%$Header$
\chapter[Solutions: \cratzeroxrefcomma{}Chapter \ref{crat0}]
{Solutions: \cratzeroxrefcomma{}Chapter \ref{crat0}, \cratzerolongtitle{}}
\label{cras0}
\vworkexercisechapterheader{}
\begin{vworkexercisesolution}{\ref{exe:crat0:sexe0:01}}
The value of the \texttt{state} variable when
evaluating the \emph{if()} clause on the
$n+1$'th invocation is
\begin{equation}
\label{eq:cras0:exe01:01}
K_1 - nK_4 + (n+1) K_2 .
\end{equation}
We require on the $n+1$'th invocation, in order for the
test of the \emph{if()} clause to fail (i.e. that the function
``\texttt{A()}'' has been run on the first $n$ invocations of
the base subroutine but is not run on the $n+1$'th invocation),
that:
\begin{equation}
\label{eq:cras0:exe01:01b}
K_1 - nK_4 + (n+1) K_2 < K_3.
\end{equation}
Solving this inequality for the smallest integral
value of $n$ yields:
\begin{equation}
\label{eq:cras0:exe01:01c}
n = \left\lceil
\frac{K_1 + K_2 - K_3 + 1}{K_4 - K_2}
\right\rceil .
\end{equation}
It can be verified using an example that
(\ref{eq:cras0:exe01:01c}). For example, with
$K_1 = 10$, $K_2 = 3$, $K_3 = 7$, and $K_4 = 5$,
(\ref{eq:cras0:exe01:01}) predicts that on the first
$\lceil 7/2 \rceil = 4$ invocations of the base subroutine
the subroutine ``\texttt{A()}'' will be run but on the 5th
invocation it will not. Tracing the algorithm with the
parameters specified reveals that at the
test in the \emph{if()} statement
on the first invocation of the
subroutine, \texttt{state}=13 (``\texttt{A()}'' executed);
on the second invocation of the
subroutine, \texttt{state}=11 (``\texttt{A()}'' executed);
on the third invocation of the
subroutine, \texttt{state}=9 (``\texttt{A()}'' executed);
on the fourth invocation of the
subroutine, \texttt{state}=7 (``\texttt{A()}'' executed);
and on the fifth invocation of the
subroutine, \texttt{state}=5 (``\texttt{A()}'' not executed).
This is in agreement with
(\ref{eq:cras0:exe01:01}).
\end{vworkexercisesolution}
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%\begin{vworkexercisesolution}{\ref{exe:cfry0:sexe0:02}}
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%\end{vworkexercisesolution}
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