I came up with this O(V+E) algorithm for calculating the chromatic number X(g) of a graph g represented by an adjacency list:

  1. Initialize an array of integers "colors" with V elements being 1
  2. Using two for loops go through each vertex and their adjacent nodes and for each of the adjacent node g[i][j] where j is adjacent to i, if j is not visited yet increment colors[g[i][j]] by 1.
  3. After doing this the maximum integer in the array "colors" is the chromatic number of the graph g(if the algorithm works).

Here is my C++ code:

#include <bits/stdc++.h>
using namespace std;

struct graph {
    vector<vector<int>> adjL;
    vector<int> colours;
    vector<bool> vis;

int chrNUM(graph& G) {
    int num = 1; 
    for(int i = 1; i < G.adjL.size(); i ++) {
        for(int j = 0; j < G.adjL[i].size(); j ++) {
            if(!G.vis[G.adjL[i][j]]) {
                G.colours[G.adjL[i][j]] ++;
                num = max(num, G.colours[G.adjL[i][j]]);
        G.vis[i] = true;
    return num;

void initGET(graph& G, int N, int M) {
    cin >> N >> M;
    G.adjL.assign(N + 1, vector<int>(0));
    G.colours.assign(N + 1, 1);
    G.vis.assign(N + 1, false);
    for(int i = 0; i < M; i ++) {
        int u,v;
        cin >> u >> v;

int main() {
    graph g;
    int n;  //number of vertices
    int m;  //number of edges
    initGET(g, n, m);
    cout << chrNUM(g);

I am wondering if there is a flaw? Maybe it works for certain graphs only? Maybe it gives X(g) for smaller graphs but a value higher than X(g) for larger graphs? I found it worked correctly for all the graphs I have tried (up to 20 vertices). I know this is an NP complete problem but I want some counterexamples for my algorithm if possible or an explanation as to why the method won't work. I have also got a recursive (DFS) solution which is a bit different but mostly similar to this. Any ideas?

Thanks in advance!

  • $\begingroup$ Although the algorithm does not work, I have upvoted the question, because it is clear, includes example code, and the OP is open to the possibility of it only working for certain graphs. $\endgroup$ Dec 28, 2019 at 14:17

1 Answer 1


Your algorithm is known as greedy coloring. Wikipedia gives an example of a bipartite graph, the crown graph, where the greedy coloring can produce a coloring using $n/2$ colors (for the worst ordering). For random $G(n,1/2)$ graphs, the greedy coloring typically produces a coloring using double the optimal number of colors.

  • 1
    $\begingroup$ Another simple example is a star graph, where a worst ordering (i.e., center node being last) will lead to a coloring using $n$ colors when the answer is just 2. $\endgroup$
    – stillanoob
    Dec 27, 2019 at 7:13

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