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Permutation Index

#======================================================================= # 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] ) )