Paths to travel each nodes using each edge (Seven Bridges of Königsberg)

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"

Recommended: Please try your approach on {IDE} first, before moving on to the 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 trough 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 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 adjcent 
// 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; 
} 

Output:

2-0  0-1  1-2  2-3  
1-0  0-2  2-3  3-1  1-0  0-2

My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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