From 7774b7754c24adfcf2d23fe13e5e4d50b861e61d Mon Sep 17 00:00:00 2001 From: Antti H S Laaksonen Date: Sun, 14 May 2017 13:51:27 +0300 Subject: [PATCH] Improve language --- chapter30.tex | 25 +++++++++++++------------ 1 file changed, 13 insertions(+), 12 deletions(-) diff --git a/chapter30.tex b/chapter30.tex index 86593a7..d4aca39 100644 --- a/chapter30.tex +++ b/chapter30.tex @@ -8,9 +8,9 @@ The idea in such algorithms is to represent an instance of the problem as a set of events that correspond to points in the plane. The events are processed in increasing order -according to their x or y coordinate. +according to their x or y coordinates. -As an example, let us consider the following problem: +As an example, consider the following problem: There is a company that has $n$ employees, and we know for each employee their arrival and leaving times on a certain day. @@ -19,8 +19,8 @@ employees that were in the office at the same time. The problem can be solved by modeling the situation so that each employee is assigned two events that -corresponds to their arrival and leaving times. -After sorting the events, we can go through them +correspond to their arrival and leaving times. +After sorting the events, we go through them and keep track of the number of people in the office. For example, the table \begin{center} @@ -122,7 +122,7 @@ The symbols $+$ and $-$ indicate whether the value of the counter increases or decreases, and the value of the counter is shown below. The maximum value of the counter is 3 -between John's arrival time and Maria's leaving time. +between John's arrival and Maria's leaving. The running time of the algorithm is $O(n \log n)$, because sorting the events takes $O(n \log n)$ time @@ -213,9 +213,9 @@ $y_1$ and $y_2$, and add this number to the total number of intersection points. To store y coordinates of horizontal segments, -we can use a binary indexed or a segment tree, +we can use a binary indexed or segment tree, possibly with index compression. -Using such a structure, processing each event +When such structures are used, processing each event takes $O(\log n)$ time, so the total running time of the algorithm is $O(n \log n)$. @@ -295,7 +295,7 @@ Thus, it suffices to only consider points that are located in those ranges, which makes the algorithm efficient. -For example, in the following picture the +For example, in the following picture, the region marked with dashed lines contains the points that can be within a distance of $d$ from the active point: @@ -332,7 +332,7 @@ from the active point: \end{center} The efficiency of the algorithm is based on the fact -that such a region always contains +that the region always contains only $O(1)$ points. We can go through those points in $O(\log n)$ time by maintaining a set of points whose x coordinate @@ -401,8 +401,9 @@ the convex hull is as follows: an easy way to construct the convex hull for a set of points in $O(n \log n)$ time. -The algorithm constructs the convex hull -in two parts: +The algorithm first locates the leftmost +and rightmost points, and then +constructs the convex hull in two parts: first the upper hull and then the lower hull. Both parts are similar, so we can focus on constructing the upper hull. @@ -414,7 +415,7 @@ add each point to the hull. Always after adding a point to the hull, we make sure that the last line segment in the hull does not turn left. -As long as this does not hold, we repeatedly remove the +As long as it turns left, we repeatedly remove the second last point from the hull. The following pictures show how