# Euler Circuit in a Directed Graph

Eulerian Path is a path in graph that visits every edge exactly once. Eulerian Circuit is an Eulerian Path which starts and ends on the same vertex.

A graph is said to be eulerian if it has a eulerian cycle. We have discussed eulerian circuit for an undirected graph. In this post, the same is discussed for a directed graph.

For example, the following graph has eulerian cycle as {1, 0, 3, 4, 0, 2, 1}

**How to check if a directed graph is eulerian?**

A directed graph has an eulerian cycle if following conditions are true (Source: Wiki)

1) All vertices with nonzero degree belong to a single strongly connected component.

2) In degree is equal to the out degree for every vertex.

We can detect singly connected component using Kosaraju’s DFS based simple algorithm.

To compare in degree and out-degree, we need to store in degree and out-degree of every vertex. Out degree can be obtained by the size of an adjacency list. In degree can be stored by creating an array of size equal to the number of vertices.

Following implementations of above approach.

## C++

// A C++ program to check if a given directed graph is Eulerian or not #include<iostream> #include <list> #define CHARS 26 using namespace std; // A class that represents an undirected graph class Graph { int V; // No. of vertices list<int> *adj; // A dynamic array of adjacency lists int *in; public: // Constructor and destructor Graph(int V); ~Graph() { delete [] adj; delete [] in; } // function to add an edge to graph void addEdge(int v, int w) { adj[v].push_back(w); (in[w])++; } // Method to check if this graph is Eulerian or not bool isEulerianCycle(); // Method to check if all non-zero degree vertices are connected bool isSC(); // Function to do DFS starting from v. Used in isConnected(); void DFSUtil(int v, bool visited[]); Graph getTranspose(); }; Graph::Graph(int V) { this->V = V; adj = new list<int>[V]; in = new int[V]; for (int i = 0; i < V; i++) in[i] = 0; } /* This function returns true if the directed graph has a eulerian cycle, otherwise returns false */ bool Graph::isEulerianCycle() { // Check if all non-zero degree vertices are connected if (isSC() == false) return false; // Check if in degree and out degree of every vertex is same for (int i = 0; i < V; i++) if (adj[i].size() != in[i]) return false; return true; } // A recursive function to do DFS starting from v 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); } // Function that returns reverse (or transpose) of this graph // This function is needed in isSC() Graph Graph::getTranspose() { Graph g(V); for (int v = 0; v < V; v++) { // Recur for all the vertices adjacent to this vertex list<int>::iterator i; for(i = adj[v].begin(); i != adj[v].end(); ++i) { g.adj[*i].push_back(v); (g.in[v])++; } } return g; } // This function returns true if all non-zero degree vertices of // graph are strongly connected (Please refer // https://www.geeksforgeeks.org/connectivity-in-a-directed-graph/ ) bool Graph::isSC() { // Mark all the vertices as not visited (For first DFS) bool visited[V]; for (int i = 0; i < V; i++) visited[i] = false; // Find the first vertex with non-zero degree int n; for (n = 0; n < V; n++) if (adj[n].size() > 0) break; // Do DFS traversal starting from first non zero degrees vertex. DFSUtil(n, visited); // If DFS traversal doesn't visit all vertices, then return false. for (int i = 0; i < V; i++) if (adj[i].size() > 0 && visited[i] == false) return false; // Create a reversed graph Graph gr = getTranspose(); // Mark all the vertices as not visited (For second DFS) for (int i = 0; i < V; i++) visited[i] = false; // Do DFS for reversed graph starting from first vertex. // Starting Vertex must be same starting point of first DFS gr.DFSUtil(n, visited); // If all vertices are not visited in second DFS, then // return false for (int i = 0; i < V; i++) if (adj[i].size() > 0 && visited[i] == false) return false; return true; } // Driver program to test above functions int main() { // Create a graph given in the above diagram Graph g(5); g.addEdge(1, 0); g.addEdge(0, 2); g.addEdge(2, 1); g.addEdge(0, 3); g.addEdge(3, 4); g.addEdge(4, 0); if (g.isEulerianCycle()) cout << "Given directed graph is eulerian n"; else cout << "Given directed graph is NOT eulerian n"; return 0; }

## Java

// A Java program to check if a given directed graph is Eulerian or not // A class that represents an undirected graph import java.io.*; import java.util.*; import java.util.LinkedList; // This class represents a directed graph using adjacency list class Graph { private int V; // No. of vertices private LinkedList<Integer> adj[];//Adjacency List private int in[]; //maintaining in degree //Constructor Graph(int v) { V = v; adj = new LinkedList[v]; in = new int[V]; for (int i=0; i<v; ++i) { adj[i] = new LinkedList(); in[i] = 0; } } //Function to add an edge into the graph void addEdge(int v,int w) { adj[v].add(w); in[w]++; } // A recursive function to print DFS starting from v void DFSUtil(int v,Boolean visited[]) { // Mark the current node as visited visited[v] = true; int n; // Recur for all the vertices adjacent to this vertex Iterator<Integer> i =adj[v].iterator(); while (i.hasNext()) { n = i.next(); if (!visited[n]) DFSUtil(n,visited); } } // Function that returns reverse (or transpose) of this graph Graph getTranspose() { Graph g = new Graph(V); for (int v = 0; v < V; v++) { // Recur for all the vertices adjacent to this vertex Iterator<Integer> i = adj[v].listIterator(); while (i.hasNext()) { g.adj[i.next()].add(v); (g.in[v])++; } } return g; } // The main function that returns true if graph is strongly // connected Boolean isSC() { // Step 1: Mark all the vertices as not visited (For // first DFS) Boolean visited[] = new Boolean[V]; for (int i = 0; i < V; i++) visited[i] = false; // Step 2: Do DFS traversal starting from the first vertex. DFSUtil(0, visited); // If DFS traversal doesn't visit all vertices, then return false. for (int i = 0; i < V; i++) if (visited[i] == false) return false; // Step 3: Create a reversed graph Graph gr = getTranspose(); // Step 4: Mark all the vertices as not visited (For second DFS) for (int i = 0; i < V; i++) visited[i] = false; // Step 5: Do DFS for reversed graph starting from first vertex. // Starting Vertex must be same starting point of first DFS gr.DFSUtil(0, visited); // If all vertices are not visited in second DFS, then // return false for (int i = 0; i < V; i++) if (visited[i] == false) return false; return true; } /* This function returns true if the directed graph has a eulerian cycle, otherwise returns false */ Boolean isEulerianCycle() { // Check if all non-zero degree vertices are connected if (isSC() == false) return false; // Check if in degree and out degree of every vertex is same for (int i = 0; i < V; i++) if (adj[i].size() != in[i]) return false; return true; } public static void main (String[] args) throws java.lang.Exception { Graph g = new Graph(5); g.addEdge(1, 0); g.addEdge(0, 2); g.addEdge(2, 1); g.addEdge(0, 3); g.addEdge(3, 4); g.addEdge(4, 0); if (g.isEulerianCycle()) System.out.println("Given directed graph is eulerian "); else System.out.println("Given directed graph is NOT eulerian "); } } //This code is contributed by Aakash Hasija

## Python3

# A Python3 program to check if a given # directed graph is Eulerian or not from collections import defaultdict class Graph(): def __init__(self, vertices): self.V = vertices self.graph = defaultdict(list) self.IN = [0] * vertices def addEdge(self, v, u): self.graph[v].append(u) self.IN[u] += 1 def DFSUtil(self, v, visited): visited[v] = True for node in self.graph[v]: if visited[node] == False: self.DFSUtil(node, visited) def getTranspose(self): gr = Graph(self.V) for node in range(self.V): for child in self.graph[node]: gr.addEdge(child, node) return gr def isSC(self): visited = [False] * self.V v = 0 for v in range(self.V): if len(self.graph[v]) > 0: break self.DFSUtil(v, visited) # If DFS traversal doesn't visit all # vertices, then return false. for i in range(self.V): if visited[i] == False: return False gr = self.getTranspose() visited = [False] * self.V gr.DFSUtil(v, visited) for i in range(self.V): if visited[i] == False: return False return True def isEulerianCycle(self): # Check if all non-zero degree vertices # are connected if self.isSC() == False: return False # Check if in degree and out degree of # every vertex is same for v in range(self.V): if len(self.graph[v]) != self.IN[v]: return False return True g = Graph(5); g.addEdge(1, 0); g.addEdge(0, 2); g.addEdge(2, 1); g.addEdge(0, 3); g.addEdge(3, 4); g.addEdge(4, 0); if g.isEulerianCycle(): print( "Given directed graph is eulerian"); else: print( "Given directed graph is NOT eulerian"); # This code is contributed by Divyanshu Mehta

## C#

// A C# program to check if a given // directed graph is Eulerian or not // A class that represents an // undirected graph using System; using System.Collections.Generic; // This class represents a directed // graph using adjacency list class Graph{ // No. of vertices public int V; // Adjacency List public List<int> []adj; // Maintaining in degree public int []init; // Constructor Graph(int v) { V = v; adj = new List<int>[v]; init = new int[V]; for(int i = 0; i < v; ++i) { adj[i] = new List<int>(); init[i] = 0; } } // Function to add an edge into the graph void addEdge(int v, int w) { adj[v].Add(w); init[w]++; } // A recursive function to print DFS // starting from v void DFSUtil(int v, Boolean []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]) { if (!visited[i]) DFSUtil(i, visited); } } // Function that returns reverse // (or transpose) of this graph Graph getTranspose() { Graph g = new Graph(V); for(int v = 0; v < V; v++) { // Recur for all the vertices // adjacent to this vertex foreach(int i in adj[v]) { g.adj[i].Add(v); (g.init[v])++; } } return g; } // The main function that returns // true if graph is strongly connected Boolean isSC() { // Step 1: Mark all the vertices // as not visited (For first DFS) Boolean []visited = new Boolean[V]; for(int i = 0; i < V; i++) visited[i] = false; // Step 2: Do DFS traversal starting // from the first vertex. DFSUtil(0, visited); // If DFS traversal doesn't visit // all vertices, then return false. for(int i = 0; i < V; i++) if (visited[i] == false) return false; // Step 3: Create a reversed graph Graph gr = getTranspose(); // Step 4: Mark all the vertices as // not visited (For second DFS) for(int i = 0; i < V; i++) visited[i] = false; // Step 5: Do DFS for reversed graph // starting from first vertex. // Staring Vertex must be same // starting point of first DFS gr.DFSUtil(0, visited); // If all vertices are not visited // in second DFS, then return false for(int i = 0; i < V; i++) if (visited[i] == false) return false; return true; } // This function returns true if the // directed graph has a eulerian // cycle, otherwise returns false Boolean isEulerianCycle() { // Check if all non-zero degree // vertices are connected if (isSC() == false) return false; // Check if in degree and out // degree of every vertex is same for(int i = 0; i < V; i++) if (adj[i].Count != init[i]) return false; return true; } // Driver code public static void Main(String[] args) { Graph g = new Graph(5); g.addEdge(1, 0); g.addEdge(0, 2); g.addEdge(2, 1); g.addEdge(0, 3); g.addEdge(3, 4); g.addEdge(4, 0); if (g.isEulerianCycle()) Console.WriteLine("Given directed " + "graph is eulerian "); else Console.WriteLine("Given directed " + "graph is NOT eulerian "); } } // This code is contributed by Princi Singh

## Javascript

<script> // A Javascript program to check if a given directed graph is Eulerian or not // This class represents a directed graph using adjacency // list representation class Graph { // Constructor constructor(v) { this.V = v; this.adj = new Array(v); this.in=new Array(v); for (let i=0; i<v; ++i) { this.adj[i] = []; this.in[i]=0; } } //Function to add an edge into the graph addEdge(v,w) { this.adj[v].push(w); this.in[w]++; } // A recursive function to print DFS starting from v DFSUtil(v,visited) { // Mark the current node as visited visited[v] = true; let n; // Recur for all the vertices adjacent to this vertex for(let i of this.adj[v]) { n = i; if (!visited[n]) this.DFSUtil(n,visited); } } // Function that returns reverse (or transpose) of this graph getTranspose() { let g = new Graph(this.V); for (let v = 0; v < this.V; v++) { // Recur for all the vertices adjacent to this vertex for(let i of this.adj[v]) { g.adj[i].push(v); (g.in[v])++; } } return g; } // The main function that returns true if graph is strongly // connected isSC() { // Step 1: Mark all the vertices as not visited (For // first DFS) let visited = new Array(this.V); for (let i = 0; i < this.V; i++) visited[i] = false; // Step 2: Do DFS traversal starting from the first vertex. this.DFSUtil(0, visited); // If DFS traversal doesn't visit all vertices, then return false. for (let i = 0; i < this.V; i++) if (visited[i] == false) return false; // Step 3: Create a reversed graph let gr = this.getTranspose(); // Step 4: Mark all the vertices as not visited (For second DFS) for (let i = 0; i < this.V; i++) visited[i] = false; // Step 5: Do DFS for reversed graph starting from first vertex. // Starting Vertex must be same starting point of first DFS gr.DFSUtil(0, visited); // If all vertices are not visited in second DFS, then // return false for (let i = 0; i < this.V; i++) if (visited[i] == false) return false; return true; } /* This function returns true if the directed graph has a eulerian cycle, otherwise returns false */ isEulerianCycle() { // Check if all non-zero degree vertices are connected if (this.isSC() == false) return false; // Check if in degree and out degree of every vertex is same for (let i = 0; i < this.V; i++) if (this.adj[i].length != this.in[i]) return false; return true; } } let g = new Graph(5); g.addEdge(1, 0); g.addEdge(0, 2); g.addEdge(2, 1); g.addEdge(0, 3); g.addEdge(3, 4); g.addEdge(4, 0); if (g.isEulerianCycle()) document.write("Given directed graph is eulerian "); else document.write("Given directed graph is NOT eulerian "); // This code is contributed by avanitrachhadiya2155 </script>

**Output: **

Given directed graph is eulerian

Time complexity of the above implementation is O(V + E) as Kosaraju’s algorithm takes O(V + E) time. After running Kosaraju’s algorithm we traverse all vertices and compare in degree with out degree which takes O(V) time.

See following as an application of this.

Find if the given array of strings can be chained to form a circle.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.