Rotating coordinates
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Antti H S Laaksonen
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luku29.tex
145
luku29.tex
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@ -675,62 +675,109 @@ from the center point, using the Euclidean and Manhattan distances:
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\end{tikzpicture}
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\end{center}
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\subsubsection{Furthest points}
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\subsubsection{Rotating coordinates}
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Some problems are easier to solve if the
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Manhattan distance is used instead of the Euclidean distance.
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For example, consider a problem where we are given
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$n$ points $(x_1,y_1),(x_2,y_2),\ldots,(x_n,y_n)$
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and our task is to calculate the maximum distance
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between any two points.
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As an example, consider a problem where we are given
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$n$ points in the two-dimensional plane
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and our task is to calculate the maximum Manhattan
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distance between any two points.
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This is a difficult problem if the Euclidean distance
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should be maximized,
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but it is easy to maximize the
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Manhattan distance,
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because it is either
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\[\max A - \min A \hspace{20px} \textrm{or} \hspace{20px} \max B - \min B,\]
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where
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\[A = \{x_i+y_i : i = 1,2,\ldots,n\}\]
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and
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\[B = \{x_i-y_i : i = 1,2,\ldots,n\}.\]
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\begin{samepage}
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The reason for this is that the Manhattan distance
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\[|x_a-x_b|+|y_a-y_b]\]
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can be written
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For example, consider the following set of points:
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\begin{center}
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\begin{tabular}{cl}
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& $\max(x_a-x_b+y_a-y_b,\,x_a-x_b-y_a+y_b)$ \\
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$=$ & $\max(x_a+y_a-(x_b+y_b),\,x_a-y_a-(x_b-y_b))$
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\end{tabular}
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\end{center}
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assuming that $x_a \ge x_b$.
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\end{samepage}
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\begin{tikzpicture}[scale=0.65]
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\draw[color=gray] (-1,-1) grid (4,4);
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\begin{samepage}
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\subsubsection{Rotating coordinates}
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\filldraw (0,2) circle (2.5pt);
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\filldraw (3,3) circle (2.5pt);
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\filldraw (1,0) circle (2.5pt);
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\filldraw (3,1) circle (2.5pt);
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A useful technique related to the Manhattan distance
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is to rotate all coordinates 45 degrees so that
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a point $(x,y)$ becomes $(a(x+y),a(y-x))$,
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where $a=1/\sqrt{2}$.
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The coefficient $a$ is chosen so that
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the distances between the points remain the same.
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After the rotation, the region within a distance of $d$
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from a point is a square with horizontal and vertical sides:
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\begin{center}
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\begin{tikzpicture}
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\draw[fill=gray!20] (0,1) -- (-1,0) -- (0,-1) -- (1,0) -- (0,1);
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\draw[fill] (0,0) circle [radius=0.05];
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\node at (2.5,0) {$\Rightarrow$};
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\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);
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\draw[fill] (5,0) circle [radius=0.05];
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\node at (0,1.5) {$A$};
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\node at (3,2.5) {$C$};
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\node at (1,-0.5) {$B$};
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\node at (3,0.5) {$D$};
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\end{tikzpicture}
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\end{center}
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\end{samepage}
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Implementing many algorithms is easier when we may assume that all
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objects are squares with horizontal and vertical sides.
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The maximum Manhattan distance is 5
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between points $B$ and $C$:
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\begin{center}
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\begin{tikzpicture}[scale=0.65]
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\draw[color=gray] (-1,-1) grid (4,4);
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\filldraw (0,2) circle (2.5pt);
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\filldraw (3,3) circle (2.5pt);
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\filldraw (1,0) circle (2.5pt);
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\filldraw (3,1) circle (2.5pt);
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\node at (0,1.5) {$A$};
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\node at (3,2.5) {$C$};
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\node at (1,-0.5) {$B$};
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\node at (3,0.5) {$D$};
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\path[draw=red,thick,line width=2pt] (1,0) -- (1,3) -- (3,3);
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\end{tikzpicture}
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\end{center}
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A useful technique related to Manhattan distances
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is to rotate all coordinates 45 degrees so that
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a point $(x,y)$ becomes $(x+y,y-x)$.
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For example, after rotating the above points,
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the result is:
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\begin{center}
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\begin{tikzpicture}[scale=0.6]
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\draw[color=gray] (0,-3) grid (7,3);
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\filldraw (2,2) circle (2.5pt);
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\filldraw (6,0) circle (2.5pt);
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\filldraw (1,-1) circle (2.5pt);
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\filldraw (4,-2) circle (2.5pt);
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\node at (2,1.5) {$A$};
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\node at (6,-0.5) {$C$};
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\node at (1,-1.5) {$B$};
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\node at (4,-2.5) {$D$};
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\end{tikzpicture}
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\end{center}
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And the maximum distance is as follows:
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\begin{center}
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\begin{tikzpicture}[scale=0.6]
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\draw[color=gray] (0,-3) grid (7,3);
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\filldraw (2,2) circle (2.5pt);
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\filldraw (6,0) circle (2.5pt);
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\filldraw (1,-1) circle (2.5pt);
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\filldraw (4,-2) circle (2.5pt);
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\node at (2,1.5) {$A$};
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\node at (6,-0.5) {$C$};
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\node at (1,-1.5) {$B$};
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\node at (4,-2.5) {$D$};
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\path[draw=red,thick,line width=2pt] (1,-1) -- (4,2) -- (6,0);
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\end{tikzpicture}
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\end{center}
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Consider two points $p_1=(x_1,y_1)$ and $p_2=(x_1,y_1)$ whose rotated
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coordinates are $p'_1=(x'_1,y'_1)$ and $p'_2=(x'_2,y'_2)$.
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The Manhattan distance is
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\[|x_1-x_2|+|y_1-y_2| = \max(|x'_1-x'_2|,|y'_1-y'_2|)\]
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For example, if $p_1=(1,0)$ and $p_2=(3,3)$,
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the rotated coordinates are $p'_1=(1,-1)$ and $p'_2=(6,0)$
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and the Manhattan distance is
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\[|1-3|+|0-3| = \max(|1-6|,|-1-0|) = 5.\]
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The rotated coordinates provide a simple way
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to operate with Manhattan distances, because we can
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consider x and y coordinates separately.
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To maximize the Manhattan distance between two points,
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we should find two points whose
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rotated coordinates maximize the value of
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\[\max(|x'_1-x'_2|,|y'_1-y'_2|).\]
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This is easy, because either the horizontal or vertical
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difference of the rotated coordinates has to be maximum.
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