CSSE 230
Data Structures and Algorithm Analysis

Written Assignment 2 (62 points)

To Be Turned In

These are to be turned in to the WA2 Drop Box. You can write solutions out by hand and scan them (there is a networked scanner in F-217), or create a file on your computer directly.   After you have submitted, click on the drop box again to verify that your submission was successful.

A single late day may be used or earned for a written assignment: you must turn in all parts early to earn a late day; if you turn in any part late, you will use a late day.

Problem numbers [in square brackets] are the numbers from the 3rd edition of the Weiss book.

1. (2 points) Weiss problem 2.7 [2.6]   (+ with ' ' and " ")
2. (6 points) Weiss problem 3.13 [3.10]  (legal and illegal references)
3. (18 points) Weiss problems 4.23-4.25 [4.21–4.23]. (generic max and min).  Assume that all of the items are of the same type T, and that this type implements the Comparable interface. (You can write these out by hand or type them. You don’t have to actually compile and run them, although you certainly may do so, if it increases your confidence that your answers are correct). If you do them in Eclipse, you should either print the code and scan it,  or copy it and paste it into your document
   To make your life a little bit simpler, I am giving you the headers for some methods.

   public static <T extends Comparable<? super T>> T min(T arg1, T arg2)

   public static <T extends Comparable<? super T>> T max(T arg1, T arg2)

   public static <T extends Comparable<? super T>> T min(T[] args)

   public static <T extends Comparable<? super T>> T[] max2(T[] args)

   Creating the Generic array for the return value of the last part is also tricky;  here is a way to do it:

   T[] result = (T[]) Array.newInstance(args[0].getClass(), 2);

4. (10 points). Use the formal definition of big-Oh (see Weiss section 5.4) to show that N² + 3N + 8 is O(N²). That is, find appropriate constants N₀ and c, and show that they work.
5. (6 points) Weiss Problem 5.3 [5.3], part a only. (T₁(N) + T₂(N)). Prove your answer, using the N₀ and c idea from the formal definition of big-Oh. [Hint: the given bounds on T₁ and T₂ tell you something about the existence of an N₀ and c for each of those functions: we could call these numbers N₀₁, c₁, N₀₂, and c₂ ]. Use those constants to come up with appropriate constants for the sum of the functions.)
6. (8 points) Weiss Problem 5.14 [5.11].   (running time if we go from size 100 to size 500).
7. (12 points) For each of the following four code fragments, similar to Weiss problem 5.20 [5.15], first give the exact number of times the "sum++" statement executes in terms of n. Then use that to give a Big-Oh running time (by ignoring the low-order terms).

   1. for (int i=0; i < n; i++)
          for (int j=0; j < n * n; j++)
              sum++;
      

   2. for (int i=0; i < n; i++)
          for (int j=i; j < n; j++)
              sum++;
      

   3. for (int i=0; i < n; i++)
          for (int j=0; j < i * n; j++)
              for (int k=0; k < n; k++)
                  sum++;
      

   4. for (int i=n; i >= 1; i = i / 2)
          sum++;
      

For Your Consideration

These problems are for you to think about and convince yourself that you could do them. It would be good practice to actually do them, but you are not required to turn them in.

• Weiss Exercise 6.5 [6.5]. (You may have to come back to it a few times before you figure out what to do.) Here is a more detailed explanation:

  The goal of this problem is to implement a collection data type that provides three operations and does them efficiently:

  push(obj); // adds obj to the collection.
  
  obj = pop(); // removes and returns the most recently-added object.
  
  obj = findMin(); // returns the smallest object (assume that all
   // of the objects in the collection are Comparable).
  

  A single stack can do the first two operations in constant time, but not the third. But if our implementation uses TWO stacks to represent a collection, it is possible to do each of the three operations in constant time. Is should be obvious that our new data structure’s push method will have to do more work (than the push operation for an ordinary stack would have to do) in order for findMin to also be a constant-time operation. All you need to do is show the code for the three operations. You may assume that there is already a Stack class that has its own push, pop, and top methods. The code for each of the three methods that you must provide is short. We hope that by making things a bit more specific, it will make it easier for you to get started on this problem.

  To further assist you in understanding this problem, here is a framework in which your answers could go: A class declaration with stubs for the three methods you are supposed to write. Each of those methods should be constant time (no loops or recursion)

  public class StackWithMin<E> {
   // TODO: Give these two stacks names indicating what they are used for:
   private Stack<E> stack1, stack2;
  
   public StackWithMin() {
   this.stack1 = new Stack<E>();
   this.stack2 = new Stack<E>();
  }
  
   public void push(E element) { /* TODO: fill in the details */ }
   public E pop() {/* TODO: fill in the details */ }
   public E findMin() {/* TODO: fill in the details */ }
  }
  

• Weiss exercises 6.26(a)(b), 6.27(a)(b), 6.28(a)(b) [6.12(a)(b), 6.13(a)(b), 6.14(a)(b)]. No implementation required; just the algorithm and analysis.