From e8563ccc65decadd22f320e057eb901784a922b8 Mon Sep 17 00:00:00 2001 From: Antti H S Laaksonen Date: Sun, 12 Feb 2017 14:39:25 +0200 Subject: [PATCH] Corrections --- luku29.tex | 144 +++++++++++++++++++++++++++-------------------------- 1 file changed, 74 insertions(+), 70 deletions(-) diff --git a/luku29.tex b/luku29.tex index 2071b86..c699e6b 100644 --- a/luku29.tex +++ b/luku29.tex @@ -2,9 +2,9 @@ \index{geometry} -In geometric problems, a usual challenge is -to realize how to approach the problem so that -the solution can be conveniently implemented +In geometric problems, it is often challenging +to find a way to approach the problem so that +the solution to the problem can be conveniently implemented and the number of special cases is small. As an example, consider a problem where @@ -51,10 +51,10 @@ $s=(a+b+c)/2$. This is a possible way to solve the problem, but there is one pitfall: how to divide the quadrilateral into triangles? -It turns out that sometimes we can't pick -arbitrary opposite vertices. -For example, in the following quadrilateral, -the division line is outside the quadrilateral: +It turns out that sometimes we cannot just pick +two arbitrary vertices. +For example, in the following situation, +the division line lies outside the quadrilateral: \begin{center} \begin{tikzpicture}[scale=0.45] @@ -80,13 +80,13 @@ However, another way to draw the line works: \draw[dashed,thick] (3,2) -- (1,1); \end{tikzpicture} \end{center} -For a human, it is clear which of the lines is the correct -choice, but for a computer the situation is difficult. +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 which is much easier to implement -and doesn't contain any special cases. -There is a general formula +another method that is much easier to implement +and does not involve any special cases. +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 whose vertices are @@ -94,9 +94,9 @@ $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ and $(x_4,y_4)$. -This formula is easy to calculate, there are no special -cases, and it turns out that we can generalize the formula -for \emph{all} polygons. +This formula is easy to implement, there are no special +cases, and it turns out that we can even generalize the formula +to \emph{all} polygons. \section{Complex numbers} @@ -106,9 +106,9 @@ for \emph{all} polygons. A \key{complex number} is a number of the form $x+y i$, where $i = \sqrt{-1}$ is the \key{imaginary unit}. -A geometric interpretation for a complex number is +A geometric interpretation of a complex number is that it represents a two-dimensional point $(x,y)$ -or a vector from the origin to point $(x,y)$. +or a vector from the origin to a point $(x,y)$. For example, $4+2i$ corresponds to the following point and vector: @@ -129,14 +129,14 @@ following point and vector: \index{complex@\texttt{complex}} The complex number class \texttt{complex} in C++ is -useful when solving geometry problems. -Using the class, we can store points and vectors +useful when solving geometric problems. +Using the class we can represent points and vectors as complex numbers, and the class also 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. -In addition, the code defines shortcuts \texttt{X} and \texttt{Y} +In addition, the code defines the macros \texttt{X} and \texttt{Y} that can be used to refer to x and y coordinates. \begin{lstlisting} @@ -164,17 +164,16 @@ P s = v+u; cout << s.X << " " << s.Y << "\n"; // 5 3 \end{lstlisting} -The appropriate type for \texttt{C} is +An appropriate coordinate type is \texttt{long long} (integer) or \texttt{long double} (real number), depending on the situation. -It is a good idea to use integers whenever possible, -because the integer calculations are exact. - -If coordinates are real numbers, -precision error should be taken into account +Integers should be used 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 -is the comparison $|a-b|<\epsilon$, +is to compare them using $|a-b|<\epsilon$, where $\epsilon$ is a small number (for example, $\epsilon=10^{-9}$). \subsubsection*{Functions} @@ -192,7 +191,7 @@ because that distance equals the length of the vector $(x_2-x_1,y_2-y_1)$. The following code calculates the distance -of points $(4,2)$ and $(3,-1)$: +between points $(4,2)$ and $(3,-1)$: \begin{lstlisting} P a = {4,2}; P b = {3,-1}; @@ -200,16 +199,16 @@ cout << abs(b-a) << "\n"; // 3.60555 \end{lstlisting} The function \texttt{arg(v)} calculates the -angle of a vector $v=(x,y)$ with respect to the x axel. +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. The angle of a vector that points to the right is 0, -and the angle decreases clockwise and increases +and angles decrease clockwise and increase counterclockwise. The function \texttt{polar(s,a)} constructs a vector -whose length is $s$ and that points to angle $a$. -In addition, a vector can be rotated by angle $a$ +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 $a$. The following code calculates the angle of @@ -229,11 +228,12 @@ cout << arg(v) << "\n"; // 0.963648 The \key{cross product} $a \times b$ of vectors $a=(x_1,y_1)$ and $b=(x_2,y_2)$ equals $x_1 y_2 - x_2 y_1$. -The cross product indicates whether the vector $b$ -turns to the left (positive value) or to the right (negative value) -when it is placed directly after the vector $a$. +The cross product tells us whether $b$ +turns left (positive value), does not turn (zero) +or turns to right (negative value) +when it is placed directly after $a$. -The following picture shows three examples: +The following picture illustrates the above cases: \begin{center} \begin{tikzpicture}[scale=0.45] @@ -264,7 +264,7 @@ The following picture shows three examples: \end{center} \noindent -For example, in the left picture +For example, in the first picture $a=(4,2)$ and $b=(1,2)$. The following code calculates the cross product using the class \texttt{complex}: @@ -275,7 +275,8 @@ P b = {1,2}; C r = (conj(a)*b).Y; // 6 \end{lstlisting} -The function \texttt{conj} negates the y coordinate +The above code works, because +the function \texttt{conj} negates the y coordinate of a vector, and when the vectors $(x_1,-y_1)$ and $(x_2,y_2)$ are multiplied together, the y coordinate @@ -304,8 +305,9 @@ $p$ is on the left side of the line: \end{tikzpicture} \end{center} -Now the cross product $(p-s_1) \times (p-s_2)$ -indicates the location of the point $p$. +In this situation, +the cross product $(p-s_1) \times (p-s_2)$ +tells us the location of the point $p$. If the cross product is positive, $p$ is located on the left side, and if the cross product is negative, @@ -369,9 +371,10 @@ intersection point is $b=c$: \end{tikzpicture} \end{center} -This case is easy to test because the -possibilities for the common intersection point -are $a=c$, $a=d$, $b=c$ and $b=d$. +This case is easy to check, because +there are only four possibilities +for the intersection point: +$a=c$, $a=d$, $b=c$ and $b=d$. \textit{Case 3:} There is exactly one intersection point @@ -410,10 +413,10 @@ using the formula where $a$, $b$ and $c$ are the vertices of the triangle. Using this formula, it is possible to calculate the -shortest distance of a point from a line. +shortest distance between a point and a line. For example, in the following picture $d$ is the -shortest distance between point $p$ and the line -that is defined by points $s_1$ and $s_2$: +shortest distance between the point $p$ and the line +that is defined by the points $s_1$ and $s_2$: \begin{center} \begin{tikzpicture}[scale=0.75] \draw (-2,-1)--(6,3); @@ -461,7 +464,7 @@ the polygon. A convenient way to solve the problem is to send a ray from the point to an arbitrary direction and calculate the number of times it touches -the border of the polygon. +the boundary of the polygon. If the number is odd, the point is inside the polygon, and if the number is even, @@ -488,10 +491,10 @@ For example, we could send the following rays: \end{samepage} The rays from $a$ touch 1 and 3 times -the border of the polygon, +the boundary of the polygon, so $a$ is inside the polygon. Correspondingly, the rays from $b$ -touch 0 and 2 times the border of the polygon, +touch 0 and 2 times the boundary of the polygon, so $b$ is outside the polygon. \section{Polygon area} @@ -500,10 +503,10 @@ 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 +where 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 +in such an order that +$p_i$ and $p_{i+1}$ are adjacent vertices on the boundary of the polygon, and the first and last vertex is the same, i.e., $p_1=p_n$. @@ -527,8 +530,8 @@ 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 $y=0$. +whose one side is a side of the polygon, +and another side lies on the horizontal line $y=0$. For example: \begin{center} \begin{tikzpicture}[scale=0.7] @@ -547,9 +550,9 @@ For example: \draw (0,0) -- (10,0); \end{tikzpicture} \end{center} -The are of such a trapezoid is +The area 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 $p_{i+1}$. +where 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}