Permutation group operation

Question

There is a well-known bijection between the permutations of n elements and the numbers 0 to n!-1 such that the lexicographic ordering of the permutations and the corresponding numbers is the same. For example, with n=3:

0 <-> (0, 1, 2)
1 <-> (0, 2, 1)
2 <-> (1, 0, 2)
3 <-> (1, 2, 0)
4 <-> (2, 0, 1)
5 <-> (2, 1, 0)

It is also well-known that the permutations of n elements form a group (the symmetric group of order n!) - so, in particular, that one permutation of n elements applied to a second permutation of n elements yields a permutation of n elements.

For example, (1, 0, 2) applied to (a, b, c) yields (b, a, c), so (1, 0, 2) applied to (2, 1, 0) yields (1, 2, 0).

Write a program which takes three integer arguments: n, p1, and p2; interprets p1 and p2 as permutations of n elements; applies the first to the second; and outputs the corresponding integer. For example:

$ ./perm.sh 3 2 5
3

Answer 1

J, 30

I like the elegance of this:

[:A.[:{/]A.~/~i.@[

or this:

13 :'A.{/(i.x)(A.)~/y'

but they work like this:

3 f 2 5
3
12 f 8 9
17

So this is the valid entry:

([:A.[:{/i.@{.A.~/}.)".}.>ARGV

Some explanations:

• 3 A. 0 1 2: gives the 3rd permutation of 0 1 2 (= 1 2 0)
• 0 1 2 (A.)~ 3: is the same but with arguments reversed
• 0 1 2 (A.)~/ 3 4 5 ... "applies" (A.)~ to 3 4 5 ..., so it gives the 3rd, 4th, 5th, ... permutation of 0 1 2.
• A. 1 2 0: gives the order of the permutation of 1 2 0 (= 3)
• i. n: gives the sequence 0 1 2 ... n-1
• 1 2 0 { 0 2 1 arranges 0 2 1 by 1 2 0 (= 2 1 0)

Answer 2

Ruby - 77 chars

n,a,b=$*.map &:to_i
l=[*[*0...n].permutation]
p l.index(l[b].values_at *l[a])

Answer 3

Python 2.6, 144 chars

import sys
from itertools import*
n,p,q=map(int,sys.argv[1:])
R=range(n)
P=list(permutations(R))
print P.index(tuple(P[q][P[p][i]] for i in R))