Due Date: October 17, 2018 [2018-10-17, Wednesday of week 08]

The following guidelines are expected for all homework submissions:

Problems for Assignment #2

This problem set has two problems from chapter 4, and three from chapter 5. You will be working with stacks, queues, and linked lists for this assignment. Your task is to complete the following problems using the programs specified in each one. As always, when you are finished and ready to submit your work, do so in your repo. Here are the problems:

  1. Write a method for the Queue class in the queue.java program (Listing 4.4) that displays the contents of the queue. Note that this does not mean simply displaying the contents of the underlying array. You should show the queue contents from the first item inserted to the last, without indicating to the viewer whether the sequence is broken by wrapping around the end of the array. Be careful that one item and no items display properly, no matter where front and rear are.
  2. Create a Deque class based on the discussion of deques (double-ended queues) in this chapter [CH 4]. It should include insertLeft(), insertRight(), removeLeft(), removeRight(), isEmpty(), and isFull() methods. It will need to support wraparound at the end of the array, as queues do.
  3. A circular list is a linked list in which the last link points back to the first link. There are many ways to design a circular list. Sometimes there is a pointer to the start of the list. However, this makes the list less like a real circle and more like an ordinary list that has its end attached to its beginning. Make a class for a singly linked circular list that has no end and no beginning. The only access to the list is a single reference, current, that can point to any link on the list. This reference can move around the list as needed. (See Programming Project 5.5 for a situation in which such a circular list is ideal.) Your list should handle insertion, searching, and deletion. You may find it convenient if these operations take place one link downstream of the link pointed to by current. (Because the upstream link is singly linked, you can't get at it without going all the way around the circle.) You should also be able to display the list (although you'll need to break the circle at some arbitrary point to print it on the screen). A step() method that moves current along to the next link might come in handy too.
  4. Implement a stack class based on the circular list of Programming Project 5.3. This exercise is not too difficult. (However, implementing a queue can be harder, unless you make the circular list doubly linked.)
  5. EXTRA CREDIT: The Josephus Problem is a famous mathematical puzzle that goes back to ancient times. There are many stories to go with the puzzle. One is that Josephus was one of a group of people who were about to be captured by the Romans. Rather than be enslaved, they chose to commit suicide. They arranged themselves in a circle and, starting at a certain person, counted off around the circle. Every nth person had to leave the circle and kill themselves. Josephus decided he didn't want to die, so he arranged the rules so that he would be the last person left. If there were (say) 20 people, and he was the seventh person from the start of the circle, what number should he tell them to use for counting? The problem is made more complicated because the circle shrinks as the counting continues.
    Create an application that uses a circular linked list (like that in Programming Project 5.3) to model this problem. Inputs are the number of people in the circle, the number used for counting off, and the number of the person where counting starts (usually 1). The output is the list of persons being eliminated. When a person drops out of the circle, counting starts again from the person who was on his left (assuming you go around clockwise).
    There are actually two ways to count this off. One way is to count by the counter value and when you land on a person at the counter value they are eleminated. Then you continue the count with the person directly adjacent to the one who was eliminated. Here's an example. There are seven people numbered 1 through 7, and you start at 1 and count off by threes. People will be eliminated in the order 4, 7, 3, 1, 6, 2. Number 5 will be left.
    The second way of counting is slightly different. You start at 1 and count by the counter value and when you land on a person at the counter value they are eliminated. However, in this case instead of continuing the count at the adjacent person, you consider that adjacent person to have TAKEN THE PLACE of the one who was eliminated, and the count begins at the next person AFTER that one. For example, in the same situation with seven people counting by threes, people will be eliminated in the order 4, 1, 6, 5, 7, 3. Number 2 is the remaining person in this case.
    If you have trouble understanding the difference, draw it out on a piece of paper or on a whiteboard and count it off to see how they are different.
    One other thing to consider is, the REAL answer to the problem is HOW NOT TO DIE, so given the number of people AND YOUR LOCATION IN THE LINE, WHAT NUMBER do you count by to ensure you are the last person left standing, so you can run away from the Romans and not either die OR be a slave.