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Fix distances

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Antti H S Laaksonen 2017-05-25 15:50:34 +03:00
parent cf7bcd21c2
commit 5a4ac55052
1 changed files with 15 additions and 15 deletions
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chapter07.tex
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@ -120,11 +120,11 @@ Of course, the same applies to coins 3 and 4.
Thus, we can use the following recursive formula Thus, we can use the following recursive formula
to calculate the minimum number of coins: to calculate the minimum number of coins:
\begin{equation*} \begin{equation*}
\begin{aligned} \begin{split}
\texttt{solve}(x) & = \min( & \texttt{solve}(x-1)+1 & , \\ \texttt{solve}(x) = \min( & \texttt{solve}(x-1)+1, \\
& & \texttt{solve}(x-3)+1 & , \\ & \texttt{solve}(x-3)+1, \\
& & \texttt{solve}(x-4)+1 & ). & \texttt{solve}(x-4)+1).
\end{aligned} \end{split}
\end{equation*} \end{equation*}
The base case of the recursion is $\texttt{solve}(0)=0$, The base case of the recursion is $\texttt{solve}(0)=0$,
because no coins are needed to form an empty sum. because no coins are needed to form an empty sum.
@ -326,11 +326,11 @@ we can form the sum $x$.
For example, if $\texttt{coins}=\{1,3,4\}$, For example, if $\texttt{coins}=\{1,3,4\}$,
then $\texttt{solve}(5)=6$ and the recursive formula is then $\texttt{solve}(5)=6$ and the recursive formula is
\begin{equation*} \begin{equation*}
\begin{aligned} \begin{split}
\texttt{solve}(x) & = & \texttt{solve}(x-1) & + \\ \texttt{solve}(x) = & \texttt{solve}(x-1) + \\
& & \texttt{solve}(x-3) & + \\ & \texttt{solve}(x-3) + \\
& & \texttt{solve}(x-4) & . & \texttt{solve}(x-4) .
\end{aligned} \end{split}
\end{equation*} \end{equation*}
In this case, the general recursive function is as follows: In this case, the general recursive function is as follows:
@ -801,11 +801,11 @@ between \texttt{x} and \texttt{y} equals $\texttt{distance}(n-1,m-1)$.
We can calculate the values of \texttt{distance} We can calculate the values of \texttt{distance}
as follows: as follows:
\begin{equation*} \begin{equation*}
\begin{aligned} \begin{split}
\texttt{distance}(a,b) & = \min(& \texttt{distance}(a,b-1)+1 & , \\ \texttt{distance}(a,b) = \min(& \texttt{distance}(a,b-1)+1, \\
& & \texttt{distance}(a-1,b)+1 & , \\ & \texttt{distance}(a-1,b)+1, \\
& & \texttt{distance}(a-1,b-1)+\texttt{cost}(a,b) & ). & \texttt{distance}(a-1,b-1)+\texttt{cost}(a,b)).
\end{aligned} \end{split}
\end{equation*} \end{equation*}
Here $\texttt{cost}(a,b)=0$ if $\texttt{x}[a]=\texttt{y}[b]$, Here $\texttt{cost}(a,b)=0$ if $\texttt{x}[a]=\texttt{y}[b]$,
and otherwise $\texttt{cost}(a,b)=1$. and otherwise $\texttt{cost}(a,b)=1$.