AP CSA 1.11 Debugging: Integer Division Inside the Formula

Unit 1: Using Objects and Methods · Lesson 1.11 · Debugging

Integer Division Inside the Formula

Math Class. This is not a blank editor: it is someone else's attempt. Find what is wrong, fix it, and submit to be graded against hidden test cases.

Back to the Lesson 1.11 page All Unit 1 lessons

Why debugging is its own skill

Math.sqrt and Math.pow work in doubles and are not the problem. The problem is the int division sitting inside the expression handed to them, which has already destroyed the fraction before the library method is ever called.

What is wrong with it

The average passed into Math.sqrt is computed with int division, so the fraction is gone before the square root sees it. Force the division to produce a double.
The hypotenuse squares its sides with a * a and b * b, which is correct, but then takes the square root of their DIFFERENCE instead of their sum. Fix the operator.
The Math.abs call is correct.

The program reads

Two positive integers a and b.

The program should print

Three lines: the square root of the average of a and b, the hypotenuse of a right triangle with legs a and b, and the absolute difference between a and b.

Worked examples

These show what the FIXED program should print. There are more cases you cannot see, and they use different values, so patching around just these numbers will fail.

Example 1 input

3 4

Example 1 output

1.8708286933869707
5.0
1

Example 2 input

6 8

Example 2 output

2.6457513110645907
10.0
2

Your answer

Main.java

import java.util.Scanner;

public class Main {
  public static void main(String[] args) {
    Scanner input = new Scanner(System.in);
    int a = input.nextInt();
    int b = input.nextInt();

    System.out.println(Math.sqrt((a + b) / 2));

    System.out.println(Math.sqrt(a * a - b * b));

    System.out.println(Math.abs(a - b));
  }
}

Input for the Run button

Run sends whatever is in the code box below, bugs and all, so you can see the crash or the wrong answer for yourself before you fix anything.

Stuck?

Hint 1

With a = 3 and b = 4 the average is 3.5, but (3 + 4) / 2 in ints is 3. Math.sqrt then dutifully computes the square root of the wrong number. Dividing by 2.0 keeps the arithmetic in doubles.

Hint 2

A method that accepts a double will happily accept an int, widening it silently. That is why nothing warns you: the damage happened before the call, in your own expression.

Hint 3

For a right triangle the hypotenuse is the square root of the SUM of the squares. Subtracting gives a negative under the root whenever b is larger than a, and Math.sqrt of a negative is NaN.

Where to go next

Rework the Lesson 1.11 page Try the from-scratch version of this lesson

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