De Bruijn sequence | Set 1

Given an integer n and a set of characters A of size k, find a string S such that every possible string on A of length n appears exactly once as a substring in S. Such a string is called de Bruijn sequence.

Examples:

Input: n = 3, k = 2, A = {0, 1)
Output: 0011101000
All possible strings of length three (000, 001, 010, 011, 100, 101, 110 and 111) appear exactly once as sub-strings in A.



Input: n = 2, k = 2, A = {0, 1)
Output: 01100

Approach:
We can solve this problem by constructing a directed graph with kn-1 nodes with each node having k outgoing edges. Each node corresponds to a string of size n-1. Every edge corresponds to one of the k characters in A and adds that character to the starting string.
For example, if n=3 and k=2, then we construct the following graph:

  • The node ’01’ is connected to node ’11’ through edge ‘1’, as adding ‘1’ to ’01’ (and removing the first character) gives us ’11’.
  • We can observe that every node in this graph has equal in-degree and out-degree, which means that a Eulerian circuit exists in this graph.
  • The Eulerian circuit will correspond to a de Bruijn sequence as every combination of a node and an outgoing edge represents a unique string of length n.
  • The de Bruijn sequence will contain the characters of the starting node and the characters of all the edges in the order they are traversed in.
  • Therefore the length of the string will be kn+n-1. We will use Hierholzer’s Algorithm to find the Eulerian circuit. The time complexity of this approach is O(kn).

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of
// the above approach
#include <bits/stdc++.h>
using namespace std;
  
unordered_set<string> seen;
vector<int> edges;
  
// Modified DFS in which no edge
// is traversed twice
void dfs(string node, int& k, string& A)
{
    for (int i = 0; i < k; ++i) {
        string str = node + A[i];
        if (seen.find(str) == seen.end()) {
            seen.insert(str);
            dfs(str.substr(1), k, A);
            edges.push_back(i);
        }
    }
}
  
// Function to find a de Bruijn sequence
// of order n on k characters
string deBruijn(int n, int k, string A)
{
  
    // Clearing global variables
    seen.clear();
    edges.clear();
  
    string startingNode = string(n - 1, A[0]);
    dfs(startingNode, k, A);
  
    string S;
  
    // Number of edges
    int l = pow(k, n);
    for (int i = 0; i < l; ++i)
        S += A[edges[i]];
    S += startingNode;
  
    return S;
}
  
// Driver code
int main()
{
    int n = 3, k = 2;
    string A = "01";
  
    cout << deBruijn(n, k, A);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of
// the above approach
import java.util.*;
  
class GFG
{
  
    static Set<String> seen = new HashSet<String>();
    static Vector<Integer> edges = new Vector<Integer>();
  
    // Modified DFS in which no edge
    // is traversed twice
    static void dfs(String node, int k, String A) 
    {
        for (int i = 0; i < k; ++i)
        {
            String str = node + A.charAt(i);
            if (!seen.contains(str))
            {
                seen.add(str);
                dfs(str.substring(1), k, A);
                edges.add(i);
            }
        }
    }
  
    // Function to find a de Bruijn sequence
    // of order n on k characters
    static String deBruijn(int n, int k, String A)
    {
  
        // Clearing global variables
        seen.clear();
        edges.clear();
  
        String startingNode = string(n - 1, A.charAt(0));
        dfs(startingNode, k, A);
  
        String S = "";
  
        // Number of edges
        int l = (int) Math.pow(k, n);
        for (int i = 0; i < l; ++i)
            S += A.charAt(edges.get(i));
        S += startingNode;
  
        return S;
    }
  
    private static String string(int n, char charAt)
    {
        String str = "";
        for (int i = 0; i < n; i++)
            str += charAt;
        return str;
    }
  
    // Driver code
    public static void main(String[] args) 
    {
        int n = 3, k = 2;
        String A = "01";
  
        System.out.print(deBruijn(n, k, A));
    }
}
  
// This code is contributed by 29AjayKumar

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of 
// the above approach 
using System;
using System.Collections.Generic;
  
class GFG 
  
    static HashSet<String> seen = new HashSet<String>(); 
    static List<int> edges = new List<int>(); 
  
    // Modified DFS in which no edge 
    // is traversed twice 
    static void dfs(String node, int k, String A) 
    
        for (int i = 0; i < k; ++i) 
        
            String str = node + A[i]; 
            if (!seen.Contains(str)) 
            
                seen.Add(str); 
                dfs(str.Substring(1), k, A); 
                edges.Add(i); 
            
        
    
  
    // Function to find a de Bruijn sequence 
    // of order n on k characters 
    static String deBruijn(int n, int k, String A) 
    
  
        // Clearing global variables 
        seen.Clear(); 
        edges.Clear(); 
  
        String startingNode = strings(n - 1, A[0]); 
        dfs(startingNode, k, A); 
  
        String S = ""
  
        // Number of edges 
        int l = (int) Math.Pow(k, n); 
        for (int i = 0; i < l; ++i) 
            S += A[edges[i]]; 
        S += startingNode; 
  
        return S; 
    
  
    private static String strings(int n, char charAt) 
    
        String str = ""
        for (int i = 0; i < n; i++) 
            str += charAt; 
        return str; 
    
  
    // Driver code 
    public static void Main(String[] args) 
    
        int n = 3, k = 2; 
        String A = "01"
  
        Console.Write(deBruijn(n, k, A)); 
    
  
// This code is contributed by 29AjayKumar

chevron_right


Output:

0011101000


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

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.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : 29AjayKumar