I am doing a course on graph algorithms in genome sequencing and the current assignment involves building an overlap graph from a bunch of k-mers. Formally the problem is as follows:

Input: A collection of Patterns of k-mers.
Output: The overlap graph in the form of an adjacency list.

I have a working algorithm that handles most cases except for self-loops. The problem is that I don't know when/why I should be adding self-loops. Any help on this would be greatly appreciated. Below are some examples illustrating my question and my current solution. Thank you!

Example 1:




Example 2:



Example 3:



The goal here is to represent a collection of kmers as an overlap graph where unique kmer will be represented as a node in the graph. Directed edges will connect nodes whose last $k-1$ characters overlap with the other node's first $k-1$ characters. Therefore, repeated kmers that overlap with a other kmer should be reflected as self loops. That's what I understood, however I am looking for a more general rule/solution.

Here is another example to illustrate this.





Current code:

 * Creates an overlap graph from a list of k-mers
 * \param patterns      a vector of string k-mers
 * \return a string containing an adjacency list representation of the
 * overlap graph as described in the problem specification
 * Output Format. The overlap graph Overlap(Patterns), in the form of an
 * adjacency list. Specifically, each line of the output represents all
 * edges (u,v) leaving a given node u in the format “u -> v”, where u and v
 * are both k-mers. If a given node u has multiple edges leaving it (e.g. v and w),
 * the destination nodes are comma-separated in any order. For example, if
 * there exist nodes ACG, CGT, and CGG, the resulting line of the adjacency
 * list would be: ACG -> CGT,CGG

#include <algorithm>
#include <iostream>
#include <map>
#include <vector>
#include <set>
#include <string>

std::string overlap_graph(const std::vector<std::string>& patterns) {
    // Overlap Graph of the k-mers collection patterns, in which a vertex exists for
    // each unique k-mer, and an edge (u,v) exists if the last k-1 characters of u are
    // equal to the first k-1 characters of v.

    if (patterns.size() == 1) return patterns[0] + "->" + patterns[0];
    // Build adjancency list O(n^2)
    std::map <std::string, std::vector<std::string>> adj;
    for (const auto& kmer_u : patterns) {
        // Add kmer as key if it doesn't exist in the map and find its neighbours.
        if (adj.find(kmer_u) == adj.end()) {
            adj[kmer_u] = {};
            for (const auto& kmer_v : patterns) {
                if ((kmer_u != kmer_v) &&
                    (kmer_u.substr(1, kmer_u.length() - 1) == kmer_v.substr(0, kmer_v.length() - 1)) && 
                    (std::find(adj[kmer_u].begin(), adj[kmer_u].end(), kmer_v) == adj[kmer_u].end())) {
        } // Otherwise continue.

    // Convert to string O(n).
    std::string ans;
    for (auto const& key_value : adj) {
        std::string kmer_u{ key_value.first };
        std::vector<std::string> kmers_v{ key_value.second };
        if (!kmers_v.empty()) {
            ans.append(kmer_u + "->" + kmers_v[0]);
            for (std::size_t kmer_v{ 1 }; kmer_v < kmers_v.size(); kmer_v++) {
                ans.append(',' + kmers_v[kmer_v]);
    return ans;

void tests() {
    /* Sample Dataset */
    std::vector<std::string> patterns{ {"AAG"}, {"AGA"}, {"ATT"}, {"CTA"}, {"CTC"}, {"GAT"},
                                       {"TAC"}, {"TCT"}, {"TCT"}, {"TTC"} };
    std::string answer{ "TTC->TCT\nCTA->TAC\nAAG->AGA\nCTC->TCT\nTCT->CTA,CTC\nAGA->GAT\nGAT->ATT\nATT->TTC\n" };
    std::cout << answer << std::endl;
    // std::cout << (overlap_graph(patterns) == answer) << std::endl;
    std::cout << overlap_graph(patterns) << std::endl;

void solution() {
    std::vector<std::string> patterns;

    std::string pattern;
    while (std::cin >> pattern) {
    std::cout << overlap_graph(patterns) << std::endl;

int main() {
  • $\begingroup$ I think this is actually a de-bruijn as opposed to an overlap graph. I believe an overlap graph has an edge whenever any positive length suffix of a node overlaps with the prefix of another node. $\endgroup$ Commented May 26, 2021 at 1:24
  • $\begingroup$ Also, since you are only concerned with length $k-1$ overlaps, you can find all of the edges in $O(n)$ time, since each node can only have 4 possible outgoing edges. I can extend my answer to include details on this if interested $\endgroup$ Commented May 26, 2021 at 4:32
  • $\begingroup$ @Throckmorton Thank you for the feedback - I really appreciate it. However, in the de Bruijn graph, the nodes are the $|k-1|$ unique k-mers whereas in the overlap graph the nodes are the unique $|k|$ k-mers. For instance, the $k=3$ de Bruijn graph for genome AGT will be AG->GT and the overlap graph is AGT->AGT (I am actually having troubles with this part - why did we need to add the self-loop here?). I agree with you that the de Bruijn graph can be solved in $O(n)$ as we don't need to check for duplicates, ie. for TTTT and $k=3$ the graph is TT->TT, TT. $\endgroup$
    – Gonzo
    Commented May 27, 2021 at 4:14
  • $\begingroup$ You can define $k$ to be whatever you want. The definition of a de-Bruin graph is a graph just all $k$-mers as nodes and an edge whenever there is a $k-1$ overlap between two nodes. This is exactly what the docstring in your question states. $\endgroup$ Commented May 27, 2021 at 18:12
  • $\begingroup$ Also, I don’t think there should be a self edge for AGT, in fact inter example the node “ACT” doesn’t have a self edge either. $\endgroup$ Commented May 27, 2021 at 18:27

2 Answers 2


You will want to create a self loop for a $k$-mer $s$ whenever $s_1,...,s_{k-1} = s_2,...,s_k$. This will only happen when all of the characters of the $k$-mer are identical, as $s_1=s_2, s_2=s_3,...,s_{k-1}=s_k$.


The usual way to avoid loops in linked lists is to keep a record of the things that have already been used / visited, and stop when an already visited thing has been seen again (or when there is nothing more to visit).


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