Assignment 4: Improving Efficiency with Sorting Solution

Assignment 4: Improving Efficiency with Sorting Solution

Type: Individual
Problem Overview
This assignment will investigate an example feature extraction problem. Feature extraction is a subproblem of pattern recognition and is also used in areas such as statistical analysis, computer vision, and image processing. For example, an image processing problem may use a feature extraction algorithm to identify
particular shapes or regions in a digitized image.
In this assignment, we’re going to focus on a very simple feature extraction problem: Given a set of points in two-dimensional space, identify every subset of four or more points that are collinear. For example, given the set of points depicted in Figure 1, your program would detect the three groups of collinear points as depicted by the line segments in Figure 2.
1 2 3 4 5 6 7 8
2 4 6 8
1 3 5 7 9
Figure 1: A set of 13 points.
1 2 3 4 5 6 7 8
2 4 6 8
1 3 5 7 9
Figure 2: Three collinear groups identified.
As always, we want our solution to be useful at scale. For example, Figure 3 plots ~100,000 points and Figure 4 shows the 34 collinear groups identified by blue line segments. Each collinear group in Figure 4 is composed of far more than four points; four is just the minimum number of points to qualify for the collinear pattern that we’re looking for. In the general problem statement we will refer to line segments instead of collinear groups, where each line segment must contain at least four points.
Problem Statement: Given a set of N distinct points in the plane, identify every line segment that connects a subset of four or more of the points. Each point will be specified by an (x, y) pair where x and y are int values in the range 0 to 32, 767. For example, the thirteen points in Figures 1 and 2 are: (1, 7), (2,
2), (2, 5), (3, 1), (4, 4), (5, 3), (5, 6), (6, 6), (7, 1), (7, 3), (7, 4), (7, 9), (8, 8).
COMP 2210

Figure 3: A set of ~100,000 points. Figure 4: 34 collinear groups identified.
You must solve this problem in terms of the classes and methods described in the following sections.
The Point class
You must create an immutable data type Point that represents a point in the plane. A shell of the Point class is provided for you, and its API is described below.
public class Point implements Comparable<Point {
public final Comparator<Point SLOPE_ORDER;
public Point(int x, int y)
public void draw()
public void drawTo(Point that)
public double slopeTo(Point that)
public int compareTo(Point that)
public String toString()
The constructor, draw method, drawto method, and toString method have been completed for you and must not be changed. You must add the bodies of the remaining methods yourself. You may add any number of private methods that you like, but you may not add any public method or constructor, nor may you change the signature of any public method or constructor.
The methods in the Point API that you must complete are described in more detail below.
The compareTo method.
This method must compare points by y-coordinates, breaking ties by x-coordinates. Thus, the invoking point (x0, y0) is less than the parameter point (x1, y1) if and only if either y0 < y1 or if y0 = y1 and x0 < x1.
For example, by this natural order of points, (0, 1) is less than (0, 2), (7, 1) is less than (5, 3), and (3, 0) is less than (4, 0).

The slopeTo method.
This method must return the slope between the invoking point (x0, y0) and the parameter point (x1, y1), which is given by the formula:
(y1 􀀀 y0)
(x1 􀀀 x0)
For example, for the point (3, 3), the slope to (1, 1) is 1.0, the slope to (4, 5) is 2.0, and the slope to (5,
2) is -0.5.
Treat the slope of a horizontal line segment as positive zero1; treat the slope of a vertical line segment as positive infinity1; treat the slope of a degenerate line segment (between a point and itself) as negative infinity1.
The SLOPE_ORDER Comparator
This field of the Point class must compare two points by the slopes they make with the invoking point (x0, y0). Thus, the point (x1, y1) is less than the point (x2, y2) if and only if
(y1 􀀀 y0)
(x1 􀀀 x0)
(y2 􀀀 y0)
(x2 􀀀 x0)
Treat horizontal, vertical, and degenerate line segments the same as in the slopeTo() method.
For example, if the invoking point is (3, 3), then (5, 2) is less than (1, 1), and (1, 1) is less than (4, 5).
You will need to write a nested inner class that implements that slope-order behavior as a Comparator<Point,
and then set the SLOPE_ORDER field to be an instance of this class.
The Line Class
This class models a line segment as a set of points.
public class Line implements Comparable<Line, Iterable<Point {
public Line()
public Line(Collection<Point l)
public void add(Point p)
public Point first()
public Point last()
public int length()
public Iterator<Point iterator()
public int compareTo(Line that)
public String toString()
The Constructors
The parameterless constructor creates a new line with no points. The second constructor creates a new
line that contains all the distinct collinear points in the Collection parameter.
1See the Java documentation of the Double class for a discussion of positive zero, positive infinity, and negative infinity.

The add method
The add method adds the given Point to the line, provided it is collinear with the existing points in the line and it isn’t already present.
The first method
The first method returns the point on the line that is least with respect to natural ordering of Point. If the line has no points, first returns null.
The last method
The last method returns the point on the line that is greatest with respect to natural ordering of Point.
If the line has no points, last returns null.
The length method
The length method returns the number of points on the line.
The iterator method
The iterator method returns an Iterator over the points on the line. The iteration order is natural
order of Point.
The compareTo method
The compareTo method compares this line with the parameter line. The natural order of Line is based
on the natural order of its first and last point. That is, given line1 and line2, line1 < line2 if line1. f irst <
line2. f irst or line1. f irst = line2. f irst and line1.last < line2.last. Two lines line1 and line2 are equal if and
only if line1. f irst = line2. f irst and line1.last = line2.last.
The Extractor Class
This class provides methods that allow clients to read in an input file of Point data and find all lines
segments of four or more collinear points in that data. The API of this class is described below, and as in the case of the Point class above, you may not modify this API in any way.
public class Extractor {
public Extractor(String filename)
public Extractor(Collection<Point c)
public void drawPoints()
public void drawLines()
public Set<Line getLinesBrute()
public Set<Line getLinesFast()

The Constructors
The first constructor for the Extractor class takes a single parameter of type String. This parameter is a filename for a file of Point data formatted as follows: The first line of the file contains a single int value N that is the number of lines Point data that follow. Each of the following N lines contains two int values
separated by one or more blanks. The first int is the x value of a Point and the second int is the y value of a Point. There may be lines of text past these first N + 1 lines of data, but they should be ignored.
A sample input file is shown below.
11000 11000
12000 10000
13000 10000
14000 10000
15000 10000
Instantiating an Extractor object with this data file would ensure that five distinct instances of the Point class are stored in a suitable data structure inside the new Extractor object.
The second constructor takes a Collection of points and creates an Extractor for this data.
The drawPoints method
This method uses the StdDraw class in the provided stdlib.jar file to graphically display all the points associated with this Extractor object. Specifically, this method should iterate over all the Point objects
in the Extractor object’s internal data structure and invoke each Point’s draw method.
This method is only to help you visualize the data that your program is processing. It will not be invoked by the grading program, and thus it will not affect your grade in any way.
The drawLines method
This method uses the StdDraw class in the provided stdlib.jar file to graphically display all the lines already identified by this Extractor object, if any. If no lines have yet been identified, this method will still raise a graphics window with no lines drawn. If lines have been identified, then this method can use the drawTo method of the Point object at one end of the line segment to draw a line from that Point to the Point object at the other end of the line segment.
This method is only to help you visualize the data that your program is processing. It will not be invoked by the grading program, and thus it will not affect your grade in any way.
The getLinesBrute method
This method implements the straight-forward, brute force approach to extracting the feature that we’re interested in. Since any combination of four distinct points that are collinear qualify as our feature, we could generate all combinations of four distinct points and check each one to see if those four points
are collinear. So, we could describe this brute force solution as a combinatoric approach to the problem:
We’re generating the combination of N things taken four at a time, and each time we generate a new combination, we’re testing it based on our feature criteria (collinearity).
For example, let’s name the points in the given sample input file p1 through p5, as shown below.
11000 11000 (p1)
12000 10000 (p2)
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13000 10000 (p3)
14000 10000 (p4)
15000 10000 (p5)
The table below shows the combinations that our code would generate, along with the result of testing
each combination for collinearity. Note that to check if four points p, q, r, and s are collinear, check
whether the slope between p and q, between p and r, and between p and s are all equal.
Combination Collinear?
p1, p2, p3, p4 no
p1, p2, p3, p5 no
p1, p2, p4, p5 no
p1, p3, p4, p5 no
p2, p3, p4, p5 yes
The advantage of this approach is that it’s fairly simple to code. Since the number of points being
selected out of the set of N total points is fixed at four, then four nested for loops should do the trick. But
of course, that’s also the problem with this approach: four nested loops each dependent on N will have
O(N4) time complexity.
In fact, since this is a combinatoric solution we can calculate exactly how many combinations of four
points will be computed for a given N.


4!(N 􀀀 4)!
For our example above, this would give 5!
4! = 5. If the input had 10 points, the brute force solution
would have to test 10!
4![1]6! = 10[1]9[1]8[1]7
4[1]3[1]2 = 210 different combinations of four points. For N = 20, the brute
force solution would generate and test 4845 combinations of four points. For N = 1000, over 41 billion
combinations of four points would be generated and tested by our program. You can see how this escalates
very quickly and the brute force solution becomes infeasible to apply for even moderately large values of
The getLinesFast method
A fundamental property of sorting is that it brings duplicates together. We can make use of this and solve
the problem much faster if we use sorting as part of our solution. Collinear points have the same slope
with respect to each other, and thus are duplicates with respect to SLOPE_ORDER.
To see if point p is part of a group of four or more collinear points we can do the following.
1. Sort the N points with respect to the slope that they make with p.
2. Scan the sorted points to find all groups of three or more consecutive points having the same slope
to p. Each such group is collinear with p and is thus, together with p, part of a line segment of at
least four points.
3. Repeat for the remaining N 􀀀 1 points.
Here’s an example of sorting the points in Figure 1 with respect to the slope they make with (7, 1).
(7, 1) (6, 6) (5, 6) (1, 7) (4, 4) (5, 3) (2, 5) (2, 2) (3, 1) (8, 8) (7, 3) (7, 4) (7, 9)
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COMP 2210 Fall 2014
Note how this sorting brings together the points of the two line segments that contain (7, 1) (underlined
in gray).
How much faster is this sort-and-scan approach? We can sort in O(N log N) time and the subsequent
scan is O(N). We have to perform these operations for all N points, so the total cost of this sort and scan
approach is N [1] (N log N + N) which is O(N2 log N). This is a significant asymptotic improvement since
O(N2 log N) ‑ O(N4), and the clock-time difference is dramatic. Problem sizes that are infeasible for the
brute force solution are solved quickly (or at least in reasonable amount of time) by this sort-and-scan
There is an extra benefit of this approach: We are not limited to identifying four-point segments. We
can now identify maximal line segments of four or more collinear points.
Notes and other requirements
Here are a couple of extra requirements plus a few things to keep in mind.
 Start this one early. There’s more reading, thinking, and up-front understanding to take care of on
this assignment. Read this handout carefully. Ask questions of your TA and of me. Ask questions
on Piazza. Start early and be proactive.
 You’ve been provided with the file stdlib.jar, which contains easy to use input/output and drawing
capabilities along with other things. This file is provided by Princeton University under the GNU
General Public License. Its documentation can be found here: http://introcs.cs.princeton.
edu/java/stdlib. You are not required to use this file at all. If you choose to do so, you will have to
add the path to this jar file to your CLASSPATH.
 You’ve been provided with shells of the Point, Line, and Extractor classes. Don’t modify the things
that have been done for you.
 Start by completing the remaining methods of the Point class. There’s no point2 in attempting the
List or Extractor class before this is complete and correct.
 Complete the List class after Point but before Extractor.
 When you begin work on the Extractor class, start with the getLinesBrute method. This is shorter
and easier to get correct quickly. After you can produce correct output with the getLinesBrute
method, turn your attention to getLinesFast.
 In getLinesFast, do not print subsegments of lines containing five or more collinear points. For
example, if the line segment p ! q ! r ! s ! t exists in the data, identify it but not any four-point
subsegment such as p ! q ! r ! s.
Assignment Submission
You must turn in the following three files toWeb-CAT for grading:,, and
While not required, it is strongly recommended that you also submit your own test cases. Note the following
rules regarding your Web-CAT submissions:
2Sorry; couldn’t resist. Here’s more:
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COMP 2210 Fall 2014
 Separate submissions for Point and Line are available on Web-CAT. You can submit an unlimited
number of times and these submissions are not counted toward your grade. This is strictly to
help you make sure Point and Line work correctly before you make submissions that include the
Extractor class.
 The feedback hints for failed test cases will be unavailable beginning one day prior to the assignment
due date.
 You can submit to Web-CAT no more than eight times for this assignment.
 The last submission that you make to Web-CAT will be used to determine your grade on the assignment,
even if its score is lower than that of an earlier submission.
 Submissions made within the 24 hour period after the published deadline will be assessed a late
penalty of 15 points.
 No submissions will be accepted more than 24 hours after the published deadline.
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