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Antti H S Laaksonen 2017-05-14 13:42:50 +03:00
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chapter29.tex
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@ -95,7 +95,7 @@ $(x_2,y_2)$,
$(x_3,y_3)$ and $(x_3,y_3)$ and
$(x_4,y_4)$. $(x_4,y_4)$.
This formula is easy to implement, there are no special This formula is easy to implement, there are no special
cases, and it turns out that we can even generalize the formula cases, and we can even generalize the formula
to \emph{all} polygons. to \emph{all} polygons.
\section{Complex numbers} \section{Complex numbers}
@ -128,7 +128,7 @@ following point and vector:
\index{complex@\texttt{complex}} \index{complex@\texttt{complex}}
The complex number class \texttt{complex} in C++ is The C++ complex number class \texttt{complex} is
useful when solving geometric problems. useful when solving geometric problems.
Using the class we can represent points and vectors Using the class we can represent points and vectors
as complex numbers, and the class contains tools as complex numbers, and the class contains tools
@ -136,7 +136,7 @@ that are useful in geometry.
In the following code, \texttt{C} is the type of In the following code, \texttt{C} is the type of
a coordinate and \texttt{P} is the type of a point or a vector. a coordinate and \texttt{P} is the type of a point or a vector.
In addition, the code defines the macros \texttt{X} and \texttt{Y} In addition, the code defines macros \texttt{X} and \texttt{Y}
that can be used to refer to x and y coordinates. that can be used to refer to x and y coordinates.
\begin{lstlisting} \begin{lstlisting}
@ -209,7 +209,7 @@ counterclockwise.
The function $\texttt{polar}(s,a)$ constructs a vector The function $\texttt{polar}(s,a)$ constructs a vector
whose length is $s$ and that points to an angle $a$. whose length is $s$ and that points to an angle $a$.
In addition, a vector can be rotated by an angle $a$ Moreover, a vector can be rotated by an angle $a$
by multiplying it by a vector with length 1 and angle $a$. by multiplying it by a vector with length 1 and angle $a$.
The following code calculates the angle of The following code calculates the angle of
@ -274,7 +274,7 @@ using the class \texttt{complex}:
\begin{lstlisting} \begin{lstlisting}
P a = {4,2}; P a = {4,2};
P b = {1,2}; P b = {1,2};
C r = (conj(a)*b).Y; // 6 C p = (conj(a)*b).Y; // 6
\end{lstlisting} \end{lstlisting}
The above code works, because The above code works, because
@ -307,8 +307,7 @@ $p$ is on the left side of the line:
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
In this situation, The cross product $(p-s_1) \times (p-s_2)$
the cross product $(p-s_1) \times (p-s_2)$
tells us the location of the point $p$. tells us the location of the point $p$.
If the cross product is positive, If the cross product is positive,
$p$ is located on the left side, $p$ is located on the left side,
@ -322,7 +321,7 @@ points $s_1$, $s_2$ and $p$ are on the same line.
\index{line segment intersection} \index{line segment intersection}
Consider the problem of checking Consider the problem of checking
whether two given line segments whether two line segments
$ab$ and $cd$ intersect. The possible cases are: $ab$ and $cd$ intersect. The possible cases are:
\textit{Case 1:} \textit{Case 1:}
@ -409,7 +408,7 @@ Hence, we can use cross products to check this.
\subsubsection{Point distance from a line} \subsubsection{Point distance from a line}
Another feature of the cross product is that Another feature of cross products is that
the area of a triangle can be calculated the area of a triangle can be calculated
using the formula using the formula
\[\frac{| (a-c) \times (b-c) |}{2},\] \[\frac{| (a-c) \times (b-c) |}{2},\]
@ -502,11 +501,11 @@ so $b$ is outside the polygon.
\section{Polygon area} \section{Polygon area}
A general formula for calculating the area A general formula for calculating the area
of a polygon\footnote{This formula is sometimes called the of a polygon, sometimes called the \key{shoelace formula},
\index{shoelace formula} \key{shoelace formula}.} is is as follows: \index{shoelace formula}
\[\frac{1}{2} |\sum_{i=1}^{n-1} (p_i \times p_{i+1})| = \[\frac{1}{2} |\sum_{i=1}^{n-1} (p_i \times p_{i+1})| =
\frac{1}{2} |\sum_{i=1}^{n-1} (x_i y_{i+1} - x_{i+1} y_i)|, \] \frac{1}{2} |\sum_{i=1}^{n-1} (x_i y_{i+1} - x_{i+1} y_i)|, \]
where the vertices are Here the vertices are
$p_1=(x_1,y_1)$, $p_2=(x_2,y_2)$, $\ldots$, $p_n=(x_n,y_n)$ $p_1=(x_1,y_1)$, $p_2=(x_2,y_2)$, $\ldots$, $p_n=(x_n,y_n)$
in such an order that in such an order that
$p_i$ and $p_{i+1}$ are adjacent vertices on the boundary $p_i$ and $p_{i+1}$ are adjacent vertices on the boundary
@ -702,7 +701,6 @@ For example, consider the following set of points:
\node at (3,0.5) {$D$}; \node at (3,0.5) {$D$};
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
The maximum Manhattan distance is 5 The maximum Manhattan distance is 5
between points $B$ and $C$: between points $B$ and $C$:
\begin{center} \begin{center}
@ -744,7 +742,6 @@ the result is:
\node at (4,-2.5) {$D$}; \node at (4,-2.5) {$D$};
\end{tikzpicture} \end{tikzpicture}
\end{center} \end{center}
And the maximum distance is as follows: And the maximum distance is as follows:
\begin{center} \begin{center}
\begin{tikzpicture}[scale=0.6] \begin{tikzpicture}[scale=0.6]