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\ihead{Errata for ``Topics on Tournaments''}
\ohead{\today}
\begin{document}

\begin{center}
\textbf{Topics on Tournaments}

\textit{John W. Moon}

\url{https://www.gutenberg.org/ebooks/42833} (version 2013)

\textbf{Errata}

\bigskip
\end{center}

\setcounter{section}{33}

\section*{Errata and comments}

All but four of the errors below were found by GPT-5.5, but verified by myself.

\begin{itemize}
\item \textbf{page 1, Section 1:} \textquotedblleft(symbolically,
$p_{i}\rightarrow_{j}$)\textquotedblright\ should be \textquotedblleft%
(symbolically, $p_{i}\rightarrow p_{j}$)\textquotedblright. It appears that
this is due to a LaTeX bug in the source:
\begin{align*}
&  \text{\texttt{%
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Dom\}[1]\{%
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to\}}}\\
\text{should be \ \ \ }  &  \text{\texttt{%
%TCIMACRO{\TEXTsymbol{\backslash}}%
%BeginExpansion
$\backslash$%
%EndExpansion
newcommand\{%
%TCIMACRO{\TEXTsymbol{\backslash}}%
%BeginExpansion
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%EndExpansion
Dom\}\{%
%TCIMACRO{\TEXTsymbol{\backslash}}%
%BeginExpansion
$\backslash$%
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to\}} \ \ \ \ (no\ \texttt{[1]}).}%
\end{align*}


\item \textbf{page 2, Section 1, Exercise 2:} \textquotedblleft the labels of
its node\textquotedblright\ should be \textquotedblleft the labels of its
nodes\textquotedblright.

\item \textbf{page 2, Section 2:} Again, replace \textquotedblleft%
$p_{i}\rightarrow_{j}$\textquotedblright\ by \textquotedblleft$p_{i}%
\rightarrow p_{j}$\textquotedblright.

\item \textbf{page 2, Section 2:} \textquotedblleft every node in $T^{\left(
j\right)  }$ dominates every node in $T^{\left(  i\right)  }$ if $1\leq i\leq
j\leq l$\textquotedblright\ is only true when $i<j$, not when $i=j$.

\item \textbf{page 6, Section 3, Exercise 1:} Replace \textquotedblleft all
the entries\textquotedblright\ by \textquotedblleft all the off-diagonal
entries\textquotedblright.

\item \textbf{page 7, Section 4, proof of Theorem 3:} \textquotedblleft
replacing the arcs $\overrightarrow{ab}$ and $\overrightarrow{be}%
$\textquotedblright\ should be \textquotedblleft replacing the arcs
$\overrightarrow{ab}$ and $\overrightarrow{bc}$\textquotedblright.

\item \textbf{page 11, Section 5, proof of Theorem 5:} \textquotedblleft To
show that $s(n,k)<n-k+1$\textquotedblright\ should be \textquotedblleft To
show that $s(n,k)\leq n-k+1$\textquotedblright.

\item \textbf{page 16, Section 6, Theorem 8:} In the formula for $\mu^{\prime
}$, replace \textquotedblleft$\dbinom{n}{8}$\textquotedblright\ by
\textquotedblleft$\dbinom{n}{2}$\textquotedblright. (This formula actually
gives the exact mean of $c\left(  m,n\right)  $, not just a limit.)

\item \textbf{page 18, Section 7, proof of Theorem 10:} \textquotedblleft and
theorem is proved\textquotedblright\ should be \textquotedblleft and the
theorem is proved\textquotedblright.

\item \textbf{page 22, Section 7, Exercise 6:} Remove the period after the
question mark.

\item \textbf{page 26, Section 9, proof of Theorem 14, equality (1):} The
upper limit of the sum should be $n$, not $m$.

\item \textbf{page 27, Section 9, proof of Theorem 14, explanation after (3):}
\textquotedblleft a term in the expansion of $\left\vert A^{T}\right\vert
$\textquotedblright\ should be \textquotedblleft a term in the expansion of
$\left\vert \overline{A}^{T}\right\vert $\textquotedblright.

\item \textbf{page 27, Section 9, proof of Theorem 14, equality (6):} Replace
the \textquotedblleft$e_{1}\subset e_{1}$\textquotedblright\ under the
summation sign by \textquotedblleft$e_{1}\subset e$\textquotedblright.

\item \textbf{page 32, Section 10, proof of Theorem 15:} \textquotedblleft If
we let $h\left(  n\right)  =\dfrac{t\left(  n\right)  }{n}$\textquotedblright%
\ should be \textquotedblleft If we let $h\left(  n\right)  =\dfrac{t\left(
n\right)  }{n!}$\textquotedblright.

\item \textbf{page 33, Section 10, Exercise 4:} \textquotedblleft
approximately $\dfrac{2\pi}{n}\left(  \dfrac{n}{2e}\right)  ^{n}%
$\textquotedblright\ should be \textquotedblleft exactly $\dfrac{\left(
n-1\right)  !}{2^{n}}$ and approximately $\sqrt{\dfrac{2\pi}{n}}\left(
\dfrac{n}{2e}\right)  ^{n}$\textquotedblright.

\item \textbf{page 41, Section 13:} \textquotedblleft if and only if there
exists a path $P\left(  p_{i},p_{j}\right)  $\textquotedblright\ should be
\textquotedblleft if and only if there exists a walk $P\left(  p_{i}%
,p_{j}\right)  $\textquotedblright, if modern graph-theoretical terminology is
to be followed. (Paths nowadays are required to have distinct vertices.) The
same change is probably needed throughout this section.

\item \textbf{page 43, Section 13:} \textquotedblleft subtournaments $T^{i}%
$\textquotedblright\ should be \textquotedblleft subtournaments $T^{\left(
i\right)  }$\textquotedblright.

\item \textbf{page 44, Section 13, Corollary:} Remove the \textquotedblleft%
$\eta\geq7$\textquotedblright\ case; it is rendered redundant by the
\textquotedblleft$\eta>4$\textquotedblright\ case.

\item \textbf{page 45, Section 13, proof of Theorem 23:} \textquotedblleft
where $T^{\left(  i\right)  }$ is\textquotedblright\ $\rightarrow$
\textquotedblleft where $T^{\left(  1\right)  }$ is\textquotedblright.

\item \textbf{page 56, Section 16, Theorem 24:} Replace \textquotedblleft%
$2^{i-1}$\textquotedblright\ by \textquotedblleft$2^{t-1}$\textquotedblright.

\item \textbf{page 56, Section 16, proof of Theorem 24:} Again, replace
\textquotedblleft$2^{i-1}$\textquotedblright\ by \textquotedblleft$2^{t-1}%
$\textquotedblright.

\item \textbf{page 59, Section 17, proof of Theorem 25:} \textquotedblleft
then $q_{1,A}\rightarrow q_{j,B}$\textquotedblright\ should be
\textquotedblleft then $q_{i,A}\rightarrow q_{j,B}$\textquotedblright.

\item \textbf{page 59, Section 17, proof of Theorem 25:} In the proof of the
$n$-universality of $H$, replace \textquotedblleft$p_{i}\in A\left(  j\right)
$\textquotedblright\ by \textquotedblleft$y_{i}\in A\left(  j\right)
$\textquotedblright. Likewise, replace \textquotedblleft$p_{i}\notin A\left(
j\right)  $\textquotedblright\ by \textquotedblleft$y_{i}\notin A\left(
j\right)  $\textquotedblright.

\item \textbf{page 62, Section 18, proof of Lemma 1:} Here you write:
\textquotedblleft 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%
\[
N=\sum_{i=1}^{t}\dbinom{s_{i}}{b}.
\]
\textquotedblright

This is imprecise: The formula $N=\sum_{i=1}^{t}\dbinom{s_{i}}{b}$ gives the
number of \textbf{pairs} $\left(  i,B\right)  $, where $i\in\left\{
1,2,\ldots,t\right\}  $ and where $B$ is a set of $b$ nodes for which $y_{i}$
dominates every node of $B$. Different pairs $\left(  i,B\right)  $ can have
the same set $B$.

\item \textbf{page 62, Section 18, Lemma 2:} Expressions of the form
\textquotedblleft$u/vw$\textquotedblright\ are inherently ambiguous; thus,
\textquotedblleft$\left(  n\log n\right)  /\left(  r+3\right)  2^{r+11}%
$\textquotedblright\ should be rewritten as \textquotedblleft$\dfrac{n\log
n}{\left(  r+3\right)  2^{r+11}}$\textquotedblright.

\item \textbf{page 64, Section 18, proof of Lemma 4:} In \textquotedblleft%
$e=e_{1}+e_{2}+\cdots+e_{i}$\textquotedblright, replace \textquotedblleft%
$e_{i}$\textquotedblright\ by \textquotedblleft$e_{t}$\textquotedblright.

\item \textbf{page 66, Section 19, McGarvey's construction:} \textquotedblleft
the orders $\left(  i,j,1,2,\ldots,n\right)  $ and $\left(  n,n-1,\ldots
,2,1,i,j\right)  $\textquotedblright\ should be \textquotedblleft the orders
$\left(  i,j,r_{1},r_{2},\ldots,r_{n-2}\right)  $ and $\left(  r_{n-2}%
,r_{n-3},\ldots,r_{1},i,j\right)  $, where $r_{1},r_{2},\ldots,r_{n-2}$ are
the elements of $\left\{  1,2,\ldots,n\right\}  \setminus\left\{  i,j\right\}
$ in increasing order\textquotedblright.

\item \textbf{page 72, Section 21, proof of Theorem 29:} The definition of the
$s_{j}^{\prime}$ would greatly profit from a picture of the Ferrers diagram of
the partitions $\left(  s_{n},s_{n-1},\ldots,s_{1}\right)  $ and $\left(
s_{n-1}^{\prime},s_{n-2}^{\prime},\ldots,s_{1}^{\prime}\right)  $ (with the
horizontal strip between the two being marked).

\item \textbf{page 73, Section 21, proof of Theorem 29:} \textquotedblleft
that $s_{1}^{\prime}=s_{i}$ for $s_{n}$ values of $i$\textquotedblright%
\ should be \textquotedblleft that $s_{i}^{\prime}=s_{i}$ for $s_{n}$ values
of $i$\textquotedblright.

\item \textbf{page 73, Section 21, proof of Theorem 29:} The inequality (3)
needs explanation.

\item \textbf{page 74, Section 21, Exercise 2:} Replace \textquotedblleft%
$\sum_{k}s_{i}$\textquotedblright\ by \textquotedblleft$\sum_{i}s_{i}%
$\textquotedblright.

\item \textbf{page 75, Section 22, Theorem 30:} The displayed equality must be
an inequality:%
\[
\sum_{i=1}^{n}\sum_{j=1}^{k_{i}}s_{ij}\geq\sum_{i=1}^{n-1}\sum_{j=i+1}%
^{n}k_{i}k_{j}.
\]


\item \textbf{page 75, Section 22, first corollary:} \textquotedblleft where
$r_{1}\leq r_{2}\cdots\leq r_{m}$\textquotedblright\ needs an extra
\textquotedblleft$\leq$\textquotedblright\ in front of the \textquotedblleft%
$\cdots$\textquotedblright. Furthermore, \textquotedblleft$\sum_{j=1}%
^{i}\left(  m-c_{j}\right)  $\textquotedblright\ should be \textquotedblleft%
$\sum_{j=1}^{l}\left(  m-c_{j}\right)  $\textquotedblright.

\item \textbf{page 78, Section 23:} In the formula%
\[
\left[  t,l\right]  ^{1}=%
\begin{cases}
1 & \text{if }t-l,\\
0 & \text{otherwise,}%
\end{cases}
\]
replace \textquotedblleft$t-l$\textquotedblright\ by \textquotedblleft%
$t=l$\textquotedblright.

\item \textbf{page 79, Section 23:} In (7), replace \textquotedblleft if
$\dfrac{1}{2}m<m\leq m$\textquotedblright\ by \textquotedblleft if $\dfrac
{1}{2}m<n\leq m$\textquotedblright\ probably.

\item \textbf{page 80, Section 23, Theorem 33:} On the right hand side,
replace \textquotedblleft$c_{1}$\textquotedblright\ by \textquotedblleft%
$c_{2}$\textquotedblright.

\item \textbf{page 80, Section 23, proof of Theorem 33:} After
\textquotedblleft we may suppose that $n$ is even, say $n=2m$%
\textquotedblright, add \textquotedblleft(since we can then handle the case of
odd $n$ by adding a new vertex that dominates all existing vertices, which
shows that $s\left(  n\right)  \leq s\left(  n+1\right)  $)\textquotedblright.

\item \textbf{page 80, Section 23, proof of Theorem 33:} \textquotedblleft the
node $p_{2m+i-1}$\textquotedblright\ should be \textquotedblleft the node
$p_{2m+1-i}$\textquotedblright.

\item \textbf{page 81, Section 23, proof of Theorem 33:} \textquotedblleft
integers $s_{1},s_{2},\ldots,s_{n}$ and $l_{1},l_{2},\ldots,l_{n}%
$\textquotedblright\ should be \textquotedblleft integers $s_{1},s_{2}%
,\ldots,s_{m}$ and $l_{1},l_{2},\ldots,l_{m}$\textquotedblright.

\item \textbf{page 82, Section 23, proof of Theorem 33:} Replace
\textquotedblleft$\left(  \dfrac{1}{m^{2}\left(  m+1\right)  ^{2}}\dbinom
{2m}{m}\right)  ^{2}m^{2}$\textquotedblright\ by \textquotedblleft$\dfrac
{1}{m^{2}}\left(  \dfrac{1}{m+1}\dbinom{2m}{m}\right)  ^{2}$\textquotedblright%
\ (it looks similar, but is a factor of $\left(  m+1\right)  ^{2}$ away).

\item \textbf{page 87, Section 25, Theorem 35:} It would be better not to
speak of score vectors (which is often sorted into nondecreasing order), but
rather be explicit about the preservation of vertices:

\begin{statement}
If $T_{n}$ and $T_{n}^{\prime}$ are two tournaments on $n$ vertices
$1,2,\ldots,n$ such that each vertex $i$ has the same score in $T_{n}$ as it
has in $T_{n}^{\prime}$, then $T_{n}$ can be transformed into $T_{n}^{\prime}$
by successively reversing the orientation of appropriate $3$-cycles.
\end{statement}

In this formulation, it is clear that the result of the transformation will be
literally $T_{n}^{\prime}$, not just a tournament isomorphic to $T_{n}%
^{\prime}$.

I don't know if your proof of Theorem 35 yields this stronger formulation, but
the proof I gave in the solution to Exercise 6 from
\href{https://www.cip.ifi.lmu.de/~grinberg/t/17s/hw2s.pdf}{Math 5707 Spring
2017 at UMN} does. (Please excuse my awkward writing...)

\item \textbf{page 95, Section 27, proof of Theorem 37:} \textquotedblleft a
permutation $\alpha$ of $G\left(  R\circ T\right)  $\textquotedblright\ should
be \textquotedblleft a permutation $\alpha$ in $G\left(  R\circ T\right)
$\textquotedblright\ or \textquotedblleft an element $\alpha$ of $G\left(
R\circ T\right)  $\textquotedblright.

\item \textbf{page 96, Section 27, Exercise 5:} Is \textquotedblleft$R\circ
T=T\circ R$\textquotedblright\ meant as a literal equality or as an isomorphism?

\item \textbf{page 107, Appendix:} The labels in the table are floating in
random places.
\end{itemize}


\end{document}