AP CSA 2.12 Debugging: Counting the Wrong Operations

Unit 2: Selection and Iteration · Lesson 2.12 · Debugging

Counting the Wrong Operations

Informal Run Time Analysis. 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 2.12 page All Unit 2 lessons

Why debugging is its own skill

Run time analysis is counting, and counting is only meaningful if the counter sits where the work happens. Put the increment one block too high and a quadratic algorithm reports linear growth, which is exactly the wrong lesson.

What is wrong with it

The comparison counter for the nested pass is incremented in the OUTER loop body instead of the inner one, so it reports n instead of n squared. Move it to where the comparison actually happens.
The single pass is supposed to report n - 1 comparisons, since comparing n items pairwise takes one fewer comparison than there are items. Its loop starts at 0 and reports n. Fix the start value.
Do not change the loop bounds of the nested pass. The point of the exercise is that its count should grow like n squared, and it already visits the right pairs.

The program reads

A single integer n of at least 2.

The program should print

Two lines: the number of comparisons a single pass makes, then the number a nested pass makes.

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

4
25

Example 2 input

Example 2 output

1
4

Your answer

Main.java

import java.util.Scanner;

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

    int single = 0;
    for (int i = 0; i < n; i++) {
      single++;
    }
    System.out.println(single);

    int nested = 0;
    for (int i = 0; i < n; i++) {
      nested++;
      for (int j = 0; j < n; j++) {
        int ignored = i + j;
      }
    }
    System.out.println(nested);
  }
}

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

Run the starter with n = 5 and then n = 10. The nested count doubles when n doubles, which is the signature of a linear algorithm. A genuinely nested loop should quadruple. That gap between the shape of the code and the shape of the number is the bug.

Hint 2

Every statement inside the inner loop runs n times per outer pass, for n * n total. Every statement in the outer body runs only n times. Where you put the increment decides which number you are measuring.

Hint 3

For the single pass, think about comparing neighbors in a list of n items: 5 items have only 4 neighboring pairs. Starting the loop at 1 instead of 0 is the whole fix.

Where to go next

Rework the Lesson 2.12 page Try the from-scratch version of this lesson

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