ANSWER IN C++ The Josephus Problem

The problem is known as the Josephus Problem (or Josephus permutation) and postulates a group of people of size N >= 1 are standing in a circle waiting to be eliminated. Counting begins at a specified point in the circle and proceeds around the circle in a specified direction. After a specified number of M >= 1 people are counted, the M^th person in the circle is eliminated. The procedure is repeated with the remaining people, starting with the next person, going in the same direction and counting the same number of people, until only one person remains.

For example, suppose that M = 3 and there are N = 5 people named A, B, C, D and E. We count three people starting at A, so that C is eliminated first. We then begin at D and count D, E and back to A, so that A is eliminated next. Then we count B, D and E, and finally B, D and B, so that D is the one who remains last.

For this computer assignment, you are to write and implement a C++ program to simulate and solve the Josephus problem. The input to the program is the number M and a list of N names, which is clockwise ordering of the circle, beginning with the person from whom the count is to start. After each removal, the program should print the names of all people in the circle until only one person remains. However, to save printing space, print the names of the remaining people only after K >= 1 eliminations, where K is also an input argument to the program. The input arguments N, M and K can be entered from stdin in the given order. (see josephus.d for values)

josephus d: 41 3 7

Programming Notes:

• Name the people in the circle in the following sequence: A1, A2 ... A9, B1, B2 ... B9, C1, C2 ..., and start counting from the person A1. Enter input values N, M and K when the program prompts for them and use a list container to store the names of N people.

• void init_vals(list &L, args &in) It reads the input values N, M and K of the struct args in when the program prompts for them. The routine prints out those values on stdout, and fills the names of people in the list L. You can find the definition of the struct args in the header file josephus.h, which is defined as:

struct args { unsigned N; unsigned M; unsigned K; }; 

• void print_list(const list &L, const unsigned &cnt) It prints out the contents of the list L at the beginning and after removing K names (each time) from the list, until only one name remains in the list, where cnt has an initial value 0 and it indicates the total number of removals so far. At the end, it also prints the name of the last remaining person. For printout, print only up to 12 names in a single line, where the names are separated by single spaces.

• The main() routine first calls the routine init_vals() and initializes cnt to 0, and then calls the print_list() to print out the names of people in circle. After that it locates the M^th person in the list, and using the member function erase(), it removes that person from the list, and by calling the print_list() prints out the current contents of the list. This process continues (in a loop) until only one person remains in the list.

  • If i (with initial value 0) indicates the position of a person in list L, then the statement: j = (i + (M 1))%L.size() returns the position of the M^th person from the position i.

  • Since the argument to the erase() function is an iterator, you can convert the index value j to an iterator by the advance(p, j) function, where p = L.begin().

• To store the names in an empty list, first change the size of the list to N, and then use the generate() function in the STL. The last argument to this function is the function object SEQ(N), which is defined in the header file josephus.h.

• Josephus.h file:

  #ifndef H_JOSEPHUS
  	#define H_JOSEPHUS
  	#include
  	#include
  	#include
  	#include
  	#include
  	using namespace std;
  	#define NO_LETS 26 // no of letters in English alphabet
  	#define NO_ITEMS 12 // no of items printed on single line
  	// struct for input arguments
  	struct args {
  	unsigned N, // no of initial people
  	M, // count to eliminate person
  	K; // frequency of printouts
  	};
  	// class to generate name tags for people
  	class SEQ {
  	private:
  	string id; // name tag for person
  	unsigned size, nd; // no of people, no of digits in name tags
  	// returns no of digits in name tags
  	unsigned find_nd ( const double& sz ) {
  	if ( ( sz / NO_LETS ) <= 1 ) return 2;
  	else return ( find_nd ( sz / NO_LETS ) + 1 );
  	}
  	public:
  	// constructor for name-tag generator
  	SEQ ( const unsigned& s = 1 ) : size ( s ) {
  	double sz = ( double ) size / 9; nd = find_nd ( sz );
  	id = string ( nd, 'A' ); id [ nd - 1 ] = '1';
  	}
  	// returns next name tag in sequence
  	string operator ( ) ( ) {
  	string tmp = id; int i = nd - 1;
  	if ( id [ i ] < '9' ) id [ i ]++;
  	else {
  	id [ i ] = '1'; bool flag = true;
  	for ( i--; i >= 0 && flag; i-- )
  	if ( id [ i ] < 'Z' ) { id [ i ]++; flag = false; }
  	else id [ i ] = 'A';
  	}
  	return tmp;
  	}
  	};
  	// reads and initializes all input arguments
  	void init_vals(list &L, args &in);
  	// prints all name tags for remaining people after elimination
  	void print_list ( const list < string >&L, const unsigned&cnt );
  	#endif

• Expected output:

  Number of people? 41 Index for elimination? 3 Index for printing? 7

  Initial group of people ----------------------- A1 A2 A3 A4 A5 A6 A7 A8 A9 B1 B2 B3 B4 B5 B6 B7 B8 B9 C1 C2 C3 C4 C5 C6 C7 C8 C9 D1 D2 D3 D4 D5 D6 D7 D8 D9 E1 E2 E3 E4 E5

  After eliminating 7th person ---------------------------- A1 A2 A4 A5 A7 A8 B1 B2 B4 B5 B7 B8 C1 C2 C4 C5 C6 C7 C8 C9 D1 D2 D3 D4 D5 D6 D7 D8 D9 E1 E2 E3 E4 E5

  After eliminating 14th person ----------------------------- A2 A4 A5 A7 A8 B1 B2 B4 B5 B7 B8 C1 C2 C4 C5 C7 C8 D1 D2 D4 D5 D7 D8 E1 E2 E4 E5

  After eliminating 21th person ----------------------------- A2 A4 A7 A8 B2 B4 B7 B8 C2 C4 C7 C8 D2 D4 D7 D8 E1 E2 E4 E5

  After eliminating 28th person ----------------------------- A2 A4 A8 B2 B7 B8 C4 C7 D2 D4 D8 E2 E4

  After eliminating 35th person ----------------------------- A2 A4 B7 C4 D4 D8

  After eliminating 40th person ----------------------------- D4