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5 files changed, 159 insertions, 8631 deletions
diff --git a/42833-pdf.zip b/42833-pdf.zip Binary files differdeleted file mode 100644 index 06be0cb..0000000 --- a/42833-pdf.zip +++ /dev/null diff --git a/42833-t.zip b/42833-t.zip Binary files differdeleted file mode 100644 index 2933cdf..0000000 --- a/42833-t.zip +++ /dev/null diff --git a/42833-t/42833-t.tex b/42833-t/42833-t.tex index 146d406..002146a 100644 --- a/42833-t/42833-t.tex +++ b/42833-t/42833-t.tex @@ -15,10 +15,11 @@ % Author: John W. Moon %
% %
% Release Date: June 5, 2013 [EBook #42833] %
+% Most recently updated: June 11, 2021 %
% %
% Language: English %
% %
-% Character set encoding: ISO-8859-1 %
+% Character set encoding: UTF-8 %
% %
% *** START OF THIS PROJECT GUTENBERG EBOOK TOPICS ON TOURNAMENTS *** %
% %
@@ -116,7 +117,7 @@ \documentclass[12pt]{book}[2005/09/16]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%% PACKAGES %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
-\usepackage[latin1]{inputenc}[2006/05/05]
+\usepackage[utf8]{inputenc}[2006/05/05]
\usepackage{ifthen}[2001/05/26] %% Logical conditionals
@@ -172,7 +173,7 @@ Minor typographical corrections have been made relative to the 1968
edition, and the proof of Theorem~34 (\Pagerefs{Thm34start}{Thm34end})
has been revised to correct an error pointed out to the author by
- B.~Bollobás in a letter dated 12~June 1980.
+ B.~Bollobás in a letter dated 12~June 1980.
\bigskip
The camera-quality files for this ebook may be downloaded at
@@ -589,10 +590,11 @@ 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: ISO-8859-1
+Character set encoding: UTF-8
*** START OF THIS PROJECT GUTENBERG EBOOK TOPICS ON TOURNAMENTS ***
\end{PGtext}
@@ -639,7 +641,7 @@ Dallas\ .\ Montreal\ .\ Toronto\ .\ London \PageSep{vi}%
\null\vfill
\begin{flushright}
-Copyright © 1968 by Holt, Rinehart and Winston, Inc. \\
+Copyright © 1968 by Holt, Rinehart and Winston, Inc. \\
All Rights Reserved
\medskip
@@ -1093,8 +1095,8 @@ proof of the theorem by induction. \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.}%
+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
@@ -1366,7 +1368,7 @@ 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).]
+can have is $[m^{2}/4] · [n^{2}/4]$. [Moon and Moser (1962a).]
\index[xauthor]{Moon, J. W.}%
\index[xauthor]{Moser, L.}%
@@ -1422,10 +1424,10 @@ If $r(ijk) = t(ijk) - \nf{1}{4}$, then where the sum is over the triples $i$,~$j$, and~$k$. The products in this expansion
are of the following types:
\[
-r(ijk) · r(ijk),\quad
-r(ijk) · r(ijw),\quad
-r(ijk) · r(ivw),\quad\text{and}\quad
-r(ijk) · r(uvw).
+r(ijk) · r(ijk),\quad
+r(ijk) · r(ijw),\quad
+r(ijk) · r(ivw),\quad\text{and}\quad
+r(ijk) · r(uvw).
\]
The variables appearing in each of the last two products are independent;
hence, the expectation of their product equals the product of their individual
@@ -1542,7 +1544,7 @@ tends to the normal distribution with zero mean and unit variance, where \Ex{1.} The quantity
\[
h = \frac{12}{n^{3} - n}
- · \sum_{i=1}^{n} \left(s_{i} - \frac{1}{2}(n - 1)\right)^{2}
+ · \sum_{i=1}^{n} \left(s_{i} - \frac{1}{2}(n - 1)\right)^{2}
\]
is called the \emph{hierarchy index} of a tournament. Show that the mean and
\index{Hierarchy index}%
@@ -1618,8 +1620,8 @@ Every tournament~$T_{n}$ ($n \geq 4$) contains at least one transitive subtourna \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.}%
+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).]
@@ -1656,8 +1658,8 @@ 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.}%
+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!]
@@ -1727,7 +1729,7 @@ proof of the theorem. \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~$3 · 2^{n-2}$.
+$T_{1}$ and~$T_{2}$, is~$3 · 2^{n-2}$.
\end{Corollary}
Let $r(n, k)$ denote the minimum number of transitive subtournaments~$T_{k}$
@@ -1738,7 +1740,7 @@ if $k > [2\log_{2} n] + 1$ and that $r(n, k) > 0$ if $k \leq [\log_{2} n] + 1$. Let
\[
\tau(n, k) = \begin{cases}
- n · \dfrac{(n - 1)}{2} · \dfrac{(n - 3)}{4} \dots \dfrac{(n - 2^{k-1} + 1)}{2^{k-1}} & \text{if $n > 2^{k-1} - 1$}, \\
+ n · \dfrac{(n - 1)}{2} · \dfrac{(n - 3)}{4} \dots \dfrac{(n - 2^{k-1} + 1)}{2^{k-1}} & \text{if $n > 2^{k-1} - 1$}, \\
0 & \text{if $n \leq 2^{k-1} - 1$}.
\end{cases}
\]
@@ -1829,8 +1831,8 @@ 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.}%
+$T_{n}$~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.
@@ -1838,7 +1840,7 @@ gave the following result. Let $w(n)$ denote the largest integer~$w$ such that every tournament~$T_{n}$
contains a set of $w$ consistent arcs. Then
\[
-w(n) \geq \left[\frac{n}{2}\right] · \left[\frac{n + 1}{2}\right]
+w(n) \geq \left[\frac{n}{2}\right] · \left[\frac{n + 1}{2}\right]
\quad\text{for all $n$ and }
w(n) \leq \frac{(1 + \eps)}{2} \binom{n}{2}
\]
@@ -1849,29 +1851,29 @@ for any positive~$\eps$ and all sufficiently large~$n$. 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
+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
\[
\left[\frac{n}{2}\right]
- + \left[\frac{n - 1}{2}\right] · \left[\frac{n}{2}\right]
- = \left[\frac{n}{2}\right] · \left[\frac{n + 1}{2}\right]
+ + \left[\frac{n - 1}{2}\right] · \left[\frac{n}{2}\right]
+ = \left[\frac{n}{2}\right] · \left[\frac{n + 1}{2}\right]
\]
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 $n! · \dbinom{N}{k}$
+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! · \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] \\
+ · \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] \\
@@ -1915,8 +1917,8 @@ tournaments. Prove that $z(n) = \left(\dfrac{1}{2} + o(1)\right) \dbinom{n}{2}$. $m$~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.}%
+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
@@ -1965,8 +1967,8 @@ that every node of~$T^{(j)}$ dominates every node of~$T^{(i)}$ if $1 \leq i < j Each of these strong subtournaments has a spanning path and these
spanning paths can be combined to yield a path spanning~$T_{n}$.
-The following theorem is a special case of a result proved by Rédei (1934)
-\index[xauthor]{Redei@Rédei, L.}%
+The following theorem is a special case of a result proved by Rédei (1934)
+\index[xauthor]{Redei@Rédei, L.}%
and generalized by Szele (1943).
\index[xauthor]{Szele, T.}%
@@ -1995,7 +1997,7 @@ The determinant of~$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
\[
-S_{k} = \sum_{1 \in e} e(A_{k}) · e'(A_{k}),
+S_{k} = \sum_{1 \in e} e(A_{k}) · e'(A_{k}),
\]
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
@@ -2045,9 +2047,9 @@ 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\} \\
+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)} \\
@@ -2063,7 +2065,7 @@ where the second summation is over all proper subsets $e_{1}$ of~$e$. (If $e$~is empty set, then both sides equal zero.) If we substitute \Eq{(5)} into~\Eq{(4)}, we find
that
\[
-h \equiv \sum_{e} \sum_{e_{1} \subset e_{1}} e'(A) · e_{1}(A) \pmod{2}.
+h \equiv \sum_{e} \sum_{e_{1} \subset e_{1}} e'(A) · e_{1}(A) \pmod{2}.
\Tag{(6)}
\]
@@ -2252,9 +2254,9 @@ 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
+&\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 \\
+&\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}
@@ -2309,8 +2311,8 @@ 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,
\[
-t(n) \leq \binom{n}{k} t(k) · t(n - k)\quad\text{or}\quad
-\frac{t(n)}{n!} \leq \frac{t(k)}{k!} · \frac{t(n - k)}{(n - k)!}
+t(n) \leq \binom{n}{k} t(k) · t(n - k)\quad\text{or}\quad
+\frac{t(n)}{n!} \leq \frac{t(k)}{k!} · \frac{t(n - k)}{(n - k)!}
\]
for all positive integers $k$ and $n$ with $k < n$. If we let $h(n) = \dfrac{t(n)}{n}$, then
\[
@@ -2369,11 +2371,11 @@ tournament having at least $\left(\dfrac{n}{3e}\right)^{n}$ spanning cycles for 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.}%
+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.}%
+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 $k = 1, 2, \dots$.}
@@ -2441,8 +2443,8 @@ 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.}%
+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)
@@ -2451,8 +2453,8 @@ n > 2^{k} k^{2} \log(2 + \eps) \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.}%
+$S(k, m)$ are similar to those used in proving the following result due to Erdös
+\index[xauthor]{Erdös, P.}%
and Moser (1964b).
\index[xauthor]{Moser, L.}%
@@ -2477,7 +2479,7 @@ depend on~$\eps$, such that if $k > K$ and $n > ck^{2}2^{k}$, then \Proof. Suppose $0 < \eps < 1$. For a particular choice of $A$~and~$B$, there are
\[
-\binom{n - k}{t} (2^{k} - 1)^{n - k - t} · 2^{\ebinom{n}{2}-k(n-k)}
+\binom{n - k}{t} (2^{k} - 1)^{n - k - t} · 2^{\ebinom{n}{2}-k(n-k)}
= 2^{\ebinom{n}{2}} \binom{n - k}{t}
\left(\frac{1}{2^{k}}\right)^{t}
\left(1 - \frac{1}{2^{k}}\right)^{n-k-t}
@@ -2555,8 +2557,8 @@ it turns out that the constant~$c$ may be any quantity greater than~$\log 2$. \Exercises
\Ex{1.} Let $T_{7}$ denote the tournament in which $p_{i} \to p_{j}$ if and only if $j - i$ is a
-quadratic residue modulo~$7$. Prove that $T_{7}$~has property $S(2, 1)$. [Erdös
-\index[xauthor]{Erdös, P.}%
+quadratic residue modulo~$7$. Prove that $T_{7}$~has property $S(2, 1)$. [Erdös
+\index[xauthor]{Erdös, P.}%
(1962).]
\PageSep{32}
@@ -2567,8 +2569,8 @@ G.~Szekeres (1965).] \Ex{3.} Prove that if a tournament~$T_{n}$ with property $S(k, 1)$ exists, then a tournament~$T_{r}$
with property $S(k, 1)$ exists whenever $r > n$. [See the remark at the
-end of the second paragraph of Erdös and Moser (1964b).]
-\index[xauthor]{Erdös, P.}%
+end of the second paragraph of Erdös and Moser (1964b).]
+\index[xauthor]{Erdös, P.}%
\index[xauthor]{Moser, L.}%
\Ex{4.} Let $A$ and $B$ denote two disjoint subsets of nodes of a tournament~$T_{n}$
@@ -2674,7 +2676,7 @@ nodes $p_{i}$ and~$p_{j}$. Then \begin{align*}
0 &\leq \frac{2^{\ebinom{n}{2}} - N(0)}{2^{\ebinom{n}{2}}}
= \frac{N(1) + N(2) + \dots + N\bigl(n(n - 1)\bigr)}{2^{\ebinom{n}{2}}} \\
- &\leq \frac{0 · N(0) + 1 · N(1) + \dots + n(n - 1) · N\bigl(n(n - 1)\bigr)}{2^{\ebinom{n}{2}}}
+ &\leq \frac{0 · N(0) + 1 · N(1) + \dots + n(n - 1) · N\bigl(n(n - 1)\bigr)}{2^{\ebinom{n}{2}}}
= E(n, \lambda).
\end{align*}
\PageSep{34}
@@ -2689,7 +2691,7 @@ and \]
Therefore,
\[
-\lim_{n \to \infty} \left(1 - N(0) · 2^{-\ebinom{n}{2}}\right) = 0,
+\lim_{n \to \infty} \left(1 - N(0) · 2^{-\ebinom{n}{2}}\right) = 0,
\]
and the corollary is proved.
@@ -3087,9 +3089,9 @@ and it matches players $k$~and~$n$ in round~$r$ if \]
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
+König (1936, p.~157) gave this construction [see also Freund (1959) and
\index[xauthor]{Freund, J.}%
-\index[xauthor]{Konig@König, D.}%
+\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),
@@ -3629,8 +3631,8 @@ disjoint copies of all the different tournaments~$T_{n}$ is $n$-universal.) \[
2^{(1/2)(n-1)} \leq \lambda(n)
\leq \begin{cases}
- n · 2^{(1/2)(n-1)} & \text{if $n$~is odd,} \\
- \dfrac{3}{2\sqrt{2}} n · 2^{(1/2)(n-1)} & \text{if $n$~is even.}
+ n · 2^{(1/2)(n-1)} & \text{if $n$~is odd,} \\
+ \dfrac{3}{2\sqrt{2}} n · 2^{(1/2)(n-1)} & \text{if $n$~is even.}
\end{cases}
\]
\end{Theorem}
@@ -3692,12 +3694,12 @@ for which \end{align*}
Hence,
\[
-\lambda(n) \leq n · 2^{\Chg{1/2}{(1/2)}(n-1)}\quad\text{if $n$ is odd,}
+\lambda(n) \leq n · 2^{\Chg{1/2}{(1/2)}(n-1)}\quad\text{if $n$ is odd,}
\]
and
\[
-\lambda(n) \leq \nf{1}{2} n · 2^{(1/2)n} + \nf{1}{2} n · 2^{(1/2)(n-2)}
- = \frac{3}{2\sqrt{2}} n · 2^{(1/2)(n-1)},\quad\text{if $n$~is even.}
+\lambda(n) \leq \nf{1}{2} n · 2^{(1/2)n} + \nf{1}{2} n · 2^{(1/2)(n-2)}
+ = \frac{3}{2\sqrt{2}} n · 2^{(1/2)(n-1)},\quad\text{if $n$~is even.}
\]
This completes the proof of the theorem.
@@ -3849,17 +3851,17 @@ 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
+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
\[
-\frac{n^{2}}{2^{2r+13/4}} - \frac{17 n^{3/2}}{2^{r}} \left[\frac{\sqrt{n}}{17 · 2^{r+5}}\right]
+\frac{n^{2}}{2^{2r+13/4}} - \frac{17 n^{3/2}}{2^{r}} \left[\frac{\sqrt{n}}{17 · 2^{r+5}}\right]
> \frac{n^{2}}{2^{2r+4}}
\]
-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
+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
\[
-\left(\left[\frac{\sqrt{n}}{17 · 2^{r+5}}\right] + 1\right)
- · [\sqrt{n}] · \left[\frac{\log n}{3(r + 3)}\right]
- > \frac{n \log n}{(r + 3) · 2^{r+11}}
+\left(\left[\frac{\sqrt{n}}{17 · 2^{r+5}}\right] + 1\right)
+ · [\sqrt{n}] · \left[\frac{\log n}{3(r + 3)}\right]
+ > \frac{n \log n}{(r + 3) · 2^{r+11}}
\]
arcs. This proves the lemma.
\PageSep{54}
@@ -3906,8 +3908,8 @@ e = e_{1} + e_{2} + \dots + e_{i} \]
or $t < 4\sqrt{e}$.
-We can now prove the following theorem due to Erdös and Moser (1964a).
-\index[xauthor]{Erdös, P.}%
+We can now prove the following theorem due to Erdös and Moser (1964a).
+\index[xauthor]{Erdös, P.}%
\index[xauthor]{Moser, L.}%
\begin{Theorem}{26.}
@@ -3935,7 +3937,7 @@ e_{i_{r}} \leq \frac{n^{2}}{2^{2r+1}}, then we shall show that
\[
i_{r+1} - i_{r}
- \leq 2^{11} · \frac{r + 3}{2^{r+1}} · \frac{n}{\log n}.
+ \leq 2^{11} · \frac{r + 3}{2^{r+1}} · \frac{n}{\log n}.
\Tag{(3)}
\]
We may suppose that
@@ -3958,7 +3960,7 @@ If we sum this inequality over all~$j$ such that $i_{r} \leq j < i_{r+1}$, we fi \[
\frac{n^{2}}{2^{2r+1}} \geq e_{i_{r}} - e_{i_{r+1}}
= \sum (e_{j} - e_{j+1})
- \geq (i_{r+1} - i_{r}) · \frac{n \log n}{(r + 3) 2^{r+11}}.
+ \geq (i_{r+1} - i_{r}) · \frac{n \log n}{(r + 3) 2^{r+11}}.
\]
This implies inequality~\Eq{(3)}.
@@ -4024,8 +4026,8 @@ 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.}%
+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.
@@ -4769,9 +4771,9 @@ support to this conjecture. \]
\end{table}
-The following bounds for~$s(n)$ are due to Erdös and Moser.\footnote
+The following bounds for~$s(n)$ are due to Erdös and Moser.\footnote
{Unpublished work.}
-\index[xauthor]{Erdös, P.}%
+\index[xauthor]{Erdös, P.}%
\index[xauthor]{Moser, L.}%
\begin{Theorem}{33.}
@@ -4904,8 +4906,8 @@ U(x) = \frac{1 - \sqrt{1 - 4x}}{2x} \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.}%
+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}}.
@@ -4953,16 +4955,16 @@ tends to one as $n$~tends to infinity for any positive~$\eps$. \Proof. If $0 < \eps < 1$, let
\[
-x_{n}^{±} = (1 ± \eps) \sqrt{2\log (n - 1)};
+x_{n}^{±} = (1 ± \eps) \sqrt{2\log (n - 1)};
\]
it follows from~\Eq{(1)} that
\[
-\Erratum{p}{P}\{w_{i} > x_{n}^{±}\}
- \sim \frac{1}{\sqrt{4\pi} (1 ± \eps) \sqrt{\log (n - 1)}}
- · (n - 1)^{-(1 ± \eps)^{2}}.
+\Erratum{p}{P}\{w_{i} > x_{n}^{±}\}
+ \sim \frac{1}{\sqrt{4\pi} (1 ± \eps) \sqrt{\log (n - 1)}}
+ · (n - 1)^{-(1 ± \eps)^{2}}.
\Tag{(2)}
\]
-Let $c_{1}$ and $c_{2}$ be positive constants such that $c_{1} < 1/\sqrt{4\pi} (1 ± \eps) < c_{2}$; if
+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
\[
@@ -4973,13 +4975,13 @@ for all sufficiently large values of~$n$. \PageSep{72}
\iffalse
-% [** Author's note: "B. Bollobás pointed out to me, in a letter dated
+% [** 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
\[
P\{s_{1} < k_{1}, \dots, s_{n} < k_{n}\}
- \leq P\{s_{1} < k_{1}\} · \dotsm · P\{s_{n} < k_{n}\}
+ \leq P\{s_{1} < k_{1}\} · \dotsm · P\{s_{n} < k_{n}\}
\Tag{(4)}
\]
for any positive numbers $k_{1}, \dots, k_{n}$. Observe that
@@ -4999,8 +5001,8 @@ 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}\}.
+ \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
@@ -5009,7 +5011,7 @@ This argument can be repeated to yield inequality~\Eq{(4)}. We now show that
\[
P\{s_{1} < k_{1}, \dots, s_{n} < k_{n}\}
- \leq P\{s_{1} < k_{1}\} · \dotsm · P\{s_{n} < k_{n}\}
+ \leq P\{s_{1} < k_{1}\} · \dotsm · P\{s_{n} < k_{n}\}
\Tag{(4)}
\]
for any positive integers $k_{1}, \dots, k_{n}$, where $s_{i} = \sum_{j \neq i} a_{ij}$,
@@ -5042,10 +5044,10 @@ P\{s_{1h} < k_{1} - c_{12}, s_{2h} < k_{2} - c_{21}, s_{3}^{(h)} < k_{3}, \dots\ \]
and
\[
-P\{b_{12} = c_{12}\} · P\{b_{21} = c_{21}\} - P\{a_{12} = c_{12}, a_{21} = c_{21}\}
+P\{b_{12} = c_{12}\} · P\{b_{21} = c_{21}\} - P\{a_{12} = c_{12}, a_{21} = c_{21}\}
\]
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
+$P\{b_{12} = c_{12}\} · P\{b_{21}=c_{21}\} = 1/4$ 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*}
@@ -5290,7 +5292,7 @@ hold for the scores of the nodes of~$T$: \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)
+ &= \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~$G$;
\[
@@ -5401,8 +5403,8 @@ Prove that $t(G) \leq n$ if $G$~is abelian. 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.}%
+identity group. [See Chvátal (1965).]
+\index[xauthor]{Chvátal, V.}%
\Section[Group of the Composition of Two Tournaments]
@@ -5420,9 +5422,9 @@ if and only if $r_{i} \to r_{j}$, or $i = j$ and $t_{k} \to t_{l}$. The composit 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
+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
\[
\alpha(x, y) = \bigl(f(x), h_{x}(y)\bigr),
\]
@@ -5433,7 +5435,7 @@ permutations~$\alpha$ of $U × V = \{(x, y) : x \in U, y \in V\}$ of the type \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
+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
@@ -5488,12 +5490,12 @@ 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
\[
-s\bigl(p(i, k)\bigr) = s(t_{k}) + b · s(r_{i}),
+s\bigl(p(i, k)\bigr) = s(t_{k}) + b · s(r_{i}),
\]
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
\[
-s(t_{k}) + b · s(r_{i}) = s(t_{l}) + b · s(r_{j})
+s(t_{k}) + b · s(r_{i}) = s(t_{l}) + b · s(r_{j})
\]
and $0 \leq s(t_{k}), s(t_{l}) < b$. Consequently,
\[
@@ -5607,7 +5609,7 @@ 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
\[
-|G| = |E(x)| · |F(x)|.
+|G| = |E(x)| · |F(x)|.
\]
\end{Theorem}
\PageSep{82}
@@ -5637,7 +5639,7 @@ 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)|,
+ = |F(x)| · |E(x)|,
\]
since each element of~$G$ is counted once and only once in the first sum.
@@ -5663,25 +5665,25 @@ If $T_{e}$ and $T_{n-e}$ denote the subtournaments determined by the nodes that \index{Subtournament}%
in~$E$ and by the nodes that are not in~$E$, then it is clear that
\[
-g(T_{n}) \leq g(T_{e}) · g(T_{n-e}) \leq g(e) · g(n - e).
+g(T_{n}) \leq g(T_{e}) · g(T_{n-e}) \leq g(e) · g(n - e).
\Tag{(3)}
\]
If $3 < e < n - 3$, then it follows from the induction hypothesis that
\[
-g(T_{n}) \leq \frac{(2.5)^{e}}{2e} · \frac{(2.5)^{n-e}}{2(n - e)}
- \leq \frac{n}{8(n - 4)} · \frac{(2.5)^{n}}{2n}
+g(T_{n}) \leq \frac{(2.5)^{e}}{2e} · \frac{(2.5)^{n-e}}{2(n - e)}
+ \leq \frac{n}{8(n - 4)} · \frac{(2.5)^{n}}{2n}
< \frac{(2.5)^{n}}{2n}.
\]
If $e = 3$ or $n - 3$, then
\[
-g(T_{n}) \leq 3 · \frac{(2.5)^{n-3}}{2(n - 3)}
+g(T_{n}) \leq 3 · \frac{(2.5)^{n-3}}{2(n - 3)}
< \frac{(2.5)^{n}}{2n},
\]
\PageSep{83}
and if $e = 1$, $2$, $n - 2$, or~$n - 1$, then
\[
-g(T_{n}) \leq 1 · \frac{2n}{5(n - 2)} · \frac{(2.5)^{n}}{2n}
+g(T_{n}) \leq 1 · \frac{2n}{5(n - 2)} · \frac{(2.5)^{n}}{2n}
< \frac{(2.5)^{n}}{2n}.
\]
A different argument must be used when $e = n$.
@@ -5704,7 +5706,7 @@ into one of the $\nf{1}{2}(n - 1)$ nodes dominated by~$p$, since $p$~is fixed. H Therefore, if $e = n$, then
\[
g(T_{n}) \leq n\left(\frac{(2.5)^{(1/2)(n-1)}}{n - 1}\right)^{2}
- = \frac{4}{5} \left(\frac{n}{n - 1}\right)^{2} · \frac{(2.5)^{n}}{2n}
+ = \frac{4}{5} \left(\frac{n}{n - 1}\right)^{2} · \frac{(2.5)^{n}}{2n}
< \frac{(2.5)^{n}}{2n}.
\]
(Notice that $\nf{1}{2} (n - 1) \geq 5$ if $n \geq 11$, so we are certainly entitled to apply
@@ -5774,7 +5776,7 @@ 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
+\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 $7 \leq n \leq 11$.
@@ -5785,7 +5787,7 @@ even. nodes that are equivalent with respect to~$G(T_{n})$. Under what circumstances
is it true that
\[
-g(T_{n}) = g(T_{a}) · g(T_{b}) \dots?
+g(T_{n}) = g(T_{a}) · g(T_{b}) \dots?
\]
\Ex{4.} Let $h(n)$ denote the order of any largest subgroup~$H$ of odd order of the
@@ -5804,10 +5806,10 @@ 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
+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.}%
+\index[xauthor]{Polya@Pólya, G.}%
summary of results on the enumeration of graphs.
\begin{Theorem}{40.}
@@ -5856,7 +5858,7 @@ of~$\pi$ contains $d_{k}$~cycles of length~$k$, for $k = 1, 2, \dots, n$, then $ \]
is of type $(2, 1, 1, 0, 0, 0, 0)$. Notice that
\[
-1 · d_{1} + 2 · d_{2} + \dots + n · d_{n} = n.
+1 · d_{1} + 2 · d_{2} + \dots + n · d_{n} = n.
\]
In the present context, we think of the permutations~$\pi$ as acting on the
@@ -5929,7 +5931,7 @@ T(n) = \sum_{(d)} \frac{2^{D}}{N}, 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
\[
-1 · d_{1} + 3 · d_{3} + 5 · d_{5} + \dots = n.
+1 · d_{1} + 3 · d_{3} + 5 · d_{5} + \dots = n.
\]
\end{Theorem}
@@ -5941,13 +5943,13 @@ calculations may be summarized as follows. \hline\Strut
d_{1} = 6 & 6! & 15 \\
\hline\Strut
-d_{1} = 3,\ d_{3} = 1 & 3 · 3! & 7 \\
+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
+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.
+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
@@ -5983,16 +5985,16 @@ $T(n) \sim \dfrac{2^{\ebinom{n}{2}}}{n!}$ as $n$~tends to infinity. \Proof. The two largest terms in the formula for~$T(n)$ are
\[
\frac{2^{\ebinom{n}{2}}}{n!}\quad\text{and}\quad
-\frac{2^{\ebinom{n}{2} - 2(n-2)}}{3 · (n - 3)!};
+\frac{2^{\ebinom{n}{2} - 2(n-2)}}{3 · (n - 3)!};
\]
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
+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,
\[
\frac{2^{\ebinom{n}{2}}}{n!} \leq T(n)
\leq \frac{2^{\ebinom{n}{2}}}{n!}
- + 2^{cn^{1/2}} · \frac{2^{\ebinom{n}{2} - 2(n-2)}}{3 · (n - 3)!}
+ + 2^{cn^{1/2}} · \frac{2^{\ebinom{n}{2} - 2(n-2)}}{3 · (n - 3)!}
= \frac{2^{\ebinom{n}{2}}}{n!} \bigl(1 + o(1)\bigr).
\]
@@ -6008,7 +6010,7 @@ number of partitions of~$n$ is less than $2^{cn^{1/2}}$, where $c = \pi(\nf{2}{3 \Ex{4.} Prove that
\[
T(n) = \frac{\Erratum{2^{\ebinom{2}{n}}}{2^{\ebinom{n}{2}}}}{n!}
- + \frac{2^{(1/2)(n^{2} - 5n + 8)}}{3 · (n - 3)!} \bigl(1 + o(1)\bigr).
+ + \frac{2^{(1/2)(n^{2} - 5n + 8)}}{3 · (n - 3)!} \bigl(1 + o(1)\bigr).
\]
\Ex{5.} Let $t(n)$ denote the number of nonisomorphic strong tournaments~$T_{n}$.
@@ -6036,14 +6038,14 @@ graphs with $n$~nodes is given by the formula \PageSep{89}
the sum is over all solutions~$(d)$ in nonnegative integers to the equation
\[
-1 · d_{1} + 2 · d_{2} + \dots + n · d_{n} = n,
+1 · d_{1} + 2 · d_{2} + \dots + n · d_{n} = n,
\]
and
\[
F = \frac{1}{2} \left\{
\sum_{k,l=1}^{n} d_{k}d_{l}(k, l)
- \sum_{k \text{ odd}} d_{k}
- - 2 · \sum_{k \text{ even}} d_{k}\right\}.
+ - 2 · \sum_{k \text{ even}} d_{k}\right\}.
\]
[See Harary (1957).]
\index[xauthor]{Harary, F.}%
@@ -6118,7 +6120,7 @@ understood that the arc is oriented from the higher node to the lower node. \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_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} \\
%
@@ -6248,8 +6250,8 @@ Acad.\ Sci.}\ Paris \Vol{249}, 2151--2152. \Bibitem J.~S.~R. Chisholm (1948). A tennis problem. \Title{Eureka} \Vol{10},~18.
\index[xauthor]{Chisholm, J. S. R.}%
-\Bibitem V.~Chvátal, (1965). On finite and countable rigid graphs and tournaments. \Title{Comment.\
-\index[xauthor]{Chvátal, V.}%
+\Bibitem V.~Chvátal, (1965). On finite and countable rigid graphs and tournaments. \Title{Comment.\
+\index[xauthor]{Chvátal, V.}%
Math.\ Univ.\ Carolinae} \Vol{6}, 429--438.
\Bibitem J.~S. Coleman (1960). The mathematical study of small groups. In \Title{Mathematical
@@ -6295,8 +6297,8 @@ Math.\ Bull.}\ \Vol{10}, 503--505. 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
+\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.
@@ -6312,9 +6314,9 @@ of orderings. \Title{Publ.\ Math.\ Inst.\ Hung.\ Acad.\ Sci.}\ \Vol{9}, 125--132 \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.|)}%
+\index[xauthor]{Erdös, P.|)}%
-\Bibitem M.~Fekete (1923). Über die Verteilung der Wurzeln bei gewissen algebraischen
+\Bibitem M.~Fekete (1923). Ãœber die Verteilung der Wurzeln bei gewissen algebraischen
\index[xauthor]{Fekete, M.}%
Gleichungen mit ganzzahligen Koeffizienten. \Title{Math.\ Zeit.}\ \Vol{17}, 228--249.
@@ -6322,8 +6324,8 @@ Gleichungen mit ganzzahligen Koeffizienten. \Title{Math.\ Zeit.}\ \Vol{17}, 228- \index[xauthor]{Feller, W.}%
York: Wiley.
-\Bibitem A.~Fernández De~Trocóniz (1966). Caminos y circuitos halmitonianos en gráfos
-\index[xauthor]{Fernández de Troconiz, A.}%
+\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.
@@ -6345,7 +6347,7 @@ 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.\
+\Bibitem G.~Frobenius (1912). Ãœber Matrizen aus nicht negativen Elementen. \Title{S.~B. Preuss.\
\index[xauthor]{Frobenius, G.}%
Akad.\ Wiss., Berlin} \Vol{23}, 456--477.
@@ -6366,7 +6368,7 @@ with multiple edges. \Title{Canad.\ J. Math.}\ \Vol{17}, 166--177. \Bibitem T.~Gallai, and A.~N. Milgram (1960). Verallgemeinerung eines graphentheoretischen
\index[xauthor]{Gallai, T.}%
\index[xauthor]{Milgram, A. N.}%
-Satzes von Rédei. \Title{Acta Sci.\ Math \(Szeged\)} \Vol{21}, 181--186.
+Satzes von Rédei. \Title{Acta Sci.\ Math \(Szeged\)} \Vol{21}, 181--186.
\PageSep{98}
\Bibitem E.~N. Gilbert (1961). Design of mixed doubles tournaments. \Title{Amer.\ Math.\ Monthly}
@@ -6415,7 +6417,7 @@ from subtournaments. \Title{Monatsh.\ Math.}\ \Vol{71}, 14--23. \index[xauthor]{Hartigan, J. A.}%
Math.\ Statist.}\ \Vol{37}, 495--503.
-\Bibitem M.~Hasse (1961). Über die Behandlung graphentheoretischer Probleme unter
+\Bibitem M.~Hasse (1961). Ãœber die Behandlung graphentheoretischer Probleme unter
\index[xauthor]{Hasse, M.}%
Verwendung der Matrizenrechnung. \Title{Wiss.\ Z.~Tech.\ Univ.\ Dresden.}\ \Vol{10}, 1313--1316.
@@ -6461,11 +6463,11 @@ Mat.\ Meh.\ Astronom.}\ \Vol{18}, 143--145. 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:
-\index[xauthor]{Konig@König, D.}%
+\Bibitem D.~König (1936). \Title{Theorie der endlichen und unendlichen Graphen.} New York:
+\index[xauthor]{Konig@König, D.}%
Chelsea.
-\Bibitem G.~Korvin (1966). On a theorem of L.~Rédei about complete oriented graphs. \Title{Acta.\
+\Bibitem G.~Korvin (1966). On a theorem of L.~Rédei about complete oriented graphs. \Title{Acta.\
\index[xauthor]{Korvin, G.}%
Sci.\ Math.}\ \Vol{27}, 99--103.
@@ -6476,7 +6478,7 @@ 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.}\
+\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.
@@ -6558,8 +6560,8 @@ comparisons. \Title{Pac.~J. Math.}\ \Vol{21}, 531--535. \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.}%
+\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.
@@ -6572,10 +6574,10 @@ und chemische Verbindungen. \Title{Acta.\ Math.}\ \Vol{68}, 145--254. \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 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
+\Bibitem M.~Reisz (1859). Ãœber eine Steinersche kombinatorische Aufgabe welche in 45sten
\index[xauthor]{Reisz, M.}%
Bande dieses Journals, Seite~181, gestellt worden ist. \Title{J.~Reine.\ Angew.\ Math.}\
\Vol{56}, 326--344.
@@ -6588,11 +6590,11 @@ Bande dieses Journals, Seite~181, gestellt worden ist. \Title{J.~Reine.\ Angew.\ \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.
+\Bibitem B.~Roy (1958). Sur quelques propriétes des graphes fortement connexes. C. R.
\index[xauthor]{Roy, B.}%
\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 \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.
\index[xauthor]{Ryser, H. J.}%
@@ -6641,22 +6643,22 @@ Vol.\ (1958/1959), Calcutta: \Title{Calcutta Math.\ Soc.}\ 323--327. \index[xauthor]{Trybula, S.}%
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},
+\Bibitem E.~and G.~Szekeres (1965). On a problem of Schütte and Erdös. \Title{Math.\ Gazette} \Vol{49},
\index[xauthor]{Szekeres, E. and G.}%
290--293,
-\Bibitem T.~Szele (1943). Kombinatorikai vizsgálatok az irányított teljes gráffal kapcsolatban.
+\Bibitem T.~Szele (1943). Kombinatorikai vizsgálatok az irányÃtott teljes gráffal kapcsolatban.
\index[xauthor]{Szele, T.}%
\Title{Mat.\ Fiz.\ Lapok.}\ \Vol{50}, 223--256. For a German translation, see Kombinatorische
\PageSep{101}
-Untersuchungen über gerichtete vollständige graphen. \Title{Publ.\ Math.\ Debrecen.}\
+Untersuchungen über gerichtete vollständige graphen. \Title{Publ.\ Math.\ Debrecen.}\
\Vol{13} (1966) 145--168.
\Bibitem G.~L. Thompson (1958). Lectures on game theory, Markov chains and related
\index[xauthor]{Thompson, G. L.}%
topics. \Title{Sandia Corporation Monograph} SCR-11.
-\Bibitem H.~Tietze (1957). Über Schachturnier-Tabellen. \Title{Math.\ Zeit.}\ \Vol{67}, 188--202.
+\Bibitem H.~Tietze (1957). Ãœber Schachturnier-Tabellen. \Title{Math.\ Zeit.}\ \Vol{67}, 188--202.
\index[xauthor]{Tietze, H.}%
\Bibitem S.~Trybula (1961). On the paradox of three random variables. \Title{Zastos.\ Mat.}\ \Vol{5},
@@ -6732,7 +6734,7 @@ Cayley, A.#Cayley 75, 96 Chisholm, J. S. R.#Chisholm 49, 96
-Chvátal, V.#Chvátal 78, 96
+Chvátal, V.#Chvátal 78, 96
Clark, A. H.#Clark 9
@@ -6752,13 +6754,13 @@ 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
+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
+Fernández de Troconiz, A.#Fernández 7, 97
Ford, L. R., Jr.#Ford 43, 44, 47, 48, 64, 97
@@ -6806,7 +6808,7 @@ 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
+Konig@König, D.#König 40, 99
Korvin, G.#Korvin 24, 28, 99
@@ -6849,7 +6851,7 @@ Palmer, E. M.#Palmer 4, 98 Plott, C. R.#Plott 57, 97
-Polya@Pólya, G.#Pólya 78, 85, 100
+Polya@Pólya, G.#Pólya 78, 85, 100
Pullman, N. J.#Pullman 35, 65, 99
@@ -6857,7 +6859,7 @@ Rado, R.#Rado 5, 51, 100 Ramanujacharyula, C.#Ramanujacharyula 46, 100
-Redei@Rédei, L.#Rédei 21, 100
+Redei@Rédei, L.#Rédei 21, 100
Reisz, M.#Reisz 39, 40, 41, 100
@@ -6877,7 +6879,7 @@ Scheid, F.#Scheid 40, 42, 100 Schreier, J.#Schreier 48, 100
-Schutte@Schütte, K.#Schütte 28
+Schutte@Schütte, K.#Schütte 28
Silverman, D. L.#Silverman 63, 100
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-
-%[** TN: Slant-style fractions in the original]
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-
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-}
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- \hyperref[eqn:#1.#2]{\ensuremath{#2}}%
-}
-
-%%%%%%%%%%%%%%%%%%%%%%%% 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]
-
-Language: English
-
-Character set encoding: ISO-8859-1
-
-*** 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}
-\small
-\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}
-\vfill
-\PageSep{vii}
-\FlushRunningHeads
-\null\vfill
-\begin{minipage}{0.85\textwidth}
-\small
-\subsection*{\centering\normalfont\scshape%
-\normalsize\MakeLowercase{\TransNote}}%
-\raggedright
-\TransNoteText
-\end{minipage}
-
-
-\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}
-
-
-\Section{}{References}
-
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-\index[xauthor]{Alspach, B.}%
-Bull.}\ \Vol{10}, 283--286.
-
-\Bibitem G.~G. Alway (1962a). Matrices and sequences. \Title{Math.\ Gazette} \Vol{46}, 208--213.
-\index[xauthor]{Alway, G. G.}%
-
-\Bibitem \Same\ (1962b). The distribution of the number of circular triads in paired comparisons.
-\Title{Biometrika} \Vol{49}, 265--269.
-
-\Bibitem A.~R. Bednarek, and A.~D. Wallace (1966). Some theorems on $P$-intersective sets.
-\index[xauthor]{Bednarek, A. R.}%
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-\Title{Acta Math.\ Acad.\ Sci.\ Hung.}\ \Vol{17}, 9--14.
-
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-\index[xauthor]{Harary, F.}%
-subtournaments. \Title{Canad.\ Math.\ Bull.}\ \Vol{8}, 491--498.
-
-\Bibitem D.~H. Bent (1964). \Title{Score Problems of Round-Robin Tournaments.} M.~Sc.\ Thesis,
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-University of Alberta.
-
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-
-\Bibitem R.~C. Bose (1956). Paired comparison designs for testing concordance between
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-
-\Bibitem A.~V. Boyd (1961). Two tournament problems. \Title{Math.\ Gazette} \Vol{45}, 213--214.
-\index[xauthor]{Boyd, A. V.}%
-
-\Bibitem R.~A. Bradley, and M.~E. Terry (1952). Rank analysis of incomplete block designs.
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-\Bibitem N.~G. De~Bruijn (1964). Polya's theory of counting. In \Title{Applied Combinatorial
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-\Title{J.~Amer.\ Statist.\ Assoc.}\ \Vol{55}, 503--520.
-
-\Bibitem H.~Buhlmann, and P.~Huber (1963). Pairwise comparisons and ranking in tournaments.
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-\index[xauthor]{Huber, P.}%
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-
-
-\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
-
-%%%%%%%%%%%%%%%%%%%%%%%%% GUTENBERG LICENSE %%%%%%%%%%%%%%%%%%%%%%%%%%
-\PGLicense
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- <./images/app_222234c.png> <./images/app_222234d.png> <./images/app_222225.png
-> <./images/app_222333a.png>] [124 <./images/app_222333b.png> <./images/app_222
-333c.png> <./images/app_222333d.png> <./images/app_222333e.png>] [125
-
-] [126] [127] [128] [129] [130] [131] [132] (./42833-t.and [133] [134
-
-] [135]) (./42833-t.ind [136
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-] [137]) [1
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-] [2] [3] [4] [5] [6] [7] [8] (./42833-t.aux)
-
- *File List*
- book.cls 2007/10/19 v1.4h Standard LaTeX document class
- bk12.clo 2007/10/19 v1.4h Standard LaTeX file (size option)
-inputenc.sty 2008/03/30 v1.1d Input encoding file
- latin1.def 2008/03/30 v1.1d Input encoding file
- ifthen.sty 2001/05/26 v1.1c Standard LaTeX ifthen package (DPC)
- amsmath.sty 2000/07/18 v2.13 AMS math features
- amstext.sty 2000/06/29 v2.01
- amsgen.sty 1999/11/30 v2.0
- amsbsy.sty 1999/11/29 v1.2d
- amsopn.sty 1999/12/14 v2.01 operator names
- amssymb.sty 2009/06/22 v3.00
-amsfonts.sty 2009/06/22 v3.00 Basic AMSFonts support
- alltt.sty 1997/06/16 v2.0g defines alltt environment
-indentfirst.sty 1995/11/23 v1.03 Indent first paragraph (DPC)
-footmisc.sty 2009/09/15 v5.5a a miscellany of footnote facilities
- array.sty 2008/09/09 v2.4c Tabular extension package (FMi)
-graphicx.sty 1999/02/16 v1.0f Enhanced LaTeX Graphics (DPC,SPQR)
- keyval.sty 1999/03/16 v1.13 key=value parser (DPC)
-graphics.sty 2009/02/05 v1.0o Standard LaTeX Graphics (DPC,SPQR)
- trig.sty 1999/03/16 v1.09 sin cos tan (DPC)
-graphics.cfg 2009/08/28 v1.8 graphics configuration of TeX Live
- pdftex.def 2009/08/25 v0.04m Graphics/color for pdfTeX
- caption.sty 2009/10/09 v3.1k Customizing captions (AR)
-caption3.sty 2009/10/09 v3.1k caption3 kernel (AR)
- makeidx.sty 2000/03/29 v1.0m Standard LaTeX package
- index.sty 2004/01/20 v4.2beta Improved index support (dmj)
-multicol.sty 2008/12/05 v1.6h multicolumn formatting (FMi)
- calc.sty 2007/08/22 v4.3 Infix arithmetic (KKT,FJ)
-fancyhdr.sty
-geometry.sty 2008/12/21 v4.2 Page Geometry
- ifpdf.sty 2009/04/10 v2.0 Provides the ifpdf switch (HO)
- ifvtex.sty 2008/11/04 v1.4 Switches for detecting VTeX and its modes (HO)
-geometry.cfg
-hyperref.sty 2009/10/09 v6.79a Hypertext links for LaTeX
- ifxetex.sty 2009/01/23 v0.5 Provides ifxetex conditional
- hycolor.sty 2009/10/02 v1.5 Code for color options of hyperref/bookmark (HO
-)
-xcolor-patch.sty 2009/10/02 xcolor patch
- pd1enc.def 2009/10/09 v6.79a Hyperref: PDFDocEncoding definition (HO)
-etexcmds.sty 2007/12/12 v1.2 Prefix for e-TeX command names (HO)
-infwarerr.sty 2007/09/09 v1.2 Providing info/warning/message (HO)
-hyperref.cfg 2002/06/06 v1.2 hyperref configuration of TeXLive
-kvoptions.sty 2009/08/13 v3.4 Keyval support for LaTeX options (HO)
-kvsetkeys.sty 2009/07/30 v1.5 Key value parser with default handler support
-(HO)
- url.sty 2006/04/12 ver 3.3 Verb mode for urls, etc.
- bitset.sty 2007/09/28 v1.0 Data type bit set (HO)
- intcalc.sty 2007/09/27 v1.1 Expandable integer calculations (HO)
-bigintcalc.sty 2007/11/11 v1.1 Expandable big integer calculations (HO)
-pdftexcmds.sty 2009/09/23 v0.6 LuaTeX support for pdfTeX utility functions (
-HO)
-ifluatex.sty 2009/04/17 v1.2 Provides the ifluatex switch (HO)
- ltxcmds.sty 2009/08/05 v1.0 Some LaTeX kernel commands for general use (HO)
-
-atbegshi.sty 2008/07/31 v1.9 At begin shipout hook (HO)
- hpdftex.def 2009/10/09 v6.79a Hyperref driver for pdfTeX
-supp-pdf.mkii
- color.sty 2005/11/14 v1.0j Standard LaTeX Color (DPC)
- color.cfg 2007/01/18 v1.5 color configuration of teTeX/TeXLive
- nameref.sty 2007/05/29 v2.31 Cross-referencing by name of section
-refcount.sty 2008/08/11 v3.1 Data extraction from references (HO)
- 42833-t.out
- 42833-t.out
- umsa.fd 2009/06/22 v3.00 AMS symbols A
- umsb.fd 2009/06/22 v3.00 AMS symbols B
- omscmr.fd 1999/05/25 v2.5h Standard LaTeX font definitions
-./images/fig1.pdf
-./images/fig2.pdf
-./images/fig3.pdf
-./images/fig4.pdf
-./images/fig5.pdf
-./images/fig6.pdf
-./images/fig7.pdf
-./images/fig8.pdf
-./images/fig9.pdf
-./images/fig10.pdf
-./images/fig11.pdf
-./images/fig12.pdf
-./images/fig13.pdf
-./images/app_0.png
-./images/app_01.png
-./images/app_012.png
-./images/app_111.png
-./images/app_0123.png
-./images/app_0222.png
-./images/app_1113.png
-./images/app_1122.png
-./images/app_01234.png
-./images/app_01333.png
-./images/app_02233.png
-./images/app_02224.png
-./images/app_11134.png
-./images/app_11224.png
-./images/app_11233a.png
-./images/app_11233b.png
-./images/app_12223a.png
-./images/app_12223b.png
-./images/app_12223c.png
-./images/app_22222.png
-./images/app_012345.png
-./images/app_012444.png
-./images/app_013335.png
-./images/app_013344.png
-./images/app_022245.png
-./images/app_022335.png
-./images/app_022344a.png
-./images/app_022344b.png
-./images/app_023334a.png
-./images/app_023334b.png
-./images/app_023334c.png
-./images/app_033333.png
-./images/app_111345.png
-./images/app_111444.png
-./images/app_112245.png
-./images/app_112335a.png
-./images/app_112335b.png
-./images/app_112344a.png
-./images/app_112344b.png
-./images/app_112344c.png
-./images/app_112344d.png
-./images/app_113334a.png
-./images/app_113334b.png
-./images/app_113334c.png
-./images/app_122235a.png
-./images/app_122235b.png
-./images/app_122235c.png
-./images/app_122244a.png
-./images/app_122244b.png
-./images/app_122244c.png
-./images/app_122334a.png
-./images/app_122334b.png
-./images/app_122334c.png
-./images/app_122334d.png
-./images/app_122334e.png
-./images/app_122334f.png
-./images/app_122334g.png
-./images/app_122334h.png
-./images/app_122334i.png
-./images/app_122334j.png
-./images/app_122334k.png
-./images/app_122334l.png
-./images/app_123333a.png
-./images/app_123333b.png
-./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:
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- 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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-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)
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diff --git a/42833-t/old/42833-t.zip b/42833-t/old/42833-t.zip Binary files differdeleted file mode 100644 index 2933cdf..0000000 --- a/42833-t/old/42833-t.zip +++ /dev/null |