732 lines
21 KiB
TeX
732 lines
21 KiB
TeX
\chapter{Geometry}
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\index{geometry}
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In geometric problems, a usual challenge is
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to realize how to approach the problem so that
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the solution can be conveniently implemented
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and the number of special cases is small.
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As an example, consider a problem where
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we are given the vertices of a quadrilateral
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(a polygon that has four vertices),
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and our task is to calculate its area.
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For example, a possible input for the problem is as follows:
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\begin{center}
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\begin{tikzpicture}[scale=0.45]
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\draw[fill] (6,2) circle [radius=0.1];
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\draw[fill] (5,6) circle [radius=0.1];
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\draw[fill] (2,5) circle [radius=0.1];
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\draw[fill] (1,1) circle [radius=0.1];
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\draw[thick] (6,2) -- (5,6) -- (2,5) -- (1,1) -- (6,2);
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\end{tikzpicture}
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\end{center}
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One way to approach the problem is to divide
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the quadrilateral into two triangles by a straight
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line between two opposite vertices:
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\begin{center}
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\begin{tikzpicture}[scale=0.45]
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\draw[fill] (6,2) circle [radius=0.1];
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\draw[fill] (5,6) circle [radius=0.1];
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\draw[fill] (2,5) circle [radius=0.1];
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\draw[fill] (1,1) circle [radius=0.1];
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\draw[thick] (6,2) -- (5,6) -- (2,5) -- (1,1) -- (6,2);
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\draw[dashed,thick] (2,5) -- (6,2);
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\end{tikzpicture}
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\end{center}
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After this, it suffices to sum the areas
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of the triangles.
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The area of a triangle can be calculated,
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for example, using \key{Heron's formula}
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\[ \sqrt{s (s-a) (s-b) (s-c)},\]
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where $a$, $b$ and $c$ are the lengths
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of the triangle's sides and
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$s=(a+b+c)/2$.
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\index{Heron's formula}
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This is a possible way to solve the problem,
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but there is one pitfall:
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how to divide the quadrilateral into triangles?
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It turns out that sometimes we can't pick
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arbitrary opposite vertices.
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For example, in the following quadrilateral,
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the division line is outside the quadrilateral:
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\begin{center}
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\begin{tikzpicture}[scale=0.45]
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\draw[fill] (6,2) circle [radius=0.1];
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\draw[fill] (3,2) circle [radius=0.1];
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\draw[fill] (2,5) circle [radius=0.1];
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\draw[fill] (1,1) circle [radius=0.1];
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\draw[thick] (6,2) -- (3,2) -- (2,5) -- (1,1) -- (6,2);
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\draw[dashed,thick] (2,5) -- (6,2);
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\end{tikzpicture}
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\end{center}
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However, another way to draw the line works:
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\begin{center}
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\begin{tikzpicture}[scale=0.45]
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\draw[fill] (6,2) circle [radius=0.1];
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\draw[fill] (3,2) circle [radius=0.1];
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\draw[fill] (2,5) circle [radius=0.1];
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\draw[fill] (1,1) circle [radius=0.1];
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\draw[thick] (6,2) -- (3,2) -- (2,5) -- (1,1) -- (6,2);
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\draw[dashed,thick] (3,2) -- (1,1);
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\end{tikzpicture}
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\end{center}
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For a human, it is clear which of the lines is the correct
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choice, but for a computer the situation is difficult.
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However, it turns out that we can solve the problem using
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another method which is much easier to implement
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and doesn't contain any special cases.
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There is a general formula
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\[x_1y_2-x_2y_1+x_2y_3-x_3y_2+x_3y_4-x_4y_3+x_4y_1-x_1y_4,\]
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that calculates the area of a quadrilateral
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whose vertices are
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$(x_1,y_1)$,
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$(x_2,y_2)$,
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$(x_3,y_3)$ and
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$(x_4,y_4)$.
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This formula is easy to calculate, there are no special
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cases, and it turns out that we can generalize the formula
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for \emph{all} polygons.
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\section{Complex numbers}
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\index{complex number}
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\index{point}
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\index{vector}
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A \key{complex number} is a number of the form $x+y i$,
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where $i = \sqrt{-1}$ is the \key{imaginary unit}.
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A geometric interpretation for a complex number is
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that it represents a two-dimensional point $(x,y)$
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or a vector from the origin to point $(x,y)$.
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For example, $4+2i$ corresponds to the
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following point and vector:
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\begin{center}
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\begin{tikzpicture}[scale=0.45]
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\draw[->,thick] (-5,0)--(5,0);
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\draw[->,thick] (0,-5)--(0,5);
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\draw[fill] (4,2) circle [radius=0.1];
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\draw[->,thick] (0,0)--(4-0.1,2-0.1);
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\node at (4,2.8) {$(4,2)$};
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\end{tikzpicture}
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\end{center}
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\index{complex@\texttt{complex}}
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The complex number class \texttt{complex} in C++ is
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useful when solving geometry problems.
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Using the class, we can store points and vectors
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as complex numbers, and the class also contains tools
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that are useful in geometry.
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In the following code, \texttt{C} is the type of
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a coordinate, and \texttt{P} is the type of a point or vector.
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In addition, the code defines shortcuts \texttt{X} and \texttt{Y}
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that can be used to refer to x and y coordinates.
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\begin{lstlisting}
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typedef long long C;
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typedef complex<C> P;
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#define X real()
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#define Y imag()
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\end{lstlisting}
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For example, the following code defines a point $p=(4,2)$
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and prints its x and y coordinates:
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\begin{lstlisting}
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P p = {4,2};
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cout << p.X << " " << p.Y << "\n"; // 4 2
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\end{lstlisting}
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The following code defines vectors $v=(3,1)$ and $u=(2,2)$,
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and after that calculates the sum $s=v+u$.
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\begin{lstlisting}
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P v = {3,1};
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P u = {2,2};
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P s = v+u;
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cout << s.X << " " << s.Y << "\n"; // 5 3
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\end{lstlisting}
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The appropriate type for \texttt{C} is
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\texttt{long long} (integer) or \texttt{long double}
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(real number), depending on the situation.
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It is a good idea to use integers whenever possible,
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because the integer calculations are exact.
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If coordinates are real numbers,
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precision error should be taken into account
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when comparing them.
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A safe way to check if numbers $a$ and $b$ are equal
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is the comparison $|a-b|<\epsilon$,
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where $\epsilon$ is a small number (for example, $\epsilon=10^{-9}$).
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\subsubsection*{Functions}
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In the following examples, the coordinate type is
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\texttt{long double}.
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The function \texttt{abs(v)} calculates the length
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$|v|$ of a vector $v=(x,y)$
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using the formula $\sqrt{x^2+y^2}$.
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The function can also be used for
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calculating the distance between points
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$(x_1,y_1)$ and $(x_2,y_2)$,
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because that distance equals the length
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of the vector $(x_2-x_1,y_2-y_1)$.
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The following code calculates the distance
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of points $(4,2)$ and $(3,-1)$:
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\begin{lstlisting}
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P a = {4,2};
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P b = {3,-1};
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cout << abs(b-a) << "\n"; // 3.60555
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\end{lstlisting}
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The function \texttt{arg(v)} calculates the
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angle of a vector $v=(x,y)$ with respect to the x axel.
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The function gives the angle in radians,
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where $r$ radians equals $180 r/\pi$ degrees.
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The angle of a vector that points to the right is 0,
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and the angle decreases clockwise and increases
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counterclockwise.
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The function \texttt{polar(s,a)} constructs a vector
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whose length is $s$ and that points to angle $a$.
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In addition, a vector can be rotated by angle $a$
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by multiplying it by a vector with length 1 and angle $a$.
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The following code calculates the angle of
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the vector $(4,2)$, rotates it $1/2$ radians
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counterclockwise, and then calculates the angle again:
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\begin{lstlisting}
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P v = {4,2};
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cout << arg(v) << "\n"; // 0.463648
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v *= polar(1.0,0.5);
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cout << arg(v) << "\n"; // 0.963648
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\end{lstlisting}
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\section{Points and lines}
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\index{cross product}
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The \key{cross product} $a \times b$ of vectors
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$a=(x_1,y_1)$ and $b=(x_2,y_2)$ equals $x_1 y_2 - x_2 y_1$.
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The cross product indicates whether the vector $b$
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turns to the left (positive value) or to the right (negative value)
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when it is placed directly after the vector $a$.
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The following picture shows three examples:
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\begin{center}
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\begin{tikzpicture}[scale=0.45]
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\draw[->,thick] (0,0)--(4,2);
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\draw[->,thick] (4,2)--(4+1,2+2);
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\node at (2.5,0.5) {$a$};
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\node at (5,2.5) {$b$};
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\node at (3,-2) {$a \times b = 6$};
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\draw[->,thick] (8+0,0)--(8+4,2);
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\draw[->,thick] (8+4,2)--(8+4+2,2+1);
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\node at (8+2.5,0.5) {$a$};
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\node at (8+5,1.5) {$b$};
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\node at (8+3,-2) {$a \times b = 0$};
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\draw[->,thick] (16+0,0)--(16+4,2);
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\draw[->,thick] (16+4,2)--(16+4+2,2-1);
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\node at (16+2.5,0.5) {$a$};
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\node at (16+5,2.5) {$b$};
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\node at (16+3,-2) {$a \times b = -8$};
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\end{tikzpicture}
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\end{center}
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\noindent
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For example, in the left picture
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$a=(4,2)$ and $b=(1,2)$.
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The following code calculates the cross product
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using the class \texttt{complex}:
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\begin{lstlisting}
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P a = {4,2};
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P b = {1,2};
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C r = (conj(a)*b).Y; // 6
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\end{lstlisting}
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The function \texttt{conj} negates the y coordinate
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of a vector,
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and when the vectors $(x_1,-y_1)$ and $(x_2,y_2)$
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are multiplied together, the y coordinate
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of the result is $x_1 y_2 - x_2 y_1$.
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\subsubsection{Point location}
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The cross product can be used for testing
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whether a point is located on the left or right
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side of a line.
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Assume that the line goes through points
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$s_1$ and $s_2$, we are looking from $s_1$
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to $s_2$ and the point is $p$.
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For example, in the following picture,
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$p$ is on the left side of the line:
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\begin{center}
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\begin{tikzpicture}[scale=0.45]
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\draw[dashed,thick,->] (0,-3)--(12,6);
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\draw[fill] (4,0) circle [radius=0.1];
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\draw[fill] (8,3) circle [radius=0.1];
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\draw[fill] (5,3) circle [radius=0.1];
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\node at (4,-1) {$s_1$};
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\node at (8,2) {$s_2$};
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\node at (5,4) {$p$};
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\end{tikzpicture}
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\end{center}
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Now the cross product $(p-s_1) \times (p-s_2)$
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indicates the location of the point $p$.
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If the cross product is positive,
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$p$ is located on the left side,
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and if the cross product is negative,
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$p$ is located on the right side.
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Finally, if the cross product is zero,
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points $s_1$, $s_2$ and $p$ are on the same line.
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\subsubsection{Line segment intersection}
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\index{line segment intersection}
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Consider a problem where we are given two line segments
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$ab$ and $cd$ and our task is to check whether they
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intersect. The possible cases are:
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\textit{Case 1:}
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The line segments are on the same line
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and they overlap each other.
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In this case, there is an infinite number of
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intersection points.
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For example, in the following picture,
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all points between $c$ and $b$ are
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intersection points:
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\begin{center}
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\begin{tikzpicture}[scale=0.9]
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\draw (1.5,1.5)--(6,3);
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\draw (0,1)--(4.5,2.5);
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\draw[fill] (0,1) circle [radius=0.05];
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\node at (0,0.5) {$a$};
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\draw[fill] (1.5,1.5) circle [radius=0.05];
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\node at (6,2.5) {$d$};
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\draw[fill] (4.5,2.5) circle [radius=0.05];
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\node at (1.5,1) {$c$};
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\draw[fill] (6,3) circle [radius=0.05];
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\node at (4.5,2) {$b$};
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\end{tikzpicture}
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\end{center}
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In this case, we can use cross products to
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check if all points are on the same line.
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After this, we can sort the points and check
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whether the line segments overlap each other.
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\textit{Case 2:}
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The line segments have a common vertex
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that is the only intersection point.
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For example, in the following picture the
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intersection point is $b=c$:
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\begin{center}
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\begin{tikzpicture}[scale=0.9]
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\draw (0,0)--(4,2);
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\draw (4,2)--(6,1);
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\draw[fill] (0,0) circle [radius=0.05];
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\draw[fill] (4,2) circle [radius=0.05];
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\draw[fill] (6,1) circle [radius=0.05];
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\node at (0,0.5) {$a$};
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\node at (4,2.5) {$b=c$};
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\node at (6,1.5) {$d$};
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\end{tikzpicture}
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\end{center}
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This case is easy to test because the
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possibilities for the common intersection point
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are $a=c$, $a=d$, $b=c$ and $b=d$.
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\textit{Case 3:}
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There is exactly one intersection point
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that is not a vertex of any line segment.
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In the following picture, the point $p$
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is the intersection point:
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\begin{center}
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\begin{tikzpicture}[scale=0.9]
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\draw (0,1)--(6,3);
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\draw (2,4)--(4,0);
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\draw[fill] (0,1) circle [radius=0.05];
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\node at (0,0.5) {$c$};
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\draw[fill] (6,3) circle [radius=0.05];
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\node at (6,2.5) {$d$};
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\draw[fill] (2,4) circle [radius=0.05];
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\node at (1.5,3.5) {$a$};
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\draw[fill] (4,0) circle [radius=0.05];
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\node at (4,-0.4) {$b$};
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\draw[fill] (3,2) circle [radius=0.05];
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\node at (3,1.5) {$p$};
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\end{tikzpicture}
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\end{center}
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In this case, the line segments intersect
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exactly when both points $c$ and $d$ are
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on different sides of a line through $a$ and $b$,
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and points $a$ and $b$ are on different
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sides of a line through $c$ and $d$.
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Hence, we can use cross products to check this.
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\subsubsection{Point distance from a line}
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The area of a triangle can be calculated
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using the formula
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\[\frac{| (a-c) \times (b-c) |}{2},\]
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where $a$, $b$ and $c$ are the vertices of the triangle.
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Using this formula, it is possible to calculate the
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shortest distance of a point from a line.
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For example, in the following picture $d$ is the
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shortest distance between point $p$ and the line
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that is defined by points $s_1$ and $s_2$:
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\begin{center}
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\begin{tikzpicture}[scale=0.75]
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\draw (-2,-1)--(6,3);
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\draw[dashed] (1,4)--(2.40,1.2);
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\node at (0,-0.5) {$s_1$};
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\node at (4,1.5) {$s_2$};
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\node at (0.5,4) {$p$};
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\node at (2,2.7) {$d$};
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\draw[fill] (0,0) circle [radius=0.05];
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\draw[fill] (4,2) circle [radius=0.05];
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\draw[fill] (1,4) circle [radius=0.05];
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\end{tikzpicture}
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\end{center}
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The area of the triangle whose vertices are
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$s_1$, $s_2$ and $p$ can be calculated in two ways:
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it is both
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$\frac{1}{2} |s_2-s_1| d$ and
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$\frac{1}{2} ((s_1-p) \times (s_2-p))$.
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Thus, the shortest distance is
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\[ d = \frac{(s_1-p) \times (s_2-p)}{|s_2-s_1|} .\]
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\subsubsection{Point inside a polygon}
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Let us now consider a problem where our task is to
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find out whether a point is located inside or outside
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a polygon.
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For example, in the following picture point $a$
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is inside the polygon and point $b$ is outside
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the polygon.
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\begin{center}
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\begin{tikzpicture}[scale=0.75]
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%\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);
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\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);
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\draw[fill] (-3,1) circle [radius=0.05];
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\node at (-3,0.5) {$a$};
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\draw[fill] (1,3) circle [radius=0.05];
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\node at (1,2.5) {$b$};
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\end{tikzpicture}
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\end{center}
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A convenient way to solve the problem is to
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send a ray from the point to an arbitrary direction
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and calculate the number of times it touches
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the border of the polygon.
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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 border of the polygon,
|
|
so $a$ is inside the polygon.
|
|
Correspondingly, the rays from $b$
|
|
touch 0 and 2 times the border of the polygon,
|
|
so $b$ is outside the polygon.
|
|
|
|
\section{Polygon are}
|
|
|
|
A general formula for calculating the area
|
|
of a polygon is
|
|
\[\frac{1}{2} |\sum_{i=1}^{n-1} (p_i \times p_{i+1})| =
|
|
\frac{1}{2} |\sum_{i=1}^{n-1} (x_i y_{i+1} - x_{i+1} y_i)|, \]
|
|
when the vertices are
|
|
$p_1=(x_1,y_1)$, $p_2=(x_2,y_2)$, $\ldots$, $p_n=(x_n,y_n)$
|
|
sorted so that
|
|
$p_i$ and $p_{i+1}$ are adjacent vertices on the border
|
|
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
|
|
\[\frac{|(2\cdot5-4\cdot5)+(5\cdot3-5\cdot7)+(7\cdot1-3\cdot4)+(4\cdot3-1\cdot4)+(4\cdot4-3\cdot2)|}{2} = 17/2.\]
|
|
|
|
The idea in the formula is to go through trapezoids
|
|
where one side is a side of the polygon,
|
|
and another side is 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 are of such a trapezoid is
|
|
\[(x_{i+1}-x_{i}) \frac{y_i+y_{i+1}}{2},\]
|
|
when 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 calculated,
|
|
because the value of the sum may be positive or negative
|
|
depending on whether we walk clockwise or counterclockwise
|
|
along the sides of the the polygon.
|
|
|
|
\subsubsection{Pick's theorem}
|
|
|
|
\index{Pick's theorem}
|
|
|
|
\key{Pick's theorem} provides another way to calculate
|
|
the area of a polygon where are vertices 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 border 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 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{Furthest points}
|
|
|
|
Some problems are easier to solve if we use the
|
|
Manhattan distance instead of the Euclidean distance.
|
|
For example, consider a problem where we are given
|
|
$n$ points $(x_1,y_1),(x_2,y_2),\ldots,(x_n,y_n)$
|
|
and our task is to calculate the maximum distance
|
|
between any two points.
|
|
|
|
This is a difficult problem if we should maximize
|
|
the Euclidean distance,
|
|
but calculating
|
|
the maximum Manhattan distance is easy
|
|
because it either
|
|
\[\max A - \min A \hspace{20px} \textrm{or} \hspace{20px} \max B - \min B,\]
|
|
where
|
|
\[A = \{x_i+y_i : i = 1,2,\ldots,n\}\]
|
|
and
|
|
\[B = \{x_i-y_i : i = 1,2,\ldots,n\}.\]
|
|
\begin{samepage}
|
|
The reason for this is that the Manhattan distance
|
|
\[|x_a-x_b|+|y_a-y_b]\]
|
|
can be written
|
|
\begin{center}
|
|
\begin{tabular}{cl}
|
|
& $\max(x_a-x_b+y_a-y_b,\,x_a-x_b-y_a+y_b)$ \\
|
|
$=$ & $\max(x_a+y_a-(x_b+y_b),\,x_a-y_a-(x_b-y_b))$
|
|
\end{tabular}
|
|
\end{center}
|
|
assuming that $x_a \ge x_b$.
|
|
\end{samepage}
|
|
|
|
\begin{samepage}
|
|
\subsubsection{Rotating coordinates}
|
|
|
|
A useful technique related to the Manhattan distance
|
|
is to rotate all coordinates 45 degrees so that
|
|
a point $(x,y)$ becomes $(a(x+y),a(y-x))$,
|
|
where $a=1/\sqrt{2}$.
|
|
The multiplier $a$ is so chosen that
|
|
the distances of the points remain the same.
|
|
|
|
After the rotation, the region within a distance of $d$
|
|
from a point is a square with horizontal and vertical sides:
|
|
\begin{center}
|
|
\begin{tikzpicture}
|
|
\draw[fill=gray!20] (0,1) -- (-1,0) -- (0,-1) -- (1,0) -- (0,1);
|
|
\draw[fill] (0,0) circle [radius=0.05];
|
|
|
|
\node at (2.5,0) {$\Rightarrow$};
|
|
|
|
\draw[fill=gray!20] (5-0.71,0.71) -- (5-0.71,-0.71) -- (5+0.71,-0.71) -- (5+0.71,0.71) -- (5-0.71,0.71);
|
|
\draw[fill] (5,0) circle [radius=0.05];
|
|
\end{tikzpicture}
|
|
\end{center}
|
|
\end{samepage}
|
|
|
|
Implementing many algorithms is easier when we may assume that all
|
|
objects are squares with horizontal and vertical sides. |