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\vskip 1cm{\LARGE\bf Zeroing the baseball indicator and the chirality of
triples
}
\vskip 1cm
\large
Christopher S. Simons and Marcus Wright \\
Department of Mathematics \\
Rowan University \\
Glassboro, New Jersey 08028 \\
USA \\
\href{mailto:simons@rowan.edu}{\tt simons@rowan.edu} \\
\href{mailto:wright@rowan.edu}{\tt wright@rowan.edu} \\
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\begin{center}
{\bf Abstract}
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\begin{quote}
Starting with a common baseball umpire indicator, we consider the zeroing
number for two-wheel indicators with states $(a,b)$ and three-wheel
indicators with states $(a,b,c)$. Elementary number theory yields formulae
for the zeroing number. The solution in the three-wheel case involves a
curiously nontrivial minimization problem whose solution determines the
chirality of the ordered triple $(a,b,c)$ of pairwise relatively prime
numbers. We prove that chirality is in fact an invariant of the unordered
triple $\{ a,b,c \}$. We also show that the chirality of Fibonacci triples
alternates between 1 and 2.
\end{quote}
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\date{\today}
\section{Introduction}
A standard three-wheel baseball umpire indicator consists of a first wheel
(for strikes) with 4 cyclic states $(0,1,2,3)$, a second wheel (for balls)
with 5 cyclic states $(0,1,2,3,4)$, and a third wheel (for outs) with 4
cyclic states $(0,1,2,3)$. The daughter of the second author asked what is
the least number of clicks required, if one alternates between advancing
the strike wheel and the ball wheel one click cyclically, to return these
two wheels to their original states. We call this the (two-wheel)
\emph{zeroing number}. After her father gave her the wrong answer of $2
\lcm (4,5) = 40$, she asked why the correct answer was in fact 31. With a
total of 40 clicks both wheels advance 20 times, and since $20\equiv 0
\pmod 4$ and $20\equiv 0 \pmod 5$ both wheels are returned to their
original state. However we can do better. A total of 31 clicks advances
the strike wheel 16 times and advances the ball wheel only 15 times.
Since $16\equiv 0 \pmod 4$ and $15\equiv 0 \pmod 5$ both wheels are
therefore zeroed. The zeroing number is 31, as 31 is the minimal such
number. However if instead one starts with balls and then moves onto
strikes, one finds that a total of only 9 clicks are required to zero the
indicator. We note that zeroing number does depend on the ordering of the
wheels and that the the sum of these two zeroing numbers gives the
incorrect solution: $31 + 9 = 40$.
Using elementary number theory we find a complete solution for the general
two-wheel indicator with cyclic states $a, b \geq 2$. The problem of the
general three-wheel indicator, with cyclic states $a, b, c \geq 2$, yields
a much more interesting result. In both cases all wheels should have at
least 2 cyclic states, since otherwise we would have wheels that could not
be clicked. For the three-wheel indicator, when $a, b, c$ are not pairwise
relatively prime a satisfactory solution is easily obtained. However when
they are pairwise relatively prime the solution involves a nontrivial
minimization problem. Of interest is whether the zeroing number modulo 3
is 1 (so that the final click is on the first wheel) or 2 (so that the
final click is on the second wheel). We call this number the {\it chirality\/} of
the triple $(a, b, c)$. While the zeroing number is highly dependent on
the ordering of the wheels, we prove that the chirality does not depend on
the ordering and is an invariant of the unordered triple $\{ a,b,c \}$. The
chirality is a mysterious quantity demonstrating many interesting patterns.
We prove one of these for the Fibonacci sequence, but we still seek more
general explanations.
\section{Two wheels} If we let $n$ be the number of times the first wheel
is advanced, then in order to obtain the zeroing number for two wheels with
a and b states we must find the minimum positive solution of
\begin{eqnarray*} n & \equiv & 0 \pmod a \\ n & \equiv & 0 \pmod b
\end{eqnarray*} or of \begin{eqnarray*} n & \equiv & 0 \pmod a \\ n -1 &
\equiv & 0 \pmod b \end{eqnarray*} depending on whether we stop after
moving the second or first wheel. The first set of equations implies
that $n = \lcm (a,b)$. The second implies \begin{eqnarray*} n & = & ak \\
ak & = & 1 \pmod b \end{eqnarray*} for the smallest positive number $k <
b$. So $n=aa^{-1}_b$ where $a^{-1}_b$ is the multiplicative inverse of $a$
in $\Z^{\ast}_b$.
Only when a and b are relatively prime does the second system have
solutions and then it provides the zeroing number. In general, the zeroing
number is
\begin{equation} f(a,b) = \left\{ \begin{array}{ll}
2 \lcm (a,b), & \mbox{if $\gcd (a,b) \neq 1$;} \\
2 a a^{-1}_b - 1, & \mbox{if $\gcd (a,b) = 1$.}
\end{array} \right. \label{eq:2wheel} \end{equation}
When a and b are relatively prime the zeroing of the wheels first occurs
when the final move is on the first wheel, as the second quantity above is
obviously smaller than the first. In fact, there is the following statement
about how much smaller it is.
\begin{theorem}
If a and b are relatively prime then
\[ f(a,b)+f(b,a) = 2ab \]
or equivalently
\[ aa^{-1}_b + bb^{-1}_a - 1 = ab. \]
\label{th:2sym}
\end{theorem}
\begin{proof}
After $f(a,b)$ advances the $(a,b)$-indicator has been zeroed for the first
time. If one continues advancing the wheels, starting with the $b$-wheel,
zeroing will first occur after another $f(b,a)$ advances. Thus the sum
represents the smallest number of advances zeroing both wheels in which
they are both advanced an equal number of times. We note that $\lcm (a,b)
= ab$ since $a,b$ are relatively prime. \end{proof}
As an aside we remark that if $a$ and $b$ are relatively prime we
have $f(a,b) \neq f(b,a)$! Otherwise by Theorem \ref{th:2sym}
$f(a,b)=ab$ and by Equation \eqref{eq:2wheel} $f(a,b)=2aa^{-1}_b-1$. The
first is a multiple of $a$, while the second is not.
We can make explicit machine computations using Mathematica
\cite{mathematica}, by using the Euler phi function
\cite{numtheory,hardywright} when $\gcd (a,b) = 1$: $a^{-1}_b \equiv
a^{\phi(b)-1} \pmod b$.
\section{Three wheels} \label{sc:3wheel}
The case of 3 or more wheels can be treated similarly, but now the
minimization problem becomes nontrivial. Again let $n$ be the number of
times the first wheel is advanced. If all three wheels are
advanced the same number of times, we must solve
\begin{eqnarray*}
n & \equiv & 0 \pmod a, \\
n & \equiv & 0 \pmod b, \\
n & \equiv & 0 \pmod c.
\end{eqnarray*}
Therefore a total of
\begin{equation*} f_0(a,b,c) = 3 \lcm (a,b,c) \label{eq:f0} \end{equation*}
advances are necessary. The other possible solutions involve unequal
numbers of advances with one or two of the wheels advanced one less than
the first. If the final move is on the first wheel then we must solve
\begin{eqnarray*}
n & \equiv & 0 \pmod a, \\
n - 1 & \equiv & 0 \pmod b, \\
n - 1 & \equiv & 0 \pmod c.
\end{eqnarray*}
Therefore a total of
\begin{equation} f_1(a,b,c) = 3 a a^{-1}_{\lcm(b,c)} - 2
\label{eq:f1} \end{equation}
advances are necessary. If the final move is on the second wheel then we
must solve
\begin{eqnarray*}
n & \equiv & 0 \pmod a, \\
n & \equiv & 0 \pmod b, \\
n -1 & \equiv & 0 \pmod c.
\end{eqnarray*} Therefore a total of
\begin{equation} f_2(a,b,c) = 3 \lcm (a,b) {\lcm (a,b)}^{-1}_c - 1
\label{eq:f2} \end{equation}
advances are necessary.
We conclude that the zeroing number is
\begin{equation} f(a,b,c) = \min \left\{ \begin{array}{ll}
3 \lcm (a,b,c), & \mbox{Case 0;} \\
3 a a^{-1}_{\lcm (b,c)} - 2, & \mbox{Case 1;} \\
3 \lcm (a,b) {\lcm (a,b)}^{-1}_c - 1, & \mbox{Case 2}
\end{array}
\right. \label{eq:3wheel} \end{equation}
where the minimization takes place over all existing cases.
Clearly $f_0(a,b,c)$ is always defined, but is larger than either of the
two remaining cases (if they exist). So this minimization problem is
nontrivial only when both $f_1(a,b,c)$ and $f_2(a,b,c)$ are defined.
For Case 1 to exist means that $a$ and $\lcm (b,c)$ are relatively
prime. For Case 2 to exist means that $\lcm (a,b)$ and $c$ are relatively
prime. Taken together, this means that only when $a,b,c$ are pairwise
relatively prime is the minimization nontrivial.
As a source of examples we provide some numerical data in Table
\ref{ta:data}.
\begin{table}
\begin{tabular}{|c|cccc|}
\hline
$(a,b,c)$ & $f_1(a,b,c)$ & $f_2(a,b,c)$ & $f(a,b,c)$ & $\chi(a,b,c)$ \\
\hline
$(2,3,5)$ & 46 & 17 & 17 & 2 \\
$(3,5,2)$ & 61 & 44 & 44 & 2 \\
$(5,2,3)$ & 73 & 29 & 29 & 2 \\
$(2,5,3)$ & 46 & 29 & 29 & 2 \\
$(3,2,5)$ & 61 & 17 & 17 & 2 \\
$(5,3,2)$ & 73 & 44 & 44 & 2 \\
\hline
\end{tabular}
\caption{Some numerical data}
\label{ta:data}
\end{table}
\section{Chirality}
In chemistry a molecule is said to be chiral if it is not superimposable
on its mirror image. Therefore such a molecule has two distinct
chiralities, left and right handedness. However in the following
definition it is more natural to denote these chiralities by 1 and 2.
\begin{definition}
The chirality of an ordered triple of pairwise relatively prime natural
numbers $\geq 2$ is the triple's zeroing number modulo 3. We write $\chi
(a,b,c) = f(a,b,c) \bmod 3$. We note that the chirality of such a triple
will always be 1 or 2.
\end{definition}
The chirality therefore corresponds to the case number from Section \ref
{sc:3wheel} that provides the zeroing number.
\begin{theorem}
The chirality of the ordered triple $(a, b, c)$ (of pairwise relatively
prime natural numbers $\geq 2$) is invariant under any permutation of the
triple. It is thus an invariant of the set $\{ a, b, c \}$.
\label{th:chir}
\end{theorem}
Recalling the origins of the problem, it can be said that the chirality
depends on the team, not on the lineup!
In order to prove the theorem we make use of some umpire indicator
identities which are valid whenever $f_1$ and $f_2$ are defined on the
given arguments.
\begin{lemma}
\begin{eqnarray}
f_1(a,b,c) + f_1(b,c,a) + f_1(c,a,b) & = & 3 \lcm (a,b,c) k_1
\label{eq:ind1} \\
f_2(a,b,c) + f_2(c,a,b) + f_2(b,c,a) & = & 3 \lcm (a,b,c) k_2
\label{eq:ind2} \\
f_2(a,b,c) + f_1(c,a,b) & = & 3 \lcm (a,b,c) \label{eq:ind3}
\end{eqnarray}
where $k_1, k_2 \in \{ 1,2 \}$
\end{lemma}
\begin{proof}
By following the advances on a three-wheel baseball umpire indicator, the
left hand sides of all three identities are easily seen to be multiples of
$3 \lcm (a,b,c)$. And since $f_1$ and $f_2$ are always strictly less than
$3 \lcm (a,b,c) $ the multiples must be as indicated.
\end{proof}
We now proceed to prove the theorem, so we assume that $a,b,c$ are pairwise
relatively prime numbers $\geq 2$. We note then that $\lcm (a,b,c) =
abc$. Adding
equations \eqref{eq:ind1} and \eqref{eq:ind2} we get
\begin{eqnarray*}
f_2(a,b,c) + f_2(c,a,b) + f_2(b,c,a)+ & & \\
f_1(a,b,c) + f_1(b,c,a) + f_1(c,b,a) & = 3 abc (k_1 + k_2).
\end{eqnarray*}
Applying the following three identities of the form of Equation
\eqref{eq:ind3}
\begin{eqnarray}
f_2(a,b,c) + f_1(c,a,b) & = & 3 abc \nonumber \\
f_2(c,a,b) + f_1(b,c,a) & = & 3 abc \label{eq:chirid} \\
f_2(b,c,a) + f_1(a,b,c) & = & 3 abc \nonumber
\end{eqnarray}
we find that $ 3abc (k_1+k_2) = 9abc $, so $k_1+k_2=3$ and $\{ k_1,k_2 \} =
\{ 1,2 \}$.
If $k_1=1$ then when we subtract Equation \eqref{eq:ind3} from Equation
\eqref{eq:ind1} we get $f_1(a,b,c) + f_1(b,c,a) - f_2(a,b,c) =0$.
Therefore $f_2(a,b,c) > f_1(a,b,c)$. So we get chirality 1.
Similarly if $k_2=1$ we get chirality 2. It follows that chirality 1 is
equivalent to $k_1=1$ (and $k_2=2$), while chirality 2 is equivalent to
$k_1=2$ (and $k_2=1$). We also note that
\begin{equation} \chi(a,b,c)=\frac{f_1(a,b,c) + f_1(b,c,a) +
f_1(c,b,a)}{3\lcm (a,b,c)}.
\label{eq:annoy} \end{equation}
Looking at Equation \eqref{eq:annoy} we see that cyclic permutations of the
triple do not change the chirality!
As an aside we note that we now have a three-wheel analog to Theorem
\ref{th:2sym}:
\begin{equation*}
f(a,b,c) + f(b,c,a) + f(c,a,b) = 3 abc
\end{equation*}
We must now prove that chirality is invariant under transposition of two
wheels of the triple. Due to the invariance under cyclic permutations, all
such transpositions are equivalent, so it suffices to check just one such
transposition. We start with the three identities in Equation
\eqref{eq:chirid}. Switching the second and third arguments of $f_1$
preserves the identities. So we have:
\begin{eqnarray*}
f_2(a,b,c) + f_1(c,b,a) & = & 3 abc \\
f_2(c,a,b) + f_1(b,a,c) & = & 3 abc \\
f_2(b,c,a) + f_1(a,c,b) & = & 3 abc
\end{eqnarray*}
Summing these three identities we get
\begin{eqnarray*}
f_2(a,b,c) + f_2(c,a,b) + f_2(b,c,a) + & & \\
f_1(a,c,b) + f_1(c,b,a) + f_1(b,a,c) & = & 9abc.
\end{eqnarray*}
Using Equation \eqref{eq:ind2} we find
\[3 abc k_2+ f_1(a,c,b)+f_1(c,b,a)+f_1(b,a,c) = 9abc\]
which along with $k_1+k_2=3$ implies that
\[f_1(a,c,b)+f_1(c,b,a)+f_1(b,a,c) = 3 abc k_1.\]
Applying Equation \eqref{eq:annoy} we see that
$\chi(a,c,b)=k_1=\chi(a,b,c)$ and we have proven Theorem \ref{th:chir}.
\section{Fibonacci triples}
Looking at the data for the three-wheel umpire indicator one notices many
intriguing patterns for the chirality of a triple of pairwise relatively
prime numbers $\geq 2$. It is difficult to analyze the data in general
since one is dealing with a 3-dimentional array, where even the existence
of chirality depends on the distribution of primes.
As an example we investigate one such pattern in detail. Let $F_n$ be the
$n$-th Fibonacci number, where $F_0 = 0$, $F_1=1$, and
$F_{n+1}=F_n+F_{n-1}$. Chirality is defined for $\{ F_n,F_{n+1},F_{n+2}
\}$ whenever $n \geq 3$. Using Equation \eqref{eq:3wheel} we found, using
Mathematica \cite{mathematica} that $\chi (2,3,5) = 2$, $\chi (3,5,8) = 1$,
$\chi (5,8,13) = 2$, $\chi (8,13,21) = 1$, \ldots . We conjectured that
this sequence continues to alternate, and then we proved:
\begin{theorem} For $n \geq 3$
\[ \chi (F_n,F_{n+1},F_{n+2}) = \left\{ \begin{array}{ll}
1, & \mbox{if $n$ is even;} \\
2, & \mbox{if $n$ is odd.}
\end{array} \right. \]
\label{th:fibonacci}
\end{theorem}
In order to prove this theorem we use the following Fibonacci identities:
\begin{lemma}[Fibonacci identities]
\begin{eqnarray*}
F_{n+1}^2-F_nF_{n+2} & = & (-1)^n \\
F_{n+1}F_{n+2} - F_nF_{n+3} & = & (-1)^n
\end{eqnarray*}
\end{lemma}
\begin{proof}
The first identity is easily proven by induction. It is a standard
exercise in elementary number theory courses.
The second identity can proven using a similar induction. However
we note that it is just a reformulation of the first identity:
\begin{eqnarray*}F_{n+1}F_{n+2} - F_nF_{n+3} & = & F_{n+1}F_{n+2} -
F_nF_{n+2} - F_nF_{n+1} \\ & = &
F_{n+1} (F_{n+2} - F_n) - F_nF_{n+2} \\ & = & F_{n+1}^2-F_nF_{n+2} \\ & = &
(-1)^n.
\end{eqnarray*}
\end{proof}
We now prove the theorem itself. First we assume that $n$ is odd. We
then must show that $\chi(F_n,F_{n+1},F_{n+2}) = 2$, but by the invariance
of chirality this is equivalent to showing that $\chi(F_n,F_{n+2},F_{n+1})
= 2$.
From Equation \eqref{eq:f1}
\[ f_1(F_n,F_{n+2},F_{n+1}) = 3F_n{F_n}^{-1}_{F_{n+2}F_{n+1}}-2 \]
But by the second Fibonacci identity $F_nF_{n+3} \equiv 1 \pmod
{F_{n+2}F_{n+1}}$, therefore
\begin{equation}
f_1(F_n,F_{n+2},F_{n+1}) = 3F_nF_{n+3}-2
\label{eq:f1odd}
\end{equation}
From Equation \eqref{eq:f2}
\[f_2(F_n,F_{n+2},F_{n+1}) = 3F_nF_{n+2}(F_nF_{n+2})^{-1}_{F_{n+1}}-1\]
But by the first Fibonacci identity $F_nF_{n+2} \equiv 1 \pmod {F_{n+1}}$,
therefore
\begin{equation}
f_2(F_n,F_{n+2},F_{n+1}) = 3F_nF_{n+2}-1.
\label{eq:f2odd}
\end{equation}
Comparing Equation~\eqref{eq:f1odd} and Equation~\eqref{eq:f2odd} we find
that $f_2$ is smaller and therefore, when $n$ is odd, $\chi
(F_n,F_{n+1},F_{n+2}) = 2$ as desired.
We now assume that $n$ is even. We must show that
$\chi(F_n,F_{n+1},F_{n+2}) = 1$, but by the invariance of chirality this is
equivalent to showing that $\chi(F_{n+1},F_{n+2},F_n) = 1$.
From Equation \eqref{eq:f1}
\[ f_1(F_{n+1},F_{n+2},F_n) = 3F_{n+1}{F_{n+1}}^{-1}_{F_{n+2}F_n}-2 \]
But by the first Fibonacci identity $F_{n+1}F_{n+1} \equiv 1 \pmod
{F_{n+2}F_n}$,
therefore
\begin{equation}
f_1(F_{n+1},F_{n+2},F_n) = 3F_{n+1}F_{n+1}-2
\label{eq:f1even}
\end{equation}
From Equation \eqref{eq:f2}
\[f_2(F_{n+1},F_{n+2},F_n) =
3F_{n+1}F_{n+2}(F_{n+1}F_{n+2})^{-1}_{F_n}-1\]
But by the second Fibonacci identity $F_{n+1}F_{n+2} \equiv 1 \pmod
{F_n}$,
therefore
\begin{equation}
f_2(F_{n+1},F_{n+2},F_n) = 3F_{n+1}F_{n+2}-1.
\label{eq:f2even}
\end{equation}
Comparing Equation~\eqref{eq:f1even} and Equation~\eqref{eq:f2even} we find
that $f_1$ is smaller and therefore, when $n$ is even, $\chi
(F_n,F_{n+1},F_{n+2}) = 1$ as desired.
This completes the proof of Theorem \ref{th:fibonacci}.
Note that we have also proven:
\begin{theorem}
If $n$ is odd \begin{equation*}
f(F_n,F_{n+2},F_{n+1}) = 3F_nF_{n+2}-1. \end{equation*}
If $n$ is even \begin{equation*}
f(F_{n+1},F_{n+2},F_n) = 3F_{n+1}^2-2. \end{equation*}
\end{theorem}
The order of the arguments of $f$ is crucial, since while chirality is an
invariant of unordered triples, the zeroing number is not.
\begin{thebibliography}{1}
\bibitem{numtheory}
David Bressoud and Stan Wagon.
\newblock {\em A Course in Computational Number Theory}.
\newblock Key College Publishing, 2000.
\bibitem{hardywright}
G.~H. Hardy and E.M. Wright.
\newblock {\em An Introduction to the Theory of Numbers}.
\newblock Oxford University Press, fifth edition, 1979.
\bibitem{mathematica}
Wolfram Research.
\newblock {\em Mathematica}.
\newblock Wolfram Research, fourth edition, 1999.
\end{thebibliography}
\vskip .5in
\bigskip
\hrule
\bigskip
\noindent 2000 {\it Mathematics Subject Classification}:
Primary 11A99; Secondary 11B39, 11B50 .
\noindent \emph{Keywords: }
chirality, Fibonacci sequence, minimization.
\bigskip
\hrule
\bigskip
\vspace*{+.1in}
\noindent
Received August 20 2003;
revised version received February 26, 2004.
Published in {\it Journal of Integer Sequences}, March 12 2004.
\bigskip
\hrule
\bigskip
\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.math.uwaterloo.ca/JIS/}.
\vskip .1in
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