector $(s_{1}, s_{2}, \dots, s_{n})$;
namely, the tournament consisting of~$T_{n-1}$ plus the node~$p_{n}$, where $p_{n}$~dominates
the $s_{n}$~nodes $p_{i}$ such that $s_{i}' = s_{i}$ and is dominated by the remaining
nodes. Therefore, we need only show that the inequality}
\[
\sum_{i=1}^{h} s_{i}' < \binom{h}{2}
\Tag{(2)}
\]
is impossible for every integer~$h$ such that $1 < h < n - 1$ in order to
complete the proof by induction.

Consider the smallest value of~$h$ for which inequality~\Eq{(2)} holds, if it ever
holds. Since
\[
\sum_{i=1}^{h-1} s_{i}' \geq \binom{h - 1}{2},
\]
it follows that $s_{h} \leq h$. Furthermore, $j \leq h$, since the first $j - 1$ scores were
unchanged. Hence,
\[
s_{h} = s_{h+1} = \dots = s_{f}
\]
if we let
\[
f = \max(h, k).
\]

Let $t$ denote the number of values of~$i$ not exceeding~$h$ such that $s_{i}' = s_{i} - 1$.
Then it must be that
\[
s_{n} \leq f - t.
\Tag{(3)}
\]
Therefore,
\begin{align*}
\binom{n}{2}
  &= \sum_{i=1}^{h} s_{i}'
      + \sum_{i=h+1}^{f} s_{i}
      + \sum_{i=f+1}^{n-1} s_{i} + s_{n} + t \\
  &< \binom{h}{2} + (f - h) s_{h} + \Erratum{\sum_{i=f+1}^{n-1} s_{i}}{(n - 1 - f) s_{n}} + f \\
  &\leq \binom{h}{2} + h(f - h) + \Erratum{\sum_{i=f+1}^{n-1} s_{i}}{f(n - 1 - f)} + f \\
  &\leq \binom{f}{2} + f(n - f)
   \leq \binom{n}{2}.
\end{align*}
Consequently, inequality~\Eq{(2)} cannot hold and the theorem is proved.


\Exercises

\Ex{1.} Show that, if the scores~$s_{i}$ of a tournament~$T_{n}$ are in nondecreasing order,
then $(i - 1)/2 \leq s_{i} \leq (n + i - 2)/2$. [Landau (1953).]
\index[xauthor]{Landau, H. G.}%
\PageSep{63}

\Ex{2.} Prove that $\sum_{k} s_{i} \leq k\bigl[n - (k + 1)/2\bigr]$, where the sum is over any $k$ scores
of a tournament~$T_{n}$. [Landau (1953).]
\index[xauthor]{Landau, H. G.}%

\Ex{3.} Prove that, in any tournament, there exists a node~$p$ such that for any
other node~$q$ either $p \to q$ or there exists a node~$r$ such that $p \to r$ and
$r \to q$. [Landau (1953).]

\Ex{4.} Prove that, if a tournament has no nodes with score zero, then it has at
\index{Score of a node}%
least three nodes~$u$ such that for any other node~$v$ either $v \to u$ or there
exists a node~$w$ such that $v \to w$ and $w \to u$. [Silverman (1961).]
\index[xauthor]{Silverman, D. L.}%

\Ex{5.} Let $l_{i}$ denote the number of nodes that dominate~$p_{i}$ in the tournament~$T_{n}$
so that $s_{i}+ l_{i} = n - 1$ for all~$i$. Prove in at least three different ways
that
\[
\sum_{i=1}^{n} s_{i}^{2} = \sum_{i=1}^{n} l_{i}^{2}.
\]

\Ex{6.} Is it possible for two nonisomorphic tournaments to have the same score
\index{Isomorphic tournaments}%
vector?

\Ex{7.} Prove that among the class of all tournaments with score vector
\index{Score vector}%
$(s_{1}, s_{2}, \dots, s_{n})$, where $s_{1} \leq s_{2} \leq \dots \leq s_{n}$, there exists a tournament~$T_{n}$ whose
matrix has only $\sum' \bigl[s_{i} - (i - 1)\bigr]$ ones above the diagonal, where the sum is
\index{Matrix!of a tournament}%
over all~$i$ such that $s_{i} > i - 1$. [Ryser (1964) and Fulkerson (\Erratum{1966}{1965}).]
\index[xauthor]{Fulkerson, D. R.}%
\index[xauthor]{Ryser, H. J.}%


\Section{22.}{Score Vectors of Generalizations of Tournaments}

An \emph{$n$-partite tournament} differs from an ordinary tournament in that there
\index{Tournament!n-partite@$n$-partite}%
are $n$~nonempty sets of nodes $P_{i} = (p_{i1}, p_{\Typo{i_{2}}{i2}}, \dots, p_{in_{i}})$ and two nodes are
joined by an oriented arc if and only if they do not belong to the same set~$P_{i}$.
The score vectors of an $n$-partite tournament are defined in the obvious
way. The following theorem was proved by Moon (1962).
\index[xauthor]{Moon, J. W.}%

\begin{Theorem}{30.}
The $n$~sets of integers $S_{i} = (s_{i1}, s_{i2}, \dots, s_{in_{i}})$, where $s_{i1} \leq s_{i2} \leq \dots \leq s_{in_{i}}$
for $i = 1, 2, \dots, n$, form the score vectors of some
$n$-partite tournament if and only if
\[
\sum_{i=1}^{n} \sum_{j=1}^{k_{i}} s_{ij}
  = \sum_{i=1}^{n-1} \sum_{j=i+1}^{n} k_{i} k_{j},
\]
for all sets of $n$~integers $k_{i}$ satisfying $0 \leq k_{i} \leq n_{i}$ with equality holding when
$k_{i} = n_{i}$ for all~$i$.
\end{Theorem}

This reduces to \ThmRef{29} when $n_{1} = n_{2} = \dots = n_{n} = 1$.

This case $n = 2$ is of special interest. There is a natural one-to-one
\index{Matrix!of zeros and ones}%
correspondence between bipartite tournaments and matrices of zeros and
\index{Tournament!bipartite}%
ones; simply let $p_{1i} \to p_{2j}$ or $p_{2j} \to p_{1i}$ according as the $(i, j)$ entry of the
\PageSep{64}
matrix is one or zero. Gale (1957) and Ryser (1957) found necessary and
\index[xauthor]{Gale, D.}%
\index[xauthor]{Ryser, H. J.}%
sufficient conditions for the existence of a matrix of zeros and ones having
prescribed row and column sums. It is not difficult to show that the following
statement is equivalent to their theorem [see also Vogel (1963)].
\index[xauthor]{Vogel, W.}%

\begin{Corollary}
There exists an $m$~by~$n$ matrix of zeros and ones with row
sums~$r_{i}$, where $r_{1} \leq r_{2} \dots \leq r_{m}$, and column sums~$c_{j}$, where $c_{1} \geq c_{2} \geq \dots \geq c_{n}$,
if and only if
\[
\sum_{i=1}^{k} r_{i} + \sum_{j=1}^{i} (m - c_{j}) \geq kl,
\]
for $k = 0, 1, \dots, m$ and $l = 0, 1, \dots, n$ with equality holding when $k = m$
and $l = n$.
\end{Corollary}

A \emph{generalized tournament} differs from an ordinary tournament in that
\index{Tournament!generalized}%
both the arcs $\Arc{p_{i}p_{j}}$ and $\Arc{p_{j}p_{i}}$ join every pair of distinct nodes $p_{i}$ and~$p_{j}$; in
addition there is a weight~$\alpha_{ij}$ associated with each arc~$\Arc{p_{i}p_{j}}$. Let $G^{*}$~denote
some set of real numbers that contains~$1$ and that forms a group with
respect to addition. We assume that the weights are nonnegative members
\index{Weights associated with arcs}%
of~$G^{*}$ and that they satisfy the equation $\alpha_{ij} + \alpha_{ji} = 1$ for all pairs of
distinct values of $i$ and~$j$. The score of~$p_{i}$ is then given by the formula
\index{Score of a node}%
\[
s_{i} = \sum_{\Typo{i}{j}=1}^{n} \alpha_{ij},
\]
if we adopt the convention that $\alpha_{ii} = 0$.

The following generalization of \ThmRef{29} is valid.

\begin{Theorem}{31.}
Let $(s_{1}, s_{2}, \dots, s_{n})$, where $s_{1} \leq s_{2} \leq \dots \leq s_{n}$, be a set of
numbers belonging to a group~$G^{*}$. Then $(s_{1}, s_{2}, \dots, s_{n})$ is the score vector
of some generalized tournament whose weights belong to~$G^{*}$ if and only if
\[
\sum_{i=1}^{k} s_{i} \geq \binom{k}{2},
\Tag{(1)}
\]
for $k = 1, 2, \dots, n$ with equality holding when $k = n$.
\end{Theorem}

Proofs of this have been given by Moon (1963) and Ryser (unpublished)
\index[xauthor]{Moon, J. W.}%
when $G^{*}$~is the group of all real numbers and by Ford and Fulkerson (\Erratum{1962}{1962; p.~41})
\index[xauthor]{Ford, L. R., Jr.}%
\index[xauthor]{Fulkerson, D. R.}%
when $G^{*}$~is the group consisting of all multiples of~$1/c$ for some integer~$c$.

Perhaps the easiest and most natural way to deal with problems of this
type is to use the theory of flows in networks. The following result, which
\index{Flows in networks}%
we state without proof, is an immediate consequence of the supply-demand
theorem due to Gale (1957). [See Ford and Fulkerson (1962) for an exposition
of the theory of flows in networks.]

\begin{Theorem}{32.}
Let $\{S_{j} : j = 1, 2, \dots, N\}$ be a family of nonempty subsets
of $T = \{1, 2, \dots, m\}$. The numbers $v_{1}$, $v_{2}$,~\dots, $v_{m}$ and $k(S_{1})$, $k(S_{2})$,~\dots, $k(S_{N})$
\PageSep{65}
all belong to a set~$G$ of real numbers that forms a group with respect
to addition. Then the system of equations
\[
\begin{aligned}
X_{ij} &= 0\quad\text{if $i \not\in S_{j}$} \\
\sum_{j=1}^{N} X_{ij} &= v_{i}\quad i = 1, 2, \dots, m \\
\sum_{i=1}^{m} X_{ij} &= k(S_{j})\quad j = 1, 2, \dots, N
\end{aligned}
\Tag{(2)}
\]
has a nonnegative solution in~$G$ if and only if for every subset~$S$ of~$T$
\[
\sum_{i \in S} v_{i} \geq \sum_{S_{j} \subset S} k(S_{j}),
\Tag{(3)}
\]
with equality holding when $S = T$.
\end{Theorem}

To deduce \ThmRef{31} from \ThmRef{32}, we let $T$ be the set of nodes of
the tournament and let the sets~$S_{j}$ be the pairs of distinct nodes; $k(S_{j}) = 1$
for all~$j$ and $v_{i} = s_{i}$ for all~$i$. Then, if the nodes $p_{i}$ and~$p_{j}$ form the subset~$S_{l}$,
we let $\alpha_{ij} = X_{il}$ and $\alpha_{ji} = X_{jl}$. It follows from~\Eq{(2)} that this will define a
generalized tournament with the score vector $(s_{1}, s_{2}, \dots, s_{n})$ and it is clear
\index{Score vector}%
that conditions \Eq{(1)} and~\Eq{(3)} are equivalent. The only difference in proving
the corresponding generalization of \ThmRef{30} is that now the sets~$S_{j}$ are
the pairs of nodes belonging to different subsets~$P_{i}$ of nodes.


\Exercises

\Ex{1.} Prove that the corollary to \ThmRef{30} is equivalent to the Gale-Ryser
\index[xauthor]{Gale, D.}%
\index[xauthor]{Ryser, H. J.}%
theorem on the existence of a $(0, 1)$~matrix with prescribed row and column
\index{Matrix!of zeros and ones}%
sums.

\Ex{2.} Deduce \ThmRef{32} from the theorem by Gale (1957).

\Ex{3.} Let $M = [\alpha_{ij}]$ be an $n$~by~$n$ matrix of nonnegative real numbers such
that $\alpha_{ij} + \alpha_{ji} = 1$ for $1 \leq i, j \leq n$. (In particular, $\alpha_{ii} = \nf{1}{2}$ for all~$i$. We
may think of~$M$ as the matrix of a generalized tournament if we wish.)
Prove that $M$~is positive semidefinite, that is, that $XMX^{T} \geq 0$ for all real
vectors $X = (x_{1}, x_{2}, \dots, x_{n})$. [N.~J. Pullman.]
\index[xauthor]{Pullman, N. J.}%


\Section{23.}{The Number of Score Vectors}

The score vectors of tournaments~$T_{n}$ may be generated by expanding the
product
\[
P_{n} = \prod_{1 \leq i < j \leq n} (a_{i} + a_{j}).
\]
\PageSep{66}
For example,
\begin{align*}
P_{3} &= (a_{1} + a_{2})(a_{1} + a_{3})(a_{2} + a_{3}) \\
  &= a_{1}^{2} a_{2} + a_{1} a_{2}^{2}
   + a_{1}^{2} a_{3} + a_{1} a_{3}^{2}
   + a_{2}^{2} a_{3} + a_{2} a_{3}^{2} + 2a_{1}a_{2}a_{3},
\end{align*}
and there are essentially two different score vectors, namely, $(0, 1, 2)$
and $(1, 1, 1)$. David (1959) used this scheme to determine the various score%
\index[xauthor]{David, H. A.}
vectors and their frequency for $n \leq 8$. Alway (1962b) extended these results
\index[xauthor]{Alway, G. G.}%
to the cases $n = 9$ and~$10$.

Let $s(n)$ denote the number of different score vectors of size~$n$, that is, the
number of sets of integers $(s_{1}, s_{2}, \dots, s_{n})$ such that
\begin{gather*}
s_{1} \leq s_{2} \leq \dots \leq s_{n} \leq n - 1,
\Tag{(1)} \\
s_{1} + s_{2} + \dots + s_{r} \geq \binom{r}{2},
\quad\text{for $r = 1, 2, \dots, n - 1$,}
\Tag{(2)} \\
\intertext{and}
s_{1} + s_{2} + \dots + s_{n} = \binom{n}{2}.
\Tag{(3)}
\end{gather*}
There seems to be no simple explicit formula for~$s(n)$. We now describe a
recursive method for determining $s(n)$ due to Bent and Narayana (1964)
\index[xauthor]{Bent, D. H.}%
\index[xauthor]{Narayana, T. V.}%
[see also Bent (1964)].

Let $[t, l]^{n}$ denote the number of sets of integers $(s_{1}, s_{2}, \dots, s_{n})$ that satisfy
\Eq{(1)},~\Eq{(2)}, and the conditions
\[
s_{1} + s_{2} + \dots + s_{n} = l
\Tag{(4)}
\]
and
\[
s_{n} = t,
\]
where $l \geq \dbinom{n}{2}$. (If $l < \dbinom{n}{2}$, then we let $[t, l]^{n} = 0$.)

Now
\[
[t, l]^{1} = \begin{cases}
  1 & \text{if $t - l$,} \\
  0 & \text{otherwise,}
\end{cases}
\]
and if $n \geq 2$, then
\[
[t, l]^{n} = \sum_{h \leq t} [h, l - t]^{n-1}.
\]
The value of $s(n)$ is given by the formula
\[
s(n) = \sum_{t} \left[t, \binom{n}{2}\right]^{n}.
\Tag{(5)}
\]
\PageSep{67}
\index[xauthor]{Narayana, T. V.}%

It is convenient to enter the values of $[t, l]^{n}$ in a table and then use the
recurrence relation to form the table of values of $[t, l]^{n+1}$. If we only want to
determine $s(n)$ for $n \leq m$, then we need only determine $[t, l]^{n}$ when
\[
\frac{n - 1}{2} \leq t \leq \frac{m + n - 2}{2}
\Tag{(6)}
\]
and
\[
\binom{n}{2} \leq l \leq \left\{
\begin{aligned}
&n\bigl[\nf{1}{2}(m - 1)\bigr]\quad\text{if $n \leq \nf{1}{2}m$,} \\
&\nf{1}{2}m\bigl[\nf{1}{2}(m - 1)\bigr] + (n - \nf{1}{2} m)[\nf{1}{2} m]
\quad\text{if $\nf{1}{2} m < m \leq m$.}
\end{aligned}
\right.
\Tag{(7)}
\]
We may assume that $[t, l]^{n} = 0$ otherwise (see Exercise~1).

The following tables illustrate the use of this method for determining
$s(n)$ when $n \leq 6$.
\begin{align*}
&\begin{array}{c|*{4}{c}}
\multicolumn{5}{c}{[t, l]^{1}} \\[6pt]
\Tvars{t}{l} & 0 & 1 & 2 & \\
\hline
\Strut
0 & 1 & 0 & 0 & \\
1 & 0 & 1 & 0 & \\
2 & 0 & 0 & 1 &
\end{array}
&
&\begin{array}{c|*{4}{c}}
\multicolumn{5}{c}{[t, l]^{2}} \\[6pt]
\Tvars{t}{l} & 1 & 2 & 3 & 4 \\
\hline
\Strut
1 & 1 & 1 & 0 & 0 \\
2 & 0 & 1 & 1 & 1 \\
3 & 0 & 0 & 1 & 1
\end{array}
&
&\begin{array}{c|*{4}{c}}
\multicolumn{5}{c}{[t, l]^{3}} \\[6pt]
\Tvars{t}{l} & 3 & 4 & 5 & 6 \\
\hline
\Strut
1 & 1 & 0 & 0 & 0 \\
2 & 1 & 2 & 1 & 1 \\
3 & 0 & 1 & 2 & 2
\end{array}
\\[12pt]
&\begin{array}{c|*{4}{c}}
\multicolumn{5}{c}{[t, l]^{4}} \\[6pt]
\Tvars{t}{l} & 6 & 7 & 8 & 9 \\
\hline
\Strut
2 & 2 & 1 & 1 & 0 \\
3 & 2 & 3 & 3 & 3 \\
4 & 0 & 2 & 3 & 3
\end{array}
&
&\begin{array}{c|*{3}{r}}
\multicolumn{4}{c}{[t, l]^{5}} \\[6pt]
\Tvars{t}{l} & 10 & 11 & 12 \\
\hline
\Strut
2 & 1 & 0 & 0 \\
3 & 4 & 4 & 3 \\
4 & 4 & 6 & 7
\end{array}
&
&\begin{array}{c|*{4}{r}}
\multicolumn{5}{c}{[t, l]^{6}} \\[6pt]
\Tvars{t}{l} & 15 &  & & \\
\hline
\Strut
3 &  3 &  & & \\
4 & 10 &  & & \\
5 &  9 &  & &
\end{array}
\end{align*}
If we sum the entries in the sixth table, we find that $s(6) = 22$.

Bent and \Typo{Naryana}{Narayana} determined the value of~$s(n)$ for $n = 1, 2, \dots, 36$,
\index[xauthor]{Bent, D. H.}%
with the aid of an electronic computer. (A slight refinement of the method
illustrated in the preceding paragraph was used when $28 \leq n \leq 36$.) The
values of~$s(n)$ for $n = 1, 2, \dots, 15$ are given in \Table{5}. They conjectured
that for $n \geq 2$ the sequence $s(n + 1)/s(n)$~is monotone increasing
with limit four. More recently, P.~Stein\footnote
  {Unpublished work.}
\index[xauthor]{Stein, P.}%
has determined the values of~$s(n)$
for $n = 1, 2, \dots, 51$ by a different method and his data gives additional
support to this conjecture.
\PageSep{68}
\begin{table}[hbt!]
\TCaption{Table 5. $s(n)$, the number of score vectors of size~$n$.}
\[
\begin{array}{r|r}
\multicolumn{1}{c}{n} & s(n) \\
\hline\Strut
 1 & 1 \\
 2 & 1 \\
 3 & 2 \\
 4 & 4 \\
 5 & 9 \\
 6 & 22 \\
 7 & 59 \\
 8 & 167 \\
 9 & 490 \\
10 & 1,486 \\
11 & 4,639 \\
12 & 14,805 \\
13 & 48,107 \\
14 & 158,808 \\
15 & 531,469
\end{array}
\]
\end{table}

The following bounds for~$s(n)$ are due to Erdös and Moser.\footnote
  {Unpublished work.}
\index[xauthor]{Erdös, P.}%
\index[xauthor]{Moser, L.}%

\begin{Theorem}{33.}
There exist constants $c_{1}$ and $c_{2}$ such that
\[
\frac{c_{1} 4^{n}}{n^{5}} < s(n) < \frac{c_{1} 4^{n}}{n^{3/2}}.
\]
\end{Theorem}

\Proof. In proving the lower bound we may suppose that $n$~is even, say
$n = 2m$. Let $l_{i}$~denote the number of nodes that dominate the node~$p_{2m+i-1}$
with the $i$th~largest score $s_{2m+1-i}$ in any tournament~$T_{2m}$, that is,
\[
l_{i} = (2m - 1) - s_{2m+1-i}.
\Tag{(8)}
\]
We now consider sets of integers $s_{1}, s_{2}, \dots, s_{n}$ and $l_{1}, l_{2}, \dots, l_{n}$, such that
\begin{gather*}
s_{1} + s_{2} + \dots + s_{m} = l_{1} + l_{2} + \dots + l_{m},
\Tag{(9)}\displaybreak[0] \\
s_{1} \leq s_{2} \leq \dots \leq s_{m} = m - 1,
\Tag{(10)}\displaybreak[0] \\
s_{i} \geq i - 1,\quad i = 1, 2, \dots , m,
\Tag{(11)}\displaybreak[0] \\
l_{1} \leq l_{2} \leq \dots \leq l_{m} = m - 1,
\Tag{(12)}\displaybreak[0] \\
\intertext{and}
l_{i} \geq i - 1,\quad i = 1, 2, \dots, m.
\Tag{(13)}
\end{gather*}

Conditions \Eq{(8)} and \Eq{(9)} imply~\Eq{(3)} and conditions \Eq{(8)},~\Eq{(10)}, and~\Eq{(12)} imply~\Eq{(1)}.
Furthermore, condition~\Eq{(11)} implies~\Eq{(2)} if $r \leq m$ and conditions \Eq{(8)},~\Eq{(11)},~\Eq{(13)},
and~\Eq{(1)} imply~\Eq{(2)} if $m < r \leq 2m - 1$, since
\PageSep{69}
\begin{align*}
&\quad s_{1} + \dots + s_{m} + s_{m+1} + \dots + s_{2m-v} \\
&= s_{1} + \dots + s_{2m} - (s_{2m-v+1} + \dots + s_{2m}) \\
&= \binom{2m}{2} - (2m - 1 - l_{1}) - \dots - (2m - 1 - l_{v}) \\
&= \binom{2m}{2} - v(2m - 1) + (l_{1} + \dots + l_{v}) \\
&\geq \binom{2m}{2} - v(2m - 1) + \binom{v}{2} = \binom{2m - v}{2}.
\end{align*}

To obtain a lower bound for~$s(2m)$ we will estimate the number of solutions
of the system \Eq{(9)}~to~\Eq{(13)}. There are $(m + 1)^{-1} \dbinom{2m}{m}$ sets of integers,
satisfying each of the systems \Eq{(10)}~and~\Eq{(11)}, and \Eq{(12)}~and~\Eq{(13)} (see Exercise~3).
Consequently, the total number of solutions of \Eq{(9)}~to~\Eq{(13)} will be
\[
\sum_{i=1}^{m^{2}} u_{i}^{2},
\]
where $u_{i}$~is the number of solutions of \Eq{(10)}~and \Eq{(11)} such that $s_{1} + s_{2} + \dots + s_{m} = i$.
[It follows from~\Eq{(10)} that $i < m^{2}$.] But
\[
\sum_{i=1}^{m^{2}} u_{i} = \frac{1}{m + 1} \binom{2m}{m},
\]
so
\begin{align*}
s(2m)
  &\geq \left(\frac{1}{m^{2}(m + 1)^{2}} \binom{2m}{m}\right)^{2} m^{2} \\
  &> \frac{c 4^{n}}{n^{5}}
\end{align*}
for some constant~$c$, by Jensen's inequality and Stirling's formula.

To obtain an upper bound for~$s(n)$, we shall estimate the number of sets of
nonnegative integers $(s_{1}, s_{2}, \dots, s_{n})$, satisfying only \Eq{(1)}~and~\Eq{(3)}. The transformation
$a_{i} = s_{i} + \Typo{1}{i}$ induces a one-to-one correspondence between these
sets and sets of integers $(a_{1}, a_{2}, \dots, a_{n})$ such that
\[
1 \leq a_{1} < a_{2} < \dots < a_{n} \leq 2n - 1
\Tag{(14)}
\]
and
\[
a_{1} + a_{2} + \dots + a_{n} = n^{2}.
\Tag{(15)}
\]

The number of solutions of~\Eq{(14)} is~$\dbinom{2n - 1}{n}$. We shall divide these
solutions into classes of $2n - 1$ solutions each, in such a way that no two
solutions in the same class have the same sum. Consequently, the number
\PageSep{70}
of solutions of both \Eq{(14)} and~\Eq{(15)} will not exceed $(2n - 1)^{-1} \dbinom{2n - 1}{n}$. The
upper bound of the theorem then follows immediately.

Place two solutions $(a_{1}, a_{2}, \dots, a_{n})$ and $(b_{1}, b_{2}, \dots, b_{n})$ in the same class
if and only if the $a$'s~and~$b$'s, in some order, differ by only a constant modulo
$2n - 1$. It is clear that each class will contain $2n - 1$ solutions. If $(a_{1}, a_{2}, \dots, a_{n})$
%[** TN: b_{1} upright in the original, presumed typo]
and $(b_{1}, b_{2}, \dots, b_{n})$ are different solutions in the same class and
the~$b$'s can be obtained from the~$a$'s by adding a positive integer~$r$ to each of
the~$a$'s modulo $2n - 1$, then
\[
a_{1} + a_{2} + \dots + a_{n} + rn \equiv b_{1} + b_{2} + \dots + b_{n}
\pmod{2n - 1}.
\]
But, $rn \equiv \nf{1}{2} r \pmod{2n - 1}$ if $r$~is even and $rn \equiv \nf{1}{2} (r - 1) + n \pmod{2n - 1}$
if $r$~is odd; in either case $rn \not\equiv 0 \pmod{2n - 1}$ for $r = 1, 2, \dots, 2n - 2$,
so $a_{1} + a_{2} + \dots + a_{n} \neq b_{1} + b_{2} + \dots + b_{n}$. This completes the
proof of the theorem.


\Exercises

\Ex{1.} Prove the correctness of the sentence containing inequalities \Eq{(6)} and~\Eq{(7)}.

\Ex{2.} Prove that
\[
[t, l]^{n} = [t - 1, l - 1]^{n} + [t, l - t]^{n-1},
\quad\text{if $t \geq 2$ and $l > \dbinom{n}{2}$}.
\]
Use this relation to extend the tables in the text and show that $s(7) = 59$.

\Ex{3.} If $u_{m}$ denotes the number of solutions of \Eq{(10)}~and~\Eq{(11)}, then show that
\[
u_{m} = u_{0} u_{m-1} + u_{1} u_{m-2} + \dots + u_{m-1} u_{0},
\]
where $u_{0} = u_{1} = 1$, so that if $U(x) = \sum_{m=0}^{\infty} u_{m} x^{m}$, then
\[
xU^{2}(x) = U(x) - 1.
\]
Deduce from this that
\[
U(x) = \frac{1 - \sqrt{1 - 4x}}{2x}
  = \sum_{m=0}^{\infty} \binom{2m}{m} \frac{x^{m}}{m + 1}.
\]

\Ex{4.} Show that of the $(m + 1)^{-1} \dbinom{2m}{m}$ solutions $(s_{1}, s_{2}, \dots, s_{m})$ of \Eq{(10)}~and
\Eq{(11)} at least $\dfrac{c_{3}}{m} \dbinom{2m}{m}$ satisfy the inequality
\[
\binom{m}{2} \leq s_{1} + s_{2} + \dots + s_{m}
  \leq \binom{m}{2} + m^{3/2}.
\]
Deduce from this that [Erdös and Moser]:
\index[xauthor]{Erdös, P.}%
\index[xauthor]{Moser, L.}%
\[
s(n) > \frac{c_{4} 4^{n}}{n^{9/2}}.
\]
\PageSep{71}

\Ex{5.} Whitworth (1878) and others have shown that there are $(m + 1)^{-1} \dbinom{2m}{m}$
\index[xauthor]{Whitworth, W. A.}%
sequences of $m + 1$'s and $m - 1$'s such that all the partial sums are
nonnegative. Describe a one-to-one correspondence between such sequences
and the solutions of \Eq{(10)}~and~\Eq{(11)}.


\Section{24.}{The Largest Score in a Tournament}
\index{Score of a node}%

If the arcs in a tournament~$T_{n}$ are oriented randomly and independently,
then the score~$s_{i}$ of each node~$p_{i}$ has a binomial distribution and the distribution
of the reduced score
\[
w_{i} = \frac{s_{i} - \nf{1}{2} (n - 1)}{\sqrt{\nf{1}{4} (n - 1)}}
\]
tends to the normal distribution with zero mean and unit variance.
Consequently,
\[
P\{w_{i} > x\} \sim \frac{1}{(2\pi)^{1/2} x} e^{-(1/2) x^{2}},
\Tag{(1)}
\]
if $x^{3}/\sqrt{\nf{1}{4} (n - 1)}$ tends to zero as $n$~and $x$ \Typo{tends}{tend} to infinity. [See Feller (1957)
\index[xauthor]{Feller, W.}%
p.~158.]

The following theorem and its proof are special cases of a somewhat more
general result due to Huber (1963a).
\index[xauthor]{Huber, P.}%

\begin{Theorem}{34.}
If $w$ denotes the largest reduced score in a random tournament~$T_{n}$,
then the probability that
\[
|w - \sqrt{2\log (n - 1)}| \leq \eps \sqrt{2\log (n - 1)}
\]
tends to one as $n$~tends to infinity for any positive~$\eps$.
\end{Theorem}

\Proof. If $0 < \eps < 1$, let
\[
x_{n}^{±} = (1 ± \eps) \sqrt{2\log (n - 1)};
\]
it follows from~\Eq{(1)} that
\[
\Erratum{p}{P}\{w_{i} > x_{n}^{±}\}
  \sim \frac{1}{\sqrt{4\pi} (1 ± \eps) \sqrt{\log (n - 1)}}
  · (n - 1)^{-(1 ± \eps)^{2}}.
\Tag{(2)}
\]
Let $c_{1}$ and $c_{2}$ be positive constants such that $c_{1} < 1/\sqrt{4\pi} (1 ± \eps) < c_{2}$; if
the constant factor in~\Eq{(2)} is replaced by $c_{1}$~or~$c_{2}$, strict inequality will hold
for all sufficiently large values of~$n$. Using Boole's inequality, $P \{\cup E_{i}\} \leq \sum P\{E_{i}\}$, we find that
\[
P\{w > x_{n}^{+}\} < c_{2} (n - 1)^{-2\eps}
\Tag{(3)}
\]
for all sufficiently large values of~$n$.
\PageSep{72}

\iffalse
% [** Author's note: "B. Bollobás pointed out to me, in a letter dated
% 12 June, 1980, that this paragraph is incorrect. It may be replaced
% by the succeeding text."]
We now show that
\[
P\{s_{1} < k_{1}, \dots, s_{n} < k_{n}\}
  \leq P\{s_{1} < k_{1}\} · \dotsm · P\{s_{n} < k_{n}\}
\Tag{(4)}
\]
for any positive numbers $k_{1}, \dots, k_{n}$. Observe that
\begin{align*}
P\{s_{1} < k_{1}, \dots, s_{n} < k_{n}\}
  &= \sum_{t<k_{1}} P\{s_{1} = t, s_{2} < k_{2}, \dots, s_{n} < k_{n}\} \\
  &= \sum_{t<k_{1}} \left(\frac{1}{2}\right)^{t} \left(\frac{1}{2}\right)^{n-1-t}
  \sum P\{s_{i_{1}} < k_{i_{1}}, \dots, s_{i_{t}} < k_{i_{t}},
          s_{i_{t+1}} < k_{i_{t+1}} - 1, \dots, s_{i_{n-1}} < k_{i_{n-1}} - 1\},
\end{align*}
where the inner sum is over the $\dbinom{n - 1}{t}$ choices of $t$~nodes to be dominated
by~$p_{1}$. But each term in the inner sum is less than, or equal to,
\[
P\{s_{2} < k_{2}, \dots, s_{n} < k_{n}\}.
\]
Therefore,
\begin{multline*}
P\{s_{1} < k_{1}, \dots, s_{n} < k_{n}\} \\
  \leq \sum_{t<k_{1}} \binom{n - 1}{t} \left(\frac{1}{2}\right)^{t}
      \left(\frac{1}{2}\right)^{n-1-t} · P\{\Typo{s_{1}}{s_{2}} < k_{2}, \dots, s_{n} < k_{n}\} \\
  = P\{s_{1} < k_{1}\} · P\{s_{2} < k_{2}, \dots, s_{n} < k_{n}\}.
\end{multline*}
This argument can be repeated to yield inequality~\Eq{(4)}.
\fi
\Pagelabel{Thm34start}%
%[** TN: Corrected text provided by the author continues to next note.]
We now show that
\[
P\{s_{1} < k_{1}, \dots, s_{n} < k_{n}\}
  \leq P\{s_{1} < k_{1}\} · \dotsm · P\{s_{n} < k_{n}\}
\Tag{(4)}
\]
for any positive integers $k_{1}, \dots, k_{n}$, where $s_{i} = \sum_{j \neq i} a_{ij}$,
for $1 \leq i, j \leq n$ and where the $a_{ij}$'s are random variables such that
\[
P\{a_{ij} = 1\} = P\{a_{ij} = 0\} = 1/2\quad\text{and}\quad
a_{ij} + a_{ji} = 1
\]
for $1 \leq i, j \leq n$ and $i \neq j$. Let $b_{ij}$, for $1 \leq i, j \leq n$ and $i \neq j$, denote
random variables with the same distribution as the~$a_{ij}$'s except that we do not impose
the dependence condition that $b_{ij} + b_{ji} = 1$.

Suppose the ordered pairs $\{(u, v) : 1 \leq u < v \leq n\}$ are assigned the
labels $1, 2, \dots, m = n(n - 1)/2$ in some convenient fashion. Let the
sequences $\{s_{i}^{(h)} : 1 \leq i \leq n\}$ be defined as follows for $0 \leq h \leq m$,
starting with $\{s_{i}^{(0)}\} = \{s_{i}\}$: if $1 \leq h \leq m$ and the label~$h$ has been
assigned to the pair~$(u, v)$, then $s_{u}^{(h)} = (s_{u}^{(h-1)} - a_{uv}) + b_{uv}$,
$s_{v}^{(h)} = (s_{v}^{(h-1)} - a_{vu}) + b_{vu}$, and $s_{i}^{(h)} = s_{i}^{(h-1)}$ if $i \neq u, v$. In other
words, $\{s_{i}^{(h)}\}$~differs from $\{s_{i}^{(h-1)}\}$ in that the dependent variables
$a_{uv}$ and~$a_{vu}$ have been replaced by the independent variables $b_{uv}$ and~$b_{vu}$.

Now let $F_{h} = P\{s_{1}^{(h)} < k_{1}, \dots , s_{n}^{(h)} < k_{n}\}$, for $0 \leq h \leq m$, and
consider the quantities $\Delta_{h} = F_{h} - F_{h-1}$ for $1 \leq h \leq m$. For notational
convenience, let us assume that the pair~$(1, 2)$ has been assigned label~$h$;
and let us write $s_{1h}$ for $s_{1}^{(h-1)} - a_{12} = s_{1}^{(h)} - b_{12}$ and $s_{2h}$ for $s_{2}^{(h-1)} - a_{21} = s_{2}^{(h)} - b_{21}$.
Then it follows from the definitions and the fact that
$b_{12}$~and~$b_{21}$ are independent, that $\Delta_{h}$~is the sum of the products of
\[
P\{s_{1h} < k_{1} - c_{12}, s_{2h} < k_{2} - c_{21}, s_{3}^{(h)} < k_{3}, \dots\}
\]
and
\[
P\{b_{12} = c_{12}\} · P\{b_{21} = c_{21}\} - P\{a_{12} = c_{12}, a_{21} = c_{21}\}
\]
over all integers $c_{12}$ and $c_{21}$ such that $0 \leq c_{12}, c_{21} \leq 1$. Now
$P\{b_{12} = c_{12}\} · P\{b_{21}=c_{21}\} = 1/4$ for all such $c_{12}$ and~$c_{21}$; and
$P\{a_{12} = c_{12}, a_{21} = c_{21}\}$ equals $\nf{1}{2}$~or $0$ according as $c_{12} + c_{21}$
does or does not equal~$1$. Therefore
\begin{align*}
\Delta_{h}
  &= (\tfrac{1}{4})\Bigl(
    P\{s_{1h} < k_{1}, s_{2h} < k_{2}, s_{3}^{(h)} < k_{3}, \dots\} \\
    &\qquad- P\{s_{1h} < k_{1} - 1, s_{2h} < k_{2}, s_{3}^{(h)} < k_{3}, \dots\} \\
    &\qquad- P\{s_{1h} < k_{1}, s_{2h} < k_{2} - 1, s_{3}^{(h)} < k_{3}, \dots\} \\
    &\qquad+ P\{s_{1h} < k_{1} - 1, s_{2h} < k_{2} - 1, s_{3}^{(h)} < k_{3}, \dots\}\Bigr) \\
  &= (\tfrac{1}{4})P\{s_{1h} = k_{1} - 1, s_{2h} = k_{2} - 1, s_{3}^{(h)} < k_{3}, \dots\} \geq 0;
\end{align*}
And the same argument shows that the analogous relations hold no matter which pair has label~$h$, for $h = 1, 2, \dots, m$. Consequently,
\[
F_{m} - F_{0} = \Delta_{1} + \dots + \Delta_{m} \geq 0,
\]
or
\[
F_{0} = P\{s_{1} < k_{1}, \dots, s_{n} < k_{n}\}
  \leq F_{m}
  = P\{s_{1}^{(m)} < k_{1}, \dots, s_{n}^{(m)} < k_{n}\}.
\]
But the variables $s_{i}^{(m)} = \sum_{j \neq i} b_{ji}$ are independent and have the
same distribution as the variables $s_{i} = \sum_{j \neq i} a_{ij}$. This implies relation~\Eq{(4)}.
\Pagelabel{Thm34end}%
%[** TN: End of corrected text provided by the author]

Using \Eq{(2)} and~\Eq{(4)}, we find that
\begin{multline*}
P\{w < x_{n}^{-}\}
  < \left(1 - \frac{c_{1} (n - 1)^{-(1-\eps)^{2}}}{\sqrt{\log (n - 1)}}\right)^{n} \\
  < \exp \left(-c_{1} \frac{(n - 1)^{2\eps - \eps^{2}}}{\sqrt{\log (n - 1)}}\right)
  < \exp \bigl(-c_{1} (n - 1)^{\eps}\bigr)
\Tag{(5)}
\end{multline*}
for all sufficiently large values of~$n$.

The theorem now follows from \Eq{(3)} and~\Eq{(5)}.

David (1959) has developed a test for determining whether the largest
\index[xauthor]{David, H. A.}%
\index{Paired comparisons, the method of}%
score in a tournament is significantly larger than the average score. Other
significance tests for use in paired comparison experiments are discussed in
David and Starks (1961), David and Trawinski (1963), and Chapter~3 of
\index[xauthor]{Starks, H. T.}%
\index[xauthor]{Trawinski, B. J.}%
David (1963).


\Exercises[Exercise]

\Ex{1.} Prove Huber's result that
\index[xauthor]{Huber, P.}%
\begin{multline*}%[** TN: Reformatted slightly from the original]
\bigl[2\log (n - 1) - (1 + \eps) \log\log (n - 1)\bigr]^{1/2} \\
  < w
  < \bigl[2\log (n - 1) - (1 - \eps) \log\log (n - 1)\bigr]^{1/2}
\end{multline*}
for almost all tournaments~$T_{n}$ and any fixed positive~$\eps$.
\PageSep{73}


\Section{25.}{A Reversal Theorem}

We saw in \SecRef{5} that the number of $3$-cycles in a tournament depends
\index{Cycles in a tournament}%
on its score vector and not on its structure, as such. If the nodes $p$,~$q$, and~$r$
\index{Score vector}%
form a $3$-cycle in a tournament, then, reversing the orientation of the arcs
of this $3$-cycle does not change any scores. The following theorem is due to
Ryser (1964); it is also a special case of a more general result due to Kotzig
\index[xauthor]{Kotzig, A.}%
\index[xauthor]{Ryser, H. J.}%
(1966).

\begin{Theorem}{35.}
If the tournaments $T_{n}$ and $T_{n}'$ have the same score vector
$(s_{1}, s_{2}, \dots, s_{n})$, where $s_{1} \leq s_{2} \leq \dots \leq s_{n}$, then $T_{n}$~can be transformed into~$T_{n}'$
by successively reversing the orientation of appropriate $3$-cycles.
\end{Theorem}

\Proof. We first state the following simple lemma.

\begin{Lemma}
If $p \to  q$ in a tournament and the score of~$p$ does not exceed the
score of~$q$, then there exists a $3$-cycle containing the arc~$\Arc{pq}$.
\end{Lemma}

It follows from the proof of \ThmRef{29} that there exists a canonical
tournament~$T_{n}^{*}$ with score vector $(s_{1}, s_{2}, \dots, s_{n})$ that enjoys the following
properties:
% [** TN: List indentation added with author's approval]
\begin{itemize}
\item[(a)] If $p_{j}^{*} \to p_{n}^{*}$ and $p_{n}^{*} \to p_{i}^{*}$, then $s_{i} \leq s_{j}$.

\item[(b)] The scores~$s_{i}'$ of the tournament~$T_{n-1}^{*}$ obtained from~$T_{n}^{*}$ by removing
$p_{n}^{*}$~and its incident arcs satisfy the inequalities $s_{1}' \leq s_{2}' \leq \dots s_{n-1}'$.

\item[(c)] The analogues of Properties (a) and~(b) hold for $T_{n-1}^{*}, T_{n-2}^{*}, \dots$.
\end{itemize}

To prove \ThmRef{35}, we may assume that $T_{n}' = T_{n}^{*}$. Let us suppose the
arc joining $p_{n}$~and~$p_{j}$ has the same orientation as the arc joining $p_{n}^{*}$~and~$p_{j}^{*}$
for $j = k + 1, k + 2, \dots, n - 1$ and that the first disagreement occurs
when $j = k$.

We treat first the case in which $p_{n} \to p_{k}$ and $p_{k}^{*} \to p_{n}^{*}$. Then there must
exist a node~$p_{i}$ such that $p_{i} \to p_{n}$ and $p_{n}^{*} \to p_{i}^{*}$. Since $i < k$, it follows that
$s_{i} \leq s_{k}$. Consequently, we may suppose that $p_{k} \to p_{i}$, for otherwise, according
to the lemma, there exists a $3$-cycle containing $\Arc{p_{i}p_{k}}$ whose orientation
could be reversed without affecting any arcs incident with~$p_{n}$. We may now
reverse the orientation of the $3$-cycle $(p_{n}, p_{k}, p_{i})$ to obtain a tournament in
which the arc joining $p_{k}$ and $p_{n}$ has the same orientation as the corresponding
arc in~$T_{n}^{*}$. It is clear that the agreement of the orientation of the arcs already
considered has not been affected.

We now treat the other possibility, namely, that $p_{k} \to p_{n}$ and $p_{k}^{*} \to p_{n}^{*}$.
As before, it follows that there exists a node~$p_{i}$ such that $p_{n} \to p_{i}$ and
$p_{i}^{*} \to p_{n}^{*}$. Now $s_{i} \leq s_{k}$, since $i < k$; but it follows from Property~(a) that
$s_{k} \leq s_{i}$. Therefore, it must be that nodes $p_{k}$ and~$p_{i}$ have the same score.
Hence, we can use the same argument as before to show that $T_{n}$~can be
\PageSep{74}
transformed, so that the arc joining $p_{k}$ and~$p_{n}$ has the same orientation as the
corresponding arc in~$T_{n}^{*}$.

By repeating this process, we can transform $T_{n}$ into a tournament in which
all arcs incident with~$p_{n}$ have the same orientation as the corresponding
arcs in~$T_{n}^{*}$. Now consider the tournament~$T_{n-1}$ obtained from this transformed
tournament by removing $p_{n}$~and its incident edges; it has the same
score vector as~$T_{n-1}^{*}$. The theorem now follows by induction, as it is clearly
true for small values of~$n$.

Analogous results for other types of graphs have been given by Ryser
\index[xauthor]{Ryser, H. J.}%
(1957) and Fulkerson, Hoffman, and McAndrew (1965).
\index[xauthor]{Fulkerson, D. R.}%
\index[xauthor]{Hoffman, A. J.}%
\index[xauthor]{MacAndrew@McAndrew, M. H.}%


\Exercises

\Ex{1.} The theorem that Ryser (1964) actually proved differs from \ThmRef{35}
in that ``$3$-cycles'' is replaced by ``$3$-cycles or $4$-cycles.'' Show that the
orientation of the arcs in any $4$-cycle may be reversed by reversing the
orientation of the arcs in two $3$-cycles.

\Ex{2.} When $n \geq 4$, would \ThmRef{35} remain valid if ``$3$-cycles'' were replaced
by ``$4$-cycles''?


\Section{26.}{Tournaments with a Given Automorphism Group}
%[** TN: Original entry points to page 44]
\index{Automorphism group of a tournament}%

Let $\alpha$ denote a dominance-preserving permutation of the nodes of a
given tournament~$T_{n}$ so that $\alpha(p) \to \alpha(q)$ if and only if $p \to q$. The set of all
such permutations forms a group, the \emph{automorphism group} $G = G(T_{n})$
of~$T_{n}$. The following theorem is due to Moon (1963).
\index[xauthor]{Moon, J. W.}%

\begin{Theorem}{36.}
A finite group~$G$ is abstractly isomorphic to the automorphism
group of some tournament if and only if the order of~$G$ is odd.
\end{Theorem}

\Proof. Suppose a tournament has a group~$G$ of even order. Then $G$~contains
at least one self-inverse element~$\alpha$ other than the identity element.
Hence, there exist two different nodes $p$ and~$q$ such that $\alpha(p) = q$ and $\alpha(q) = p$.
We may assume that $p \to q$; but then $\alpha(q) \to \alpha(p)$ and this contradicts
the definition of~$G$. Thus, if a finite group~$G$ is to be isomorphic to the
group of a tournament, a necessary condition is that the order of~$G$ be odd.
We now show that this condition is also sufficient.

Let $G$~be a group of odd order whose elements are $g_{1}, g_{2}, \dots, g_{n}$. Suppose
that $g_{1}, g_{2}, \dots, g_{h}$ forms a minimal set of generators for~$G$; that is, every
element of~$G$ can be expressed as a finite product of powers of these $h$
elements, and no smaller set has this property. It is very easy to construct a
\PageSep{75}
tournament whose group is a cyclic group of given odd order (see Exercise~1)
so we shall assume that $h \geq 2$ henceforth.

In constructing a tournament whose group is isomorphic to~$G$, we begin
by forming what is essentially the Cayley color graph~$T^{*}$ of~$G$ [see Cayley
\index[xauthor]{Cayley, A.}%
(1878)]. The nodes of~$T^{*}$ correspond to the elements of~$G$. For convenience,
we use the same symbol for a node and its corresponding group element.
With each generator~$g_{j}$ we associate a certain set of arcs in~$T^{*}$ that are said
to have color~$j$. There is an arc of color~$j$ ($j = 1, 2, \dots, h$) going from $p$ to~$q$
in~$T^{*}$ if and only if $pg_{j} = q$. At each node, there is now one incoming and
one outgoing arc for each generator. No node is joined to itself by an arc,
since the identity element is not one of the generators. No two nodes are
joined by two arcs, one oriented in each direction, since the colors of these
arcs would correspond to two group elements that were the inverses of
each other; a minimal set of generators would not contain both of these
elements.

If two distinct nodes $p$ and~$q$ are not joined by an arc in this procedure,
we introduce one of the $0$th~color oriented toward $q$~or~$p$ according as the
element $p^{-1}q$~or $q^{-1}p$ has the larger subscript in the original listing of the
elements of~$G$. (These products are not equal, since $G$~contains no self-inverse
elements other than the identity.) If an arc of the $0$th~color goes
from $p$ to~$q$, then an arc of the $0$th~color also goes from $q' = pq^{-1}p$ to~$p$. Each
pair of distinct nodes of~$T^{*}$ is now joined by a colored arc and the orientations
are such that the score~$s(g)$ of each node~$g$ is equal to $\nf{1}{2} (n - 1)$.
(If $G$~is the direct product of two cyclic groups of order three, generated by
$r$~and~$s$, then $T^{*}$~is the graph illustrated in \Figure{11}; the arcs of the $0$th~color
are omitted.)
\begin{figure}[hbt!]
\centering
\Graphic[3.6in]{fig11}
\FCaption{Figure 11}
\end{figure}

The group of dominance and color-preserving automorphisms of~$T^{*}$ is
isomorphic to~$G$. Our object is to maintain this property while transforming~$T^{*}$
into a tournament~$T$ whose arcs all have the same color, or rather none
at all. We accomplish this by introducing $j$~new nodes for each arc of color~$j$
($j = 0, 1, \dots, h$).
\PageSep{76}

The new nodes are labeled $x(i, j, k)$ where $i = 1, 2, \dots, n$; $j = 1, 2, \dots, h$;
and $k = 1, 2, \dots, j$. Consider any node~$g_{i}$; if there is an arc of color~$j$
from $q$ to~$g_{i}$ in~$T^{*}$, that is, if $q = g_{i}g_{j}^{-1}$, then $q \to x(i, j, k)$ and $x(i, j, k) \to g_{i}$
in~$T$ if $2 \leq k \leq j$ but $x(i, j, 1)$~dominates both $q$ and~$g_{i}$. All the colored arcs
in~$T^{*}$ are replaced by the corresponding uncolored arcs in~$T$. Any node of the
type $x(i, j, k)$ is said to belong to~$g_{i}$ (we also say that $g_{i}$~belongs to itself).
If the nodes $x$ and~$y$ belong to the nodes $p$ and~$q$ where $p \neq q$, then $x \to y$
if and only if $p \to q$. Finally, $x(i, j, k) \to x(i, l, m)$ if and only if $j > l$ or
$j = l$ and $k > m$. This completes the definition of the tournament~$T$.
(If $T^{*}$~is the graph in \Figure{11}, then a portion of~$T$ is illustrated in \Figure{12}.)
\begin{figure}[hbt!]
\centering
\Graphic[3in]{fig12}
\FCaption{Figure 12}
\end{figure}

It is not difficult to verify, appealing to the definition of~$T$ and recalling
the exceptional character of the nodes $x(i, j, 1)$, that the following formulas
hold for the scores of the nodes of~$T$:
\index{Score of a node}%
\begin{align*}
s(g) &= (1 + 2 + \dots + h)h
  + \nf{1}{2} (n - 2h - 1) \bigl[(1 + 2 + \dots + h) + 1\bigr] \\
  &= \nf{1}{2} (n - 1) · \binom{h + 1}{2} + \nf{1}{2} (n - 2h - 1)
\end{align*}
for each node~$g$ associated with an element of~$G$;
\[
s\bigl(x(i, j, k)\bigr) = s(g) + \binom{j}{2} + h + k,
\quad\text{if $2 \leq k \leq j$,}
\]
and
\[
s\bigl(x(i, j, 1)\bigr) = s\bigl(x(i, j, 2)\bigr),
\quad\text{for all $i$ and~$j$.}
\]

The important fact here is that $s(x) > s(g)$ for every new node~$x$ and that
\[
s\bigl(x(i, j, k)\bigr) = s\bigl(x(t, l, m)\bigr)
\]
if and only if $j = l$, and $k = m$ or $km = 2$. Since $s(p) = s\bigl(\alpha(p)\bigr)$ for any
admissible automorphism~$\alpha$ of~$T$, it follows, in particular, that the identity
of the sets of original nodes~$g$ and new nodes~$x$ is preserved by~$\alpha$.

{\Loosen We now prove that the automorphism group of~$T$ is isomorphic to~$G$.
Let $\alpha$~be any automorphism that leaves some node~$g_{i}$ fixed, that is, $\alpha(g_{i}) = g_{i}$.
We first show that this implies that $\alpha$~leaves every node of~$T$ fixed.}
\PageSep{77}

Consider any node $x = x(i, j, k)$ where $k \geq 2$ that belongs to~$g_{i}$. At least
one such node exists since $h \geq 2$. Recall that $x \to g_{i}$; hence $\alpha(x)$~could
belong to a node~$g_{i}$ in~$\alpha(T)$ such that $g_{i} \to g_{l}$ only if
\[
\alpha(x) = x(l, j, 1).
\]
If this happens, there exists a node~$y$ (take $y = x(l, 2, 2)$ if $j = 1$ or $y = x(l, j, 2)$
if $j \neq 1$) such that $g_{i} \to y$ and $y \to \alpha(x)$ in~$\alpha(T)$. But there exists no
such corresponding path from $g_{i}$ to~$x$ in~$T$, so this possibility is excluded.
If $\alpha(x)$~belongs to a node~$g_{l}$ in~$\alpha(T)$ such that $g_{l} \to g_{i}$, then $\alpha(x) \to  g_{l}$
and $g_{l} \to  g_{i}$. But there is no corresponding path of length two from $x$ to~$g_{i}$
in~$T$ whose intermediate node is one of the nodes~$g$. The only alternative
is that $\alpha(x)$~belongs to $g_{i}$ itself. Since $s(x) = s\bigl(\alpha(x)\bigr)$, it follows that $\alpha(x) = x$
except, possibly, when $k = 2$ and $\alpha(x) = x(i, j, 1)$. This last alternative can
be ruled out by considering the types of paths of length two from $x$ to~$g_{i}$.

Now consider any remaining node
\[
x = x(i, j, 1)
\]
that belongs to~$g_{i}$. It follows by the same argument as before that $\alpha(x)$~cannot
belong to a node~$g_{i}$ in~$\alpha(T)$ such that $g_{i} \to g_{l}$. But if $\alpha(x)$~belongs to~$g_{l}$
where $g_{l} \to g_{i}$, then
\[
\alpha(x) \to \alpha\bigl(x(i, h, 2)\bigr) = x(i, h, 2).
\]
This contradicts the fact that $\alpha$~preserves the orientation of the arc going
from $x(i, h, 2)$ to~$x$. Hence, $\alpha(x)$~belongs to~$g_{i}$. The only conclusion compatible
with these results and the condition that $s(x) = s\bigl(\alpha(x)\bigr)$ is that, if
$\alpha(g_{i}) = g_{i}$, then $\alpha(x) = x$ for every node~$x$ that belongs to~$g_{i}$.

Now consider any node $g = g_{i}g_{j}^{-1}$ ($1 \leq j \leq h$). Since $x(i, j, 1) \to g$ and
$g \to g_{i}$, it must be that $x(i, j, 1) \to \alpha(g)$ and $\alpha(g) \to g_{i}$ because $x(i, j, 1)$ and~$g_{i}$
are fixed under~$\alpha$. But $g$~is the only node associated with an element of~$G$
that has this property. (This is where the exceptional property of $x(i, j, 1)$
is used.) Hence, $\alpha(g) = g$ for each such~$g$.

The tournament~$T$ is strongly connected; this follows from the way $T$~is
\index{Strong tournament}%
constructed from~$T^{*}$ which itself is strongly connected (see Exercise~3).
Hence, by repeating the above argument as often as is necessary we eventually
conclude that, if $\alpha(g_{i}) = g_{i}$ for any node~$g_{i}$, then $\alpha(p) = p$ for every
node~$p$ in~$T$.

We have shown that, if $\alpha$~is not the identity element, then $\alpha(g) \neq g$ for
every node~$g$ associated with an element of~$G$. It follows from this that,
for any two such nodes $g_{u}$ and~$g_{v}$, there exists at most one automorphism~$\alpha$
of~$T$ such that $\alpha(g_{u}) = g_{v}$, (see Exercise~4). But, the group element $\alpha = g_{v}g_{u}^{-1}$
induces such an automorphism defined as follows: $\alpha(g) = \alpha g$ for all nodes~$g$
associated with an element of~$G$, and $\alpha\bigl(x(i, j, k)\bigr) = x(l, j, k)$ if $g_{l} = \alpha(g_{i})$,
for all $i$,~$j$, and~$k$. It follows from these results that the group of the tournament~$T$
\PageSep{78}
is abstractly isomorphic to the group~$G$. This completes the proof of
the theorem.


\Exercises

\Ex{1.} Construct a tournament~$T_{n}$ whose group is the cyclic group~$C_{n}$ when $n$~is
odd.

\Ex{2.} Verify the formulas given for the scores of the nodes of~$T$.

\Ex{3.} Prove that the tournaments $T^{*}$ and~$T$ are strongly connected.

\Ex{4.} Prove that there exists at most one automorphism~$\alpha$ of~$T$ such that
\index{Automorphism group of a tournament}%
$\alpha(g_{u}) = g_{v}$ where $g_{u}$ and~$g_{v}$ are any two nodes of~$T$ that are associated
with elements of~$G$.

\Ex{5.} Prove that, if $G$~is a finite group of odd order, then there exist infinitely
many strong tournaments whose group is isomorphic to~$G$.

\Ex{6.} Let $t(G)$ denote the number of nodes in the smallest tournament whose
group is isomorphic to the group~$G$. If $G$~has odd order~$n$ and is generated
by $h$~of its elements, then the tournament~$T$ shows that
\[
t(G) \leq n \binom{h + 1}{2} + n.
\]
Prove that $t(G) \leq n$ if $G$~is abelian.

\Ex{7.} What is the smallest odd integer~$n$ ($n \geq 3$) for which there exists a regular
tournament~$T_{n}$ such that $G(T_{n})$~is the identity group?

\Ex{8.} Construct a nontransitive infinite tournament~$T$ such that $G(T)$~is the
identity group. [See Chvátal (1965).]
\index[xauthor]{Chvátal, V.}%


\Section[Group of the Composition of Two Tournaments]
{27.}{The Group of the Composition of Two Tournaments}
\index{Automorphism group of a tournament}%
\index{Composition!of two groups}%
\index{Composition!of two tournaments}%

Let $R$ and~$T$ denote two tournaments with nodes $r_{1}, r_{2}, \dots, r_{a}$ and
$t_{1}, t_{2}, \dots, t_{b}$. The \emph{composition} of $R$ with~$T$ is the tournament $R \circ T$ obtained
by replacing each node~$r_{i}$ of~$R$ by a copy~$T(i)$ of~$T$ so that, if $r_{i} \to r_{j}$ in~$R$,
then every node of~$T(i)$ dominates every node of~$T(j)$ in~$R \circ T$. More precisely,
there are $ab$~nodes $p(i, k)$ in~$R \circ T$ ($1 \leq i \leq a$, $1 \leq k \leq b$) and $p(i, k) \to  p(j, l)$
if and only if $r_{i} \to r_{j}$, or $i = j$ and $t_{k} \to t_{l}$. The composition of two $3$-cycles is
illustrated in \Figure{13}.

Let $F$ and~$H$ denote two permutation groups with object sets $U$ and~$V$.
The \emph{composition} [see Pólya (1937)] of $F$ with~$H$ is the group $F \circ H$ of all
\index[xauthor]{Polya@Pólya, G.}%
permutations~$\alpha$ of $U × V = \{(x, y) : x \in U, y \in V\}$ of the type
\[
\alpha(x, y) = \bigl(f(x), h_{x}(y)\bigr),
\]
\PageSep{79}
\begin{figure}[hbt!]
\centering
\Graphic[2.25in]{fig13}
\FCaption{Figure 13}
\end{figure}
where $f$~is any element of~$F$ and $h_{x}$, for each~$x$, is any element of~$H$. If
the objects of $U × V$ are arranged in a matrix so that the rows and columns
correspond to the objects of $U$~and $V$, respectively, then $F \circ H$ is the
group of permutations obtained by permuting the objects in each row
according to some element of~$H$ (not necessarily the same element for
every row) and then permuting the rows themselves according to some
element of~$F$. If $F$~and~$H$ have order $m$~and~$n$ and degree $u$~and~$v$, then $F \circ H$
has order~$mn^{u}$ and degree~$uv$.

We now prove that the group of the composition of two tournaments is
equal to the composition of their groups.

\begin{Theorem}{37.}
$G(R \circ T) = G(R) \circ G(T)$.
\end{Theorem}

\Proof. The nodes $p(i, k)$ in each copy~$T(i)$ may be permuted according to
any element of~$G(T)$ and the copies themselves may be permuted according
to any element of~$G(R)$, so $G(R) \circ G(T)$ is certainly a subgroup of~$G(R \circ T)$.
To prove the groups are the same, it will suffice to show that, if a permutation~$\alpha$
of~$G(R \circ T)$ takes any node of~$T(i)$ into a node of~$T(j)$, then $\alpha$~takes
every node of~$T(i)$ into~$T(j)$.

Suppose, on the contrary, that there exists a permutation~$\alpha$ of~$G(R \circ T)$
such that the nodes of~$\alpha\bigl(T(i)\bigr)$ do not all belong to~$T(j)$. Let $X$~denote the set
of all nodes~$p(i, k)$ of~$T(i)$ such that $\alpha\bigl(p(i, k)\bigr)$ is in~$T(j)$, where $X$~and $T(i) - X$
are nonempty. There is no loss of generality (see Exercise~1) if we assume
that $i \neq j$ and that every node of~$X$ dominates every node of~$\alpha(X)$ (symbolically,
$X \to \alpha\bigl(X)$). If any node~$p$ in~$T(i) - X$ both dominates and is
dominated by nodes of~$X$, then $\alpha(p)$~must be in~$T(j)$ (see Exercise~2), contrary
to the definition of~$X$. Hence, one of the following alternatives holds with
respect to~$T(i)$.

(1a) There exists a node~$p$ in~$T(i) - X$ such that $p \to X$.

(1b) Every node in~$X$ dominates every node in~$T(i) - X$.

Similarly, one of the following alternatives holds with respect to~$T(j)$.

(2a) There exists a node~$q$ in~$T(j) - \alpha(X)$ such that $q \to \alpha(X)$.

(2b) Every node in~$\alpha(X)$ dominates every node in~$T(j) - \alpha(X)$.
\PageSep{80}

We first suppose that alternatives (1a) and (2a) hold. Since $p \to X$,
it follows that $\alpha(p) \to \alpha(X)$. Since $\alpha(p)$~is not in~$T(j)$, it must be that $\alpha(p) \to T(j)$,
appealing to the definition of~$R \circ T$; in particular, $\alpha(p) \to q$. Similarly,
$\alpha^{-1}(q) \to X$ since $q \to \alpha(X)$, and $a^{-1}(q) \to T(i)$ since $\alpha^{-1}(q)$~is not in~$T(i)$; in
particular, $a^{-1}(q) \to p$. Thus there exist two nodes, $\alpha^{-1}(q)$ and~$p$, such that
$\alpha^{-1}(q) \to p$ and $\alpha(p) \to \alpha\bigl(\alpha^{-1}(q)\bigr) = q$. This contradicts the fact that $\alpha$~is
dominance-preserving.

We next suppose that alternatives (1a) and (2b) hold. The score of any
\index{Score of a node}%
node~$p(i, k)$ in~$R \circ T$ is given by the formula
\[
s\bigl(p(i, k)\bigr) = s(t_{k}) + b · s(r_{i}),
\]
where $s(t_{k})$ and~$s(r_{i})$ denote the scores of $t_{k}$ and~$r_{i}$ in $T$~and~$R$. It follows
that if $\alpha\bigl(p(i, k)\bigr) - p(j, l)$ then $s(t_{k}) = s(t_{l})$, since
\[
s(t_{k}) + b · s(r_{i}) = s(t_{l}) + b · s(r_{j})
\]
and $0 \leq s(t_{k}), s(t_{l}) < b$. Consequently,
\[
\sum_{p(i, k) \in X} s(t_{k}) = \sum_{p(j, l) \in \alpha(X)} s(t_{l}).
\]
But if there are $m$~nodes in~$X$ and in~$\alpha(X)$, then
\[
\sum_{p(i, k) \in X} s(t_{k}) \leq \binom{m}{2} + m(b - 1 - m),
\]
by (1a), and
\[
\sum_{p(j, l) \in \alpha(X)} s(t_{l}) = \binom{m}{2} + m(b - m),
\]
by~(2b). Therefore, the two sums cannot be equal and we have again reached
a contradiction.

It remains to treat the cases when alternatives (1b) and (2a)~or (2b) hold.
We shall omit the arguments for these cases, since they are similar to the
arguments already used (see Exercise~3).

It follows from these arguments that every permutation of~$G(R \circ T)$
preserves the identity of the various copies~$T(i)$. Since every permutation
with this property belongs to $G(R) \circ G(T)$, the theorem is proved.

Sabidussi (1961) and Hemminger (1966) have proved an analogous result
\index[xauthor]{Hemminger, R. L.}%
\index[xauthor]{Sabidussi, G.}%
for ordinary graphs; there the problem is complicated by the fact that there
are certain exceptional cases in which the result does not hold.


\Exercises

\Ex{1.} Why may we assume that $X \to \alpha(X)$ in the proof of \ThmRef{37}?

\Ex{2.} Why may we assume that $\alpha(p)$~is in~$T(j)$ if $p$~both dominates and is
dominated by nodes of~$X$?

\Ex{3.} Supply the arguments omitted in the proof of \ThmRef{37}.
\PageSep{81}

\Ex{4.} Let $C_{1}$ denote a $3$-cycle and let $C_{k} = C_{1} \circ C_{k-1}$ for $k \geq 2$. Show that the
\index{Automorphism group of a tournament}%
group of~$C_{k}$ has order $3^{(1/2)(3^{k}-1)}$.

\Ex{5.} For what tournaments $R$ and $T$ is it true that $R \circ T = T \circ R$?


\Section{28.}{The Maximum Order of the Group of a Tournament}

The groups $G(T_{n})$ of the tournaments~$T_{n}$ are specified in the appendix
for $n \leq 6$. If $g(T_{n})$~denotes the order of the group~$G(T_{n})$, let $g(n)$~denote the
maximum of~$g(T_{n})$ taken over all tournaments~$T_{n}$. The entries in \Table{6}
were given by Goldberg (1966).
\index[xauthor]{Goldberg, M.}%
\begin{table}[hbt!]
\TCaption{Table 6. $g(n)$, the maximum possible order of the group
of a tournament~$T_{n}$.}
\[
\begin{array}[t]{r>{\quad}r>{\quad}c}
\multicolumn{1}{c}{n} & g(n) & g(n)^{1/(n-1)} \\
\hline\Strut
 1 &   1  & \text{---} \\
 2 &   1  & 1\phantom{.999} \\
 3 &   3  & 1.732 \\
 4 &   3  & 1.442 \\
 5 &   5  & 1.495 \\
 6 &   9  & 1.552 \\
 7 &  21  & 1.662 \\
 8 &  21  & 1.545 \\
 9 &  81  & 1.732 \\
10 &  81  & 1.629 \\
11 &  81  & 1.552 \\
12 & 243  & 1.647 \\
13 & 243  & 1.581 \\
14 & 441  & 1.597
\end{array}
\quad
\begin{array}[t]{r>{\quad}r>{\quad}c}
\multicolumn{1}{c}{n} & g(n)\quad\null & g(n)^{1/(n-1)} \\
\hline\Strut
15 &     1,215 & 1.661 \\
16 &     1,701 & 1.643 \\
17 &     1,701 & 1.592 \\
18 &     6,561 & 1.677 \\
19 &     6,561 & 1.629 \\
20 &     6,561 & 1.588 \\
21 &    45,927 & 1.710 \\
22 &    45,927 & 1.667 \\
23 &    45,927 & 1.629 \\
24 &   137,781 & 1.673 \\
25 &   137,781 & 1.637 \\
26 &   229,635 & 1.639 \\
27 & 1,594,323 & 1.732
\end{array}
\]
\end{table}

Goldberg and Moon (1966) obtained bounds for~$g(n)$. Their argument is
\index[xauthor]{Moon, J. W.}%
based on a simple result of group theory [see Burnside (1911, pp.\ 170~and
\index[xauthor]{Burnside, W.}%
185)].

Let $G$~be a group of permutations that act on the elements of a finite set~$X$.
Two elements $x$~and~$y$ are \emph{equivalent} (with respect to~$G$) if and only if there
\index{Equivalent elements in a set}%
exists a permutation~$\alpha$ in~$G$ such that $\alpha(x) = y$. Then $X$~can be partitioned
into classes of equivalent elements.

\begin{Theorem}{38.}
If $x \in X$, let $E(x) = \{\alpha(x) : \alpha \in G\}$ and $F(x) = \{\gamma : \gamma \in G \text{ and } \gamma(x) = x\}$. Then
\[
|G| = |E(x)| · |F(x)|.
\]
\end{Theorem}
\PageSep{82}

\Proof. If $x$ and $y$ are any two elements in the same equivalence class of~$X$
with respect to~$G$, let
\[
g(x, y) = \big|\{\alpha : \alpha \in G \text{ and } \alpha(x) = y\}\big|.
\]
We shall show that
\[
g(x, y) = g(y, y).
\Tag{(1)}
\]

Suppose the equation $\alpha_{i}(x) = y$ holds if and only if $i = 1, 2, \dots, h$. Then
$\beta_{i}(y) = y$, where $\beta_{i} = \alpha_{i}\alpha_{1}^{-1}$ for $i = 1, 2, \dots, h$; the $\beta_{i}$'s are clearly distinct.
If $\beta(y) = y$; then $\beta\alpha_{1}(x) = y$; consequently, $\beta\alpha_{1} = \alpha_{i}$ for some~$i$ such that
$1 \leq i \leq h$ and $\beta = \alpha_{i}\alpha_{1}^{-1}$. This completes the proof of~\Eq{(1)}.

Since $g(x, y) = g(y, x)$, it follows from~\Eq{(1)} that
\[
g(x, x) = g(y, y)
\]
for all elements~$y$ in $E = E(x)$. Therefore,
\[
|G| = \sum_{y \in E} g(x, y)
    = \sum_{y \in E} g(y, y)
    = g(x, x) \sum_{y \in E} 1
    = |F(x)| · |E(x)|,
\]
since each element of~$G$ is counted once and only once in the first sum.

\begin{Theorem}{39.}
The limit of~$g(n)^{1/n}$ as $n$~tends to infinity exists and lies
between $\sqrt{3}$ and~$2.5$, inclusive.
\end{Theorem}

\Proof. We shall first prove by induction that
\[
g(n) \leq \frac{(2.5)^{n}}{2n}
\Tag{(2)}
\]
if $n \geq 4$. The exact values of~$g(n)$ in \Table{6} can be used to verify that this
inequality holds when $4 \leq n \leq 9$.

Consider any node~$p$ of an arbitrary tournament~$T_{n}$, where $n \geq 10$.
Let $e$~denote the number of different nodes in the set
\[
E = \{\alpha(p) : \alpha \in G(T_{n})\}.
\]
If $T_{e}$ and $T_{n-e}$ denote the subtournaments determined by the nodes that are
\index{Subtournament}%
in~$E$ and by the nodes that are not in~$E$, then it is clear that
\[
g(T_{n}) \leq g(T_{e}) · g(T_{n-e}) \leq g(e) · g(n - e).
\Tag{(3)}
\]

If $3 < e < n - 3$, then it follows from the induction hypothesis that
\[
g(T_{n}) \leq \frac{(2.5)^{e}}{2e} · \frac{(2.5)^{n-e}}{2(n - e)}
  \leq \frac{n}{8(n - 4)} · \frac{(2.5)^{n}}{2n}
  < \frac{(2.5)^{n}}{2n}.
\]
If $e = 3$ or $n - 3$, then
\[
g(T_{n}) \leq 3 · \frac{(2.5)^{n-3}}{2(n - 3)}
  < \frac{(2.5)^{n}}{2n},
\]
\PageSep{83}
and if $e = 1$, $2$, $n - 2$, or~$n - 1$, then
\[
g(T_{n}) \leq 1 · \frac{2n}{5(n - 2)} · \frac{(2.5)^{n}}{2n}
  < \frac{(2.5)^{n}}{2n}.
\]
A different argument must be used when $e = n$.

There are $\nf{1}{2} n(n - 1)$ arcs in the tournaments~$T_{n}$. Hence, if $e = n$ and
the nodes of~$T_{n}$ are all similar to each other with respect to the group~$G(T_{n})$, it must be that each node has score $\nf{1}{2} (n - 1)$. This can happen only
\index{Score of a node}%
when $n$~is odd.

Consider the subgroup~$F$ of automorphisms~$\alpha$ of~$G(T_{n})$ such that $\alpha(p) = p$.
It follows from \ThmRef{38} that, if $e = n$, then
\[
g(T_{n}) = n|F|.
\]
No element of~$F$ can transform one of the $\nf{1}{2} (n - 1)$ nodes that dominate~$p$
into one of the $\nf{1}{2}(n - 1)$ nodes dominated by~$p$, since $p$~is fixed. Hence,
\[
|F| \leq \left(g\left(\frac{n - 1}{2}\right)\right)^{2}.
\]
Therefore, if $e = n$, then
\[
g(T_{n}) \leq n\left(\frac{(2.5)^{(1/2)(n-1)}}{n - 1}\right)^{2}
  = \frac{4}{5} \left(\frac{n}{n - 1}\right)^{2} · \frac{(2.5)^{n}}{2n}
  < \frac{(2.5)^{n}}{2n}.
\]
(Notice that $\nf{1}{2} (n - 1) \geq 5$ if $n \geq 11$, so we are certainly entitled to apply
the induction hypothesis to $g\bigl((n - 1)/2\bigr)$.) This completes the proof of inequality~\Eq{(2)}.

An immediate consequence of~\Eq{(2)} is that
\[
\limsup g(n)^{1/n} < 2.5.
\Tag{(4)}
\]

If $T_{a} \circ T_{b}$ denotes the composition of $T_{a}$ with~$T_{b}$, then $g(T_{a} \circ T_{b}) = g(T_{a}) \bigl[g(T_{b})\bigr]^{a}$,
\index{Composition!of two tournaments}%
by \ThmRef{37}. Therefore,
\[
g(ab) \geq g(a) g(b)^{a}
\Tag{(5)}
\]
for all integers $a$ and~$b$. Since $g(3) = 3$, it follows by induction that
\[
g(n) \geq \sqrt{3}^{n-1}
\Tag{(6)}
\]
if $n$~is a power of three (see Exercise~27.4). Hence,
\[
\limsup g(n)^{1/n} \geq \sqrt{3}.
\Tag{(7)}
\]

{\Loosen We now use inequality~\Eq{(5)} to prove the following assertion: If $g(m)^{1/m} > \gamma$,
then $g(n)^{1/n} > \gamma - \eps$ for any positive~$\eps$ and all sufficiently large~$n$.}

We may suppose that $\gamma > 1$. Let $l$~be the least integer such that $\gamma^{-1/l} >  1 - \eps/\gamma$.
Every sufficiently large integer~$n$ can be written in the form
$n = km + t$, where $k > l$ and $0 \leq t < m$. Then,
\PageSep{84}
\begin{align*}
g(n)^{1/n}
  &= g(km + t)^{1/km+t} \geq g(km)^{1/m(k+1)} \\
  &\geq \bigl[g(m)^{1/m}\bigr]^{k/k+1} > \gamma^{k/k+1} \\
  &> \gamma^{l/l+1} > \gamma^{1 - 1/l} \\
  &> \gamma \left(1 - \dfrac{\eps}{\gamma}\right) = \gamma - \eps.
\end{align*}

Let $\beta = \limsup g(n)^{1/n}$. (We know that $\sqrt{3} \leq \beta \leq 2 \nf{1}{2}$.) For every positive~$\eps$
there exists an integer~$m$ such that
\[
g(m)^{1/m} > \beta - \eps.
\]
But then,
\[
g(n)^{1/n} > \beta - 2\eps
\]
for all sufficiently large~$n$. Hence,
\[
\liminf g(n)^{1/n} > \beta - 2\eps
\]
for every positive~$\eps$. Therefore,
\[
\liminf g(n)^{1/n} = \limsup g(n)^{1/n}.
\Tag{(8)}
\]

\ThmRef{39} now follows from \Eq{(4)},~\Eq{(7)}, and~\Eq{(8)}. (The actual value of the
limit is~$\sqrt{3}$; see Exercise~5.)


\Exercises

\Ex{1.} Prove that $g(n) = \max \{g(d) · g(n - d)\}$, $d = 1, 3, 5, \dots, n - 1$, if $n$~is
even.

\Ex{2.} Verify the entries in \Table{6} when $7 \leq n \leq 11$.

\Ex{3.} Let $T_{a}, T_{b}, \dots$ denote the subtournaments determined by the classes of
\index{Isomorphic tournaments}%
\index{Subtournament}%
nodes that are equivalent with respect to~$G(T_{n})$. Under what circumstances
is it true that
\[
g(T_{n}) = g(T_{a}) · g(T_{b}) \dots?
\]

\Ex{4.} Let $h(n)$ denote the order of any largest subgroup~$H$ of odd order of the
symmetric group on $n$~objects. Prove that $h(n) = g(n)$. [Hint: Construct a
tournament~$T_{n}$ such that $H$~is a subgroup of~$G(T_{n})$.]

\Ex{5.} Prove that $g(n) \leq \sqrt{3}^{n-1}$ with equality holding if and only if $n$~is a
power of three. [Dixon (1967).]
\index[xauthor]{Dixon, J. D.}%


\Section{29.}{The Number of Nonisomorphic Tournaments}

Before determining the number of nonisomorphic tournaments~$T_{n}$, we
derive a result due to Burnside (1911, p.~191) that is used in dealing with a
\index[xauthor]{Burnside, W.}%
\PageSep{85}
general class of enumeration problems. A theory of enumeration has been
developed by Pólya (1937) and de~Bruijn (1964); Harary (1964) has given a
\index[xauthor]{Debruijn@de Bruijn, N. G.}%
\index[xauthor]{Harary, F.}%
\index[xauthor]{Polya@Pólya, G.}%
summary of results on the enumeration of graphs.

\begin{Theorem}{40.}
Let $G$ denote a permutation group that acts on a finite set~$X$
and let $f(\alpha)$~denote the number of elements of~$X$ left fixed by the permutation~$\alpha$.
Then the number~$m$ of equivalence classes of~$X$ with respect to~$G$ is given
\index{Equivalent elements in a set}%
by the formula
\[
m = \frac{1}{|G|} \sum_{\alpha \in G} f(\alpha).
\]
\end{Theorem}

\Proof. We showed in proving \ThmRef{38} that
\[
\sum_{x \in E} g(x, x) = |G|,
\]
where the sum is over the elements in any equivalence class~$E$ of~$X$ and
where $g(x, x)$~denotes the number of permutations of~$G$ that leave $x$ \Typo{fiixed}{fixed}.
Therefore,
\begin{align*}
\sum_{\alpha \in G} f(\alpha)
  &= \sum_{x \in X} g(x, x) \\
  &= \sum_{E} \left(\sum_{x \in E} g(x,x)\right)
   = \sum_{E} |G| \\
  &= m|G|.
\end{align*}
This proves the theorem.

Davis (1953) used \ThmRef{40} to enumerate various relations on a finite
\index[xauthor]{Davis, R. L.}%
set; later, in (1954), he dealt specifically with the problem of determining~$T(n)$,
the number of nonisomorphic tournaments~$T_{n}$. We now derive his
formula for~$T(n)$.

Any permutation~$\pi$ that belongs to the symmetric group~$S_{n}$ can be
expressed as a product of disjoint cycles. If the disjoint cycle representation
of~$\pi$ contains $d_{k}$~cycles of length~$k$, for $k = 1, 2, \dots, n$, then $\pi$~is said to be of
\index{Cycle type of a permutation}%
(\emph{cycle}) \emph{type} $(d) = (d_{1}, d_{2}, \dots, d_{n})$. For example, the permutation
\[
\pi = \left(\begin{array}{@{}*{7}{c}@{}}
1 & 2 & 3 & 4 & 5 & 6 & 7 \\
1 & 2 & 4 & 3 & 6 & 7 & 5
\end{array}\right) = (1)(2)(34)(567)
\]
is of type $(2, 1, 1, 0, 0, 0, 0)$. Notice that
\[
1 · d_{1} + 2 · d_{2} + \dots + n · d_{n} = n.
\]

In the present context, we think of the permutations~$\pi$ as acting on the
nodes of tournaments~$T_{n}$; these permutations~$\pi$ then, in effect, define
permutations among the tournaments themselves. We must determine~$f(\pi)$,
the number of tournaments~$T_{n}$ such that $\pi(T_{n}) = T_{n}$. (Recall that
$T_{n} = T_{n}'$ if and only if the arc joining $p_{i}$~and~$p_{j}$ has the same orientation in
both tournaments for all pairs of distinct values of $i$~and~$j$.)
\PageSep{86}

\begin{Lemma}
If the permutation~$\pi$ is of type $(d_{1}, d_{2}, \dots, d_{n})$, then $f(\pi) = 0$ if $\pi$~has
any cycles of even length; otherwise $f(\pi) = 2^{D}$, where
\[
D = \frac{1}{2} \left\{
  \sum_{k,l=1}^{n} d_{k}d_{l}(k, l) - \sum_{k=1}^{n} d_{k}
\right\}
\Tag{(1)}
\]
and $(k, l)$ denotes the greatest common divisor of $k$~and~$l$.
\end{Lemma}

\Proof. We shall give the proof only for the case when all the cycles of~$\pi$
have odd length (see Exercise~1). Let $T_{n}$~be any tournament such that $\pi(T_{n}) = T_{n}$.
Then $T_{n}$~can be partitioned into subtournaments $T^{(1)}, T^{(2)}, \dots$ in such
a way that two nodes $p_{i}$ and~$p_{j}$ belong to the same subtournament if and only
\index{Subtournament}%
if $i$~and~$j$ belong to the same cycle of~$\pi$. Among the subtournaments $T^{(1)}, T^{(2)}, \dots$,
there are $d_{1}$~that contain a single node, $d_{3}$~that contain three
nodes, and so forth.

Let $p_{1}, p_{2}, \dots, p_{k}$ and $p_{k+1}, p_{k+2}, \dots, p_{k+l}$ denote the nodes of two of these
subtournaments, $T^{(1)}$~and~$T^{(2)}$ say. We may suppose the permutation~$\pi$
contains the cycles $(1, 2, \dots, k)$ and $(k + 1, k + 2, \dots, k + l)$. If the
$\nf{1}{2} (k - 1)$ arcs joining $p_{1}$ to~$p_{i}$ ($i = 2, 3, \dots, \nf{1}{2}(k + 1)$) are oriented
arbitrarily, then the orientations of the remaining arcs of~$T^{(1)}$ are determined
uniquely by the condition that $\pi(T_{n}) = T_{n}$. Furthermore, if the $(k, l)$ arcs
joining $p_{1}$ to~$p_{k+i}$ ($i = 1, 2, \dots, (k, l)$) are oriented arbitrarily, then the
orientations of the remaining arcs joining $T^{(1)}$~and~$T^{(2)}$ also are determined
uniquely by the condition that $\pi(T_{n}) = T_{n}$. This argument can be repeated
as often as necessary. Therefore, in constructing tournaments~$T_{n}$ such that
$\pi(T_{n}) = T_{n}$, we are free to orient arbitrarily only
\begin{multline*}
\sum_{k=1}^{n} \frac{1}{2} d_{k}(k - 1)
  + \sum_{k=1}^{n} \binom{d_{k}}{2} (k, k)
  + \sum_{k<l} d_{k}d_{l}(k, l) \\
  = \frac{1}{2} \left\{\sum_{k,l=1}^{n} d_{k}d_{l}(k, l)
    - \sum_{k=1}^{n} d_{k}\right\}
\end{multline*}
arcs, the orientations of the remaining arcs being determined by the
orientations of these. The lemma now follows.

We have seen that, if $\pi$~is a permutation of degree~$n$, then $f(\pi)$~depends
only on the cycle type of~$\pi$. There are
\[
\frac{n!}{N}
  = \frac{n!}{1^{d_{1}} d_{1}!\, 2^{d_{2}} d_{2}! \dotsm n^{d_{n}} d_{n}!}
\Tag{(2)}
\]
permutations~$\pi$ of type $(d_{1}, d_{2}, \dots, d_{n})$. (See Exercise~2.) The preceding
lemma and \ThmRef{40} imply the following result.

\begin{Theorem}{41.}
If $T(n)$~denotes the number of nonisomorphic tournaments~$T_{n}$,
then
\[
T(n) = \sum_{(d)} \frac{2^{D}}{N},
\]
\PageSep{87}
where $D$ and~$N$ are as defined in \Eq{(1)} and~\Eq{(2)} and where the sum is over all
solutions~$(d)$ in nonnegative integers of the equation
\[
1 · d_{1} + 3 · d_{3} + 5 · d_{5} + \dots = n.
\]
\end{Theorem}

We illustrate the use of \ThmRef{41} by determining~$T(6)$. The necessary
calculations may be summarized as follows.
\begin{gather*}
\begin{array}{c>{\qquad}c>{\qquad}r}
(d) & N & D \\
\hline\Strut
d_{1} = 6 & 6! & 15 \\
\hline\Strut
d_{1} = 3,\ d_{3} = 1 &  3 · 3! & 7 \\
\hline\Strut
d_{1} = 1,\ d_{5} = 1 & 5 & 3 \\
\hline\Strut
d_{3} = 2 & 2 · 3^{2} & 5
\end{array}\displaybreak[0] \\
T(6) = \frac{2^{15}}{6!} + \frac{2^{7}}{3 · 3!} + \frac{2^{3}}{5} + \frac{2^{5}}{2 · 3!} = 56.
\end{gather*}

The values of~$T(n)$ for $n = 1, 2, \dots, 12$ are given in \Table{7}; the first eight
of these values were given by Davis (1954).
\index[xauthor]{Davis, R. L.}%
\begin{table}[hbt!]
\TCaption{Table 7. $T(n)$, the number of nonisomorphic tournaments~$T_{n}$.}
\[
\begin{array}{r>{\qquad}r}
 n &       T(n) \\
 1 &               1 \\
 2 &               1 \\
 3 &               2 \\
 4 &               4 \\
 5 &              12 \\
 6 &              56 \\
 7 &             456 \\
 8 &           6,880 \\
 9 &         191,536 \\
10 &       9,733,056 \\
11 &     903,753,248 \\
12 & 154,108,311,168
\end{array}
\]
\end{table}
\PageSep{88}
\index{Automorphism group of a tournament}%

\begin{Theorem}{42.}
$T(n) \sim \dfrac{2^{\ebinom{n}{2}}}{n!}$ as $n$~tends to infinity.
\end{Theorem}

\Proof. The two largest terms in the formula for~$T(n)$ are
\[
\frac{2^{\ebinom{n}{2}}}{n!}\quad\text{and}\quad
\frac{2^{\ebinom{n}{2} - 2(n-2)}}{3 · (n - 3)!};
\]
they arise from permutations of type $(n, 0, \dots, 0)$ and $(n - 3, 1, 0, \dots, 0)$.
(See Exercise~3.) The number of terms in the formula is equal to the number
of partitions of~$n$ into odd integers. Erdös (1942)\index[xauthor]{Erdös, P.} has shown that the total
number of partitions of~$n$ is less than $2^{cn^{1/2}}$, where $c = \pi(\nf{2}{3})^{1/2} \log_{2} e$. Hence,
\[
\frac{2^{\ebinom{n}{2}}}{n!} \leq T(n)
  \leq \frac{2^{\ebinom{n}{2}}}{n!}
    + 2^{cn^{1/2}} · \frac{2^{\ebinom{n}{2} - 2(n-2)}}{3 · (n - 3)!}
  = \frac{2^{\ebinom{n}{2}}}{n!} \bigl(1 + o(1)\bigr).
\]


\Exercises

\Ex{1.} Prove that $f(\pi) = 0$ if the permutation~$\pi$ has any even cycles.

\Ex{2.} Prove the assertion involving expression~\Eq{(2)}.

\Ex{3.} Prove the first assertion in the proof of \ThmRef{42}.

\Ex{4.} Prove that
\[
T(n) = \frac{\Erratum{2^{\ebinom{2}{n}}}{2^{\ebinom{n}{2}}}}{n!}
     + \frac{2^{(1/2)(n^{2} - 5n + 8)}}{3 · (n - 3)!} \bigl(1 + o(1)\bigr).
\]

\Ex{5.} Let $t(n)$ denote the number of nonisomorphic strong tournaments~$T_{n}$.
\index{Isomorphic tournaments}%
\index{Strong tournament}%
Show that if
\[
T(x) = \sum_{n=1}^{\infty} T(n)x^{n}\quad\text{and}\quad
t(x) = \sum_{n=1}^{\infty} t(n)x^{n},
\]
then $t(x) = T(x)/\bigl[1 + T(x)\bigr]$. Use this result to determine~$t(n)$ for $1 \leq n \leq 6$.

\Ex{6.} Is $T(n)$ always even when $n \geq 3$?

\Ex{7.} Determine the number of nonisomorphic tournaments~$T_{n}$ such that
$T_{n}$~is not transitive but can be transformed into a transitive tournament by
reversing the orientation of one arc.

\Ex{8.} Use \ThmRef{41} to show that the number of nonisomorphic oriented
\index{Oriented graph}%
graphs with $n$~nodes is given by the formula
\[
\sum_{(d)} \frac{3^{F}}{N};
\]
\PageSep{89}
the sum is over all solutions~$(d)$ in nonnegative integers to the equation
\[
1 · d_{1} + 2 · d_{2} + \dots +  n · d_{n} = n,
\]
and
\[
F = \frac{1}{2} \left\{
  \sum_{k,l=1}^{n} d_{k}d_{l}(k, l)
  - \sum_{k \text{ odd}} d_{k}
  - 2 · \sum_{k \text{ even}} d_{k}\right\}.
\]
[See Harary (1957).]
\index[xauthor]{Harary, F.}%

\Ex{9.} Determine the number of tournaments~$T_{n}$ that are isomorphic to their
\index{Isomorphic tournaments}%
\index{Self-complementary tournament}%
complement.
\index{Complement of a tournament}%

\Ex{10.} Kotzig (1964) has raised the problem of determining the number of
\index[xauthor]{Kotzig, A.}%
nonisomorphic regular tournaments~$T_{n}$.
\index{Regular tournament}%
\PageSep{90}
\PageSep{91}
\BackMatter


\Section{}{Appendix}

The following drawings illustrate the nonisomorphic tournaments~$T_{n}$
\index{Isomorphic tournaments}%
\index{Score vector}%
($n \leq 6$), their score vectors, the number of ways of labeling their nodes,
and their automorphism groups. Most of the material is taken from the
\index{Automorphism group of a tournament}%
thesis of Goldberg (1966). Not all of the arcs have been included in the
\index[xauthor]{Goldberg, M.}%
drawings; if an arc joining two nodes has not been drawn, then it is to be
understood that the arc is oriented from the higher node to the lower node.

\noindent
\AppFig{app_0}{(0)}{1}{I}
\AppFig{app_01}{(0, 1)}{2}{I}
\AppFig{app_012}{(0, 1, 2)}{6}{I}
\AppFig{app_111}{(1, 1, 1)}{2}{C_{3}} \\
%
\AppFig{app_0123}{(0, 1, 2, 3)}{24}{I}
\AppFig{app_0222}{(0, 2, 2, 2)}{8}{C_{3}}
\AppFig{app_1113}{(1, 1, 1, 3)}{8}{C_{3}}
\AppFig{app_1122}{(1, 1, 2, 2)}{24}{I} \\
%
\AppFig{app_01234}{(0, 1, 2, 3, 4)}{120}{I}
\AppFig{app_01333}{(0, 1, 3, 3, 3)}{40}{C_{3}}
\AppFig{app_02233}{(0, 2, 2, 3, 3)}{120}{I}
\AppFig{app_02224}{(0, 2, 2, 2, 4)}{40}{C_{3}} \\
\PageSep{92}
\AppFig{app_11134}{(1, 1, 1, 3, 4)}{40}{C_{3}}
\AppFig{app_11224}{(1, 1, 2, 2, 4)}{120}{I}
\AppFig{app_11233a}{(1, 1, 2, 3, 3)}{120}{I}
\AppFig{app_11233b}{(1, 1, 2, 3, 3)}{120}{I} \\
%
\AppFig{app_12223a}{(1, 2, 2, 2, 3)}{120}{I}
\AppFig{app_12223b}{(1, 2, 2, 2, 3)}{120}{I}
\AppFig{app_12223c}{(1, 2, 2, 2, 3)}{40}{C_{3}}
\AppFig{app_22222}{(2, 2, 2, 2, 2)}{24}{C_{5}} \\
%
\AppFig{app_012345}{(0, 1, 2, 3, 4, 5)}{720}{I}
\AppFig{app_012444}{(0, 1, 2, 4, 4, 4)}{240}{C_{3}}
\AppFig{app_013335}{(0, 1, 3, 3, 3, 5)}{240}{C_{3}}
\AppFig{app_013344}{(0, 1, 3, 3, 4, 4)}{720}{I} \\
%
\AppFig{app_022245}{(0, 2, 2, 2, 4, 5)}{240}{C_{3}}
\AppFig{app_022335}{(0, 2, 2, 3, 3, 5)}{720}{I}
\AppFig{app_022344a}{(0, 2, 2, 3, 4, 4)}{720}{I}
\AppFig{app_022344b}{(0, 2, 2, 3, 4, 4)}{720}{I} \\
\PageSep{93}
\AppFig{app_023334a}{(0, 2, 3, 3, 3, 4)}{720}{I}
\AppFig{app_023334b}{(0, 2, 3, 3, 3, 4)}{720}{I}
\AppFig{app_023334c}{(0, 2, 3, 3, 3, 4)}{240}{C_{3}}
\AppFig{app_033333}{(0, 3, 3, 3, 3, 3)}{144}{C_{5}} \\
%
\AppFig{app_111345}{(1, 1, 1, 3, 4, 5)}{240}{C_{3}}
\AppFig{app_111444}{(1, 1, 1, 4, 4, 4)}{80}{C_{3} × C_{3}}
\AppFig{app_112245}{(1, 1, 2, 2, 4, 5)}{720}{I}
\AppFig{app_112335a}{(1, 1, 2, 3, 3, 5)}{720}{I} \\
%
\AppFig{app_112335b}{(1, 1, 2, 3, 3, 5)}{720}{I}
\AppFig{app_112344a}{(1, 1, 2, 3, 4, 4)}{720}{I}
\AppFig{app_112344b}{(1, 1, 2, 3, 4, 4)}{720}{I}
\AppFig{app_112344c}{(1, 1, 2, 3, 4, 4)}{720}{I} \\
%
\AppFig{app_112344d}{(1, 1, 2, 3, 4, 4)}{720}{I}
\AppFig{app_113334a}{(1, 1, 3, 3, 3, 4)}{720}{I}
\AppFig{app_113334b}{(1, 1, 3, 3, 3, 4)}{720}{I}
\AppFig{app_113334c}{(1, 1, 3, 3, 3, 4)}{240}{C_{3}} \\
\PageSep{94}
\AppFig{app_122235a}{(1, 2, 2, 2, 3, 5)}{720}{I}
\AppFig{app_122235b}{(1, 2, 2, 2, 3, 5)}{720}{I}
\AppFig{app_122235c}{(1, 2, 2, 2, 3, 5)}{240}{C_{3}}
\AppFig{app_122244a}{(1, 2, 2, 2, 4, 4)}{720}{I} \\
%
\AppFig{app_122244b}{(1, 2, 2, 2, 4, 4)}{720}{I}
\AppFig{app_122244c}{(1, 2, 2, 2, 4, 4)}{240}{C_{3}}
\AppFig{app_122334a}{(1, 2, 2, 3, 3, 4)}{720}{I}
\AppFig{app_122334b}{(1, 2, 2, 3, 3, 4)}{720}{I} \\
%
\AppFig{app_122334c}{(1, 2, 2, 3, 3, 4)}{720}{I}
\AppFig{app_122334d}{(1, 2, 2, 3, 3, 4)}{720}{I}
\AppFig{app_122334e}{(1, 2, 2, 3, 3, 4)}{720}{I}
\AppFig{app_122334f}{(1, 2, 2, 3, 3, 4)}{720}{I} \\
%
\AppFig{app_122334g}{(1, 2, 2, 3, 3, 4)}{720}{I}
\AppFig{app_122334h}{(1, 2, 2, 3, 3, 4)}{720}{I}
\AppFig{app_122334i}{(1, 2, 2, 3, 3, 4)}{720}{I}
\AppFig{app_122334j}{(1, 2, 2, 3, 3, 4)}{720}{I} \\
\PageSep{95}
\AppFig{app_122334k}{(1, 2, 2, 3, 3, 4)}{720}{I}
\AppFig{app_122334l}{(1, 2, 2, 3, 3, 4)}{720}{I}
\AppFig{app_123333a}{(1, 2, 3, 3, 3, 3)}{720}{I}
\AppFig{app_123333b}{(1, 2, 3, 3, 3, 3)}{720}{I} \\
%
\AppFig{app_123333c}{(1, 2, 3, 3, 3, 3)}{240}{C_{3}}
\AppFig{app_123333d}{(1, 2, 3, 3, 3, 3)}{720}{I}
\AppFig{app_222234a}{(2, 2, 2, 2, 3, 4)}{720}{I}
\AppFig{app_222234b}{(2, 2, 2, 2, 3, 4)}{240}{C_{3}} \\
%
\AppFig{app_222234c}{(2, 2, 2, 2, 3, 4)}{720}{I}
\AppFig{app_222234d}{(2, 2, 2, 2, 3, 4)}{720}{I}
\AppFig{app_222225}{(2, 2, 2, 2, 2, 5)}{144}{C_{5}}
\AppFig{app_222333a}{(2, 2, 2, 3, 3, 3)}{720}{I} \\
%
\AppFig{app_222333b}{(2, 2, 2, 3, 3, 3)}{720}{I}
\AppFig{app_222333c}{(2, 2, 2, 3, 3, 3)}{720}{I}
\AppFig{app_222333d}{(2, 2, 2, 3, 3, 3)}{240}{C_{3}}
\AppFig{app_222333e}{(2, 2, 2, 3, 3, 3)}{240}{C_{3}}
\PageSep{96}


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Thinking in the Measurement of Behavior} (H.~Solomon, ed.). Glencoe: The
Free Press, 1--149.

\Bibitem U.~Colombo (1964). Sui circuiti nei grafi completi. \Title{Boll.\ Un.\ Mat.\ Ital.}\ \Vol{19}, 153--170.
\index[xauthor]{Colombo, U.}%

\Bibitem H.~A. David (1959). Tournaments and paired comparisons. \Title{Biometrika} \Vol{46}, 139--149.
\index[xauthor]{David, H. A.}%

\Bibitem \Same\ (1963). \Title{The Method of Paired Comparisons.} London: Griffin.

\Bibitem \Same, and H.~T. Starks (1961). Significance tests for paired comparison experiments.
\index[xauthor]{Starks, H. T.}%
\Title{Biometrika} \Vol{48}, 95--108.
\PageSep{97}

\Bibitem \Same, and B.~J. Trawinski (1963). Selection of the best treatment in a paired-comparison
\index[xauthor]{Trawinski, B. J.}%
experiment. \Title{Ann.\ Math.\ Statist.}\ \Vol{34}, 75--91.

\Bibitem R.~L. Davis (1953). The number of structures of finite relations. \Title{Proc.\ Amer.\ Math.\
\index[xauthor]{Davis, R. L.}%
Soc.}\ \Vol{4}, 486--495.

\Bibitem \Same\ (1954). Structures of dominance relations. \Title{Bull.\ Math.\ Biophys.}\ \Vol{16}, 131--140.

\Bibitem F.~R. DeMeyer, and C.~R. Plott (1967). The probability of a cyclical majority.
\index[xauthor]{DeMeyer, F. R.}%
\index[xauthor]{Plott, C. R.}%
\Title{Notices Amer.\ Math.\ Soc.}\ \Vol{14}, 151.

\Bibitem J.~D. Dixon (1967). The maximum order of the group of a tournament. \Title{Canad.\
\index[xauthor]{Dixon, J. D.}%
Math.\ Bull.}\ \Vol{10}, 503--505.

\Bibitem A.~L. Dulmage, and N.~S. Mendelsohn (1965). Graphs and matrices. University
\index[xauthor]{Dulmage, A. L.}%
\index[xauthor]{Mendelsohn, N. S.}%
of Alberta Preprint Series, Edmonton. In \Title{Graph Theory and Theoretical Physics}
(F.~Harary, ed.). London: Academic Press, 1967, 167--227.

\Bibitem \index[xauthor]{Erdös, P.|(}%
P.~Erdös (1942). On an elementary proof of some asymptotic formulas in the theory
of partitions. \Title{Ann.\ of Math.}\ \Vol{43}, 437--450.

\Bibitem \Same\ (1963). On a problem in graph theory. \Title{Math.\ Gazette} \Vol{47}, 220--223.

\Bibitem \Same, and L.~Moser (1964a). On the representation of directed graphs as unions
\index[xauthor]{Moser, L.|(}%
of orderings. \Title{Publ.\ Math.\ Inst.\ Hung.\ Acad.\ Sci.}\ \Vol{9}, 125--132.

\Bibitem \Same, and \Same\ (1964b). A problem on tournaments. \Title{Canad.\ Math.\ Bull.}\
\Vol{7}, 351--356.
\index[xauthor]{Moser, L.|)}%

\Bibitem \Same, and J.~W. Moon (1965). On sets of consistent arcs in a tournament,
\index[xauthor]{Moon, J. W.}%
\Title{Canad.\ Math.\ Bull.}\ \Vol{8}, 269--271.
\index[xauthor]{Erdös, P.|)}%

\Bibitem M.~Fekete (1923). Über die Verteilung der Wurzeln bei gewissen algebraischen
\index[xauthor]{Fekete, M.}%
Gleichungen mit ganzzahligen Koeffizienten. \Title{Math.\ Zeit.}\ \Vol{17}, 228--249.

\Bibitem W.~Feller (1957). \Title{An Introduction to Probability Theory and its Applications.} New
\index[xauthor]{Feller, W.}%
York: Wiley.

\Bibitem A.~Fernández De~Trocóniz (1966). Caminos y circuitos halmitonianos en gráfos
\index[xauthor]{Fernández de Troconiz, A.}%
fuertemente conexos. \Title{Trabajos Estadist.}\ \Vol{17}, 17--32.

\Bibitem L.~R. Ford,~Jr. (1957). Solution of a ranking problem from binary comparisons.
\index[xauthor]{Ford, L. R., Jr.}%
\Title{Amer.\ Math.\ Monthly} \Vol{64}, 28--33.

\Bibitem \Same, and S.~M. Johnson (1959). A tournament problem. \Title{Amer.\ Math.\ Monthly}
\index[xauthor]{Johnson, S. M.}%
\Vol{66}, 387--389.

\Bibitem \Same, and D.~R. Fulkerson (1962). \Title{Flows in Networks.} Princeton University
\index[xauthor]{Fulkerson, D. R.}%
Press.

\Bibitem J.~D. Foulkes (1960). Directed graphs and assembly schedules. \Title{Proc.\ Sympos.\ Appl.\
\index[xauthor]{Foulkes, J. D.}%
Math.\ Providence, Amer.\ Math.\ Soc.}\ \Vol{10}, 231--289.

\Bibitem J.~Freund (1959). Round robin mathematics. \Title{Amer.\ Math.\ Monthly} \Vol{63}, 112--114.
\index[xauthor]{Freund, J.}%

\Bibitem G.~Frobenius (1912). Über Matrizen aus nicht negativen Elementen. \Title{S.~B. Preuss.\
\index[xauthor]{Frobenius, G.}%
Akad.\ Wiss., Berlin} \Vol{23}, 456--477.

\Bibitem D.~R. Fulkerson (1965). Upsets in round robin tournaments. \Title{Canad.\ J. Math.}\
\Vol{17}, 957--969.

\Bibitem \Same, A.~J. Hoffman, and M.~H. McAndrew (1965). Some properties of graphs
\index[xauthor]{Hoffman, A. J.}%
\index[xauthor]{MacAndrew@McAndrew, M. H.}%
with multiple edges. \Title{Canad.\ J. Math.}\ \Vol{17}, 166--177.

\Bibitem D.~Gale (1957). A theorem on flows in networks. \Title{Pac.\ J. Math.}\ \Vol{1}, 1073--1082.
\index[xauthor]{Gale, D.}%

\Bibitem \Same\ (1964). On the number of faces of a convex polygon. \Title{Canad.\ J. Math.}\
\Vol{16}, 12--17.

\Bibitem T.~Gallai, and A.~N. Milgram (1960). Verallgemeinerung eines graphentheoretischen
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\index[xauthor]{Milgram, A. N.}%
Satzes von Rédei. \Title{Acta Sci.\ Math \(Szeged\)} \Vol{21}, 181--186.
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\Bibitem E.~N. Gilbert (1961). Design of mixed doubles tournaments. \Title{Amer.\ Math.\ Monthly}
\index[xauthor]{Gilbert, E. N.}%
\Vol{68}, 124--131.

\Bibitem M.~Goldberg (1966). \Title{Results on the Automorphism Group of a Graph.} M.~Sc. Thesis,
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University of Alberta.

\Bibitem \Same, and J.~W. Moon (1966). On the maximum order of the group of a tournament.
\index[xauthor]{Moon, J. W.}%
\Title{Canad.\ Math.\ Bull.}\ \Vol{9}, 563--569.

\Bibitem I.~J. Good (1955). On the marking of chess-players. \Title{Math.\ Gazette} \Vol{39}, 292--296.
\index[xauthor]{Good, I. J.}%

\Bibitem N.~T. Gridgeman (1963). Significance and adjustment in paired comparisons.
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\Title{Biometrics} \Vol{19}, 213--228.

\Bibitem F.~Harary (1957). The number of oriented graphs. \Title{Mich.\ Math.~J.} \Vol{4}, 221--224.
\index[xauthor]{Harary, F.}%

\Bibitem \Same\ (1960). On the consistency of precedence matrices. \Title{J.~Assoc.\ Comput.\
Mach.}\ \Vol{7}, 255--259.

\Bibitem \Same\ (1964). Combinatorial problems in graphical enumeration. In \Title{Applied
Combinatorial Mathematics} (E.~Beckenbach, ed.). New York: Wiley, 185--217.

\Bibitem \Same, R.~Z. Norman, and D.~Cartwright (1965). \Title{Structural Models: An Introduction
\index[xauthor]{Cartwright, D.}%
\index[xauthor]{Norman, R. Z.}%
to the Theory of Directed Graphs.} New York: Wiley.

\Bibitem \Same, and L.~Moser (1966). The theory of round robin tournaments. \Title{Amer.\
\index[xauthor]{Moser, L.}%
Math.\ Monthly} \Vol{73}, 231--246.

\Bibitem \Same, and E.~Palmer (1967). On the problem of reconstructing a tournament
\index[xauthor]{Palmer, E. M.}%
from subtournaments. \Title{Monatsh.\ Math.}\ \Vol{71}, 14--23.

%[** TN: [sic] "knockout" (no hyphen)]
\Bibitem J.~A. Hartigan (1966). Probabilistic completion of a knockout tournament. \Title{Ann.\
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Math.\ Statist.}\ \Vol{37}, 495--503.

\Bibitem M.~Hasse (1961). Über die Behandlung graphentheoretischer Probleme unter
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Verwendung der Matrizenrechnung. \Title{Wiss.\ Z.~Tech.\ Univ.\ Dresden.}\ \Vol{10}, 1313--1316.

\Bibitem R.~L. Hemminger (1966). The lexicographic product of graphs. \Title{Duke Math.~J.} \Vol{33},
\index[xauthor]{Hemminger, R. L.}%
499--501.

\Bibitem P.~J. Huber (1963a). A remark on a paper of Trawinski and David entitled: Selection
\index[xauthor]{Huber, P.}%
of the best treatment in a paired comparison experiment. \Title{Ann.\ Math.\ Statist.}\
\Vol{34}, 92--94.

\Bibitem \Same\ (1963b). Pairwise comparison and ranking: optimum properties of the
row sum procedure. \Title{Ann.\ Math.\ Statist.}\ \Vol{34}, 511--520.

\Bibitem J.~B. Kadane (1966). Some equivalence classes in paired comparisons. \Title{Ann.\ Math.\
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Statist.}\ \Vol{37}, 488--494.

\Bibitem L.~Katz (1953). A new status index derived from sociometric analysis. \Title{Psychometrika}
\index[xauthor]{Katz, L.}%
\Vol{18}, 39--43.

\Bibitem M.~G. Kendall (1955). Further contributions to the theory of paired comparisons.
\index[xauthor]{Kendall, M. G.}%
\Title{Biometrics} \Vol{11}, 43--62.

\Bibitem \Same\ (1962). \Title{Rank Correlation Methods}, 3\textsuperscript{d}~ed. New York: Hafner.

\Bibitem \Same, and B.~Babington Smith (1940). On the method of paired comparisons.
\index[xauthor]{Smith, B. B.}%
\Title{Biometrika} \Erratum{\Vol{33}, 239--251}{\Vol{31}, 324--345}.

\Bibitem \Same, and A.~Stuart (1958). \Title{The Advanced Theory of Statistics.} London: Griffin.
\index[xauthor]{Stuart, A.}%

\Bibitem S.~S. Kislicyn (1963). An improved bound for the least mean number of comparisons
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needed for the complete ordering of a finite set. \Title{Vestnik Leningrad.\ Univ.\ Ser.\
Mat.\ Meh.\ Astronom.}\ \Vol{18}, 143--145.

\Bibitem \Same\ (1964). On the selection of the $k$-th~element of an ordered set by pairwise
comparison. \Title{Sibirsk.\ Mat.\ \v{Z}.} \Vol{5}, 557--564.
\PageSep{99}

\Bibitem D.~König (1936). \Title{Theorie der endlichen und unendlichen Graphen.} New York:
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Chelsea.

\Bibitem G.~Korvin (1966). On a theorem of L.~Rédei about complete oriented graphs. \Title{Acta.\
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Sci.\ Math.}\ \Vol{27}, 99--103.

\Bibitem A.~Kotzig (1964). \Title{Theory of Graphs and its Applications} (M.~Fiedler, ed.). New
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York: Academic Press,~162.

\Bibitem \Same\ (1966). Cycles in a complete graph oriented in equilibrium. \Title{Mat.-Fyz.\
\v{C}asopis.}\ \Vol{16}, 175--182.

\Bibitem \Same\ (1966). O $\delta$-transformáciách antisymetrických grafov. \Title{Mat.-Fyz.\ \v{C}asopis.}\
\Vol{16}, 353--356.

\Bibitem M.~Kraitchik (1954). \Title{Mathematical Recreations.} New York, Dover.
\index[xauthor]{Kraitchik, M.}%

\Bibitem H.~G. Landau (1951a). On dominance relations and the structure of animal societies.
\index[xauthor]{Landau, H. G.}%
I.~Effect of inherent characteristics. \Title{Bull.\ Math.\ Biophys.}\ \Vol{13}, 1--19.

\Bibitem \Same\ (1951b). On dominance relations and the structure of animal societies.
II.~Some effects of possible social factors. \Title{Bull.\ Math.\ Biophys.}\ \Vol{13}, 245--262.

\Bibitem \Same\ (1953). On dominance relations and the structure of animal societies.
III.~The condition for a score structure. \Title{Bull.\ Math.\ Biophys.}\ \Vol{15}, 143--148.

\Bibitem \index[xauthor]{Lockwood, E. H.|(}%
E.~H. Lockwood (1936). American tournament. \Title{Math.\ Gazette} \Vol{20},~333.

\Bibitem \Same\ (1962). Tournament problems. \Title{Math.\ Gazette} \Vol{46},~220.
\index[xauthor]{Lockwood, E. H.|)}%

\Bibitem R.~B. Marimont (1959). A new method of checking the consistency of precedence
\index[xauthor]{Marimont, R. B.}%
matrices. \Title{J.~Assoc.\ Comput.\ Mach.}\ \Vol{6}, 164--171.

\Bibitem D.~S. McGarvey (1953). A theorem on the construction of voting paradoxes.
\index[xauthor]{MacGarvey@McGarvey, D. S.}%
\Title{Econometrica} \Vol{21}, 608--610.

\Bibitem J.~H. McKay (1966). The William Lowell \Typo{Putman}{Putnam} Mathematical Competition.
\index[xauthor]{MacKay@McKay, J. H.}%
\Title{Amer.\ Math.\ Monthly} \Vol{73}, 726--732.

\Bibitem N.~S. Mendelsohn (1949). Problem~4346. \Title{Amer.\ Math.\ Monthly} \Vol{56},~343.
\index[xauthor]{Mendelsohn, N. S.}%

\Bibitem \Same\ (1953). Problem~62. \Title{Pi Mu Epsilon~J.} \Vol{1}, 365--366. Solution by K.~A.
Vrons (1955). Same~\Title{J.} \Vol{2}, 84--85.
\index[xauthor]{Vrons, K. A.}%

\Bibitem J. W. Moon (1962). On the score sequence of an $n$-partite tournament. \Title{Canad.\
\index[xauthor]{Moon, J. W.}%
Math.\ Bull.}\ \Vol{5}, 51--58.

\Bibitem \Same\ (1963). An extension of Landau's theorem on tournaments. \Title{Pac.\ J. Math.}\
\Vol{13}, 1343--1345.

\Bibitem \Same\ (1964). Tournaments with a given automorphism group. \Title{Canad.\ J. Math.}\
\Vol{16}, 485--489.

\Bibitem \Same\ (1965). On minimal $n$-universal graphs. \Title{Proc.\ Glasgow Math.\ Soc.}\ \Vol{7},
32--33.

\Bibitem \Same\ (1966). On subtournaments of a tournament. \Title{Canad.\ Math.\ Bull.}\ \Vol{9},
297--301.

\Bibitem \Same, and L.~Moser (1962a). On the distribution of $4$-cycles in random bipartite
\index[xauthor]{Moser, L.}%
tournaments. \Title{Canad.\ Math.\ Bull.}\ \Vol{5}, 5--12.

\Bibitem \Same, and \Same\ (1962b). Almost all tournaments are irreducible. \Title{Canad.\
Math.\ Bull.}\ \Vol{5}, 61--65.

\Bibitem \Same, and \Same\ (1966). Almost all $(0, 1)$~matrices are primitive. \Title{Stud.\ Sci.\
Math.\ Hung.}\ \Vol{1}, 153--156.

\Bibitem \Same\ and \Same\ (1967). Generating oriented graphs by means of team
comparisons. \Title{Pac.~J. Math.}\ \Vol{21}, 531--535.

\Bibitem \Same, and N.~J. Pullman (1967). On the powers of tournament matrices.
\index[xauthor]{Pullman, N. J.}%
\Title{J.~Comb.\ Theory} \Vol{3}, 1--9.
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\Bibitem P.~A.~P. Moran (1947). On the method of paired comparisons. \Title{Biometrika} \Vol{34},
\index[xauthor]{Moran, P. A. P.}%
363--365.

\Bibitem O.~Ore (1963). \Title{Graphs and Their Uses.} New York: Random House.
\index[xauthor]{Ore, O.}%

\Bibitem G.~Pólya (1937). Kombinatorische Anzahlbestimmungen für Gruppen, Graphen
\index[xauthor]{Polya@Pólya, G.}%
und chemische Verbindungen. \Title{Acta.\ Math.}\ \Vol{68}, 145--254.

\Bibitem R.~Rado (1943). Theorems on linear combinatorial topology and general measure.
\index[xauthor]{Rado, R.}%
\Title{Ann.\ of Math.}\ \Vol{44}, 228--270.

\Bibitem \Same\ (1964). Universal graphs and universal functions. \Title{Acta.\ Arith.}\ \Vol{9}, 331--340.

\Bibitem C.~Ramanujacharyula (1964). Analysis of preferential experiments. \Title{Psychometrika}
\index[xauthor]{Ramanujacharyula, C.}%
\Vol{29}, 257--261.

\Bibitem L.~Rédei (1934). Ein kombinatorischer Satz. \Title{Acta.\ Litt.\ Szeged.}\ \Vol{7}, 39--43.
\index[xauthor]{Redei@Rédei, L.}%

\Bibitem M.~Reisz (1859). Über eine Steinersche kombinatorische Aufgabe welche in 45sten
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Bande dieses Journals, Seite~181, gestellt worden ist. \Title{J.~Reine.\ Angew.\ Math.}\
\Vol{56}, 326--344.

\Bibitem R.~Remage, Jr., and W.~A. Thompson, Jr. (1964). Rankings from paired comparisons.
\index[xauthor]{Remage, R., Jr.}%
\index[xauthor]{Thompson, W. A., Jr.}%
\Title{Ann.\ Math.\ Statist.}\ \Vol{35}, 739--747.

\Bibitem \Same, and \Same\ (1966). Maximum likelihood paired comparison rankings.
\Title{Biometrika} \Vol{53}, 143--149.

\Bibitem B.~Roy (1958). Sur quelques propriétes des graphes fortement connexes. C. R.
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\Title{Acad.\ Sci.\ Paris} \Vol{247}, 399--401.

\Bibitem \Same\ (1959). Transitivité et connexité. \Title{C.~R. Acad.\ Sci.\ Paris} \Vol{249}, 216--218.

\Bibitem H.~J. Ryser (1957). Combinatorial properties of matrices of zeros and ones. \Title{Canad.~J.
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Math.}\ \Vol{9}, 371--377.

\Bibitem \Same\ (1964). Matrices of zeros and ones in combinatorial mathematics. \Title{Recent
Advances in Matrix Theory.} Madison: Univ.\ Wisconsin Press, 103--124.

\Bibitem G.~Sabidussi (1961). The lexicographic product of graphs. \Title{Duke Math.~J.} \Vol{28},
\index[xauthor]{Sabidussi, G.}%
573--578.

\Bibitem H.~Sachs (1965). Bemerkung zur Konstruktion zyklischer selbstkomplementarer
\index[xauthor]{Sachs, H.}%
gerichteter Graphen. \Title{Wiss.\ Z.~Techn.\ Hochsch.\ Ilmenau.}\ \Vol{11}, 161--162.

\Bibitem F.~Scheid (1960). A tournament problem. \Title{Amer.\ Math.\ Monthly} \Vol{67}, 39--41.
\index[xauthor]{Scheid, F.}%

\Bibitem J.~Schreier (1932). O systemach eliminacji no tumiejach. \Title{Mathesis Polska} \Vol{7},
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\Bibitem D.~L. Silverman (1961). Problem~463. \Title{Math.\ Mag.}\ \Vol{34},~425. Solution by J.~W. Moon
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\index[xauthor]{Silverman, D. L.}%
(1962). \Title{Math.\ Mag.}\ \Vol{35},~189.

\Bibitem P.~Slater (1961). Inconsistencies in a schedule of paired comparisons. \Title{Biometrika}
\index[xauthor]{Slater, P.}%
\Vol{48}, 303--312.

\Bibitem J.~Slupecki (1951). On the systems of tournaments. \Title{Colloq.\ Math.}\ \Vol{2}, 286--290.
\index[xauthor]{Slupecki, J.}%

\Bibitem R.~Stearns (1959). The voting problem. \Title{Amer.\ Math.\ Monthly} \Vol{66}, 761--763.
\index[xauthor]{Stearns, R.}%

\Bibitem H.~Steinhaus (1950). \Title{Mathematical Snapshots.} Oxford University Press.
\index[xauthor]{Steinhaus, H.}%

\Bibitem \Same\ (1963). Some remarks about tournaments. Golden Jubilee Commemoration
Vol.\ (1958/1959), Calcutta: \Title{Calcutta Math.\ Soc.}\ 323--327.

\Bibitem \Same, and S.~Trybula (1959). On a paradox in applied probabilities. \Title{Bull.\
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Acad.\ Polon.\ Sci.}\ \Vol{7}, 67--69.

\Bibitem E.~and G.~Szekeres (1965). On a problem of Schütte and Erdös. \Title{Math.\ Gazette} \Vol{49},
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290--293,

\Bibitem T.~Szele (1943). Kombinatorikai vizsgálatok az irányított teljes gráffal kapcsolatban.
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Untersuchungen über gerichtete vollständige graphen. \Title{Publ.\ Math.\ Debrecen.}\
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\Bibitem G.~L. Thompson (1958). Lectures on game theory, Markov chains and related
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topics. \Title{Sandia Corporation Monograph} SCR-11.

\Bibitem H.~Tietze (1957). Über Schachturnier-Tabellen. \Title{Math.\ Zeit.}\ \Vol{67}, 188--202.
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\Bibitem S.~Trybula (1961). On the paradox of three random variables. \Title{Zastos.\ Mat.}\ \Vol{5},
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321--332.

\Bibitem Z.~Usiskin (1964). Max-min probabilities in the voting paradox. \Title{Ann.\ Math.\ Statist.}\
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\Bibitem W.~Vogel (1963). Bemerkungen zur Theorie der Matrizen aus Nullen und Einsen.
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\Bibitem G.~L. Watson (1954). Bridge problem. \Title{Math.\ Gazette} \Vol{38}, 129--130.
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\Bibitem T.~H. Wei (1952). \Title{The algebraic foundations of ranking theory.} Ph.~D. Thesis,
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Cambridge University.

\Bibitem W.~A. Whitworth (1878). Arrangements of $m$~things of one sort and $m$~things of
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another sort under certain conditions of priority. \Title{Messenger of Math.}\ \Vol{8}, 105--114.

\Bibitem H.~Wielandt (1950). Unzerlegbare, nicht negative Matrizen. \Title{Math.\ Zeit.}\ \Vol{52}, 642--648.
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\Bibitem C.~C. Yalavigi (1963). A tournament problem. \Title{Math.\ Student.}\ \Vol{31}, 51--64.
\index[xauthor]{Yalavigi, C. C.}%

\Bibitem E.~Zermelo (1929). Die Berechnung der Tumier-Ergebnisse als ein Maximalproblem
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der Wahrscheinlichkeitsrechnung. \Title{Math.\ Zeit.}\ \Vol{29}, 436--460.
\PageSep{102}


\AuthorIndex

\iffalse
%[** TN: All entries pointing to References section off by one in the original]
Author Index

Alspach, B.#Alspach 7, 96

Alway, G. G.#Alway 11, 61, 66, 96

Bednarek, A. R.#Bednarek 5, 96

Beineke, L. W.#Beineke 4, 8, 11, 96

Bent, D. H.#Bent 16, 66, 67, 96

Bose, R. C.#Bose 40, 96

Boyd, A. V.#Boyd 41, 96

Bradley, R. A.#Bradley 43, 96

Brunk, H. D.#Brunk 46, 96

\Typo{Buhlman}{Buhlmann}, N. H.#Buhlman 46, 96

Burnside, W.#Burnside 81, 84, 96

Bush, L. E.#Bush 18, 96

Camion, P.#Camion 6, 18, 96

Cartwright, D.#Cartwright 10, 98

Cayley, A.#Cayley 75, 96

Chisholm, J. S. R.#Chisholm 49, 96

Chvátal, V.#Chvátal 78, 96

Clark, A. H.#Clark 9

Coleman, J. S.#Coleman 1, 96

Colombo, U.#Colombo 9, 96

David, H. A.#David 1, 40, 44, 46, 48, 66, 72, 96, 97

Davis, R. L.#Davis 85, 87, 97

Debruijn@de Bruijn, N. G.#de Bruijn 51, 85, 96

DeMeyer, F. R.#DeMeyer 57, 97

Dixon, J. D.#Dixon 84, 97

Dulmage, A. L.#Dulmage 35, 97

Erdös, P.#Erdös 15, 16, 19, 21, 28, 29, 30, 31, 32, 54, 56, 68, 70, 88, 97

Fekete, M.#Fekete 27, 97

Feller, W.#Feller 33, 71, 97

Fernández de Troconiz, A.#Fernández 7, 97

Ford, L. R., Jr.#Ford 43, 44, 47, 48, 64, 97

Foulkes, J. D.#Foulkes 7, 97

Freund, J.#Freund 40, 97

Frobenius, G.#Frobenius 45, 97

Fulkerson, D. R.#Fulkerson 61, 63, 64, 74, 97

Gale, D.#Gale 9, 64, 65, 97

Gallai, T.#Gallai 25, 97

Gilbert, E. N.#Gilbert 40, 98

Goldberg, M.#Goldberg 81, 91, 98

Good, I. J.#Good 46, 98

Gridgeman, N. T.#Gridgeman 46, 98

Harary, F.#Harary 4, 6, 8, 10, 18, 85, 89, 96, 98

Hartigan, J. A.#Hartigan 48, 98

Hasse, M.#Hasse 46, 98

Hemminger, R. L.#Hemminger 80, 98

Hoffman, A. J.#Hoffman 74, 97

Huber, P.#Huber 46, 71, 72, 96, 98

Johnson, S. M.#Johnson 47, 48, 97

Kadane, J. B.#Kadane 46, 98

Katz, L.#Katz 46, 98

Kelly, P. J.#Kelly 7

Kendall, M. G.#Kendall 1, 9, 11, 12, 40, 44, 45, 46, 98

Kislicyn, S. S.#Kislicyn 48, 98

Konig@König, D.#König 40, 99

Korvin, G.#Korvin 24, 28, 99

Kotzig, A.#Kotzig 7, 73, 89, 99

Kraitchik, M.#Kraitchik 40, 41, 99

Landau, H. G.#Landau 14, 61, 62, 63, 99

Lockwood, E. H.#Lockwood 40, 99

MacAndrew@McAndrew, M. H.#McAndrew 74, 97

MacGarvey@McGarvey, D. S.#McGarvey 56, 99

MacKay@McKay, J. H.#McKay 42, 99

MacLeod, R. A.#MacLeod 3

Marimont, R. B.#Marimont 18, 99

Mendelsohn, N. S.#Mendelsohn 35, 42, 49, 97, 99

Milgram, A. N.#Milgram 25, 97

Moon, J. W.#Moon 3, 10, 11, 14, 16, 19, 21, 32, 35, 51, 58, 63, 64, 74, 81, 97, 98, 99, 100

Moran, P. A. P.#Moran 11, 14, 100

Moser, L.#Moser 3, 4, 6, 11, 14, 15, 16, 18, 28, 30, 32, 54, 56, 58, 68, 70, 97, 98, 99

Narayana, T. V.#Narayana 66, 67, 96

Norman, R. Z.#Norman 10, 98

Ore, O.#Ore 40, 100
%\PageSep{103}

Palmer, E. M.#Palmer 4, 98

Plott, C. R.#Plott 57, 97

Polya@Pólya, G.#Pólya 78, 85, 100

Pullman, N. J.#Pullman 35, 65, 99

Rado, R.#Rado 5, 51, 100

Ramanujacharyula, C.#Ramanujacharyula 46, 100

Redei@Rédei, L.#Rédei 21, 100

Reisz, M.#Reisz 39, 40, 41, 100

Remage, R., Jr.#Remage 21, 46, 100

Robbins, H.#Robbins 57

Roy, B.#Roy 5, 100

Ryser, H. J.#Ryser 61, 63, 64, 65, 73, 74, 100

Sabidussi, G.#Sabidussi 80, 100

Sachs, H.#Sachs 2, 100

Scheid, F.#Scheid 40, 42, 100

Schreier, J.#Schreier 48, 100

Schutte@Schütte, K.#Schütte 28

Silverman, D. L.#Silverman 63, 100

Slater, P.#Slater 21, 46, 100

Slupecki, J.#Slupecki 48, 100

Smith, B. B.#Smith 9, 11, 98

Starks, H. T.#Starks 72, 96

Stearns, R.#Stearns 15, 56, 57, 100

Stein, P.#Stein 67

Steinhaus, H.#Steinhaus 47, 48, 58, 60, 100

Stuart, A.#Stuart 12, 98

Szekeres, E. and G.#Szekeres 28, 32, 100

Szele, T.#Szele 9, 18, 21, 24, 26, 28, 100

Terry, M. E.#Terry 43, 96

Thompson, G. L.#Thompson 35, 45, 46, 101

Thompson, W. A., Jr.#Thompson 21, 46, 100

Tietze, H.#Tietze 40, 42, 101

Trawinski, B. J.#Trawinski 72, 97

Trybula, S.#Trybula 58, 60, 100, 101

Usiskin, Z.#Usiskin 60, 101

Vogel, W.#Vogel 64, 101

Vrons, K. A.#Vrons 99

Wallace, A. D.#Wallace 5, 96

Watson, G. L.#Watson 40, 42, 101

Wei, T. H.#Wei 40, 46, 101

Whitworth, W. A.#Whitworth 71, 101

Wielandt, H.#Wielandt 34, 101

Yalavigi, C. C.#Yalavigi 40, 101

Zermelo, E.#Zermelo 42, 44, 101
\PageSep{104}
\fi


\SubjectIndex

\iffalse

Subject Index

Automorphism group of a tournament#Automorphism 44, 78, 81, 88, 91

Bilevel graph 51, 57, 59

Complement of a tournament#Complement 1, 89

Components of a bilevel graph#Components 51, 57, 59

Composition
  of two groups 78
  of two tournaments 78, 83

Consistent arcs#Consistent 19

Cycle type of a permutation#Cycle type 85

Cycles in a tournament#Cycles 5, 8, 11, 14, 28, 35, 73

Degree of a node#Degree 51

Diameter of a tournament#Diameter 32, 35

Directed graph 34, 51

Dominance relation#Dominance 1, 47

Equivalent elements in a set#Equivalent 81, 85

Exponent
  of a matrix 34
  of a tournament 35

Flows in networks#Flows 64

Hierarchy index#Hierarchy 14

Index
  of convergence of a matrix 38
  of maximum density of a matrix 36

Irreducible tournament#Irreducible 2, 4, 6, 7, 35, 43, 46

Isomorphic tournaments#Isomorphic 1, 4, 49, 63, 84, 88, 89, 91

Length of a path or cycle#Length 5

Matrix
  of a tournament 2, 5, 15, 22, 34, 45, 63
  of zeros and ones 34, 63, 65

Maximal transitive subtournament#Maximal 18

Maximum likelihood principle#Maximum likelihood 43

Net score 58

Oriented graph#Oriented 21, 24, 51, 56, 57, 88

Paired comparisons, the method of#Paired comparisons 1, 11, 40, 72

Paths in a tournament#Paths 5, 21, 26, 32, 35

Period of a matrix#Period 38

Preference matrix 56

Primitive
  matrix 34, 45
  tournament 35

Property $S(k, m)$ 28

Ranking problems 42, 46, 47

Reducible tournament#Reducible 2, 7, 46

Regular tournament#Regular 7, 8, 11, 50, 89

Score of a node#Score 1, 7, 15, 19, 34, 39, 63, 64, 71, 76, 80, 83

Score vector 1, 2, 8, 15, 61, 63, 65, 73, 91

Self-complementary tournament#Self-complementary 1, 89

Spanning path or cycle#Spanning 5, 21, 26, 28

Strength of a player#Strength 43, 58

Strong tournament#Strong 4, 8, 16, 21, 32, 77, 88

Subtournament 7, 8, 15, 21, 26, 36, 49, 82, 84, 86

Teams 41, 57, 61

Tournament
  bipartite 11, 14, 51, 63
  generalized 64
  knock-out 48
  n-partite@$n$-partite 63
  round-robin 1, 39, 44

Transitive tournament#Transitive 14, 21, 25

Universal tournament#Universal 49

Upsets in a tournament#Upsets 46

Voters 56

Weights associated with arcs#Weights 64
\fi

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> <./images/app_023334a.png> <./images/app_023334b.png> <./images/app_023334c.p
ng> <./images/app_033333.png> <./images/app_111345.png> <./images/app_111444.pn
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] [126] [127] [128] [129] [130] [131] [132] (./42833-t.and [133] [134

] [135]) (./42833-t.ind [136



] [137]) [1



] [2] [3] [4] [5] [6] [7] [8] (./42833-t.aux)

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 42833-t.out
 42833-t.out
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./images/app_123333c.png
./images/app_123333d.png
./images/app_222234a.png
./images/app_222234b.png
./images/app_222234c.png
./images/app_222234d.png
./images/app_222225.png
./images/app_222333a.png
./images/app_222333b.png
./images/app_222333c.png
./images/app_222333d.png
./images/app_222333e.png
 42833-t.and
 42833-t.ind
 ***********

 ) 
Here is how much of TeX's memory you used:
 8539 strings out of 493848
 126773 string characters out of 1152823
 246540 words of memory out of 3000000
 10861 multiletter control sequences out of 15000+50000
 17344 words of font info for 67 fonts, out of 3000000 for 9000
 716 hyphenation exceptions out of 8191
 37i,20n,46p,330b,590s stack positions out of 5000i,500n,10000p,200000b,50000s
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texlive/fonts/type1/public/amsfonts/cm/cmcsc10.pfb></usr/share/texmf-texlive/fo
nts/type1/public/amsfonts/cm/cmex10.pfb></usr/share/texmf-texlive/fonts/type1/p
ublic/amsfonts/cmextra/cmex8.pfb></usr/share/texmf-texlive/fonts/type1/public/a
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/usr/share/texmf-texlive/fonts/type1/public/amsfonts/cm/cmmi8.pfb></usr/share/t
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/fonts/type1/public/amsfonts/cm/cmr12.pfb></usr/share/texmf-texlive/fonts/type1
/public/amsfonts/cm/cmr17.pfb></usr/share/texmf-texlive/fonts/type1/public/amsf
onts/cm/cmr6.pfb></usr/share/texmf-texlive/fonts/type1/public/amsfonts/cm/cmr8.
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f-texlive/fonts/type1/public/amsfonts/cm/cmsy6.pfb></usr/share/texmf-texlive/fo
nts/type1/public/amsfonts/cm/cmsy7.pfb></usr/share/texmf-texlive/fonts/type1/pu
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s/cm/cmti12.pfb></usr/share/texmf-texlive/fonts/type1/public/amsfonts/cm/cmtt10
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hare/texmf-texlive/fonts/type1/public/amsfonts/symbols/msam10.pfb>
Output written on 42833-t.pdf (152 pages, 1198135 bytes).
PDF statistics:
 2594 PDF objects out of 2984 (max. 8388607)
 725 named destinations out of 1000 (max. 500000)
 766 words of extra memory for PDF output out of 10000 (max. 10000000)

