Concrete mathematics : a foundation for computer science

80 INTEGER FUNCTIONS
We’ve got more tools to work with than we had in Chapter 1, so let’s
consider the more authentic Josephus problem in which every third person is
eliminated, instead of every second. If we apply the methods that worked in
Chapter 1 to this more difficult problem, we wind up with a recurrence like
J3(n) = [iJ3(Ljnl) + a,] modn+ 1,
where ‘mod’ is a function that we will be studying shortly, and where we have
a,, = -2, +1 , or -i according as n mod 3 = 0, 1, or 2. But this recurrence
is too horrible to pursue.
There’s another approach to the Josephus problem that gives a much
better setup. Whenever a person is passed over, we can assign a new number.
Thus, 1 and 2 become n + 1 and n + 2, then 3 is executed; 4 and 5 become
n + 3 and n + 4, then 6 is executed; . . . ; 3kSl and 3k+2 become n+2k+ 1
and n + 2k + 2, then 3k + 3 is executed; . . . then 3n is executed (or left to
survive). For example, when n = 10 the numbers are
1 2 3 4 5 6 7 8 9 10
11 12 13 14 15 16 17
18 19 20 21 22
23 24 25
26 27
28
29
30
The kth person eliminated ends up with number 3k. So we can figure out who
the survivor is if we can figure out the original number of person number 3n.
If N > n, person number N must have had a previous number, and we
can find it as follows: We have N = n + 2k + 1 or N = n + 2k + 2, hence
k = [(N - n - 1)/2J ; the previous number was 3k + 1 or 3k + 2, respectively.
That is, it was 3k + (N - n - 2k) = k + N - n. Hence we can calculate the
survivor’s number J3 (n) as follows:
N := 3n;
while N>n do N:= [“-r-‘] +N-n;
J3(n) := N.
This is not a closed form for Jj(n); it’s not even a recurrence. But at least it
“Not too slow,
not too fast,”
tells us how to calculate the answer reasonably fast, if n is large. -L. Amstrong