786 lines
22 KiB
TeX
786 lines
22 KiB
TeX
\chapter{Geometry}
|
|
|
|
\index{geometry}
|
|
|
|
In geometric problems, it is often challenging
|
|
to find a way to approach the problem so that
|
|
the solution to the problem can be conveniently implemented
|
|
and the number of special cases is small.
|
|
|
|
As an example, consider a problem where
|
|
we are given the vertices of a quadrilateral
|
|
(a polygon that has four vertices),
|
|
and our task is to calculate its area.
|
|
For example, a possible input for the problem is as follows:
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.45]
|
|
|
|
\draw[fill] (6,2) circle [radius=0.1];
|
|
\draw[fill] (5,6) circle [radius=0.1];
|
|
\draw[fill] (2,5) circle [radius=0.1];
|
|
\draw[fill] (1,1) circle [radius=0.1];
|
|
\draw[thick] (6,2) -- (5,6) -- (2,5) -- (1,1) -- (6,2);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
One way to approach the problem is to divide
|
|
the quadrilateral into two triangles by a straight
|
|
line between two opposite vertices:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.45]
|
|
|
|
\draw[fill] (6,2) circle [radius=0.1];
|
|
\draw[fill] (5,6) circle [radius=0.1];
|
|
\draw[fill] (2,5) circle [radius=0.1];
|
|
\draw[fill] (1,1) circle [radius=0.1];
|
|
|
|
\draw[thick] (6,2) -- (5,6) -- (2,5) -- (1,1) -- (6,2);
|
|
\draw[dashed,thick] (2,5) -- (6,2);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
After this, it suffices to sum the areas
|
|
of the triangles.
|
|
The area of a triangle can be calculated,
|
|
for example, using \key{Heron's formula}
|
|
%\footnote{Heron of Alexandria (c. 10--70) was a Greek mathematician.}
|
|
\[ \sqrt{s (s-a) (s-b) (s-c)},\]
|
|
where $a$, $b$ and $c$ are the lengths
|
|
of the triangle's sides and
|
|
$s=(a+b+c)/2$.
|
|
\index{Heron's formula}
|
|
|
|
This is a possible way to solve the problem,
|
|
but there is one pitfall:
|
|
how to divide the quadrilateral into triangles?
|
|
It turns out that sometimes we cannot just pick
|
|
two arbitrary opposite vertices.
|
|
For example, in the following situation,
|
|
the division line is outside the quadrilateral:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.45]
|
|
|
|
\draw[fill] (6,2) circle [radius=0.1];
|
|
\draw[fill] (3,2) circle [radius=0.1];
|
|
\draw[fill] (2,5) circle [radius=0.1];
|
|
\draw[fill] (1,1) circle [radius=0.1];
|
|
\draw[thick] (6,2) -- (3,2) -- (2,5) -- (1,1) -- (6,2);
|
|
|
|
\draw[dashed,thick] (2,5) -- (6,2);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
However, another way to draw the line works:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.45]
|
|
|
|
\draw[fill] (6,2) circle [radius=0.1];
|
|
\draw[fill] (3,2) circle [radius=0.1];
|
|
\draw[fill] (2,5) circle [radius=0.1];
|
|
\draw[fill] (1,1) circle [radius=0.1];
|
|
\draw[thick] (6,2) -- (3,2) -- (2,5) -- (1,1) -- (6,2);
|
|
|
|
\draw[dashed,thick] (3,2) -- (1,1);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
It is clear for a human which of the lines is the correct
|
|
choice, but the situation is difficult for a computer.
|
|
|
|
However, it turns out that we can solve the problem using
|
|
another method that is much easier to implement
|
|
and does not involve any special cases.
|
|
Namely, there is a general formula
|
|
\[x_1y_2-x_2y_1+x_2y_3-x_3y_2+x_3y_4-x_4y_3+x_4y_1-x_1y_4,\]
|
|
that calculates the area of a quadrilateral
|
|
whose vertices are
|
|
$(x_1,y_1)$,
|
|
$(x_2,y_2)$,
|
|
$(x_3,y_3)$ and
|
|
$(x_4,y_4)$.
|
|
This formula is easy to implement, there are no special
|
|
cases, and it turns out that we can even generalize the formula
|
|
to \emph{all} polygons.
|
|
|
|
\section{Complex numbers}
|
|
|
|
\index{complex number}
|
|
\index{point}
|
|
\index{vector}
|
|
|
|
A \key{complex number} is a number of the form $x+y i$,
|
|
where $i = \sqrt{-1}$ is the \key{imaginary unit}.
|
|
A geometric interpretation of a complex number is
|
|
that it represents a two-dimensional point $(x,y)$
|
|
or a vector from the origin to a point $(x,y)$.
|
|
|
|
For example, $4+2i$ corresponds to the
|
|
following point and vector:
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.45]
|
|
|
|
\draw[->,thick] (-5,0)--(5,0);
|
|
\draw[->,thick] (0,-5)--(0,5);
|
|
|
|
\draw[fill] (4,2) circle [radius=0.1];
|
|
\draw[->,thick] (0,0)--(4-0.1,2-0.1);
|
|
|
|
\node at (4,2.8) {$(4,2)$};
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
\index{complex@\texttt{complex}}
|
|
|
|
The complex number class \texttt{complex} in C++ is
|
|
useful when solving geometric problems.
|
|
Using the class we can represent points and vectors
|
|
as complex numbers, and the class also contains tools
|
|
that are useful in geometry.
|
|
|
|
In the following code, \texttt{C} is the type of
|
|
a coordinate and \texttt{P} is the type of a point or vector.
|
|
In addition, the code defines the macros \texttt{X} and \texttt{Y}
|
|
that can be used to refer to x and y coordinates.
|
|
|
|
\begin{lstlisting}
|
|
typedef long long C;
|
|
typedef complex<C> P;
|
|
#define X real()
|
|
#define Y imag()
|
|
\end{lstlisting}
|
|
|
|
For example, the following code defines a point $p=(4,2)$
|
|
and prints its x and y coordinates:
|
|
|
|
\begin{lstlisting}
|
|
P p = {4,2};
|
|
cout << p.X << " " << p.Y << "\n"; // 4 2
|
|
\end{lstlisting}
|
|
|
|
The following code defines vectors $v=(3,1)$ and $u=(2,2)$,
|
|
and after that calculates the sum $s=v+u$.
|
|
|
|
\begin{lstlisting}
|
|
P v = {3,1};
|
|
P u = {2,2};
|
|
P s = v+u;
|
|
cout << s.X << " " << s.Y << "\n"; // 5 3
|
|
\end{lstlisting}
|
|
|
|
An appropriate coordinate type is
|
|
\texttt{long long} (integer) or \texttt{long double}
|
|
(real number), depending on the situation.
|
|
Integers should be used whenever possible,
|
|
because calculations with integers are exact.
|
|
If real numbers are needed,
|
|
precision errors should be taken into account
|
|
when comparing them.
|
|
A safe way to check if numbers $a$ and $b$ are equal
|
|
is to compare them using $|a-b|<\epsilon$,
|
|
where $\epsilon$ is a small number (for example, $\epsilon=10^{-9}$).
|
|
|
|
\subsubsection*{Functions}
|
|
|
|
In the following examples, the coordinate type is
|
|
\texttt{long double}.
|
|
|
|
The function \texttt{abs(v)} calculates the length
|
|
$|v|$ of a vector $v=(x,y)$
|
|
using the formula $\sqrt{x^2+y^2}$.
|
|
The function can also be used for
|
|
calculating the distance between points
|
|
$(x_1,y_1)$ and $(x_2,y_2)$,
|
|
because that distance equals the length
|
|
of the vector $(x_2-x_1,y_2-y_1)$.
|
|
|
|
The following code calculates the distance
|
|
between points $(4,2)$ and $(3,-1)$:
|
|
\begin{lstlisting}
|
|
P a = {4,2};
|
|
P b = {3,-1};
|
|
cout << abs(b-a) << "\n"; // 3.16228
|
|
\end{lstlisting}
|
|
|
|
The function \texttt{arg(v)} calculates the
|
|
angle of a vector $v=(x,y)$ with respect to the x axis.
|
|
The function gives the angle in radians,
|
|
where $r$ radians equals $180 r/\pi$ degrees.
|
|
The angle of a vector that points to the right is 0,
|
|
and angles decrease clockwise and increase
|
|
counterclockwise.
|
|
|
|
The function \texttt{polar(s,a)} constructs a vector
|
|
whose length is $s$ and that points to an angle $a$.
|
|
In addition, a vector can be rotated by an angle $a$
|
|
by multiplying it by a vector with length 1 and angle $a$.
|
|
|
|
The following code calculates the angle of
|
|
the vector $(4,2)$, rotates it $1/2$ radians
|
|
counterclockwise, and then calculates the angle again:
|
|
|
|
\begin{lstlisting}
|
|
P v = {4,2};
|
|
cout << arg(v) << "\n"; // 0.463648
|
|
v *= polar(1.0,0.5);
|
|
cout << arg(v) << "\n"; // 0.963648
|
|
\end{lstlisting}
|
|
|
|
\section{Points and lines}
|
|
|
|
\index{cross product}
|
|
|
|
The \key{cross product} $a \times b$ of vectors
|
|
$a=(x_1,y_1)$ and $b=(x_2,y_2)$ equals $x_1 y_2 - x_2 y_1$.
|
|
The cross product tells us whether $b$
|
|
turns left (positive value), does not turn (zero)
|
|
or turns right (negative value)
|
|
when it is placed directly after $a$.
|
|
|
|
The following picture illustrates the above cases:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.45]
|
|
|
|
\draw[->,thick] (0,0)--(4,2);
|
|
\draw[->,thick] (4,2)--(4+1,2+2);
|
|
|
|
\node at (2.5,0.5) {$a$};
|
|
\node at (5,2.5) {$b$};
|
|
|
|
\node at (3,-2) {$a \times b = 6$};
|
|
|
|
\draw[->,thick] (8+0,0)--(8+4,2);
|
|
\draw[->,thick] (8+4,2)--(8+4+2,2+1);
|
|
|
|
\node at (8+2.5,0.5) {$a$};
|
|
\node at (8+5,1.5) {$b$};
|
|
|
|
\node at (8+3,-2) {$a \times b = 0$};
|
|
|
|
\draw[->,thick] (16+0,0)--(16+4,2);
|
|
\draw[->,thick] (16+4,2)--(16+4+2,2-1);
|
|
|
|
\node at (16+2.5,0.5) {$a$};
|
|
\node at (16+5,2.5) {$b$};
|
|
|
|
\node at (16+3,-2) {$a \times b = -8$};
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
\noindent
|
|
For example, in the first picture
|
|
$a=(4,2)$ and $b=(1,2)$.
|
|
The following code calculates the cross product
|
|
using the class \texttt{complex}:
|
|
|
|
\begin{lstlisting}
|
|
P a = {4,2};
|
|
P b = {1,2};
|
|
C r = (conj(a)*b).Y; // 6
|
|
\end{lstlisting}
|
|
|
|
The above code works, because
|
|
the function \texttt{conj} negates the y coordinate
|
|
of a vector,
|
|
and when the vectors $(x_1,-y_1)$ and $(x_2,y_2)$
|
|
are multiplied together, the y coordinate
|
|
of the result is $x_1 y_2 - x_2 y_1$.
|
|
|
|
\subsubsection{Point location}
|
|
|
|
Cross products can be used for testing
|
|
whether a point is located on the left or right
|
|
side of a line.
|
|
Assume that the line goes through points
|
|
$s_1$ and $s_2$, we are looking from $s_1$
|
|
to $s_2$ and the point is $p$.
|
|
|
|
For example, in the following picture,
|
|
$p$ is on the left side of the line:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.45]
|
|
\draw[dashed,thick,->] (0,-3)--(12,6);
|
|
\draw[fill] (4,0) circle [radius=0.1];
|
|
\draw[fill] (8,3) circle [radius=0.1];
|
|
\draw[fill] (5,3) circle [radius=0.1];
|
|
\node at (4,-1) {$s_1$};
|
|
\node at (8,2) {$s_2$};
|
|
\node at (5,4) {$p$};
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
In this situation,
|
|
the cross product $(p-s_1) \times (p-s_2)$
|
|
tells us the location of the point $p$.
|
|
If the cross product is positive,
|
|
$p$ is located on the left side,
|
|
and if the cross product is negative,
|
|
$p$ is located on the right side.
|
|
Finally, if the cross product is zero,
|
|
points $s_1$, $s_2$ and $p$ are on the same line.
|
|
|
|
\subsubsection{Line segment intersection}
|
|
|
|
\index{line segment intersection}
|
|
|
|
Consider a problem where we are given two line segments
|
|
$ab$ and $cd$ and our task is to check whether they
|
|
intersect. The possible cases are:
|
|
|
|
\textit{Case 1:}
|
|
The line segments are on the same line
|
|
and they overlap each other.
|
|
In this case, there is an infinite number of
|
|
intersection points.
|
|
For example, in the following picture,
|
|
all points between $c$ and $b$ are
|
|
intersection points:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.9]
|
|
\draw (1.5,1.5)--(6,3);
|
|
\draw (0,1)--(4.5,2.5);
|
|
\draw[fill] (0,1) circle [radius=0.05];
|
|
\node at (0,0.5) {$a$};
|
|
\draw[fill] (1.5,1.5) circle [radius=0.05];
|
|
\node at (6,2.5) {$d$};
|
|
\draw[fill] (4.5,2.5) circle [radius=0.05];
|
|
\node at (1.5,1) {$c$};
|
|
\draw[fill] (6,3) circle [radius=0.05];
|
|
\node at (4.5,2) {$b$};
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
In this case, we can use cross products to
|
|
check if all points are on the same line.
|
|
After this, we can sort the points and check
|
|
whether the line segments overlap each other.
|
|
|
|
\textit{Case 2:}
|
|
The line segments have a common vertex
|
|
that is the only intersection point.
|
|
For example, in the following picture the
|
|
intersection point is $b=c$:
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.9]
|
|
\draw (0,0)--(4,2);
|
|
\draw (4,2)--(6,1);
|
|
\draw[fill] (0,0) circle [radius=0.05];
|
|
\draw[fill] (4,2) circle [radius=0.05];
|
|
\draw[fill] (6,1) circle [radius=0.05];
|
|
|
|
\node at (0,0.5) {$a$};
|
|
\node at (4,2.5) {$b=c$};
|
|
\node at (6,1.5) {$d$};
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
This case is easy to check, because
|
|
there are only four possibilities
|
|
for the intersection point:
|
|
$a=c$, $a=d$, $b=c$ and $b=d$.
|
|
|
|
\textit{Case 3:}
|
|
There is exactly one intersection point
|
|
that is not a vertex of any line segment.
|
|
In the following picture, the point $p$
|
|
is the intersection point:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.9]
|
|
\draw (0,1)--(6,3);
|
|
\draw (2,4)--(4,0);
|
|
\draw[fill] (0,1) circle [radius=0.05];
|
|
\node at (0,0.5) {$c$};
|
|
\draw[fill] (6,3) circle [radius=0.05];
|
|
\node at (6,2.5) {$d$};
|
|
\draw[fill] (2,4) circle [radius=0.05];
|
|
\node at (1.5,3.5) {$a$};
|
|
\draw[fill] (4,0) circle [radius=0.05];
|
|
\node at (4,-0.4) {$b$};
|
|
\draw[fill] (3,2) circle [radius=0.05];
|
|
\node at (3,1.5) {$p$};
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
In this case, the line segments intersect
|
|
exactly when both points $c$ and $d$ are
|
|
on different sides of a line through $a$ and $b$,
|
|
and points $a$ and $b$ are on different
|
|
sides of a line through $c$ and $d$.
|
|
Hence, we can use cross products to check this.
|
|
|
|
\subsubsection{Point distance from a line}
|
|
|
|
The area of a triangle can be calculated
|
|
using the formula
|
|
\[\frac{| (a-c) \times (b-c) |}{2},\]
|
|
where $a$, $b$ and $c$ are the vertices of the triangle.
|
|
|
|
Using this formula, it is possible to calculate the
|
|
shortest distance between a point and a line.
|
|
For example, in the following picture $d$ is the
|
|
shortest distance between the point $p$ and the line
|
|
that is defined by the points $s_1$ and $s_2$:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.75]
|
|
\draw (-2,-1)--(6,3);
|
|
\draw[dashed] (1,4)--(2.40,1.2);
|
|
\node at (0,-0.5) {$s_1$};
|
|
\node at (4,1.5) {$s_2$};
|
|
\node at (0.5,4) {$p$};
|
|
\node at (2,2.7) {$d$};
|
|
\draw[fill] (0,0) circle [radius=0.05];
|
|
\draw[fill] (4,2) circle [radius=0.05];
|
|
\draw[fill] (1,4) circle [radius=0.05];
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
The area of the triangle whose vertices are
|
|
$s_1$, $s_2$ and $p$ can be calculated in two ways:
|
|
it is both
|
|
$\frac{1}{2} |s_2-s_1| d$ and
|
|
$\frac{1}{2} ((s_1-p) \times (s_2-p))$.
|
|
Thus, the shortest distance is
|
|
|
|
\[ d = \frac{(s_1-p) \times (s_2-p)}{|s_2-s_1|} .\]
|
|
|
|
\subsubsection{Point inside a polygon}
|
|
|
|
Let us now consider a problem where our task is to
|
|
find out whether a point is located inside or outside
|
|
a polygon.
|
|
For example, in the following picture point $a$
|
|
is inside the polygon and point $b$ is outside
|
|
the polygon.
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.75]
|
|
%\draw (0,0)--(2,-2)--(3,1)--(5,1)--(2,3)--(1,2)--(-1,2)--(1,4)--(-2,4)--(-2,1)--(-3,3)--(-4,0)--(0,0);
|
|
\draw (0,0)--(2,2)--(5,1)--(2,3)--(1,2)--(-1,2)--(1,4)--(-2,4)--(-2,1)--(-3,3)--(-4,0)--(0,0);
|
|
|
|
\draw[fill] (-3,1) circle [radius=0.05];
|
|
\node at (-3,0.5) {$a$};
|
|
\draw[fill] (1,3) circle [radius=0.05];
|
|
\node at (1,2.5) {$b$};
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
A convenient way to solve the problem is to
|
|
send a ray from the point to an arbitrary direction
|
|
and calculate the number of times it touches
|
|
the boundary of the polygon.
|
|
If the number is odd,
|
|
the point is inside the polygon,
|
|
and if the number is even,
|
|
the point is outside the polygon.
|
|
|
|
\begin{samepage}
|
|
For example, we could send the following rays:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.75]
|
|
\draw (0,0)--(2,2)--(5,1)--(2,3)--(1,2)--(-1,2)--(1,4)--(-2,4)--(-2,1)--(-3,3)--(-4,0)--(0,0);
|
|
|
|
\draw[fill] (-3,1) circle [radius=0.05];
|
|
\node at (-3,0.5) {$a$};
|
|
\draw[fill] (1,3) circle [radius=0.05];
|
|
\node at (1,2.5) {$b$};
|
|
|
|
\draw[dashed,->] (-3,1)--(-6,0);
|
|
\draw[dashed,->] (-3,1)--(0,5);
|
|
|
|
\draw[dashed,->] (1,3)--(3.5,0);
|
|
\draw[dashed,->] (1,3)--(3,4);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
\end{samepage}
|
|
|
|
The rays from $a$ touch 1 and 3 times
|
|
the boundary of the polygon,
|
|
so $a$ is inside the polygon.
|
|
Correspondingly, the rays from $b$
|
|
touch 0 and 2 times the boundary of the polygon,
|
|
so $b$ is outside the polygon.
|
|
|
|
\section{Polygon area}
|
|
|
|
A general formula for calculating the area
|
|
of a polygon\footnote{This formula is sometimes called the
|
|
\index{shoelace formula} \key{shoelace formula}.} is
|
|
\[\frac{1}{2} |\sum_{i=1}^{n-1} (p_i \times p_{i+1})| =
|
|
\frac{1}{2} |\sum_{i=1}^{n-1} (x_i y_{i+1} - x_{i+1} y_i)|, \]
|
|
where the vertices are
|
|
$p_1=(x_1,y_1)$, $p_2=(x_2,y_2)$, $\ldots$, $p_n=(x_n,y_n)$
|
|
in such an order that
|
|
$p_i$ and $p_{i+1}$ are adjacent vertices on the boundary
|
|
of the polygon,
|
|
and the first and last vertex is the same, i.e., $p_1=p_n$.
|
|
|
|
For example, the area of the polygon
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.7]
|
|
\filldraw (4,1.4) circle (2pt);
|
|
\filldraw (7,3.4) circle (2pt);
|
|
\filldraw (5,5.4) circle (2pt);
|
|
\filldraw (2,4.4) circle (2pt);
|
|
\filldraw (4,3.4) circle (2pt);
|
|
\node (1) at (4,1) {(4,1)};
|
|
\node (2) at (7.2,3) {(7,3)};
|
|
\node (3) at (5,5.8) {(5,5)};
|
|
\node (4) at (2,4) {(2,4)};
|
|
\node (5) at (3.5,3) {(4,3)};
|
|
\path[draw] (4,1.4) -- (7,3.4) -- (5,5.4) -- (2,4.4) -- (4,3.4) -- (4,1.4);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
is
|
|
\[\frac{|(2\cdot5-5\cdot4)+(5\cdot3-7\cdot5)+(7\cdot1-4\cdot3)+(4\cdot3-4\cdot1)+(4\cdot4-2\cdot3)|}{2} = 17/2.\]
|
|
|
|
The idea in the formula is to go through trapezoids
|
|
whose one side is a side of the polygon,
|
|
and another side lies on the horizontal line $y=0$.
|
|
For example:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.7]
|
|
\path[draw,fill=lightgray] (5,5.4) -- (7,3.4) -- (7,0) -- (5,0) -- (5,5.4);
|
|
\filldraw (4,1.4) circle (2pt);
|
|
\filldraw (7,3.4) circle (2pt);
|
|
\filldraw (5,5.4) circle (2pt);
|
|
\filldraw (2,4.4) circle (2pt);
|
|
\filldraw (4,3.4) circle (2pt);
|
|
\node (1) at (4,1) {(4,1)};
|
|
\node (2) at (7.2,3) {(7,3)};
|
|
\node (3) at (5,5.8) {(5,5)};
|
|
\node (4) at (2,4) {(2,4)};
|
|
\node (5) at (3.5,3) {(4,3)};
|
|
\path[draw] (4,1.4) -- (7,3.4) -- (5,5.4) -- (2,4.4) -- (4,3.4) -- (4,1.4);
|
|
\draw (0,0) -- (10,0);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
The area of such a trapezoid is
|
|
\[(x_{i+1}-x_{i}) \frac{y_i+y_{i+1}}{2},\]
|
|
where the vertices of the polygon are $p_i$ and $p_{i+1}$.
|
|
If $x_{i+1}>x_{i}$, the area is positive,
|
|
and if $x_{i+1}<x_{i}$, the area is negative.
|
|
|
|
The area of the polygon is the sum of areas of
|
|
all such trapezoids, which yields the formula
|
|
|
|
\[|\sum_{i=1}^{n-1} (x_{i+1}-x_{i}) \frac{y_i+y_{i+1}}{2}| =
|
|
\frac{1}{2} |\sum_{i=1}^{n-1} (x_i y_{i+1} - x_{i+1} y_i)|.\]
|
|
|
|
Note that the absolute value of the sum is taken,
|
|
because the value of the sum may be positive or negative
|
|
depending on whether we walk clockwise or counterclockwise
|
|
along the boundary of the polygon.
|
|
|
|
\subsubsection{Pick's theorem}
|
|
|
|
\index{Pick's theorem}
|
|
|
|
\key{Pick's theorem} provides another way to calculate
|
|
the area of a polygon provided that all vertices
|
|
of the polygon have integer coordinates.
|
|
According to Pick's theorem, the area of the polygon is
|
|
\[ a + b/2 -1,\]
|
|
where $a$ is the number of integer points inside the polygon
|
|
and $b$ is the number of integer points on the boundary of the polygon.
|
|
|
|
For example, the area of the polygon
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.7]
|
|
\filldraw (4,1.4) circle (2pt);
|
|
\filldraw (7,3.4) circle (2pt);
|
|
\filldraw (5,5.4) circle (2pt);
|
|
\filldraw (2,4.4) circle (2pt);
|
|
\filldraw (4,3.4) circle (2pt);
|
|
\node (1) at (4,1) {(4,1)};
|
|
\node (2) at (7.2,3) {(7,3)};
|
|
\node (3) at (5,5.8) {(5,5)};
|
|
\node (4) at (2,4) {(2,4)};
|
|
\node (5) at (3.5,3) {(4,3)};
|
|
\path[draw] (4,1.4) -- (7,3.4) -- (5,5.4) -- (2,4.4) -- (4,3.4) -- (4,1.4);
|
|
|
|
\filldraw (2,4.4) circle (2pt);
|
|
\filldraw (3,4.4) circle (2pt);
|
|
\filldraw (4,4.4) circle (2pt);
|
|
\filldraw (5,4.4) circle (2pt);
|
|
\filldraw (6,4.4) circle (2pt);
|
|
|
|
\filldraw (4,3.4) circle (2pt);
|
|
\filldraw (5,3.4) circle (2pt);
|
|
\filldraw (6,3.4) circle (2pt);
|
|
\filldraw (7,3.4) circle (2pt);
|
|
|
|
\filldraw (4,2.4) circle (2pt);
|
|
\filldraw (5,2.4) circle (2pt);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
is $6+7/2-1=17/2$.
|
|
|
|
\section{Distance functions}
|
|
|
|
\index{distance function}
|
|
\index{Euklidean distance}
|
|
\index{Manhattan distance}
|
|
|
|
A \key{distance function} defines the distance between
|
|
two points.
|
|
The usual distance function in geometry is the
|
|
\key{Euclidean distance} where the distance between
|
|
points $(x_1,y_1)$ and $(x_2,y_2)$ is
|
|
\[\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.\]
|
|
An alternative distance function is the
|
|
\key{Manhattan distance}
|
|
where the distance between points
|
|
$(x_1,y_1)$ and $(x_2,y_2)$ is
|
|
\[|x_1-x_2|+|y_1-y_2|.\]
|
|
\begin{samepage}
|
|
For example, consider the following picture:
|
|
\begin{center}
|
|
\begin{tikzpicture}
|
|
|
|
\draw[fill] (2,1) circle [radius=0.05];
|
|
\draw[fill] (5,2) circle [radius=0.05];
|
|
|
|
\node at (2,0.5) {$(2,1)$};
|
|
\node at (5,1.5) {$(5,2)$};
|
|
|
|
\draw[dashed] (2,1) -- (5,2);
|
|
|
|
\draw[fill] (5+2,1) circle [radius=0.05];
|
|
\draw[fill] (5+5,2) circle [radius=0.05];
|
|
|
|
\node at (5+2,0.5) {$(2,1)$};
|
|
\node at (5+5,1.5) {$(5,2)$};
|
|
|
|
\draw[dashed] (5+2,1) -- (5+2,2);
|
|
\draw[dashed] (5+2,2) -- (5+5,2);
|
|
|
|
\node at (3.5,-0.5) {Euclidean distance};
|
|
\node at (5+3.5,-0.5) {Manhattan distance};
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
\end{samepage}
|
|
The Euclidean distance between the points is
|
|
\[\sqrt{(5-2)^2+(2-1)^2}=\sqrt{10}\]
|
|
and the Manhattan distance is
|
|
\[|5-2|+|2-1|=4.\]
|
|
The following picture shows regions that are within a distance of 1
|
|
from the center point, using the Euclidean and Manhattan distances:
|
|
\begin{center}
|
|
\begin{tikzpicture}
|
|
|
|
\draw[fill=gray!20] (0,0) circle [radius=1];
|
|
\draw[fill] (0,0) circle [radius=0.05];
|
|
|
|
\node at (0,-1.5) {Euclidean distance};
|
|
|
|
\draw[fill=gray!20] (5+0,1) -- (5-1,0) -- (5+0,-1) -- (5+1,0) -- (5+0,1);
|
|
\draw[fill] (5,0) circle [radius=0.05];
|
|
\node at (5,-1.5) {Manhattan distance};
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
\subsubsection{Rotating coordinates}
|
|
|
|
Some problems are easier to solve if the
|
|
Manhattan distance is used instead of the Euclidean distance.
|
|
As an example, consider a problem where we are given
|
|
$n$ points in the two-dimensional plane
|
|
and our task is to calculate the maximum Manhattan
|
|
distance between any two points.
|
|
|
|
For example, consider the following set of points:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.65]
|
|
\draw[color=gray] (-1,-1) grid (4,4);
|
|
|
|
\filldraw (0,2) circle (2.5pt);
|
|
\filldraw (3,3) circle (2.5pt);
|
|
\filldraw (1,0) circle (2.5pt);
|
|
\filldraw (3,1) circle (2.5pt);
|
|
|
|
\node at (0,1.5) {$A$};
|
|
\node at (3,2.5) {$C$};
|
|
\node at (1,-0.5) {$B$};
|
|
\node at (3,0.5) {$D$};
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
The maximum Manhattan distance is 5
|
|
between points $B$ and $C$:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.65]
|
|
\draw[color=gray] (-1,-1) grid (4,4);
|
|
|
|
\filldraw (0,2) circle (2.5pt);
|
|
\filldraw (3,3) circle (2.5pt);
|
|
\filldraw (1,0) circle (2.5pt);
|
|
\filldraw (3,1) circle (2.5pt);
|
|
|
|
\node at (0,1.5) {$A$};
|
|
\node at (3,2.5) {$C$};
|
|
\node at (1,-0.5) {$B$};
|
|
\node at (3,0.5) {$D$};
|
|
|
|
\path[draw=red,thick,line width=2pt] (1,0) -- (1,3) -- (3,3);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
A useful technique related to Manhattan distances
|
|
is to rotate all coordinates 45 degrees so that
|
|
a point $(x,y)$ becomes $(x+y,y-x)$.
|
|
For example, after rotating the above points,
|
|
the result is:
|
|
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.6]
|
|
\draw[color=gray] (0,-3) grid (7,3);
|
|
|
|
\filldraw (2,2) circle (2.5pt);
|
|
\filldraw (6,0) circle (2.5pt);
|
|
\filldraw (1,-1) circle (2.5pt);
|
|
\filldraw (4,-2) circle (2.5pt);
|
|
|
|
\node at (2,1.5) {$A$};
|
|
\node at (6,-0.5) {$C$};
|
|
\node at (1,-1.5) {$B$};
|
|
\node at (4,-2.5) {$D$};
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
And the maximum distance is as follows:
|
|
\begin{center}
|
|
\begin{tikzpicture}[scale=0.6]
|
|
\draw[color=gray] (0,-3) grid (7,3);
|
|
|
|
\filldraw (2,2) circle (2.5pt);
|
|
\filldraw (6,0) circle (2.5pt);
|
|
\filldraw (1,-1) circle (2.5pt);
|
|
\filldraw (4,-2) circle (2.5pt);
|
|
|
|
\node at (2,1.5) {$A$};
|
|
\node at (6,-0.5) {$C$};
|
|
\node at (1,-1.5) {$B$};
|
|
\node at (4,-2.5) {$D$};
|
|
|
|
\path[draw=red,thick,line width=2pt] (1,-1) -- (4,2) -- (6,0);
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
|
|
Consider two points $p_1=(x_1,y_1)$ and $p_2=(x_1,y_1)$ whose rotated
|
|
coordinates are $p'_1=(x'_1,y'_1)$ and $p'_2=(x'_2,y'_2)$.
|
|
The Manhattan distance is
|
|
\[|x_1-x_2|+|y_1-y_2| = \max(|x'_1-x'_2|,|y'_1-y'_2|)\]
|
|
|
|
For example, if $p_1=(1,0)$ and $p_2=(3,3)$,
|
|
the rotated coordinates are $p'_1=(1,-1)$ and $p'_2=(6,0)$
|
|
and the Manhattan distance is
|
|
\[|1-3|+|0-3| = \max(|1-6|,|-1-0|) = 5.\]
|
|
|
|
The rotated coordinates provide a simple way
|
|
to operate with Manhattan distances, because we can
|
|
consider x and y coordinates separately.
|
|
To maximize the Manhattan distance between two points,
|
|
we should find two points whose
|
|
rotated coordinates maximize the value of
|
|
\[\max(|x'_1-x'_2|,|y'_1-y'_2|).\]
|
|
This is easy, because either the horizontal or vertical
|
|
difference of the rotated coordinates has to be maximum.
|