Eulerian Path is a path in a graph that visits every edge exactly once. Eulerian Circuit is an Eulerian Path that starts and ends on the same vertex.
How to find whether a given graph is Eulerian or not?
The problem is same as following question. “Is it possible to draw a given graph without lifting pencil from the paper and without tracing any of the edges more than once”.
A graph is called Eulerian if it has an Eulerian Cycle and called Semi-Eulerian if it has an Eulerian Path. The problem seems similar to Hamiltonian Path which is NP complete problem for a general graph. Fortunately, we can find whether a given graph has a Eulerian Path or not in polynomial time. In fact, we can find it in O(V+E) time.
Following are some interesting properties of undirected graphs with an Eulerian path and cycle. We can use these properties to find whether a graph is Eulerian or not.
Eulerian Cycle: An undirected graph has Eulerian cycle if following two conditions are true.
- All vertices with non-zero degree are connected. We don’t care about vertices with zero degree because they don’t belong to Eulerian Cycle or Path (we only consider all edges).
- All vertices have even degree.
Eulerian Path: An undirected graph has Eulerian Path if following two conditions are true.
- Same as condition (a) for Eulerian Cycle.
- If zero or two vertices have odd degree and all other vertices have even degree. Note that only one vertex with odd degree is not possible in an undirected graph (sum of all degrees is always even in an undirected graph)
Note that a graph with no edges is considered Eulerian because there are no edges to traverse.
How does this work?
In Eulerian path, each time we visit a vertex v, we walk through two unvisited edges with one end point as v. Therefore, all middle vertices in Eulerian Path must have even degree. For Eulerian Cycle, any vertex can be middle vertex, therefore all vertices must have even degree.
Implementation:
C++
// A C++ program to check if a given graph is Eulerian or not#include<iostream>#include <list>using namespace std;// A class that represents an undirected graphclass Graph{ int V; // No. of vertices list<int> *adj; // A dynamic array of adjacency listspublic: // Constructor and destructor Graph(int V) {this->V = V; adj = new list<int>[V]; } ~Graph() { delete [] adj; } // To avoid memory leak // function to add an edge to graph void addEdge(int v, int w); // Method to check if this graph is Eulerian or not int isEulerian(); // Method to check if all non-zero degree vertices are connected bool isConnected(); // Function to do DFS starting from v. Used in isConnected(); void DFSUtil(int v, bool visited[]);};void Graph::addEdge(int v, int w){ adj[v].push_back(w); adj[w].push_back(v); // Note: the graph is undirected}void Graph::DFSUtil(int v, bool visited[]){ // Mark the current node as visited and print it visited[v] = true; // Recur for all the vertices adjacent to this vertex list<int>::iterator i; for (i = adj[v].begin(); i != adj[v].end(); ++i) if (!visited[*i]) DFSUtil(*i, visited);}// Method to check if all non-zero degree vertices are connected.// It mainly does DFS traversal starting frombool Graph::isConnected(){ // Mark all the vertices as not visited bool visited[V]; int i; for (i = 0; i < V; i++) visited[i] = false; // Find a vertex with non-zero degree for (i = 0; i < V; i++) if (adj[i].size() != 0) break; // If there are no edges in the graph, return true if (i == V) return true; // Start DFS traversal from a vertex with non-zero degree DFSUtil(i, visited); // Check if all non-zero degree vertices are visited for (i = 0; i < V; i++) if (visited[i] == false && adj[i].size() > 0) return false; return true;}/* The function returns one of the following values 0 --> If graph is not Eulerian 1 --> If graph has an Euler path (Semi-Eulerian) 2 --> If graph has an Euler Circuit (Eulerian) */int Graph::isEulerian(){ // Check if all non-zero degree vertices are connected if (isConnected() == false) return 0; // Count vertices with odd degree int odd = 0; for (int i = 0; i < V; i++) if (adj[i].size() & 1) odd++; // If count is more than 2, then graph is not Eulerian if (odd > 2) return 0; // If odd count is 2, then semi-eulerian. // If odd count is 0, then eulerian // Note that odd count can never be 1 for undirected graph return (odd)? 1 : 2;}// Function to run test casesvoid test(Graph &g){ int res = g.isEulerian(); if (res == 0) cout << "graph is not Eulerian\n"; else if (res == 1) cout << "graph has a Euler path\n"; else cout << "graph has a Euler cycle\n";}// Driver program to test above functionint main(){ // Let us create and test graphs shown in above figures Graph g1(5); g1.addEdge(1, 0); g1.addEdge(0, 2); g1.addEdge(2, 1); g1.addEdge(0, 3); g1.addEdge(3, 4); test(g1); Graph g2(5); g2.addEdge(1, 0); g2.addEdge(0, 2); g2.addEdge(2, 1); g2.addEdge(0, 3); g2.addEdge(3, 4); g2.addEdge(4, 0); test(g2); Graph g3(5); g3.addEdge(1, 0); g3.addEdge(0, 2); g3.addEdge(2, 1); g3.addEdge(0, 3); g3.addEdge(3, 4); g3.addEdge(1, 3); test(g3); // Let us create a graph with 3 vertices // connected in the form of cycle Graph g4(3); g4.addEdge(0, 1); g4.addEdge(1, 2); g4.addEdge(2, 0); test(g4); // Let us create a graph with all vertices // with zero degree Graph g5(3); test(g5); return 0;} |
Java
// A Java program to check if a given graph is Eulerian or notimport java.io.*;import java.util.*;import java.util.LinkedList;// This class represents an undirected graph using adjacency list// representationclass Graph{ private int V; // No. of vertices // Array of lists for Adjacency List Representation private LinkedList<Integer> adj[]; // Constructor Graph(int v) { V = v; adj = new LinkedList[v]; for (int i=0; i<v; ++i) adj[i] = new LinkedList(); } //Function to add an edge into the graph void addEdge(int v, int w) { adj[v].add(w);// Add w to v's list. adj[w].add(v); //The graph is undirected } // A function used by DFS void DFSUtil(int v,boolean visited[]) { // Mark the current node as visited visited[v] = true; // Recur for all the vertices adjacent to this vertex Iterator<Integer> i = adj[v].listIterator(); while (i.hasNext()) { int n = i.next(); if (!visited[n]) DFSUtil(n, visited); } } // Method to check if all non-zero degree vertices are // connected. It mainly does DFS traversal starting from boolean isConnected() { // Mark all the vertices as not visited boolean visited[] = new boolean[V]; int i; for (i = 0; i < V; i++) visited[i] = false; // Find a vertex with non-zero degree for (i = 0; i < V; i++) if (adj[i].size() != 0) break; // If there are no edges in the graph, return true if (i == V) return true; // Start DFS traversal from a vertex with non-zero degree DFSUtil(i, visited); // Check if all non-zero degree vertices are visited for (i = 0; i < V; i++) if (visited[i] == false && adj[i].size() > 0) return false; return true; } /* The function returns one of the following values 0 --> If graph is not Eulerian 1 --> If graph has an Euler path (Semi-Eulerian) 2 --> If graph has an Euler Circuit (Eulerian) */ int isEulerian() { // Check if all non-zero degree vertices are connected if (isConnected() == false) return 0; // Count vertices with odd degree int odd = 0; for (int i = 0; i < V; i++) if (adj[i].size()%2!=0) odd++; // If count is more than 2, then graph is not Eulerian if (odd > 2) return 0; // If odd count is 2, then semi-eulerian. // If odd count is 0, then eulerian // Note that odd count can never be 1 for undirected graph return (odd==2)? 1 : 2; } // Function to run test cases void test() { int res = isEulerian(); if (res == 0) System.out.println("graph is not Eulerian"); else if (res == 1) System.out.println("graph has a Euler path"); else System.out.println("graph has a Euler cycle"); } // Driver method public static void main(String args[]) { // Let us create and test graphs shown in above figures Graph g1 = new Graph(5); g1.addEdge(1, 0); g1.addEdge(0, 2); g1.addEdge(2, 1); g1.addEdge(0, 3); g1.addEdge(3, 4); g1.test(); Graph g2 = new Graph(5); g2.addEdge(1, 0); g2.addEdge(0, 2); g2.addEdge(2, 1); g2.addEdge(0, 3); g2.addEdge(3, 4); g2.addEdge(4, 0); g2.test(); Graph g3 = new Graph(5); g3.addEdge(1, 0); g3.addEdge(0, 2); g3.addEdge(2, 1); g3.addEdge(0, 3); g3.addEdge(3, 4); g3.addEdge(1, 3); g3.test(); // Let us create a graph with 3 vertices // connected in the form of cycle Graph g4 = new Graph(3); g4.addEdge(0, 1); g4.addEdge(1, 2); g4.addEdge(2, 0); g4.test(); // Let us create a graph with all vertices // with zero degree Graph g5 = new Graph(3); g5.test(); }}// This code is contributed by Aakash Hasija |
Python3
# Python program to check if a given graph is Eulerian or not#Complexity : O(V+E)from collections import defaultdict# This class represents a undirected graph using adjacency list representationclass Graph: def __init__(self, vertices): self.V = vertices # No. of vertices self.graph = defaultdict(list) # default dictionary to store graph # function to add an edge to graph def addEdge(self, u, v): self.graph[u].append(v) self.graph[v].append(u) # A function used by isConnected def DFSUtil(self, v, visited): # Mark the current node as visited visited[v] = True # Recur for all the vertices adjacent to this vertex for i in self.graph[v]: if visited[i] == False: self.DFSUtil(i, visited) '''Method to check if all non-zero degree vertices are connected. It mainly does DFS traversal starting from node with non-zero degree''' def isConnected(self): # Mark all the vertices as not visited visited = [False]*(self.V) # Find a vertex with non-zero degree for i in range(self.V): if len(self.graph[i]) != 0: break # If there are no edges in the graph, return true if i == self.V-1: return True # Start DFS traversal from a vertex with non-zero degree self.DFSUtil(i, visited) # Check if all non-zero degree vertices are visited for i in range(self.V): if visited[i] == False and len(self.graph[i]) > 0: return False return True '''The function returns one of the following values 0 --> If graph is not Eulerian 1 --> If graph has an Euler path (Semi-Eulerian) 2 --> If graph has an Euler Circuit (Eulerian) ''' def isEulerian(self): # Check if all non-zero degree vertices are connected if self.isConnected() == False: return 0 else: # Count vertices with odd degree odd = 0 for i in range(self.V): if len(self.graph[i]) % 2 != 0: odd += 1 '''If odd count is 2, then semi-eulerian. If odd count is 0, then eulerian If count is more than 2, then graph is not Eulerian Note that odd count can never be 1 for undirected graph''' if odd == 0: return 2 elif odd == 2: return 1 elif odd > 2: return 0 # Function to run test cases def test(self): res = self.isEulerian() if res == 0: print("graph is not Eulerian") elif res == 1: print("graph has a Euler path") else: print("graph has a Euler cycle")# Let us create and test graphs shown in above figuresg1 = Graph(5)g1.addEdge(1, 0)g1.addEdge(0, 2)g1.addEdge(2, 1)g1.addEdge(0, 3)g1.addEdge(3, 4)g1.test()g2 = Graph(5)g2.addEdge(1, 0)g2.addEdge(0, 2)g2.addEdge(2, 1)g2.addEdge(0, 3)g2.addEdge(3, 4)g2.addEdge(4, 0)g2.test()g3 = Graph(5)g3.addEdge(1, 0)g3.addEdge(0, 2)g3.addEdge(2, 1)g3.addEdge(0, 3)g3.addEdge(3, 4)g3.addEdge(1, 3)g3.test()# Let us create a graph with 3 vertices# connected in the form of cycleg4 = Graph(3)g4.addEdge(0, 1)g4.addEdge(1, 2)g4.addEdge(2, 0)g4.test()# Let us create a graph with all vertices# with zero degreeg5 = Graph(3)g5.test()# This code is contributed by Neelam Yadav |
C#
// A C# program to check if a given graph is Eulerian or notusing System;using System.Collections.Generic; // This class represents an undirected graph using adjacency list// representationpublic class Graph{ private int V; // No. of vertices // Array of lists for Adjacency List Representation private List<int> []adj; // Constructor Graph(int v) { V = v; adj = new List<int>[v]; for (int i=0; i<v; ++i) adj[i] = new List<int>(); } //Function to add an edge into the graph void addEdge(int v, int w) { adj[v].Add(w);// Add w to v's list. adj[w].Add(v); //The graph is undirected } // A function used by DFS void DFSUtil(int v,bool []visited) { // Mark the current node as visited visited[v] = true; // Recur for all the vertices adjacent to this vertex foreach(int i in adj[v]){ int n = i; if (!visited[n]) DFSUtil(n, visited); } } // Method to check if all non-zero degree vertices are // connected. It mainly does DFS traversal starting from bool isConnected() { // Mark all the vertices as not visited bool []visited = new bool[V]; int i; for (i = 0; i < V; i++) visited[i] = false; // Find a vertex with non-zero degree for (i = 0; i < V; i++) if (adj[i].Count != 0) break; // If there are no edges in the graph, return true if (i == V) return true; // Start DFS traversal from a vertex with non-zero degree DFSUtil(i, visited); // Check if all non-zero degree vertices are visited for (i = 0; i < V; i++) if (visited[i] == false && adj[i].Count > 0) return false; return true; } /* The function returns one of the following values 0 --> If graph is not Eulerian 1 --> If graph has an Euler path (Semi-Eulerian) 2 --> If graph has an Euler Circuit (Eulerian) */ int isEulerian() { // Check if all non-zero degree vertices are connected if (isConnected() == false) return 0; // Count vertices with odd degree int odd = 0; for (int i = 0; i < V; i++) if (adj[i].Count%2!=0) odd++; // If count is more than 2, then graph is not Eulerian if (odd > 2) return 0; // If odd count is 2, then semi-eulerian. // If odd count is 0, then eulerian // Note that odd count can never be 1 for undirected graph return (odd==2)? 1 : 2; } // Function to run test cases void test() { int res = isEulerian(); if (res == 0) Console.WriteLine("graph is not Eulerian"); else if (res == 1) Console.WriteLine("graph has a Euler path"); else Console.WriteLine("graph has a Euler cycle"); } // Driver method public static void Main(String []args) { // Let us create and test graphs shown in above figures Graph g1 = new Graph(5); g1.addEdge(1, 0); g1.addEdge(0, 2); g1.addEdge(2, 1); g1.addEdge(0, 3); g1.addEdge(3, 4); g1.test(); Graph g2 = new Graph(5); g2.addEdge(1, 0); g2.addEdge(0, 2); g2.addEdge(2, 1); g2.addEdge(0, 3); g2.addEdge(3, 4); g2.addEdge(4, 0); g2.test(); Graph g3 = new Graph(5); g3.addEdge(1, 0); g3.addEdge(0, 2); g3.addEdge(2, 1); g3.addEdge(0, 3); g3.addEdge(3, 4); g3.addEdge(1, 3); g3.test(); // Let us create a graph with 3 vertices // connected in the form of cycle Graph g4 = new Graph(3); g4.addEdge(0, 1); g4.addEdge(1, 2); g4.addEdge(2, 0); g4.test(); // Let us create a graph with all vertices // with zero degree Graph g5 = new Graph(3); g5.test(); }}// This code contributed by PrinciRaj1992 |
Javascript
<script>// A Javascript program to check if a given graph is Eulerian or not// This class represents an undirected graph using adjacency list// representationclass Graph{ // Constructor constructor(v) { this.V = v; this.adj = new Array(v); for (let i = 0; i < v; ++i) this.adj[i] = []; } // Function to add an edge into the graph addEdge(v,w) { this.adj[v].push(w);// Add w to v's list. this.adj[w].push(v); //The graph is undirected } // A function used by DFS DFSUtil(v,visited) { // Mark the current node as visited visited[v] = true; // Recur for all the vertices adjacent to this vertex for(let i of this.adj[v]) { let n = i; if (!visited[n]) this.DFSUtil(n, visited); } } // Method to check if all non-zero degree vertices are // connected. It mainly does DFS traversal starting from isConnected() { // Mark all the vertices as not visited let visited = new Array(this.V); let i; for (i = 0; i < this.V; i++) visited[i] = false; // Find a vertex with non-zero degree for (i = 0; i < this.V; i++) if (this.adj[i].length != 0) break; // If there are no edges in the graph, return true if (i == this.V) return true; // Start DFS traversal from a vertex with non-zero degree this.DFSUtil(i, visited); // Check if all non-zero degree vertices are visited for (i = 0; i < this.V; i++) if (visited[i] == false && this.adj[i].length > 0) return false; return true; } /* The function returns one of the following values 0 --> If graph is not Eulerian 1 --> If graph has an Euler path (Semi-Eulerian) 2 --> If graph has an Euler Circuit (Eulerian) */ isEulerian() { // Check if all non-zero degree vertices are connected if (this.isConnected() == false) return 0; // Count vertices with odd degree let odd = 0; for (let i = 0; i < this.V; i++) if (this.adj[i].length%2!=0) odd++; // If count is more than 2, then graph is not Eulerian if (odd > 2) return 0; // If odd count is 2, then semi-eulerian. // If odd count is 0, then eulerian // Note that odd count can never be 1 for undirected graph return (odd==2)? 1 : 2; } // Function to run test cases test() { let res = this.isEulerian(); if (res == 0) document.write("graph is not Eulerian<br>"); else if (res == 1) document.write("graph has a Euler path<br>"); else document.write("graph has a Euler cycle<br>"); }}// Driver method// Let us create and test graphs shown in above figureslet g1 = new Graph(5);g1.addEdge(1, 0);g1.addEdge(0, 2);g1.addEdge(2, 1);g1.addEdge(0, 3);g1.addEdge(3, 4);g1.test();let g2 = new Graph(5);g2.addEdge(1, 0);g2.addEdge(0, 2);g2.addEdge(2, 1);g2.addEdge(0, 3);g2.addEdge(3, 4);g2.addEdge(4, 0);g2.test();let g3 = new Graph(5);g3.addEdge(1, 0);g3.addEdge(0, 2);g3.addEdge(2, 1);g3.addEdge(0, 3);g3.addEdge(3, 4);g3.addEdge(1, 3);g3.test();// Let us create a graph with 3 vertices// connected in the form of cyclelet g4 = new Graph(3);g4.addEdge(0, 1);g4.addEdge(1, 2);g4.addEdge(2, 0);g4.test();// Let us create a graph with all vertices// with zero degreelet g5 = new Graph(3);g5.test();// This code is contributed by avanitrachhadiya2155</script> |
Output
graph has a Euler path graph has a Euler cycle graph is not Eulerian graph has a Euler cycle graph has a Euler cycle
Time Complexity: O(V+E)
Space Complexity: O(V+E)
Next Articles:
Eulerian Path and Circuit for a Directed Graphs.
Fleury’s Algorithm to print a Eulerian Path or Circuit?
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