AP CSA 2.6 FRQ Practice: De Morgan in Practice

Unit 2: Selection and Iteration · Lesson 2.6 · FRQ Practice

De Morgan in Practice

A free response question in the shape the exam uses: a stated contract, four rubric parts, and no main method handed to you. Worth 4 points.

Back to the Lesson 2.6 pageAll Unit 2 lessons

Why this question is worth four points

Two boolean expressions are equivalent when they agree on EVERY input, and the exam tests that with the one rule students reliably get backwards: not (A and B) is (not A) OR (not B), never (not A) and (not B). This question asks for the same rule written both ways so they can be compared directly.

What you are given

Nothing is given. Write a class named Logic containing exactly the four static methods described below. Do not write a main method.

The question

  1. (a) public static boolean bothOpen(boolean a, boolean b) returns whether both are true.
  2. (b) public static boolean notBothOpen(boolean a, boolean b) returns the opposite of part (a), written WITHOUT calling part (a).
  3. (c) public static boolean neitherOpen(boolean a, boolean b) returns whether both are false.
  4. (d) public static boolean onlyFirstOpen(boolean a, boolean b) returns whether a is true and b is false.

What the reader is looking for

  1. Write class Logic with the four static methods exactly as described. No main method.
  2. Parts (b) and (c) are DIFFERENT rules and the driver will show you where they differ. "Not both" is true when one is open; "neither" is not.
  3. Part (d) is true only for the first door. It is deliberately not symmetric: onlyFirstOpen(true, false) and onlyFirstOpen(false, true) must give different answers.

Worked examples

These show what a correct answer prints. There are more cases you cannot see, and they use different values, so an answer built around just these numbers will fail.

Example 1 input
true true
Example 1 output
true
false
false
false
Example 2 input
true false
Example 2 output
false
true
false
true

Your answer

Main.java

Input for the Run button

How this is scored: your answer runs against every test case, and the fraction it passes becomes your score out of 4. That is not how a human AP reader marks a rubric, so treat the score as a check on whether your code works, and the rubric above as the thing you are actually practising.


  

  
Stuck?

    
Hint 1
Distributing a not across an and flips the connector too. !(a && b) is !a || !b. Writing !a && !b gives you "neither", which is a different rule that happens to agree when both are false.

    
Hint 2
Part (d) needs both halves: a must be true AND b must be false. Returning a alone is right on three of the four rows, which is exactly the kind of nearly correct that a full truth table catches.

    
Hint 3
Fill in the four row truth table for your parts (b) and (c) before submitting. They must disagree on two of the four rows.



  
Before you submit

  
5 mistake(s) that lose points on this question
    
Each of these is a real error the grader catches. Check your answer against them before you submit, not instead of trying.

    
    
      
  • part (b) distributes the not without flipping the connector, becoming neither
    • 
      
  • part (c) is written as not both instead of neither
    • 
      
  • part (d) ignores the second door and returns a alone
    • 
      
  • part (d) is written symmetrically, so it cannot tell the two doors apart
    • 
      
  • part (a) uses || instead of &&
    • 
    

  


  
Where to go next

  

    Rework the Lesson 2.6 page
    Try the from-scratch exercise for this lesson
    Try the debugging exercise for this lesson
  


  
  






  

















  
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