A Peterson Graph Problem

The following graph G is called a Petersen graph and its vertices have been numbered from 0 to 9. Some letters have also been assigned to vertices of G, as can be seen from the following picture:
Let’s consider a walk W in graph G, which consists of L vertices W1, W2, …, WL. A string S of L letters 'A' – 'E' is realized by walk W if the sequence of letters written along W is equal to S. Vertices can be visited multiple times while walking along W.

For example, S = 'ABBECCD' is realized by W = (0, 1, 6, 9, 7, 2, 3). Determine whether there is a walk W which realizes a given string S in graph G, and if so then find the lexicographically least such walk. The only line of input contains one string S. If there is no walk W which realizes S, then output -1 otherwise, you should output the least lexicographical walk W which realizes S.

Examples:

Input : s = 'ABB'
Output: 016
Explanation: As we can see in the graph
             the path from ABB is 016.
Input : s = 'AABE'
Output :-1
Explanation: As there is no path that
             exists, hence output is -1.



We apply breadth first search to visit each vertex of the graph.

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// C++ program to find the
// path in Peterson graph
#include <bits/stdc++.h>
using namespace std;
  
// path to be checked 
char S[100005]; 
  
// adjacency matrix. 
bool adj[10][10];
  
// resulted path - way 
char result[100005];
  
// we are applying breadth first search
// here
bool findthepath(char* S, int v)
{
    result[0] = v + '0';
    for (int i = 1; S[i]; i++) {
          
        // first traverse the outer graph
        if (adj[v][S[i] - 'A'] || adj[S[i] -
                              'A'][v]) {
            v = S[i] - 'A';
        }
  
        // then traverse the inner graph
        else if (adj[v][S[i] - 'A' + 5] || 
                 adj[S[i] - 'A' + 5][v]) {
            v = S[i] - 'A' + 5;
        }
  
        // if the condition failed to satisfy
        // return false
        else 
            return false;
  
        result[i] = v + '0';
    }
  
    return true;
}
  
// driver code
int main()
{
    // here we have used adjacency matrix to make
    // connections between the connected nodes
    adj[0][1] = adj[1][2] = adj[2][3] = adj[3][4] = 
    adj[4][0] = adj[0][5] = adj[1][6] = adj[2][7] =
    adj[3][8] = adj[4][9] = adj[5][7] = adj[7][9] =
    adj[9][6] = adj[6][8] = adj[8][5] = true;
      
    // path to be checked
    char S[] = "ABB";
      
    if (findthepath(S, S[0] - 'A') || 
        findthepath(S, S[0] - 'A' + 5)) {
        cout << result;
    } else {
        cout << "-1";
    }
    return 0;
}

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Output:

016

This article is contributed by Sunidhi Chaudhary. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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