Do problem 5.5 on pages 248-249 using Java (one file only--multiple classes ok).

 The input will be a sequence of 3 ints entered from the keyboard (separated only by one or more blanks), such as 7 1 3 as in problem 5.5. This kind of input can be entered again and again until the user enters stop. 7 1 3 means that there are seven items: 1 2 3 4 5 6 7 (the numbering always starts at 1) and that the holder starts at 1 and that the passing is 3 so that means that the first one eliminated is 4 creating a new list 1 2 3 5 6 7 which now starts holding at 5 and the next one eliminated is 1 creating a new list 2 3 5 6 7 which now starts holding at 2 and the next one eliminated is 6 creating a new list 2 3 5 7 which now starts holding at 7 and the next one eliminated is 5 creating a new list 2 3 7 which now starts holding at 7 and the next one eliminated is 7 creating a new list 2 3 which now starts holding at 2 and the next one eliminated is 3 creating the final list 2 IN ALL SITUATIONS THE FIRST NUMBER IN THE LIST IS 1. When the user wants to stop the input, it should just be: stop The user should be able to continue providing input after the problem is solved. The output should be sent to the console (lower pane on jGrasp.) Please note also that the first number could be an extremely large int. It might be fun to try a circular linked list in one class , an iterator in another, and then the application itself, but the choice is yours. Try to learn from the concepts of the chapter! JUST PRINT THE FINAL LIST OF A SINGLE int. 1. turnin an electronic version of "bigone" (the file with the classes--unsharred) using $ turnin ONLY turnin the .java file...do not turnin any runs. Name your file your n number.java. DO NOT USE shar...

Problem - 5.5:

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 Jews 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, started counting off around the circle. Every nth person had to leave the circle and commit suicide. Josephus decided he didnt want to die, so he arranged the rules so 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 off? The problem is made much 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). Heres 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, 1, 6, 5, 7, 3. Number 2 will be left.

Programming Project 5.3:

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 ideally suited.) 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 cant get at it without going all the way around the circle.) You should also be able to display the list (although youll need to break the circle at some arbi- trary point to print it on the screen). A step() method that moves currentalong to the next link might come in handy too.