Parameterized Eulerian Strong Component Arc Deletion Problem on Tournaments

A digraph D is strongly connected if for every x, y ∈ V (D) there is a directed path in D from x to y. In particular, the digraph consisting of just one vertex is strongly connected. A strong component of D is a maximal induced subgraph C in D that is strongly connected. D is Eulerian if there is a directed closed trail in D that traverses every vertex of D and uses every arc in A(D) once. Recall that D is Eulerian if and only if D is strongly connected and balanced. For X, Y ⊆ V (D) let X ⇒ Y denote the fact that all arcs between X and Y go from a vertex in X to a vertex in Y . In particular, we can write X ⇒ Y if there are no arcs between X and Y . If Y is a set of vertices in D or a subgraph of D, for x ∈ V (D) we let d + Y (x) denote the number of arcs of the form xy, for some vertex y in Y , and define d - Y (x) similarly. For a positive integer n, let [n] = {1, . . . , n}. For more information on digraphs, see [1].

If A ′ is an arc-set in D then D -A ′ denotes the subgraph of D obtained by deleting all arcs of A ′ from D. A set A ′ ⊆ A(D) is said to be a DESC-set (standing for Delete in order to obtain Eulerian Strong Components) if all strong components in D -A ′ are Eulerian in D -A ′ . The size of a smallest DESC-set in D is denoted by desc(D). We can now define the problem Min-DESC.

Instance: A digraph D and a nonnegative integer k. Question: Decide whether desc(D) ≤ k. [4] introduced Min-DESC in the context of housing markets, and asked if the problem is fixed-parameter tractable when parameterized by k.

Cygan et al. [6] study a number of related problems; in particular, the problem of finding a set of arcs A ′ such that D -A ′ is balanced, and the problem of finding a set of arcs A ′ such that D -A ′ is Eulerian. They proved that the first problem is polynomial-time solvable and the second problem is fixed-parameter tractable when the parameter is k. (Cygan et al. also studied related problems with graphs instead of digraphs, and deleting vertices instead of arcs or edges).

Note that unlike these problems, in Min-DESC we do not require that D -A ′ is balanced, since we allow arcs between strong components. This makes it more difficult to say anything about what D -A ′ must look like locally, as whether a given arc is between two strong components or not depends on the rest of the digraph.

The complexity of Min-DESC parameterized by k remains open. In this paper, we consider Min-DESC in the special case when D is a tournament. This problem is still NP-hard as proved in [5]. Since the proof is very long and of relatively little interest to the reader, we have decided to omit it from the paper. In this paper we prove the following theorem.

Theorem 1. Min-DESC for tournaments parameterized by k is fixedparameter tractable and has a kernel with at most 4k(4k + 2) vertices. 1) where f is a function of the parameter k only [7,9,10]. Given a parameterized problem L, a kernelization of L is a polynomial-time algorithm that maps an instance (x, k) to an instance

for some functions h and g. The function g(k) is called the size of the kernel. It is well-known [7,9,10] that a decidable parameterized problem L is fixed-parameter tractable if and only if it has a kernel. Polynomial-size kernels are of main interest due to applications [7,9,10], but unfortunately many fixed-parameter problems do not have such kernels unless coNP⊆NP/poly, see, e.g., [2,3,8].

In this section we will prove Theorem 1. To prove this result, consider a tournament T , and assume desc(T ) ≤ k. Let A ′ ⊆ A(T ) be a DESC-set for T of size at most k and let T ′ = T -A ′ . Let ≻ be a linear ordering on the strong components of T ′ , such that if A ≻ B then A ⇒ B. For sets of vertices X, Y , we let X ≻ Y denote the fact that A ≻ B, for any strong components A, B in

We firstly prove some properties of T, T ′ and A ′ . For sets of vertices X and Y , let d * (X, Y ) be the number of arcs in A ′ with one end-vertex in X and the other in Y (in either direction). Let d * (X) = d * (X, X), d * (x, Y ) = d * ({x}, Y ) and d * (x) = d * (x, V (T )).

Proof. Let Z be the maximal set of vertices for which B ≻ Z, and let R be the maximal set of vertices for which R ≻ A.

Note that if X is the vertex set of a strong component in X). Also note that if X ≻ Y , then the arcs between X and Y in A ′ will be exactly the arcs from Y to X. Hence, for all a ∈ A, b ∈ B we have

The last inequality holds since d * (a, A)

as there are at most k deleted arcs, with each deleted arc counted at most once, with a possible exception of an arc from a to b, which would be counted twice.

Lemma 2. Suppose there exists X ⊆ V (T ) such that |X| ≥ 4k + 3 and

Proof. For the sake of contradiction assume that the lemma does not hold. Let A, B, W be disjoint sets of vertices such that Proof. Let H be the strong component of T containing x and y, and let V = V (T ) and U = V (H). Form a digraph Q from T ′ by reinserting deleted arcs between V and V \ U and then reorienti

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