A Sub-exponential FPT Algorithm and a Polynomial Kernel for Minimum Directed Bisection on Semicomplete Digraphs

Abstract

Given an n-vertex digraph D and a non-negative integer k, the Minimum Directed Bisection problem asks if the vertices of D can be partitioned into two parts, say L and R, such that ${\vert {L} \vert }$ and ${\vert {R} \vert }$ differ by at most 1 and the number of arcs from R to L is at most k. This problem is known to be NP-hard even when $k = 0$. We investigate the parameterized complexity of this problem on semicomplete digraphs. We show that Minimum Directed Bisection admits a sub-exponential time fixed-parameter tractable algorithm on semicomplete digraphs. We also show that Minimum Directed Bisection admits a polynomial kernel on semicomplete digraphs. To design the kernel, we use $(n,k,k^2)$-splitters, which, to the best of our knowledge, have never been used before in the design of kernels. We also prove that Minimum Directed Bisection is NP-hard on semicomplete digraphs, but polynomial time solvable on tournaments.

Appendix: Proof of Observation 3.3

It is easy to see that Maximum Directed Bisection belongs to the class NP. Now, to see that the problem is NP-hard on DAGs, we consider the Directed Max-Cut problem on DAGs. Here, the input consists of a DAG G and an integer k, and the task is to determine whether V(G) can be partitioned into two parts X and Y such that ${\vert {A(Y,X)} \vert } \ge k$. Now, given an instance (G, k) of Directed Max-Cut where G is a DAG on n vertices, we construct an instance (D, k) of Minimum Directed Bisection as follows. The DAG D is constructed from G by simply adding n new isolated vertices to G. That is, $V(D) = V(G) \cup Z$ where Z is a set of vertices such that $V(G) \cap Z =\emptyset$, and ${\vert {Z} \vert }=n$, and $A(D)=A(G)$. Now, if (X, Y) is a cut of G of size at least k, then define a bisection (L, R) of D as follows. Take $L=X \cup Z_1$ and $R=Y \cup Z_2$, where $Z_1$ is any subset of Z of size $n-{\vert {X} \vert }$ and $Z_2=Z \setminus Z_1$. Then, note that ${\vert {L} \vert }={\vert {X} \vert }+(n-{\vert {X} \vert })=n$. Moreover, we have ${\vert {Y} \vert }=n-{\vert {X} \vert }$ and ${\vert {Z_2} \vert }={\vert {X} \vert }$, and therefore ${\vert {R} \vert }={\vert {Y} \vert }+{\vert {Z_2} \vert }=(n-{\vert {X} \vert })+{\vert {X} \vert }=n$. Thus, (L, R) is indeed a bisection of D and ${\vert {A(R,L)} \vert }={\vert {A(Y,X)} \vert } \ge k$. Conversely, if $(L',R')$ is a bisection of D of size at least k, then because Z is a set of isolated vertices, we get that $(L'\setminus Z, R'\setminus Z)$ is a cut of G such that ${\vert {A(R'\setminus Z, L'\setminus Z)} \vert } \ge k$.