-
Notifications
You must be signed in to change notification settings - Fork 68
/
JoeSurvivor.rb
43 lines (36 loc) · 1.33 KB
/
JoeSurvivor.rb
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
31
32
33
34
35
36
37
38
39
40
41
42
43
=begin
In this kata you have to correctly return who is the "survivor", ie: the last
element of a Josephus permutation.
Basically you have to assume that n people are put into a circle and that they
are eliminated in steps of k elements, like this:
josephus_survivor(7,3) => means 7 people in a circle;
one every 3 is eliminated until one remains
[1,2,3,4,5,6,7] - initial sequence
[1,2,4,5,6,7] => 3 is counted out
[1,2,4,5,7] => 6 is counted out
[1,4,5,7] => 2 is counted out
[1,4,5] => 7 is counted out
[1,4] => 5 is counted out
[4] => 1 counted out, 4 is the last element - the survivor!
The above link about the "base" kata description will give you a more thorough
insight about the origin of this kind of permutation, but basically that's all
that there is to know to solve this kata.
Notes and tips: using the solution to the other kata to check your function may
be helpful, but as much larger numbers will be used, using an array/list to
compute the number of the survivor may be too slow; you may assume that both n
and k will always be >=1.
=end
# My Solution
def josephus_survivor(n,k)
people = []
0.upto(n-1) {|x| people << x+1}
0.upto(n-2) do |x|
people = people.rotate(k-1)
people.shift
end
people[0]
end
# Better Solution
def josephus_survivor(n, k)
n < 1 ? 1 : (josephus_survivor(n - 1, k) + k - 1) % n + 1;
end