Symmetric Relation on a Set
Last Updated :
02 Jan, 2023
A relation is a subset of the cartesian product of a set with another set. A relation contains ordered pairs of elements of the set it is defined on. To learn more about relations refer to the article on “Relation and their types“.
What is a Symmetric Relation?
A relation R on a set A is called symmetric relation if and only if
∀ a, b ∈ A, if (a, b) ∈ R then (b, a) ∈ R and vice versa i.e.,
∀ a, b ∈ A, (a, b) ∈ R 
(b, a) ∈ R,
where R is a subset of (A x A), i.e. the cartesian product of set A with itself.
This means if an ordered pair of elements “a” to “b” (aRb) is present in relation R, then an ordered pair of elements “b” to “a” (bRa) should also be present in relation R. If any such bRa is not present for any aRb in R then R is not a symmetric relation.
Example:
Consider set A = {a, b}
then R = {(a, b), (b, a)} is symmetric relation but
R = { (a, b), (a, a) } is not a symmetric relation as for (a, b) tuple, (b, a) tuple is not present.
Properties of Symmetric Relation
- Empty relation on any set is always symmetric.
- Universal relations are always symmetric.
- If the relation is reflexive/irreflexive then it need not be symmetric.
How to verify Symmetric Relation?
To verify a symmetric relation do the following:
- Manually check for the existence of every bRa tuple for every aRb tuple in the relation.
- If any of the tuples does not exist then the relation is not symmetric else it is symmetric.
Follow the below illustration for a better understanding
Illustration:
Consider set A = { 1, 2, 3, 4 } and a relation R = { (1, 2), (1, 3), (2, 1), (3, 4), (3, 1) }
For the pair (1, 2) in R:
=> The reversed pair (2, 1) is present in the relation.
=> This pair satisfies the condition
For the pair (1, 3) in R:
=> The reversed pair (3, 1) is present in the relation.
=> This pair satisfies the condition
For the pair (2, 1) in R:
=> The reversed pair (1, 2) is present in the relation.
=> This pair satisfies the condition
For the pair (3, 4) in R:
=> The reversed pair (4, 3) is not present in the relation.
=> This pair does not satisfy the condition
For the pair (3, 1) in R:
=> The reversed pair (1, 3) is present in the relation.
=> This pair satisfies the condition
As the pair (3, 4) does not satisfy the condition, the relation is not symmetric.
Below is the code implementation of the idea:
C++
#include <bits/stdc++.h>
using namespace std;
class Relation {
public:
bool checkSymmetric(set<pair<int, int> > R)
{
if (R.size() == 0) {
return true;
}
for (auto i = R.begin(); i != R.end(); i++) {
auto temp = make_pair(i->second, i->first);
if (R.find(temp) == R.end()) {
return false;
}
}
return true;
}
};
int main()
{
set<pair<int, int> > R;
R.insert(make_pair(1, 1));
R.insert(make_pair(1, 2));
R.insert(make_pair(2, 1));
R.insert(make_pair(2, 3));
R.insert(make_pair(3, 2));
R.insert(make_pair(3, 4));
Relation obj;
if (obj.checkSymmetric(R)) {
cout << "Symmetric Relation" << endl;
}
else {
cout << "Not a Symmetric Relation" << endl;
}
return 0;
}
|
Java
import java.io.*;
import java.util.*;
class pair {
int first, second;
pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
class GFG {
static class Relation {
boolean checkSymmetric(Set<pair> R)
{
if (R.size() == 0) {
return true;
}
for (var i : R) {
pair temp = new pair(i.second, i.first);
if (!R.contains(temp)) {
return false;
}
}
return true;
}
}
public static void main(String[] args)
{
Set<pair> R = new HashSet<>();
R.add(new pair(1, 1));
R.add(new pair(1, 2));
R.add(new pair(2, 1));
R.add(new pair(2, 3));
R.add(new pair(3, 2));
R.add(new pair(3, 4));
Relation obj = new Relation();
if (obj.checkSymmetric(R)) {
System.out.println("Symmetric Relation");
}
else {
System.out.println("Not a Symmetric Relation");
}
}
}
|
Python3
class Relation:
def checkSymmetric(self, R):
if len(R) == 0:
return True
for i in R:
if (i[1], i[0]) not in R:
return False
return True
if __name__ == '__main__':
R = {(1, 1), (1, 2), (2, 1), (2, 3), (3, 2), (3, 4)}
obj = Relation()
if obj.checkSymmetric(R):
print("Symmetric Relation")
else:
print("Not a Symmetric Relation")
|
C#
using System;
using System.Collections.Generic;
class pair {
public int first, second;
public pair(int first, int second)
{
this.first = first;
this.second = second;
}
}
public class GFG {
class Relation {
public bool checkSymmetric(HashSet<pair> R)
{
if (R.Count == 0) {
return true;
}
foreach(var i in R)
{
pair temp = new pair(i.second, i.first);
if (!R.Contains(temp)) {
return false;
}
}
return true;
}
}
static public void Main()
{
HashSet<pair> R = new HashSet<pair>();
R.Add(new pair(1, 1));
R.Add(new pair(1, 2));
R.Add(new pair(2, 1));
R.Add(new pair(2, 3));
R.Add(new pair(3, 2));
R.Add(new pair(3, 4));
Relation obj = new Relation();
if (obj.checkSymmetric(R)) {
Console.WriteLine("Symmetric Relation");
}
else {
Console.WriteLine("Not a Symmetric Relation");
}
}
}
|
Javascript
class Relation {
constructor() {}
checkSymmetric(R) {
if (R.size === 0) {
return true;
}
for (const i of R) {
const temp = [i[1], i[0]];
if (!R.has(temp)) {
return false;
}
}
return true;
}
}
function main() {
const R = new Set();
R.add([1, 1]);
R.add([1, 2]);
R.add([2, 1]);
R.add([2, 3]);
R.add([3, 2]);
R.add([3, 4]);
const obj = new Relation();
if (obj.checkSymmetric(R)) {
console.log("Symmetric Relation");
} else {
console.log("Not a Symmetric Relation");
}
}
main();
|
Output
Not a Symmetric Relation
Time Complexity: O(N * log N) where N is the number of tuples in the relation
Auxiliary Space: O(1)
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