References etc.

chapter29.tex

@@ -14,7 +14,7 @@ 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]
+\begin{tikzpicture}[scale=0.44]
 
 \draw[fill] (6,2) circle [radius=0.1];
 \draw[fill] (5,6) circle [radius=0.1];
@@ -27,7 +27,7 @@ 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]
+\begin{tikzpicture}[scale=0.44]
 
 \draw[fill] (6,2) circle [radius=0.1];
 \draw[fill] (5,6) circle [radius=0.1];
@@ -41,7 +41,8 @@ line between two opposite vertices:
 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}
+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
@@ -56,7 +57,7 @@ two arbitrary opposite vertices.
 For example, in the following situation,
 the division line is outside the quadrilateral:
 \begin{center}
-\begin{tikzpicture}[scale=0.45]
+\begin{tikzpicture}[scale=0.44]
 
 \draw[fill] (6,2) circle [radius=0.1];
 \draw[fill] (3,2) circle [radius=0.1];
@@ -69,7 +70,7 @@ the division line is outside the quadrilateral:
 \end{center}
 However, another way to draw the line works:
 \begin{center}
-\begin{tikzpicture}[scale=0.45]
+\begin{tikzpicture}[scale=0.44]
 
 \draw[fill] (6,2) circle [radius=0.1];
 \draw[fill] (3,2) circle [radius=0.1];
@@ -500,7 +501,8 @@ so $b$ is outside the polygon.
 \section{Polygon area}
 
 A general formula for calculating the area
-of a polygon is
+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
@@ -571,7 +573,7 @@ along the boundary of the polygon.
 
 \index{Pick's theorem}
 
-\key{Pick's theorem} provides another way to calculate
+\key{Pick's theorem} \cite{pic99} 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

chapter30.tex

@@ -270,7 +270,11 @@ we should find the following points:
 
 This is another example of a problem
 that can be solved in $O(n \log n)$ time
-using a sweep line algorithm.
+using a sweep line algorithm\footnote{Besides this approach,
+there is also an
+$O(n \log n)$ time divide-and-conquer algorithm \cite{sha75}
+that divides the points into two sets and recursively
+solves the problem for both sets.}.
 We go through the points from left to right
 and maintain a value $d$: the minimum distance
 between two points seen so far.

list.tex

@@ -206,6 +206,12 @@
   Where to use and how not to use polynomial string hashing.
   \emph{Olympiads in Informatics}, 7(1):90--100, 2013.
 
+\bibitem{pic99}
+  G. Pick.
+  Geometrisches zur Zahlenlehre.
+  \emph{Sitzungsberichte des deutschen naturwissenschaftlich-medicinischen Vereines
+  für Böhmen "Lotos" in Prag. (Neue Folge)}, 19:311--319, 1899.
+
 \bibitem{pri57}
   R. C. Prim.
   Shortest connection networks and some generalizations.
@@ -215,6 +221,11 @@
   27-Queens Puzzle: Massively Parallel Enumeration and Solution Counting.
   \url{https://github.com/preusser/q27}
 
+\bibitem{sha75}
+  M. I. Shamos and D. Hoey.
+  Closest-point problems.
+  \emph{16th Annual Symposium on Foundations of Computer Science}, 151--162, 1975.
+
 \bibitem{sha81}
   M. Sharir.
   A strong-connectivity algorithm and its applications in data flow analysis.