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Problem: Give a linear-time algorithm to find an odd-length (directed) cycle in a directed graph. (Exercise 3.21 of Algorithms by S. Dasgupta, C. Papadimitriou, and U. Vazirani.)

The related post@cs.stackexchange asks for the existence of an odd-length directed cycle in a digraph, which is solved by the following theorem@algs4.cs.princeton.edu.

Theorem: A directed graph has an odd-length directed cycle if and only if one (or more) of its strong components is non-bipartite (when treated as an undirected graph).

Thus, we can assume that the digraph is strongly connected.

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hengxin
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2 Answers2

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Let $G$ be strongly connected. Run BFS (!) from an arbitrary vertex $s$. BFS creates a leveled tree where level of a vertex $v$ is it's directed distance from $s$. If while running BFS you have never seen an edge $(u,v)$ that goes between levels of the same parity then $G$ is bipartite. The parity of the level of $u$ defines the partition that $u$ belongs to. Note that since every vertex is reachable from $s$, BFS will explore all the edges, so all of the edges would be going between the two blocks of the partition. Otherwise, you find an edge $(u,v)$ that goes between levels of the same parity. Suppose level of $u$ is even and level of $v$ is even. Then run DFS (or BFS) from $v$ to get a path to $s$. If this path is of even length then you get an odd directed cycle $s \rightsquigarrow u \rightarrow v \rightsquigarrow s$. If this path is of odd length then you get an odd directed cycle $s \rightsquigarrow v \rightsquigarrow s$. Analogous thing holds when level of $u$ is odd and level of $v$ is odd. Since this algorithm simply does 2 runs of BFS/DFS, it runs in linear time.

Denis Pankratov
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There actually is the answer to your question on a page you have linked: algs4.cs.princeton.edu: Directed Graphs. It's under Creative Problems (41.) and the trick is that you can construct a directed odd-length cycle from an undirected odd-length cycle in a strongly connected component.

  1. Odd-length directed cycle. Design a linear-time algorithm to determine whether a digraph has an odd-length directed cycle.

Solution. We claim that a digraph G has an odd-length directed cycle if and only if one (or more) of its strong components is nonbipartite (when treated as an undirected graph).

If the digraph G has an odd-length directed cycle, then this cycle will be entirely contained in one of the strong components. When the strong component is treated as an undirected graph, the odd-length directed cycle becomes an odd-length cycle. Recall that an undirected graph is bipartite if and only if it has no odd-length cycle.

Suppose a strong component of G is nonbipartite (when treated as an undirected graph). This means that there is an odd-length cycle C in the strong component, ignoring direction. If C is a directed cycle, then we are done. Otherwise, if an edge v->w is pointing in the "wrong" direction, we can replace it with an odd-length path that is pointing in the opposite direction (which preserves the parity of the number of edges in the cycle). To see how, note that there exists a directed path P from w to v because v and w are in the same strong component. If P has odd length, then we replace edge v->w by P; if P has even length, then this path P combined with v->w is an odd-length cycle.

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Philip Becker
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