Maximum Disc Coverage

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Topic Coverage

Problem Description

In this problem, you are given a set of points on a 2D plane. We want to place a unit disc (i.e., a circle of radius 1) so that it covers as many points as possible. Note that this is different from previous exercises where one of the points must be at the centre of the disc.

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What is the maximum number of points that we can cover with the disc at any one time?

To help you along, here are some observations:

The algorithm

The high level algorithm for finding the maximum discoverage coverage is as follows:

  1. For every pair of distinct points of distance no more than 2, create a unit circle such that the pair of points lie on that circle's perimeter.
  2. For each unit circle created in the previous step, count the number of points that it covers.
  3. Return the maximum coverage.

The most interesting part of this task lies in creating the unit circle with two given points, such that they lie on the perimeter of the circle. Some elementary trigonometry can guide us as to how to do this:

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Given points p and q, we can

  1. Find the midpoint of p and q, denoted as point m.
  2. Find the angle of the line pq with respect to the x axis, denoted as θ
  3. Find the length of line mc, denoted as d, which can be obtained using the Pythagorean Theorem
  4. Move point m by distance d, at the angle of line mc with respect to the x-axis, which is θ + π/2. The new location of point m is the centre of the circle, c.

The Task

Given a set of points, go through every pair of points, and for each circle that passes through them, count how many points are covered by each circle. Finally, return the maximum count.

You may assume that there are always at least two points with a positive distance less than 2 between them.

This task is divided into five levels. You may submit as soon as you complete one level. There is no restriction on the number of attempts.

In the spirit of incremental coding and testing, you will need to complete ALL levels to get full credit for this lab.

Level 1

Represent a Point

The first thing we can do is to group x and y coordinates into a single data structure or class.

Design a class Point to represent a point object with each pair of x- and y- coordinates.

class Point {
    // include the properties of a Point

    // complete the constructor
    Point(double x, double y) {
    }

    // include a toString method
}

Define an overriding toString method to output each point. In addition, the output of each double value, say d, is to be formatted with String.format("%.3f", d); 

jshell> /open Point.java

jshell> new Point(0.0, 1.0)
$.. ==> point (0.000, 1.000)

jshell> /exit

Level 2

Find the mid-point and angle of line pq

Find the mid-point between two consecutive points p and q. In addition, find the angle (in radians) of line pq using the atan or atan2 Math function (refer to the Java API specifications).

Complete the midPoint and angleTo methods and include them in the Point class. You may define other helper methods to further abstract away lower-level functionalities. 

class Point {
    ...

    Point midPoint(Point q) {
        ...
    }

    double angleTo(Point q) {
        ...
    }
}

Inconsistencies between your output and the actual output involving -0.000 and 0.000 can be ignored. 

jshell> /open Point.java

jshell> new Point(0.0, 0.0).midPoint(new Point(1.0, 1.0))
$.. ==> point (0.500, 0.500)

jshell> new Point(0.0, 0.0).angleTo(new Point(1.0, 1.0))
$.. ==> 0.7853981633974483

jshell> new Point(0, 0).angleTo(new Point(-1, -1))
$.. ==> -2.356194490192345

jshell> Point p = new Point(1, 1)
p ==> point (1.000, 1.000)

jshell> p.midPoint(new Point(2,2))
$.. ==> point (1.500, 1.500)

jshell> p
p ==> point (1.000, 1.000)

jshell> /exit

Level 3

Moving the point

Now we need to allow points to be "moved", so that we can move point m to c. However, in the spirit of immutability, the original point is not moved. Instead, a new point is returned, which reflects the new position of the original point if it has been moved by angle θ and distance d.

Hint: if a point, say m, is at (x, y), then moving m at an angle θ and distance d, would result in the new position having the coordinates (x + d cosθ, y + d sinθ)

class Point {
    ...

    Point moveTo(double theta, double d) {
        ...
        return new Point...;
    }
}

jshell> /open Point.java

jshell> Point p = new Point(0.0, 0.0)
p ==> point (0.000, 0.000)

jshell> p.moveTo(Math.PI / 2, 1.0)
$.. ==> point (0.000, 1.000)

jshell> p
p ==> point (0.000, 0.000)

jshell> /exit

Level 4

Creating the Circle

Define the Circle class to allow for the creation of circle objects with a given centre and radius. You may assume that all circles created are valid. 

jshell> /open Point.java 

jshell> /open Circle.java 

jshell> new Circle(new Point(0.0, 0.0), 1.0)
$.. ==> circle of radius 1.0 centered at point (0.000, 0.000)

jshell> new Circle(new Point(1, 1), 2)
$.. ==> circle of radius 2.0 centered at point (1.000, 1.000)

We are now ready to construct unit circles whose perimeter passes through two given points.

By using the mid-point and angle of line pq, move the mid-point to the centre of the circle of radius r whose perimeter coincides with points p and q. You will need to work out the respective angle and distance values.

Create a jshell script maxDiscCoverage.jsh and define a createUnitCircle method that takes in two points p and q and returns the unit circle created. 

Circle createUnitCircle(Point p, Point q) {
    ...
    return new Circle...;
}
Notice that if the distance between points p and q is larger than 2 units, then there is no unit circle whose perimeter coincides with the points. In this level, you may assume that the two points will form a unit circle. 

jshell> /open maxDiscCoverage.jsh 

jshell> createUnitCircle(new Point(0, 0), new Point(1, 0))
$.. ==> circle of radius 1.0 centered at point (0.500, 0.866)

jshell> createUnitCircle(new Point(0, 0), new Point(2, 0))
$.. ==> circle of radius 1.0 centered at point (1.000, 0.000)

jshell> /exit

Level 5

Maximum Disc Coverage

You are now ready to find the maximum unit disc coverage.

Within maxDiscCoverage.jsh, further define a method findMaxDiscCoverage that takes in an array of Point objects and finds the maximum coverage among them. You may make use of other helper methods. 

Circle createUnitCircle(Point p, Point q) {
    ...
}

int findMaxDiscCoverage(Point[] points) {
    int maxDiscCoverage = 0;

    for (int i = 0; i < points.length - 1; i++) {
        for (int j = i + 1; j < points.length; j++) {
            // find coverage with (points[i], points[j])
            ...
        }
    }
    return maxDiscCoverage;
}

Once again, keep in mind that if the distance between points p and q is larger than 2 units, then there is no unit circle whose perimeter coincides with the points.

jshell> /open Point.java 

jshell> /open Circle.java 

jshell> /open maxDiscCoverage.jsh 

jshell> Point[] points = new Point[]{new Point(0, 0), new Point(1, 0)}
points ==> Point[2] { point (0.000, 0.000), point (1.000, 0.000) }

jshell> findMaxDiscCoverage(points)
$.. ==> 2

jshell> points = new Point[]{new Point(0, -1), new Point(1, 0),
   ...> new Point(0, 1), new Point(-1, 0)}
points ==> Point[4] { point (0.000, -1.000), point (1.000, 0 ... ), point (-1.000, 0.000) }

jshell> findMaxDiscCoverage(points)
$.. ==> 4

jshell> /exit