My question is the directed version of this one. (I know the results and proofs about feedback vertex set in undirected graphs or undirected planar graphs; so I am concern about the directed feedback vertex set.)

In wiki, it says that directed Feedback vertex set (DFVS) problem is $\mathsf{NP}$-hard if the maximum in-degree $\Delta_{in} = 2$ and maximum out-degree $\Delta_{out} = 2$; and planar directed Feedback vertex set problem is $\mathsf{NP}$-hard if the maximum in-degree $\Delta_{in} = 3$ and maximum out-degree $\Delta_{out} = 3$.

Here is what I know:

  1. When $(\Delta_{in}, \Delta_{out}) = (1, d)$, DFVS is polynominal-time solvable.
  2. When $(\Delta_{in}, \Delta_{out}) \geq (2, 2)$, DFVS is $\mathsf{NP}$-hard.
  3. When $(\Delta_{in}, \Delta_{out}) \geq (3, 3)$, Planar-DFVS is $\mathsf{NP}$-hard.

  • Item 1 is trivial I think.
  • Item 2 and 3 are claimed by Garey and Johnson's book.
  • For Item 3, Vertex Cover in $3$-degree planar graphs can reduce to Planar-DFVS with $(\Delta_{in}, \Delta_{out}) = (3, 3)$.
  • I do not find the proof about item 2.
  • Is Planar-DFVS in $\mathsf{P}$ if $\Delta_{in}$ (or $\Delta_{out}$) is no more than $2$?

2 Answers 2



Warning: this does not completely solve your question because the gadget is not planar. But since I could not find a proof of this, I think it might be worth posting.

Speckenmeyer proved that the undirected feedback vertex set problem is NP-hard on planar graphs with maximum degree of four. I try to adapt his gadgets to directed graphs. However, I cannot keep the planarity. But it satisfies the degree bounds.

We replace a vertex $x$ by this gadget and assign the entering arcs to $a_1$ and $a_2$, and the leaving arcs to $d_1$ and $d_2$. (It does not matter who receives two.)

In the gray box, there are four directed triangles. We need to remove at least two vertices to break them. The only choice is $\{b_1, b_2\}$. This corresponds to a solution not containing $x$. On the other hand, a solution containing $x$ can be replaced by $\{o, c_1, c_2\}$.

  • $\begingroup$ After a second thought, I guess it's very unlikely the same approach can be applied to the planar case. The essential obstacle is that $K_{3,3}$ is not planar. We probably need a different source problem instead of degree reductions. $\endgroup$
    – Yixin Cao
    Mar 3, 2023 at 8:33
  • $\begingroup$ Your construction of the gadget is much helpful. Perhaps, this problem would have been settled if my reduction for the planar version is correct. $\endgroup$
    – Blanco
    Mar 6, 2023 at 4:43

Thanks to Prof. @Yixin Cao's construction of the reduction, I think I have found a way to show that Planar-DFVS remains $\textsf{NP}$-hard when $(\Delta_{in}, \Delta_{out}) = (2, 2)$.

We use the Vertex Cover in planar $3$-regular Hamiltonian graphs to reduce.

Let $G$ be an instance of Vertex Cover in planar $3$-regular Hamiltonian graphs. $G$ is bridge-less, since $G$ is Hamiltonian. According to Peterson's theorem, $G$ has a perfect matching, and let $M$ be a perfect matching of $G$.


For each edge $e$ in $G$, we add a new $2$-degree vertex $w_{e}$ connecting the endpoints of $e$.

If $e$ is in $M$, we orient the triangle containing $w_{e}$ in the anticlockwise direction; otherwise, we orient the triangle in the clockwise direction.

Let $x$ be a vertex in $G$. Without loss of generality, suppose there are one clockwise directed triangle and two anticlockwise directed triangles containing $x$.

enter image description here

Now, the in-degree and out-degree of every vertex are exactly three.


Based on Speckenmeyer's method, we use a gadget to replace each vertex $x$.

enter image description here

As Prof. Cao's statements, in the dotted box, there are four directed triangles. To Break all of them by using two vertices, we have a unique choice $\{u_{1},u_{5}\}$. This corresponds to a solution not containing $x$. On the other hand, a solution containing $x$ can always be replaced by $\{v_{1}, u_{3}, v_{3}\}$ or $\{v_{2}, u_{3}, v_{4}\}$.

Therefore, we obtain a planar digraph with $(\Delta_{in}, \Delta_{out}) = (2, 2)$.

  • $\begingroup$ What about a cycle entering from $u_1$ and leaving from $v_2$? $\endgroup$
    – Yixin Cao
    Mar 4, 2023 at 0:32

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