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was the sentence of death to the person next to the one who got the potato Josephus rigged the game to get the last lot and convinced the remaining intended victim that the two of them should surrender That is how we know about this game; in effect, Josephus cheated' If M = 0, the players are eliminated in order, and the last player always wins For other values of M, things are not so obvious Figure 141 shows that if N = 5 and M = I, the players are eliminated in the order 2, 4, I , 5 In this case, player 3 wins The steps are as follows

1 At the start, the potato is at player 1 After one pass it is at player 2 2 Player 2 is eliminated Player 3 picks up the potato, and after one pass, it is at player 4 3 Player 4 is eliminated Player 5 picks up the potato and passes it to player I 4 Player I is eliminated Player 3 picks up the potato and passes it to player 5 5 Player 5 is eliminated, so player 3 wins

(12.34)

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First, we write a program that simulates, pass for pass, a game for any values of N and M The running time of the simulation is O(MN), which is acceptable if the number of passes is small Each step takes O(M) time because it performs M passes We then show how to implement each step in O(log N) time, regardless of the number of passes performed The running time of the simulation becomes O(N log N)

1 Thanks to David Teague for relaying this story The version that we solve differs from the historical description In Exercise 1412 you are asked to solve the historical version

(1.1 )

The passing stage in the Josephus problem suggests that we represent the players in a linked list We create a linked list in which the elements 1, 2, , N are inserted in order We then set an iterator to the front element Each pass of the potato corresponds to a + + operation on the iterator At the last player (currently remaining) in the list we implement the pass by resetting the iterator to the first element This action mimics the circle When we have finished passing, we remove the element on which the iterator has landed An implementation is shown in Figure 142 The linked list and iterator are declared at lines 8 and 9, respectively We construct the initial list by using the loop at lines 14 and 15 In Figure 142, the code at lines 20 to 33 plays one step of the algorithm by passing the potato (lines 20 to 25) and then eliminating a player (lines 30-33) This procedure is repeated until the test at line 18 tells us that only one player remains At that point we return the player's number at line 36 The running time of this routine is O ( M N ) because that is exactly the number of passes that occur during the algorithm For small M, this running time is acceptable, although we should mention that the case M = 0 does not yield a running time of O ( 0 ) ;obviously the running time is O(N)We do not merely multiply by zero when trying to interpret a Big-Oh expression

EXERCISE 12.8 Derive (12.34) by performing the dw integration on the righthand side (see Exercise 6.7).

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