Efficiently determine the parity of a permutation
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Efficiently Determine the Parity of a Permutation in Java
Learn various algorithms to calculate the parity (even or odd) of a permutation, focusing on efficient Java implementations and their underlying mathematical principles.
Determining the parity of a permutation is a fundamental problem in combinatorics and abstract algebra, with applications in areas like cryptography, puzzle solving, and algorithm analysis. A permutation's parity indicates whether it can be expressed as an even or odd number of transpositions (swaps of two elements). This article explores different methods to calculate permutation parity in Java, emphasizing efficiency and clarity.
Understanding Permutation Parity
The parity of a permutation is a property that classifies it as either 'even' or 'odd'. This classification is based on the minimum number of swaps (transpositions) required to transform the permutation back into the identity permutation (sorted order). If an even number of swaps is needed, the permutation is even; if an odd number, it's odd.
Two primary methods are commonly used to determine parity:
- Counting Inversions: An inversion occurs when two elements are in the 'wrong' order relative to each other. The parity of a permutation is the same as the parity of its number of inversions.
- Cycle Decomposition: Any permutation can be decomposed into disjoint cycles. The parity of a permutation is determined by the number of cycles and the length of the permutation. Specifically, a permutation is even if
n - cis even, wherenis the length of the permutation andcis the number of cycles.
flowchart TD
A[Start: Permutation P] --> B{Choose Method}
B -->|Method 1: Inversions| C[Count Inversions]
C --> D{Is Inversion Count Even?}
D -->|Yes| E[Parity: Even]
D -->|No| F[Parity: Odd]
B -->|Method 2: Cycle Decomposition| G[Decompose into Cycles]
G --> H[Count Cycles (c) and Length (n)]
H --> I{Is (n - c) Even?}
I -->|Yes| E
I -->|No| FFlowchart illustrating two methods for determining permutation parity.
Method 1: Counting Inversions
The inversion counting method is straightforward: iterate through all pairs of elements in the permutation and count how many are out of order. For a permutation P = [p_0, p_1, ..., p_{n-1}], an inversion is a pair (i, j) such that i < j but p_i > p_j. The total number of inversions directly determines the parity: if the count is even, the permutation is even; if odd, the permutation is odd.
While conceptually simple, a naive implementation of counting inversions has a time complexity of O(n^2), where n is the length of the permutation. For larger permutations, more efficient algorithms like merge sort can count inversions in O(n log n) time.
public class PermutationParity {
// O(n^2) approach for counting inversions
public static boolean isEvenParityByInversionsNaive(int[] p) {
int inversions = 0;
for (int i = 0; i < p.length - 1; i++) {
for (int j = i + 1; j < p.length; j++) {
if (p[i] > p[j]) {
inversions++;
}
}
}
return inversions % 2 == 0;
}
// O(n log n) approach using a modified merge sort
public static boolean isEvenParityByInversionsMergeSort(int[] p) {
int[] temp = new int[p.length];
long inversions = mergeSortAndCount(p, temp, 0, p.length - 1);
return inversions % 2 == 0;
}
private static long mergeSortAndCount(int[] arr, int[] temp, int left, int right) {
long invCount = 0;
if (right > left) {
int mid = (left + right) / 2;
invCount += mergeSortAndCount(arr, temp, left, mid);
invCount += mergeSortAndCount(arr, temp, mid + 1, right);
invCount += mergeAndCount(arr, temp, left, mid + 1, right);
}
return invCount;
}
private static long mergeAndCount(int[] arr, int[] temp, int left, int mid, int right) {
int i = left; // Initial index for left subarray
int j = mid; // Initial index for right subarray
int k = left; // Initial index for merged subarray
long invCount = 0;
while ((i <= mid - 1) && (j <= right)) {
if (arr[i] <= arr[j]) {
temp[k++] = arr[i++];
} else {
temp[k++] = arr[j++];
// This is where inversions are counted
// If arr[i] > arr[j], then all elements from arr[i] to arr[mid-1]
// are greater than arr[j]
invCount += (mid - i);
}
}
while (i <= mid - 1) {
temp[k++] = arr[i++];
}
while (j <= right) {
temp[k++] = arr[j++];
}
for (i = left; i <= right; i++) {
arr[i] = temp[i];
}
return invCount;
}
public static void main(String[] args) {
int[] p1 = {1, 0, 3, 2}; // 2 inversions (1,0), (3,2) -> Even
int[] p2 = {0, 2, 1, 3}; // 1 inversion (2,1) -> Odd
int[] p3 = {3, 2, 1, 0}; // 6 inversions (3,2), (3,1), (3,0), (2,1), (2,0), (1,0) -> Even
System.out.println("Permutation {1, 0, 3, 2} (Naive): " + (isEvenParityByInversionsNaive(p1) ? "Even" : "Odd"));
System.out.println("Permutation {0, 2, 1, 3} (Naive): " + (isEvenParityByInversionsNaive(p2) ? "Even" : "Odd"));
System.out.println("Permutation {3, 2, 1, 0} (Naive): " + (isEvenParityByInversionsNaive(p3) ? "Even" : "Odd"));
System.out.println("Permutation {1, 0, 3, 2} (Merge Sort): " + (isEvenParityByInversionsMergeSort(p1) ? "Even" : "Odd"));
System.out.println("Permutation {0, 2, 1, 3} (Merge Sort): " + (isEvenParityByInversionsMergeSort(p2) ? "Even" : "Odd"));
System.out.println("Permutation {3, 2, 1, 0} (Merge Sort): " + (isEvenParityByInversionsMergeSort(p3) ? "Even" : "Odd"));
}
}
Java implementation for counting inversions using both naive O(n^2) and merge sort O(n log n) approaches.
Method 2: Cycle Decomposition
Every permutation can be uniquely decomposed into a product of disjoint cycles. For example, the permutation [0, 2, 1, 3] (mapping 0->0, 1->2, 2->1, 3->3) can be decomposed into (0)(1 2)(3). The parity of a permutation is determined by the formula (-1)^(n - c), where n is the length of the permutation and c is the number of disjoint cycles. If n - c is even, the permutation is even; if n - c is odd, the permutation is odd.
To implement this, we can iterate through the elements. If an element hasn't been visited, start a new cycle and follow the chain until we return to the starting element, marking all elements in the cycle as visited. This process continues until all elements have been visited. This method has a time complexity of O(n) because each element is visited and processed exactly once.
import java.util.Arrays;
public class PermutationParityCycles {
public static boolean isEvenParityByCycles(int[] p) {
int n = p.length;
boolean[] visited = new boolean[n];
int numCycles = 0;
for (int i = 0; i < n; i++) {
if (!visited[i]) {
numCycles++;
int current = i;
while (!visited[current]) {
visited[current] = true;
current = p[current];
}
}
}
// Parity is even if (n - numCycles) is even
return (n - numCycles) % 2 == 0;
}
public static void main(String[] args) {
// Permutation p = [p_0, p_1, ..., p_{n-1}]
// where p_i is the element at index i, meaning i maps to p_i
int[] p1 = {1, 0, 3, 2}; // (0 1)(2 3) -> n=4, c=2. n-c = 2 (Even)
int[] p2 = {0, 2, 1, 3}; // (0)(1 2)(3) -> n=4, c=3. n-c = 1 (Odd)
int[] p3 = {3, 2, 1, 0}; // (0 3)(1 2) -> n=4, c=2. n-c = 2 (Even)
int[] p4 = {0, 1, 2, 3}; // (0)(1)(2)(3) -> n=4, c=4. n-c = 0 (Even - Identity)
int[] p5 = {1, 2, 3, 0}; // (0 1 2 3) -> n=4, c=1. n-c = 3 (Odd)
System.out.println("Permutation {1, 0, 3, 2} (Cycles): " + (isEvenParityByCycles(p1) ? "Even" : "Odd"));
System.out.println("Permutation {0, 2, 1, 3} (Cycles): " + (isEvenParityByCycles(p2) ? "Even" : "Odd"));
System.out.println("Permutation {3, 2, 1, 0} (Cycles): " + (isEvenParityByCycles(p3) ? "Even" : "Odd"));
System.out.println("Permutation {0, 1, 2, 3} (Cycles): " + (isEvenParityByCycles(p4) ? "Even" : "Odd"));
System.out.println("Permutation {1, 2, 3, 0} (Cycles): " + (isEvenParityByCycles(p5) ? "Even" : "Odd"));
}
}
Java implementation for determining permutation parity using cycle decomposition.