Due Date: Monday, October 29, or Tuesday, October 30, 2007, before the beginning of your lab section.
Objective: The main objective of this assignment is to learn to design user-defined classes and ADTs.
Programming assignment: an interval [a, b] is characterized by its lower endpoint a and its upper endpoint b. A midpoint of the interval is defined as (a + b)/2, and its width is defined as b - a.
Arithmetic operations on the intervals are defined based on the following logic. For example, for addition, if I have between 5 and 8 dollars with me, and my friend has between 10 and 30, then the smallest amount we have together is 5 + 10 = 15, and the largest is 8 + 30 = 38.
For subtraction, if I have between 10 and 30 dollars, and I spend between 5 and 8 for lunch, then the smallest amount that will remain is when I had the smallest amount and spend the largest amount, i.e., 10 - 8 = 2. The largest remaining amount is when I had most money and spent the least amount of money, i.e., 30 - 5 = 25.
As a result, the arithmetic operations with intervals are defined as follows:
[a, b] + [c, d] = [a + c, b + d] [a, b] - [c, d] = [a - d, b - c] [a, b] * [c, d] = [min(a*c, a*d, b*c, b*d), max(a*c, a*d, b*c, b*d)] 1/[a, b] = [1/b, 1/a] if 0 is not in the interval [a, b] (and undefined otherwise) [a, b]/[c, d] = [a, b] * (1/[c, d])
Describe a class whose elements are intervals, and methods include:
Solution:
public class Interval {
//endpoints
private double low;
private double up;
//constructor
public (double lower, double upper)
{low = lower; up = upper;}
//assessor methods
public double getLower()
{return low;}
public double getUpper()
{return up;}
//mutator methods
public void setLower(double newLower)
{low = newLower;}
public void setUpper(double newUpper)
{up = newUpper;}
//midpoint and width
public double midpoint()
{return (low + up)/2.0;}
public double width()
{return up - low;}
//arithmetic operations
public Interval add(Interval secondInterval)
{
double newLower = low + secondInterval.getLower();
double newUpper = up + secondInterval.getUpper();
return new Interval(newLower, newUpper);
}
public Interval subtract(Interval secondInterval)
{
double newLower = low - secondInterval.getUpper();
double newUpper = up - secondInterval.getLower();
return new Interval(newLower, newUpper);
}
public Interval multiply(Interval secondInterval)
{
double product1 = low * secondInterval.getLower();
double product2 = low * secondInterval.getUpper();
double product3 = up * secondInterval.getLower();
double product4 = up * secondInterval.getUpper();
//computing the lower endpoint
double newLower = product1;
newLower = Math.min(newLower, product2);
newLower = Math.min(newLower, product3);
newLower = Math.min(newLeft, product4);
//computing the upper endpoint
double newUpper = product1;
newUpper = Math.max(newUpper, product2);
newUpper = Math.max(newUpper, product3);
newUpper = Math.max(newUpper, product4);
return new Interval(newLower, newUpper);
}
public Interval inverse()
{
if (low <= 0 && 0 <= up)
{System.out.println("Inverse not defined");
else
{
double newLower = 1/up;
double newUpper = 1/low;
return new Interval(newLower, newUpper);
}
}
public Interval divide(Interval secondInterval)
{
return self.multiply(secondInterval.inverse());
}
}
public class Assignment8
{
public static void main(String[] args)
{
Interval int1 = new Interval(5.0, 8.0);
Interval int2 = new Interval(10.0, 30.0);
Interval sum = int1.add(int2);
System.out.println("The sum of [5, 8] and [10, 30] is ["+
sum.getLower() +", " + sum.getUpper() + "]");
Interval difference = int2.subtract(int1);
System.out.println("The difference of [10, 30] and [5, 8] is ["+
difference.getLower() +", " + difference.getUpper() + "]");
}
}
Homework assignment: on a separate sheet of paper,
solve Ex. 6 from p. 520 and Ex. 16 from p. 522. Deliverables: as instructed by your TA.