Graph Coloring | Set 2 (Greedy Algorithm) - GeeksforGeeks

Unfortunately, there is no efficient algorithm available for coloring a graph with minimum number of colors as the problem is a known NP Complete problem. There are approximate algorithms to solve the problem though. Following is the basic Greedy Algorithm to assign colors. It doesn’t guarantee to use minimum colors, but it guarantees an upper bound on the number of colors. The basic algorithm never uses more than d+1 colors where d is the maximum degree of a vertex in the given graph.

Basic Greedy Coloring Algorithm:

1. Color first vertex with first color.
2. Do following for remaining V-1 vertices.
          a) Consider the currently picked vertex and color it with the 
             lowest numbered color that has not been used on any previously
             colored vertices adjacent to it. If all previously used colors 
             appear on vertices adjacent to v, assign a new color to it.

void Graph::greedyColoring()

{

    int result[V];

    // Assign the first color to first vertex

    result[0]  = 0;

    // Initialize remaining V-1 vertices as unassigned

    for (int u = 1; u < V; u++)

        result[u] = -1;  // no color is assigned to u

    // A temporary array to store the available colors. True

    // value of available[cr] would mean that the color cr is

    // assigned to one of its adjacent vertices

    bool available[V];

    for (int cr = 0; cr < V; cr++)

        available[cr] = false;

    // Assign colors to remaining V-1 vertices

    for (int u = 1; u < V; u++)

    {

        // Process all adjacent vertices and flag their colors

        // as unavailable

        list<int>::iterator i;

        for (i = adj[u].begin(); i != adj[u].end(); ++i)

            if (result[*i] != -1)

                available[result[*i]] = true;

        // Find the first available color

        int cr;

        for (cr = 0; cr < V; cr++)

            if (available[cr] == false)

                break;

        result[u] = cr; // Assign the found color

        // Reset the values back to false for the next iteration

        for (i = adj[u].begin(); i != adj[u].end(); ++i)

            if (result[*i] != -1)

                available[result[*i]] = false;

    }

    // print the result

    for (int u = 0; u < V; u++)

        cout << "Vertex " << u << " --->  Color "

             << result[u] << endl;

}

Time Complexity: O(V^2 + E) in worst case.

Analysis of Basic Algorithm
The above algorithm doesn’t always use minimum number of colors. Also, the number of colors used sometime depend on the order in which vertices are processed. For example, consider the following two graphs. Note that in graph on right side, vertices 3 and 4 are swapped. If we consider the vertices 0, 1, 2, 3, 4 in left graph, we can color the graph using 3 colors. But if we consider the vertices 0, 1, 2, 3, 4 in right graph, we need 4 colors.

So the order in which the vertices are picked is important. Many people have suggested different ways to find an ordering that work better than the basic algorithm on average. The most common isWelsh–Powell Algorithm which considers vertices in descending order of degrees.

How does the basic algorithm guarantee an upper bound of d+1?
Here d is the maximum degree in the given graph. Since d is maximum degree, a vertex cannot be attached to more than d vertices. When we color a vertex, at most d colors could have already been used by its adjacent. To color this vertex, we need to pick the smallest numbered color that is not used by the adjacent vertices. If colors are numbered like 1, 2, …., then the value of such smallest number must be between 1 to d+1 (Note that d numbers are already picked by adjacent vertices).
This can also be proved using induction. See this video lecture for proof.

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