From 480f58d386ea70111852e85674e872ca9d35db0f Mon Sep 17 00:00:00 2001 From: Antti H S Laaksonen Date: Sat, 22 Apr 2017 15:13:42 +0300 Subject: [PATCH] Small improvements --- chapter29.tex | 65 ++++++++++++++++++++++++++------------------------- 1 file changed, 33 insertions(+), 32 deletions(-) diff --git a/chapter29.tex b/chapter29.tex index 78a3759..67c6c57 100644 --- a/chapter29.tex +++ b/chapter29.tex @@ -55,7 +55,7 @@ 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: +the division line is \emph{outside} the quadrilateral: \begin{center} \begin{tikzpicture}[scale=0.45] @@ -85,8 +85,7 @@ 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. +another method that is much easier to use. 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 @@ -132,11 +131,11 @@ following point and vector: 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 +as complex numbers, and the class 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. +a coordinate and \texttt{P} is the type of a point or a vector. In addition, the code defines the macros \texttt{X} and \texttt{Y} that can be used to refer to x and y coordinates. @@ -165,15 +164,16 @@ P s = v+u; cout << s.X << " " << s.Y << "\n"; // 5 3 \end{lstlisting} -An appropriate coordinate type is +In practice, +an appropriate coordinate type is usually \texttt{long long} (integer) or \texttt{long double} -(real number), depending on the situation. -Integers should be used whenever possible, +(real number). +It is a good idea to use integer 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 +when comparing numbers. +A safe way to check if real numbers $a$ and $b$ are equal is to compare them using $|a-b|<\epsilon$, where $\epsilon$ is a small number (for example, $\epsilon=10^{-9}$). @@ -182,7 +182,7 @@ where $\epsilon$ is a small number (for example, $\epsilon=10^{-9}$). In the following examples, the coordinate type is \texttt{long double}. -The function \texttt{abs(v)} calculates the length +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 @@ -199,7 +199,7 @@ P b = {3,-1}; cout << abs(b-a) << "\n"; // 3.16228 \end{lstlisting} -The function \texttt{arg(v)} calculates the +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. @@ -207,7 +207,7 @@ 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 +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 $a$. @@ -228,7 +228,8 @@ cout << arg(v) << "\n"; // 0.963648 \index{cross product} The \key{cross product} $a \times b$ of vectors -$a=(x_1,y_1)$ and $b=(x_2,y_2)$ equals $x_1 y_2 - x_2 y_1$. +$a=(x_1,y_1)$ and $b=(x_2,y_2)$ is calculated +using the formula $x_1 y_2 - x_2 y_1$. The cross product tells us whether $b$ turns left (positive value), does not turn (zero) or turns right (negative value) @@ -265,7 +266,7 @@ The following picture illustrates the above cases: \end{center} \noindent -For example, in the first picture +For example, in the first case $a=(4,2)$ and $b=(1,2)$. The following code calculates the cross product using the class \texttt{complex}: @@ -285,7 +286,7 @@ of the result is $x_1 y_2 - x_2 y_1$. \subsubsection{Point location} -Cross products can be used for testing +Cross products can be used to test whether a point is located on the left or right side of a line. Assume that the line goes through points @@ -320,9 +321,9 @@ points $s_1$, $s_2$ and $p$ are on the same line. \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: +Consider the problem of checking +whether two given line segments +$ab$ and $cd$ intersect. The possible cases are: \textit{Case 1:} The line segments are on the same line @@ -408,7 +409,8 @@ Hence, we can use cross products to check this. \subsubsection{Point distance from a line} -The area of a triangle can be calculated +Another feature of the cross product is that +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. @@ -438,13 +440,12 @@ 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 +Let us now consider the problem of +testing 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 @@ -463,7 +464,7 @@ the polygon. \end{center} A convenient way to solve the problem is to -send a ray from the point to an arbitrary direction +send a \emph{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, @@ -531,7 +532,7 @@ For example, the area of the polygon 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 +The idea of 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: @@ -560,12 +561,11 @@ and if $x_{i+1}