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When looking at the feedback arc set problem, given some graph $G = (V, A)$ where the weights are on the arcs, I'm hoping to find an approximation algorithm that provides a guarantee when the number of arcs is $O(n \log n)$ or some other commonly assumed number of arcs for a sparse graph.

There have been results published for the tournament graph in this setting interestingly enough, but I haven't seen any on a sparse graph.

Do you think solving this exactly would be the best route to go?

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Feedback Arc Set is NP-complete even on graphs with constant degree, so sparsity does help that much. Indeed, the opposite seems to be the case: for very dense graph, the problem (Feedback Arc Set on Tournaments) is slightly easier. The approximation landscape of feedback arc set is quite open.

In Combinatorial algorithms for feedback problems in directed graphs, Demetrescu and Finocchi (2003) give an algorithm running in time $O(m \cdot n)$ that computes a minimal FAS with approximation equal to the length of the longest cycle.

Do you think solving this exactly would be the best route to go?

It depends on what you want. There are really fast algorithms for giving really good solutions, but there are no guarantees, unless you go for exact algorithms.

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According to the article FPT-approximation for FPT Problems

Theorem 1.1. Directed Feedback Vertex Set, Subset Directed Feedback Vertex Set, Directed Odd Cycle Transversal (DOCT), and Multicut have $2^{\mathrm O(k)}n^{\mathrm O(1)}$-time factor-$2$ approximation algorithms.

Note that the Directed Feedback Vertex Set (DFVS) problem can be linearly reduced to the (Directed) Feedback Arc Set (DFAS) by vertex subdivision, while DFAS can be reduced to DFVS with factor $k$ for the number of vertices and arcs. So the theorem above applies to DFAS too. (Here $k$ is the cardinality of feedback set, and $n$ is the order of graph.)

Note that the best known exact solution of DFVS takes $\mathrm O(k!4^k k^5 (n + m))$ of time, where $m$ is the number of arcs. That's why such result for $2$-approximation makes sense.

However there exists a strongly polynomial approximation scheme of DFVS with approximation factor $\mathrm O(\log n \log \log n)$.

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