[Solved] CS2030s Lab 1-Maximum Disc Coverage

Level 1

Represent a PointThe 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.javajshell> new Point(0.0, 1.0)$.. ==> point (0.000, 1.000)jshell> /exit

Level 2

Find the mid-point and angle of line pqFind 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.javajshell> new Point(0.0, 0.0).midPoint(new Point(1.0, 1.0))$.. ==> point (0.500, 0.500)jshell> new Point(0.0, 0.0).angleTo(new Point(1.0, 1.0))$.. ==> 0.7853981633974483jshell> new Point(0, 0).angleTo(new Point(-1, -1))$.. ==> -2.356194490192345jshell> Point p = new Point(1, 1)p ==> point (1.000, 1.000)jshell> p.midPoint(new Point(2,2))$.. ==> point (1.500, 1.500)jshell> pp ==> point (1.000, 1.000)jshell> /exit

Level 3

Moving the pointNow 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 and distance d.Hint: if a point, say m, is at (x, y), then moving m at an angle and distance d, would result in the new position having the coordinates (x + d cos, y + d sin)

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

jshell> /open Point.javajshell> 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> pp ==> point (0.000, 0.000)jshell> /exit

Level 4

Creating the CircleDefine 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 CoverageYou 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)$.. ==> 2jshell> points = new Point[]{new Point(0, -1), new Point(1, 0),   ...> new Point(0, 1), new Point(-1, 0)}points ==> Point[4] { point (0.000, -1.000), point (1.000, 0 ... ), point (-1.000, 0.000) }jshell> findMaxDiscCoverage(points)$.. ==> 4jshell> /exit