There are n nodes and m bridges in between these nodes. Print the possible path through each node using each edges (if possible), traveling through each edges only once.
Examples :
Input : [[0, 1, 0, 0, 1], [1, 0, 1, 1, 0], [0, 1, 0, 1, 0], [0, 1, 1, 0, 0], [1, 0, 0, 0, 0]] Output : 5 -> 1 -> 2 -> 4 -> 3 -> 2 Input : [[0, 1, 0, 1, 1], [1, 0, 1, 0, 1], [0, 1, 0, 1, 1], [1, 1, 1, 0, 0], [1, 0, 1, 0, 0]] Output : "No Solution"
It is one of the famous problems in Graph Theory and known as problem of “Seven Bridges of Königsberg”. This problem was solved by famous mathematician Leonhard Euler in 1735. This problem is also considered as the beginning of Graph Theory.
The problem back then was that: There was 7 bridges connecting 4 lands around the city of Königsberg in Prussia. Was there any way to start from any of the land and go through each of the bridges once and only once? Please see these wikipedia images for more clarity.
Euler first introduced graph theory to solve this problem. He considered each of the lands as a node of a graph and each bridge in between as an edge in between. Now he calculated if there is any Eulerian Path in that graph. If there is an Eulerian path then there is a solution otherwise not.
Problem here, is a generalized version of the problem in 1735.
Below is the implementation :
// A C++ program print Eulerian Trail in a // given Eulerian or Semi-Eulerian Graph #include <iostream> #include <string.h> #include <algorithm> #include <list> using namespace std;
// A class that represents an undirected graph class Graph
{ // No. of vertices int V;
// A dynamic array of adjacency lists
list< int > *adj;
public :
// Constructor and destructor
Graph( int V)
{
this ->V = V;
adj = new list< int >[V];
}
~Graph()
{
delete [] adj;
}
// functions to add and remove edge
void addEdge( int u, int v)
{
adj[u].push_back(v);
adj[v].push_back(u);
}
void rmvEdge( int u, int v);
// Methods to print Eulerian tour
void printEulerTour();
void printEulerUtil( int s);
// This function returns count of vertices
// reachable from v. It does DFS
int DFSCount( int v, bool visited[]);
// Utility function to check if edge u-v
// is a valid next edge in Eulerian trail or circuit
bool isValidNextEdge( int u, int v);
}; /* The main function that print Eulerian Trail. It first finds an odd degree vertex (if there is any) and then calls printEulerUtil() to print the path */ void Graph::printEulerTour()
{ // Find a vertex with odd degree
int u = 0;
for ( int i = 0; i < V; i++)
if (adj[i].size() & 1)
{
u = i;
break ;
}
// Print tour starting from oddv
printEulerUtil(u);
cout << endl;
} // Print Euler tour starting from vertex u void Graph::printEulerUtil( int u)
{ // Recur for all the vertices adjacent to
// this vertex
list< int >::iterator i;
for (i = adj[u].begin(); i != adj[u].end(); ++i)
{
int v = *i;
// If edge u-v is not removed and it's a
// valid next edge
if (v != -1 && isValidNextEdge(u, v))
{
cout << u << "-" << v << " " ;
rmvEdge(u, v);
printEulerUtil(v);
}
}
} // The function to check if edge u-v can be considered // as next edge in Euler Tout bool Graph::isValidNextEdge( int u, int v)
{ // The edge u-v is valid in one of the following
// two cases:
// 1) If v is the only adjacent vertex of u
int count = 0; // To store count of adjacent vertices
list< int >::iterator i;
for (i = adj[u].begin(); i != adj[u].end(); ++i)
if (*i != -1)
count++;
if (count == 1)
return true ;
// 2) If there are multiple adjacents, then u-v
// is not a bridge
// Do following steps to check if u-v is a bridge
// 2.a) count of vertices reachable from u
bool visited[V];
memset (visited, false , V);
int count1 = DFSCount(u, visited);
// 2.b) Remove edge (u, v) and after removing
// the edge, count vertices reachable from u
rmvEdge(u, v);
memset (visited, false , V);
int count2 = DFSCount(u, visited);
// 2.c) Add the edge back to the graph
addEdge(u, v);
// 2.d) If count1 is greater, then edge (u, v)
// is a bridge
return (count1 > count2)? false : true ;
} // This function removes edge u-v from graph. // It removes the edge by replacing adjacent // vertex value with -1. void Graph::rmvEdge( int u, int v)
{ // Find v in adjacency list of u and replace
// it with -1
list< int >::iterator iv = find(adj[u].begin(),
adj[u].end(), v);
*iv = -1;
// Find u in adjacency list of v and replace
// it with -1
list< int >::iterator iu = find(adj[v].begin(),
adj[v].end(), u);
*iu = -1;
} // A DFS based function to count reachable // vertices from v int Graph::DFSCount( int v, bool visited[])
{ // Mark the current node as visited
visited[v] = true ;
int count = 1;
// Recur for all vertices adjacent to this vertex
list< int >::iterator i;
for (i = adj[v].begin(); i != adj[v].end(); ++i)
if (*i != -1 && !visited[*i])
count += DFSCount(*i, visited);
return count;
} // Driver program to test above function int main()
{ // Let us first create and test
// graphs shown in above figure
Graph g1(4);
g1.addEdge(0, 1);
g1.addEdge(0, 2);
g1.addEdge(1, 2);
g1.addEdge(2, 3);
g1.printEulerTour();
Graph g3(4);
g3.addEdge(0, 1);
g3.addEdge(1, 0);
g3.addEdge(0, 2);
g3.addEdge(2, 0);
g3.addEdge(2, 3);
g3.addEdge(3, 1);
// comment out this line and you will see that
// it gives TLE because there is no possible
// output g3.addEdge(0, 3);
g3.printEulerTour();
return 0;
} |
// A java program print Eulerian Trail in a // given Eulerian or Semi-Eulerian Graph import java.util.*;
public class GFG{
// A class that represents an undirected graph
static class Graph
{
// No. of vertices
int V;
// A dynamic array of adjacency lists
ArrayList<ArrayList<Integer>> adj;
// Constructor
Graph( int V)
{
this .V = V;
adj = new ArrayList<ArrayList<Integer>>();
for ( int i= 0 ; i<V; i++){
adj.add( new ArrayList<Integer>());
}
}
// functions to add and remove edge
void addEdge( int u, int v)
{
adj.get(u).add(v);
adj.get(v).add(u);
}
// This function removes edge u-v from graph.
// It removes the edge by replacing adjacent
// vertex value with -1.
void rmvEdge( int u, int v)
{
// Find v in adjacency list of u and replace
// it with -1
int iv = find(adj.get(u), v);
adj.get(u).set(iv, - 1 );
// Find u in adjacency list of v and replace
// it with -1
int iu = find(adj.get(v), u);
adj.get(v).set(iu, - 1 );
}
int find(ArrayList<Integer> al, int v){
for ( int i= 0 ; i<al.size(); i++){
if (al.get(i) == v){
return i;
}
}
return - 1 ;
}
// Methods to print Eulerian tour
/* The main function that print Eulerian Trail.
It first finds an odd degree vertex (if there is any)
and then calls printEulerUtil() to print the path */
void printEulerTour()
{
// Find a vertex with odd degree
int u = 0 ;
for ( int i = 0 ; i < V; i++){
if (adj.get(i).size() % 2 == 1 )
{
u = i;
break ;
}
}
// Print tour starting from oddv
printEulerUtil(u);
System.out.println();
}
// Print Euler tour starting from vertex u
void printEulerUtil( int u)
{
// Recur for all the vertices adjacent to
// this vertex
for ( int i = 0 ; i<adj.get(u).size(); ++i)
{
int v = adj.get(u).get(i);
// If edge u-v is not removed and it's a
// valid next edge
if (v != - 1 && isValidNextEdge(u, v))
{
System.out.print(u + "-" + v + " " );
rmvEdge(u, v);
printEulerUtil(v);
}
}
}
// This function returns count of vertices
// reachable from v. It does DFS
// A DFS based function to count reachable
// vertices from v
int DFSCount( int v, boolean visited[])
{
// Mark the current node as visited
visited[v] = true ;
int count = 1 ;
// Recur for all vertices adjacent to this vertex
for ( int i = 0 ; i<adj.get(v).size(); ++i){
int u = adj.get(v).get(i);
if (u != - 1 && !visited[u]){
count += DFSCount(u, visited);
}
}
return count;
}
// Utility function to check if edge u-v
// is a valid next edge in Eulerian trail or circuit
// The function to check if edge u-v can be considered
// as next edge in Euler Tout
boolean isValidNextEdge( int u, int v)
{
// The edge u-v is valid in one of the following
// two cases:
// 1) If v is the only adjacent vertex of u
int count = 0 ; // To store count of adjacent vertices
for ( int i = 0 ; i<adj.get(u).size(); ++i)
if (adj.get(u).get(i) != - 1 )
count++;
if (count == 1 )
return true ;
// 2) If there are multiple adjacents, then u-v
// is not a bridge
// Do following steps to check if u-v is a bridge
// 2.a) count of vertices reachable from u
boolean visited[] = new boolean [V];
Arrays.fill(visited, false );
int count1 = DFSCount(u, visited);
// 2.b) Remove edge (u, v) and after removing
// the edge, count vertices reachable from u
rmvEdge(u, v);
Arrays.fill(visited, false );
int count2 = DFSCount(u, visited);
// 2.c) Add the edge back to the graph
addEdge(u, v);
// 2.d) If count1 is greater, then edge (u, v)
// is a bridge
return (count1 > count2)? false : true ;
}
}
// Driver program to test above function
public static void main(String args[])
{
// Let us first create and test
// graphs shown in above figure
Graph g1 = new Graph( 4 );
g1.addEdge( 0 , 1 );
g1.addEdge( 0 , 2 );
g1.addEdge( 1 , 2 );
g1.addEdge( 2 , 3 );
g1.printEulerTour();
Graph g3 = new Graph( 4 );
g3.addEdge( 0 , 1 );
g3.addEdge( 1 , 0 );
g3.addEdge( 0 , 2 );
g3.addEdge( 2 , 0 );
g3.addEdge( 2 , 3 );
g3.addEdge( 3 , 1 );
// comment out this line and you will see that
// it gives TLE because there is no possible
// output g3.addEdge(0, 3);
g3.printEulerTour();
}
} // This code is contributed by adityapande88. |
# A Python program to print Eulerian trail in a # given Eulerian or Semi-Eulerian Graph from collections import defaultdict
class Graph:
# Constructor and destructor
def __init__( self , V):
self .V = V
self .adj = defaultdict( list )
# functions to add and remove edge
def addEdge( self , u, v):
self .adj[u].append(v)
self .adj[v].append(u)
def rmvEdge( self , u, v):
self .adj[u].remove(v)
self .adj[v].remove(u)
# Methods to print Eulerian tour
def printEulerTour( self ):
# Find a vertex with odd degree
u = 0
for i in range ( self .V):
if len ( self .adj[i]) % 2 = = 1 :
u = i
break
# Print tour starting from oddv
self .printEulerUtil(u)
print ()
def printEulerUtil( self , u):
# Recur for all the vertices adjacent to this vertex
for v in self .adj[u]:
# If edge u-v is not removed and it's a valid next edge
if v ! = - 1 and self .isValidNextEdge(u, v):
print (u, "-" , v, " " , end = "")
self .rmvEdge(u, v)
self .printEulerUtil(v)
# The function to check if edge u-v can be considered
# as next edge in Euler Tout
def isValidNextEdge( self , u, v):
# The edge u-v is valid in one of the following
# two cases:
# 1) If v is the only adjacent vertex of u
count = 0 # To store count of adjacent vertices
for i in self .adj[u]:
if i ! = - 1 :
count + = 1
if count = = 1 :
return True
# 2) If there are multiple adjacents, then u-v is not a bridge
# Do following steps to check if u-v is a bridge
# 2.a) count of vertices reachable from u
visited = [ False ] * ( self .V)
count1 = self .DFSCount(u, visited)
# 2.b) Remove edge (u, v) and after removing
# the edge, count vertices reachable from u
self .rmvEdge(u, v)
visited = [ False ] * ( self .V)
count2 = self .DFSCount(u, visited)
# 2.c) Add the edge back to the graph
self .addEdge(u, v)
# 2.d) If count1 is greater, then edge (u, v) is a bridge
return False if count1 > count2 else True
# A DFS based function to count reachable vertices from v
def DFSCount( self , v, visited):
# Mark the current node as visited
visited[v] = True
count = 1
# Recur for all the vertices adjacent to this vertex
for i in self .adj[v]:
if not visited[i]:
count + = self .DFSCount(i, visited)
return count
# utility function to form edge between two vertices
# source and dest
def makeEdge(src, dest):
graph.addEdge(src, dest)
# Driver program to test above functions def main():
# Let us first create and test
# graphs shown in above figure
g1 = Graph( 4 )
g1.addEdge( 0 , 1 )
g1.addEdge( 0 , 2 )
g1.addEdge( 1 , 2 )
g1.addEdge( 2 , 3 )
g1.printEulerTour()
g3 = Graph( 4 )
g3.addEdge( 0 , 1 )
g3.addEdge( 1 , 0 )
g3.addEdge( 0 , 2 )
g3.addEdge( 2 , 0 )
g3.addEdge( 2 , 3 )
g3.addEdge( 3 , 1 )
# comment out this line and you will see that
# it gives TLE because there is no possible
# output g3.addEdge(0, 3);
g3.printEulerTour()
if __name__ = = "__main__" :
main()
# This code is contributed by vikramshirsath177
|
// A Javascript program print Eulerian Trail in a given // Eulerian or Semi-Eulerian Graph // A class that represents an undirected graph
class Graph {
constructor(V) {
this .V = V; // No. of vertices
//A dynamic array of adjacency lists
this .adj = Array.from(Array(V), () => new Array());
}
// functions to add and remove edge
addEdge(u, v) {
this .adj[u].push(v);
this .adj[v].push(u);
}
// This function removes edge u-v from graph.
// It removes the edge by replacing adjacent
// vertex value with -1.
rmvEdge(u, v) {
// Find v in adjacency list of u and replace
// it with -1
for (let i = 0; i < this .adj[u].length; i++) {
if ( this .adj[u][i] == v) {
this .adj[u][i] = -1;
break ;
}
}
// Find u in adjacency list of v and replace
// it with -1
for (let i = 0; i < this .adj[v].length; i++) {
if ( this .adj[v][i] == u) {
this .adj[v][i] = -1;
break ;
}
}
}
// Methods to print Eulerian tour
printEulerTour() {
// Find a vertex with odd degree
let u = 0;
for (let i = 0; i < this .V; i++)
if ( this .adj[i].length & 1) {
u = i;
break ;
}
// Print tour starting from oddv
this .printEulerUtil(u);
console.log( " " );
}
//Print Euler tour starting from vertex u
printEulerUtil(u) {
{
// Recur for all the vertices adjacent to
// this vertex
for (let j in this .adj[u]) {
let v = this .adj[u][j];
// If edge u-v is not removed and it's a
// valid next edge
if (v != -1 && this .isValidNextEdge(u, v)) {
console.log(u + "-" + v);
this .rmvEdge(u, v);
this .printEulerUtil(v);
}
}
}
}
// This function returns count of vertices
// reachable from v. It does DFS
DFSCount(v, visited) {
// Mark the current node as visited
visited[v] = true ;
let count = 1;
// Recur for all vertices adjacent to this vertex
for (let j in this .adj[v]) {
let i = this .adj[v][j];
if (i != -1 && !visited[i]) count += this .DFSCount(i, visited);
}
return count;
}
// The function to check if edge u-v can be considered
// as next edge in Euler Tout
isValidNextEdge(u, v) {
// The edge u-v is valid in one of the following
// two cases:
// 1) If v is the only adjacent vertex of u
let count = 0; // To store count of adjacent vertices
for (let j in this .adj[u]) {
let i = this .adj[u][j];
if (i != -1) count++;
}
if (count == 1) return true ;
// 2) If there are multiple adjacents, then u-v
// is not a bridge
// Do following steps to check if u-v is a bridge
// 2.a) count of vertices reachable from u
let visited = new Array( this .V);
visited.fill( false );
let count1 = this .DFSCount(u, visited);
// 2.b) Remove edge (u, v) and after removing
// the edge, count vertices reachable from u
this .rmvEdge(u, v);
visited.fill( false );
let count2 = this .DFSCount(u, visited);
// 2.c) Add the edge back to the graph
this .addEdge(u, v);
// 2.d) If count1 is greater, then edge (u, v)
// is a bridge
return count1 > count2 ? false : true ;
}
}
// Driver program to test above function
// Let us first create and test
// graphs shown in above figure
let g1 = new Graph(4);
g1.addEdge(0, 1);
g1.addEdge(0, 2);
g1.addEdge(1, 2);
g1.addEdge(2, 3);
g1.printEulerTour();
let g3 = new Graph(4);
g3.addEdge(0, 1);
g3.addEdge(1, 0);
g3.addEdge(0, 2);
g3.addEdge(2, 0);
g3.addEdge(2, 3);
g3.addEdge(3, 1);
// comment out this line and you will see that
// it gives TLE because there is no possible
// output g3.addEdge(0, 3);
g3.printEulerTour();
|
// A C# program print Eulerian Trail in a // given Eulerian or Semi-Eulerian Graph using System;
using System.Collections.Generic;
class GFG {
// A class that represents an undirected graph
class Graph {
// No. of vertices
int V;
// A dynamic array of adjacency lists
List<List< int > > adj;
// Constructor
public Graph( int V)
{
this .V = V;
adj = new List<List< int > >();
for ( int i = 0; i < V; i++) {
adj.Add( new List< int >());
}
}
public void addEdge( int u, int v)
{
adj[u].Add(v);
adj[v].Add(u);
}
// This function removes edge u-v from graph.
// It removes the edge by replacing adjacent
// vertex value with -1.
public void rmvEdge( int u, int v)
{
// Find v in adjacency list of u and replace
// it with -1
int iv = adj[u].IndexOf(v);
adj[u][iv] = -1;
// Find u in adjacency list of v and replace
// it with -1
int iu = adj[v].IndexOf(u);
adj[v][iu] = -1;
}
int find(List< int > al, int v)
{
for ( int i = 0; i < al.Count; i++) {
if (al[i] == v) {
return i;
}
}
return -1;
}
// Methods to print Eulerian tour
/* The main function that print Eulerian Trail.
It first finds an odd degree vertex (if there is any)
and then calls printEulerUtil() to print the path */
public void printEulerTour()
{
// Find a vertex with odd degree
int u = 0;
for ( int i = 0; i < V; i++) {
if (adj[i].Count % 2 == 1) {
u = i;
break ;
}
}
// Print tour starting from oddv
printEulerUtil(u);
Console.WriteLine();
}
// Print Euler tour starting from vertex u
void printEulerUtil( int u)
{
// Recur for all the vertices adjacent to
// this vertex
for ( int i = 0; i < adj[u].Count; ++i) {
int v = adj[u][i];
// If edge u-v is not removed and it's a
// valid next edge
if (v != -1 && isValidNextEdge(u, v)) {
Console.Write(u + "-" + v + " " );
rmvEdge(u, v);
printEulerUtil(v);
}
}
}
// This function returns count of vertices
// reachable from v. It does DFS
// A DFS based function to count reachable
// vertices from v
int DFSCount( int v, bool [] visited)
{
// Mark the current node as visited
visited[v] = true ;
int count = 1;
// Recur for all vertices adjacent to this vertex
for ( int i = 0; i < adj[v].Count; ++i) {
int u = adj[v][i];
if (u != -1 && !visited[u]) {
count += DFSCount(u, visited);
}
}
return count;
}
// Utility function to check if edge u-v
// is a valid next edge in Eulerian trail or circuit
// The function to check if edge u-v can be considered
// as next edge in Euler Tout
bool isValidNextEdge( int u, int v)
{
// The edge u-v is valid in one of the following
// two cases:
// 1) If v is the only adjacent vertex of u
int count = 0; // To store count of adjacent vertices
for ( int i = 0; i < adj[u].Count; ++i)
if (adj[u][i] != -1)
count++;
if (count == 1)
return true ;
// 2) If there are multiple adjacents, then u-v
// is not a bridge
// Do following steps to check if u-v is a bridge
// 2.a) count of vertices reachable from u
bool [] visited = new bool [V];
Array.Fill(visited, false );
int count1 = DFSCount(u, visited);
// 2.b) Remove edge (u, v) and after removing
// the edge, count vertices reachable from u
rmvEdge(u, v);
Array.Fill(visited, false );
int count2 = DFSCount(u, visited);
// 2.c) Add the edge back to the graph
addEdge(u, v);
// 2.d) If count1 is greater, then edge (u, v)
// is a bridge
return (count1 > count2) ? false : true ;
}
}
// Driver program to test above function
public static void Main()
{
// Let us first create and test
// graphs shown in above figure
Graph g1 = new Graph(4);
g1.addEdge(0, 1);
g1.addEdge(0, 2);
g1.addEdge(1, 2);
g1.addEdge(2, 3);
g1.printEulerTour();
Graph g3 = new Graph(4);
g3.addEdge(0, 1);
g3.addEdge(1, 0);
g3.addEdge(0, 2);
g3.addEdge(2, 0);
g3.addEdge(2, 3);
g3.addEdge(3, 1);
// comment out this line and you will see that
// it gives TLE because there is no possible
// output g3.addEdge(0, 3);
g3.printEulerTour();
}
} // This code is contributed by Prajwal Kandekar |
2-0 0-1 1-2 2-3 1-0 0-2 2-3 3-1 1-0 0-2
Time Complexity: O(V+E)
The time complexity of the above algorithm is O(V+E) where V is the number of vertices and E is the number of edges in the graph.
Space Complexity: O(V+E)
The space complexity of the above algorithm is O(V+E) where V is the number of vertices and E is the number of edges in the graph.