diff --git a/luku29.tex b/luku29.tex
index dcf606a..a4914f6 100644
--- a/luku29.tex
+++ b/luku29.tex
@@ -675,62 +675,109 @@ from the center point, using the Euclidean and Manhattan distances:
 \end{tikzpicture}
 \end{center}
 
-\subsubsection{Furthest points}
+\subsubsection{Rotating coordinates}
 
 Some problems are easier to solve if the
 Manhattan distance is used 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.
+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.
 
-This is a difficult problem if the Euclidean distance
-should be maximized,
-but it is easy to maximize the
-Manhattan distance,
-because it is 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
+For example, consider the following set of points:
 \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{tikzpicture}[scale=0.65]
+\draw[color=gray] (-1,-1) grid (4,4);
 
-\begin{samepage}
-\subsubsection{Rotating coordinates}
+\filldraw (0,2) circle (2.5pt);
+\filldraw (3,3) circle (2.5pt);
+\filldraw (1,0) circle (2.5pt);
+\filldraw (3,1) circle (2.5pt);
 
-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 coefficient $a$ is chosen so that
-the distances between 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];
+\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}
-\end{samepage}
 
-Implementing many algorithms is easier when we may assume that all
-objects are squares with horizontal and vertical sides.
\ No newline at end of file
+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.
\ No newline at end of file