about Josephus problem

Problem: Josephus problem is a classical problem which can be solved by DP. The problem assume n persons in a circle, then start from the 1st person, we eliminate a person after skipping a fix number of persons. The process stops when there is only one person left. We want to know the last person's index in the original circle. For example, if n = 5 and k = 2, at the beginning, we have 1 2 3 4 5, then we eliminate 2 and 4, then we eliminate 1 and 5, so the last person's index is 3.

Solution: For the general case where k is an arbitrary number, we use F(n, k) to denote the solution for n persons. Then we have F(n, k) = (F(n-1, k) + k ) mod n. Why? we can think the problem as follow:

• The first person we need to eliminate is the kth person, after he is removed, we have n-1 persons now.
• However, we can not directly resort to the subproblem F(n-1, k), since now our process will start at the k+1th person at the original circle. Therefore we need to remap the indices. Since the process is applied on a circle, we can move the first k-1 persons (1st, 2rd, ... k-1th person at the original circle) to the end of the circle. Thus, assume the new index in F(n-1, k) is i, the corresponding index in F(n, k) is (i + k) mod n.

The above approach will take O(N) time. For smaller k, actually O(logN) solution exists. For example, when k = 2, we have F(2n) = 2*F(n) -1 and F(2n+1) = 2*F(n) +1. Actually for k = 2, even an O(1) solution exists: if n = 2^m + l where  0 =< l  < 2^m, the last person will be 2*l + 1.