AP CSA 4.16 Debugging: The Base Case That Returns Zero

Unit 4: Data Collections · Lesson 4.16 · Debugging

The Base Case That Returns Zero

Recursion. 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 4.16 page All Unit 4 lessons

Why debugging is its own skill

The recursion terminates, the structure is textbook, and every answer is zero. A base case is not just a stopping point: it is the value the entire chain of multiplications is built on top of, and returning 0 from it annihilates everything the recursion computed on the way down.

What is wrong with it

The factorial base case returns 0. Every result is then multiplied by that 0 as the calls unwind, so the answer is always 0. Return the value that leaves a product unchanged.
The sum-to-n method recurses on n instead of n - 1, so it never approaches its base case. Change the recursive call so the problem gets smaller each time.
The countdown method is correct. Notice that it does reduce its argument, and use it as the model.

The program reads

A single integer n from 1 to 12.

The program should print

The countdown from n to 1 one per line, then n factorial, then the sum of 1 through n.

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

Example 1 output

Example 2 input

Example 2 output

1
1
1

Your answer

Main.java

import java.util.Scanner;

public class Main {
  public static void countdown(int n) {
    if (n <= 0) {
      return;
    }
    System.out.println(n);
    countdown(n - 1);
  }

  public static int factorial(int n) {
    if (n <= 1) {
      return 0;
    }
    return n * factorial(n - 1);
  }

  public static int sumTo(int n) {
    if (n <= 0) {
      return 0;
    }
    return n + sumTo(n);
  }

  public static void main(String[] args) {
    Scanner input = new Scanner(System.in);
    int n = input.nextInt();
    countdown(n);
    System.out.println(factorial(n));
    System.out.println(sumTo(n));
  }
}

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

Expand factorial(3) by hand: 3 * factorial(2), which is 3 * 2 * factorial(1), which is 3 * 2 * 0. Anything multiplied by zero is zero, so the base case silently destroys the whole product. The identity for multiplication is 1.

Hint 2

A recursive call must make the problem strictly smaller or it never reaches the base case. sumTo(n) calling sumTo(n) is the same problem again, forever.

Hint 3

Compare the three methods. countdown passes n - 1 and factorial passes n - 1. Only sumTo passes n unchanged, which makes it the odd one out.

Where to go next

Rework the Lesson 4.16 page Try the from-scratch version of this lesson

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