Please corect me if this is not true. The topological sort may not be unique i.e. My union find code only get 38/47 test cases passed. While I was searching for it I came across different techniques like DFS and topological sorting to detect cycle in a directed graph. So, initially all vertices are white. And the answer is: If no vertex has indegree 0, we can find a cycle by tracing backwards through vertices with positive indegree; since every vertex on the trace back has a positive indegree, we eventually reach a vertex twice, and the cycle has been found. I was trying to write code for detecting a cycle in a directed graph and if there is no cycle then return a topological order of the same. What about undirected graphs? Topological Sort / Cycle Detection. The answer is that both approaches will work. I can determine the topological sort of a directed graph using DFS algorithm. It seems like your question is the following: can you use depth-first search to detect cycles in an undirected graph, or should you use topological sort instead? "White color" means that the vertex hasn't been visited yet. If the given graph contains a cycle, then there is at least one node which is a parent as well as a child so this will break Topological Order. Is "topological sort of an undirected graph… Thus, the above file defines a directed graph. Here's a little code for topological sort and cycle detection. And if the graph contains cycle then it does not form a topological sort, because no node of the cycle can appear before the other nodes of the cycle in the ordering. Detect cycle in a directed graph using topological sort. "Gray" means that we've visited the vertex but haven't visited all vertices in its subtree. Before going into them, whenever you are dealing with representing graphs in files, you have to decide how you are going to format them. Use the following approach: consider we have three colors, and each vertex should be painted with one of these colors. Detect cycle in Directed Graph using Topological Sort , In Topological Sort, the idea is to visit the parent node followed by the child node. We have discussed a DFS based solution to detect cycle in a directed graph.In this post, BFS based solution is discussed. If there is a cycle, I assume the topological order is useless. If there are no cycles, I assume the topological order I found is valid. An acyclic graph always has a topological sort. Am I correct so far? The idea is to simply use Kahn’s algorithm for Topological Sorting. (Don’t use depth first search, we want just a modification of the basic topological sort.) If there is a cycle in a directed graph, then you can detect this by running a depth-first search over the graph. Topological Sort: TS can be only used to directed graph. 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