/***************************************************************************
* 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]));
}
}
}
