EECS2030E Test 4

Version D

You have 80 minutes to complete this test. This is a closed book test.

GETTING STARTED

  1. Start eclipse.
  2. Download this project file.
  3. Import the test project by doing the following:
    1. Under the File menu choose Import...
    2. Under General choose Existing Projects into Workspace and press Next
    3. Click the Select archive file radio button, and click the Browse... button.
    4. Navigate to your home directory (the file chooser is probably in the workspace directory).
    5. Select the file test4D.zip and click OK
    6. Click Finish.
  4. All of the files you need for this test should now appear in eclipse.
  5. Open a terminal. You will use this terminal to submit your work.
  6. Copy and paste the command cd workspace/Test4D/src/test4 into the terminal and press enter.

Resources

Question 1 (18 marks total)

Implement the class described by this API. You do not have to include javadoc comments.

submit 2030L secEtest4D Test4D.java

Question 2 (12 marks total)

Consider the following method: 

 /**
  * Re-orders the elements of the argument list t so that
  * all of the negative elements of t are in the front part
  * of the list and all of the positive elements of t are
  * in the back part of the list. There is no guarantee of
  * the ordering of the positive elements or of the negative 
  * elements.
  * 
  * Zero is considered to be a positive number for the purposes
  * of this method.
  * 
  * For example, a possible solution results in the following: 
  * 
  * if t is [-1, 2]               then partition(t) makes t [-1, 2]
  * if t is [2, -1]               then partition(t) makes t [-1, 2]
  * if t is [8, -2, 1]            then partition(t) makes t [-2, 1, 8]  
  * if t is [1, -5, -9, 4]        then partition(t) makes t [-5, -9, 4, 1] 
  * if t is [-3, 7, -1, 3, -5]    then partition(t) makes t [-3, -1, -5, 7, 3]
  * 
  * @param t a list of integers
  */
 public static void partition(List<Integer> t) {
     if (t.size() < 2) {                         // base case
         return;
     }
     int first = t.get(0);
     if (first < 0) {                            // recursive case 1
         partition(t.subList(1, t.size()));
     }
     else {                                      // recursive case 2
         t.add(first);
         t.remove(0);
         partition(t.subList(1, t.size() - 1));
     }
 }
A. (1 marks)

Explain why the base case is correct when the size of t is equal to 0.


B. (1 marks)

Explain why the base case is correct when the size of t is equal to 1.


C. (2 marks)

In recursive case 1, what should we assume about the recursive call partition(t.subList(1, t.size())) if we wanted to prove that the recursive case 1 is correct?


D. (2 marks)

Using your assumption from Part C prove that the recursive case 1 is correct.


E. (2 marks)

Define the size of the problem solved by partition(t).


F. (2 marks)

Using your definition from Part E prove that the method terminates.


G. (2 marks)

Recursive case 2 contains an error; what is it?


submit 2030L secEtest4D answers.txt