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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. **************************************************************************/ public int permutationIndex(int[] permutation) { int index = 0; int position = 2;// position 1 is paired with factor 0 and so is skipped int factor = 1; for (int p = permutation.length - 2; p >= 0; p--) { int successors = 0; for (int q = p + 1; q < permutation.length; q++) { if (permutation[p] > permutation[q]) { successors++; } } index += (successors * factor); factor *= position; position++; } return index; }

import org.junit.Test; import static org.junit.Assert.*; public class NumbersTest { /** * Test of permutationIndex method, of class Numbers. */ @Test public void testPermutationIndex() { System.out.println("permutationIndex"); int[][] three = {{1, 2, 3}, {1, 3, 2}, {2, 1, 3}, {2, 3, 1}, {3, 1, 2}, {3, 2, 1}}; Numbers instance = new Numbers(); for (int i = 0; i < three.length; i++) { assertEquals(i, instance.permutationIndex(three[i])); } int[][] 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 (int i = 0; i < four.length; i++) { assertEquals(i, instance.permutationIndex(four[i])); } } }