#=======================================================================
# Author: Isai Damier
# Title: Permutation Index
# Project: geekviewpoint
# Package: algorithms
#
# Statement:
# Given a permutation of a set, return the index of the permutation.
#
# Sample Input: [3, 1, 2]
# Sample Output: 4
#
# Time Complexity of Solution:
# Best = Average = Worst = O(n^2)
# Space Complexity of Solution:
# Best = Average = Worst = O(1)
#
# Details:
# A permutation is an ordering of the elements of a set. So for a set
# constituted of the elements 1,2,3; then [1,2,3] and [2,1,3] are two
# different permutations of the set. A set of length n has n!
# permutations. For example, if a set contains 3 elements, it has
# 3! = 3*2*1 = 6 permutations. The following algorithm uses the
# relation between permutation and factorial to find the index of a
# given permutation of a set.
#
# Illustrating by manually getting the index of [2, 4, 3, 1].
# Since this is a 4-element set, we know there are 4! permutations
# (4! = 4*3*2*1). If the set only had 3 elements, we would have 3*2*1
# permutations. If the set only had 2 elements, we would have 2!=2*1
# permutations; and so on.
#
# ASIDE: The decimal system of counting is a positional system.
# A 3-element decimal number, for instance, has the following three
# positional weights: hundred, ten, unit. Hence, we know the value of
# the number 472 because we understand: 4*hundred + 7*ten + 2*unit.
#
# If we treat our 4-element set as a positional system, then we get
# the following positional weights: 3!, 2!, 1!, 0. So that the index
# of [2, 4, 3, 1] is: x*3!+y*2!+z*1!+w*0. Presently it suffices to
# find the values of x,y,z to calculate the index (we ignore w because
# it is paired with 0). x,y,z are counters: the number of succeeding
# elements less than the element being considered. For example, in
# [2, 4, 3, 1], there are two succeeding elements less than 4
# (namely 3 and 1). For 2 it's 1 (1); for 4 it's 2 (3 and 1); for 3
# it's 1 (1); for 1 it's 0.
#
# Now we can calculate the index of [2, 4, 3, 1] as: x=1, y=2, z=1:
# x*3!+y*2!+z*1!+w*0 = 1*3! + 2*2! + 1*1! = 6 + 4 + 1 = 11.
#
# Now that we have our algorithm, the trick is implementing it. As you
# may imagine there are a number of possible implementations. The
# presented implementation focuses on using constant auxiliary memory:
# memory = O(1) and time = O(n^2). The code is written for readability.
#=======================================================================
def permutationIndex( permutation ):
index = 0
position = 2 # position 1 is paired with factor 0 and so is skipped
factor = 1
for p in range( len( permutation ) - 2, -1, -1 ):
successors = 0
for q in range( p + 1, len( permutation ) ):
if permutation[p] > permutation[q]:
successors += 1
index += ( successors * factor )
factor *= position
position += 1
return index
import unittest
from algorithms import numbers as algorithm
class Test( unittest.TestCase ):
def testPermutationIndex( self ):
three = [[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1],
[3, 1, 2], [3, 2, 1]]
for i in range( len( three ) ) :
self.assertEquals( i, algorithm.permutationIndex( three[i] ) )
four = [[1, 2, 3, 4], [1, 2, 4, 3], [1, 3, 2, 4], [1, 3, 4, 2],
[1, 4, 2, 3], [1, 4, 3, 2], [2, 1, 3, 4], [2, 1, 4, 3],
[2, 3, 1, 4], [2, 3, 4, 1], [2, 4, 1, 3], [2, 4, 3, 1],
[3, 1, 2, 4], [3, 1, 4, 2], [3, 2, 1, 4], [3, 2, 4, 1],
[3, 4, 1, 2], [3, 4, 2, 1], [4, 1, 2, 3], [4, 1, 3, 2],
[4, 2, 1, 3], [4, 2, 3, 1], [4, 3, 1, 2], [4, 3, 2, 1]]
for i in range( len( four ) ) :
self.assertEquals( i, algorithm.permutationIndex( four[i] ) )
