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[Solved]: Coloring a Unicyclic Graph

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Problem Detail: 

I am trying to design an efficienct algorithm to color a unicyclic graph. I know if a graph does not contain any cycles (it's a tree) then it is 2-colorable. But cycles are either 2 (is even number of vertices) or 3 (is odd number of vertices) colorable. So if there contains just one cycle then the chromatic number of this unicyclic graph should be 3 (if the cycle contains an odd number of vertices).

So enough of that. on to the algorithm.

I would start at a vertex and color it color1. Now I would perform a BSF from the originator. Then each vertex that is adjacent to the originator vertex i would color it color 2, and continue this process of switching from color1 to color 2 unless if one of the vertices has an adjacent vertex that is already colored the same color. That means i found the cycle and will color that vertex color 3. If it had an adajcent vertex that was a different color than what i was going to color it then its an even number cycle and it would be 2-colorable.

So that means, each step of the BFS I would check each vertex adjacent to the vertex selected for that step of the BFS. Therefore $O(|E|)$?

is there a more efficient algorithm to color a unicyclic graph and is my time complexity correct?

Asked By : gprime

Answered By : Paresh

Yes, your algorithm is correct. The maximum number of edges that a graph can have without forming a cycle is when it is a tree. Lesser number of edges is only when the graph is disconnected. We will assume that the graph is connected, without loss of generality. In such a case, there would be $|V| - 1$ edges in the graph. Any other edge will create one cycle. So, any connected graph with exactly one cycle will have $|V|$ edges.

In such a case, you are right in doing a BFS to color alternatively the nodes alternatively. What you are doing is essentially starting from a root, and all nodes at odd depth have a different color than the root, and all nodes at an even depth have the same color as the root. If you encounter any node which has 2 colored neighbours (parent is always colored, but one of the children is also colored), then you have found a cycle. As you rightly note, if the second vertex is of a different colour than what you planned, it is an even cycle, and you can just go ahead - you will get a 2-coloring. Otherwise, just assign this vertex the third color, and go on with the alternating between two colors for the rest. This gives a 3-coloring.

As for the complexity, you are doing one BFS, checking for colored neighbors while putting them in the queue. Thus, complexity is $O(|V| + |E|)$, as you are visiting every edge and vertex once. However, since $|E| = |V|$, the overall complexity is $O(|V|)$. To color $|V|$ vertices, you need at least $\Omega(|V|)$ time, and so the algorithm is asymptotically optimal (no asymptotically faster algorithm is possible). This algorithm is $\Theta(|V|)$.

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