    # The Parameterized Complexity of Packing Arc-Disjoint Cycles in Tournaments

Given a directed graph D on n vertices and a positive integer k, the Arc-Disjoint Cycle Packing problem is to determine whether D has k arc-disjoint cycles. This problem is known to be W-hard in general directed graphs. In this paper, we initiate a systematic study on the parameterized complexity of the problem restricted to tournaments. We show that the problem is fixed-parameter tractable and admits a polynomial kernel when parameterized by the solution size k. In particular, we show that it can be solved in 2^O(k k) n^O(1) time and has a kernel with O(k) vertices. The primary ingredient in both these results is a min-max theorem that states that every tournament either contains k arc-disjoint triangles or has a feedback arc set of size at most 6k. Our belief is that this combinatorial result is of independent interest and could be useful in other problems related to cycles in tournaments.

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## 1 Introduction

Given a (directed or undirected) graph and a positive integer , the Disjoint Cycle Packing problem is to determine whether has (vertex or arc/edge) disjoint cycles. Packing disjoint cycles is a fundamental problem in Graph Theory and Algorithm Design with applications in several areas. Since the publication of the classic Erdös-Pósa theorem in 1965 , this problem has received significant scientific attention in various algorithmic realms. In particular, Vertex-Disjoint Cycle Packing in undirected graphs is one of the first problems studied in the framework of parameterized complexity. In this framework, each problem instance is associated with a non-negative integer called parameter, and a problem is said to be fixed-parameter tractable () if it can be solved in time for some function , where is the input size. For convenience, the running time where grows super-polynomially with is denoted as . A kernelization algorithm is a polynomial-time algorithm that transforms an arbitrary instance of the problem to an equivalent instance of the same problem whose size is bounded by some computable function of the parameter of the original instance. The resulting instance is called a kernel and if is a polynomial function, then it is called a polynomial kernel and we say that the problem admits a polynomial kernel. Kernelization typically involves applying a set of rules (called reduction rules) to the given instance to produce another instance. A reduction rule is said to be safe

if it is sound and complete, i.e., applying it to the given instance produces an equivalent instance. In order to classify parameterized problems as being  or not, the -hierarchy is defined:

. It is believed that the subset relations in this sequence are all strict, and a parameterized problem that is hard for some complexity class above  in this hierarchy is said to be fixed-parameter intractable. Further details on parameterized algorithms can be found in [15, 16, 19].

Vertex-Disjoint Cycle Packing in undirected graphs is  with respect to the solution size [10, 25] but has no polynomial kernel unless   co/poly . In contrast, Edge-Disjoint Cycle Packing in undirected graphs admits a kernel with vertices (and is therefore ) . On directed graphs, Vertex-Disjoint Cycle Packing and Arc-Disjoint Cycle Packing are equivalent and turn out to be -hard [23, 30]. Therefore, studying these problems on a subclass of directed graphs is a natural direction of research. Tournaments form a mathematically rich subclass of directed graphs with interesting structural and algorithmic properties [5, 26]. A tournament is a directed graph in which there is a single arc between every pair of distinct vertices. Tournaments have several applications in modeling round-robin tournaments and in the study of voting theory and social choice theory. Further, the combinatorics of inclusion relations of tournaments is reasonably well-understood . A seminal result in the theory of undirected graphs is the Graph Minor Theorem (also known as the Robertson and Seymour theorem) that states that undirected graphs are well-quasi-ordered under the minor relation . Developing a similar theory of inclusion relations of directed graphs has been a long-standing research challenge. However, there is such a result known for tournaments that states that tournaments are well-quasi-ordered under the strong immersion relation . In fact, this result also holds for a superclass of tournaments, namely, semicomplete digraphs . A semicomplete digraph is a directed graph in which there is at least one arc between every pair of distinct vertices. Many results (including some of the ones described in this work) for tournaments straightaway hold for semicomplete digraphs too. This is another reason why tournaments is one of the most well-studied classes of directed graphs.

Feedback Vertex Set and Feedback Arc Set are two well-explored algorithmic problems on tournaments. A feedback vertex (arc) set is a set of vertices (arcs) whose deletion results in an acyclic graph. Given a directed graph and a positive integer , Feedback Arc (Vertex) Set is the problem of determining if the graph has a set of at most arcs (vertices) whose deletion results in an acyclic graph. These problems are the dual problems of Vertex-Disjoint Cycle Packing and Arc-Disjoint Cycle Packing, respectively. They are known to be -hard on tournaments [2, 12, 14, 31] but are  when parameterized by [3, 22, 20, 28]. Further, Feedback Arc Set in Tournaments has a kernel with vertices  and Feedback Vertex Set in Tournaments has a kernel with vertices . Though Feedback Arc Set and Feedback Vertex Set are intensively studied in tournaments, their duals have surprisingly not been considered in the literature until recently [8, 24]. Any tournament that has a cycle also has a triangle . Therefore, if a tournament has vertex-disjoint cycles, then it also has vertex-disjoint triangles. Thus, Vertex-Disjoint Cycle Packing in tournaments is just packing vertex-disjoint triangles. A straightforward application of the color coding technique shows that this problem is  and a kernel with vertices is an immediate consequence of the quadratic element kernel known for 3-Set Packing . Recently, a kernel with vertices was shown for this problem using interesting variants and generalizations of the popular expansion lemma .

Focusing on Arc-Disjoint Cycle Packing in tournaments, it is easy to verify that a tournament that has arc-disjoint cycles need not necessarily have arc-disjoint triangles. This observation hints that packing arc-disjoint cycles could be significantly harder than packing vertex-disjoint cycles. This is the starting point of our study. In this paper, we investigate the parameterized complexity of Arc-Disjoint Cycle Packing in tournaments.

Arc-Disjoint Cycle Packing in Tournaments (ACT) Parameter: Input: A tournament and a positive integer . Question: Do there exist arc-disjoint cycles in ?

We show that ACT is  and admits a polynomial kernel. En route, we discover an interesting min-max relation analogous to the classical Erdös-Pósa theorem. In particular, we show the following results.

• A tournament has arc-disjoint cycles if and only if has arc-disjoint cycles each of length at most (Lemma 1).

• Every tournament either contains arc-disjoint triangles or has a feedback arc set of size at most (Theorem 4).

• ACT admits a kernel with vertices (Theorem 13).

• ACT can be solved in time (Theorem 14).

The paper is organized as follows. In Section 2, we give some definitions related to directed graphs, cycles and tournaments. In Section 3, we show the first two combinatorial results about tournaments. In Section 4, we show that ACT is  and admits a polynomial kernel. In Section 5, we describe an improved  algorithm and a linear vertex kernel for ACT. Finally, we conclude with some remarks in Section 6.

## 2 Preliminaries

The set is denoted by . A directed graph (or digraph) is a pair consisting of a set of vertices and a set

of arcs. An arc is specified as an ordered pair of vertices. We will consider only simple unweighted digraphs. For a digraph

, and denote the set of its vertices and the set of its arcs, respectively. Two vertices , are said to be adjacent in if or . For an arc , denotes and denotes . For a vertex , its out-neighborhood, denoted by , is the set and its in-neighborhood, denoted by , is the set . For a set of arcs , denotes the union of the sets of endpoints of arcs in . For a set , denotes the digraph obtained from by deleting .

A path in is a sequence of distinct vertices such that for each , . The set is denoted by and the set is denoted by . A path is called an induced (or chordless) path if there is no arc in that is between two non-consecutive vertices of . A cycle in is a sequence of distinct vertices such that is a path and . The set is denoted by and the set is denoted by . A cycle is called an induced (or chordless) cycle if there is no arc in that is between two non-consecutive vertices of with the exception of the arc . The length of a path or cycle is the number of vertices in it and is denoted by . For a set of paths or cycles, denotes the set . A cycle on three vertices is called a triangle. A digraph is said to be triangle-free if it has no triangles.

A digraph is called a directed acyclic graph if it has no cycles. Any directed acyclic graph has an ordering called topological ordering of its vertices such that for each , holds. A feedback vertex (arc) set is a set of vertices (arcs) whose deletion results in an acyclic graph. For a digraph , let denote the size of a minimum feedback arc set of . A tournament is a digraph in which for every pair of distinct vertices either or but not both. A tournament is called transitive if it is a directed acyclic graph and a transitive tournament has a unique topological ordering.

## 3 An Erdös-Pósa Type Theorem

The classical Erdös-Pósa theorem for cycles in undirected graphs states that there exists a function such that for each non-negative integer , every undirected graph either contains vertex-disjoint cycles or has a feedback vertex set consisting of vertices . An interesting consequence of this theorem is that it leads to an  algorithm for Vertex-Disjoint Cycle Packing. It is well known that the treewidth () of a graph is not larger than the size of its feedback vertex set, and that a naive dynamic programming scheme solves Vertex-Disjoint Cycle Packing in time and exponential space (see, e.g., ). Thus, the existence of a time algorithm that uses exponential space can be viewed as a direct consequence of the Erdös-Pósa theorem (see  for more details).

In this section, we show that that there exists a function such that for each non-negative integer , every tournament either contains arc-disjoint cycles or has a feedback arc set consisting of arcs. First, we show that is and then improve it to . The following result is crucial in proving the former.

###### Lemma 1.

Let and be positive integers such that . If a tournament contains a set of arc-disjoint cycles, then it also contains a set of arc-disjoint cycles each of length at most .

###### Proof.

Let be a set of arc-disjoint cycles in that minimizes . If every cycle in is a triangle, then the claim trivially holds. Otherwise, let be a longest cycle in and let denote its length. Let be a pair of non-consecutive vertices in . Then, either or . In any case, the arc between and along with forms a cycle of length less than with . By our choice of , this implies that is an arc in some other cycle . This property is true for the arc between any pair of non-consecutive vertices in . Therefore, we have leading to . ∎

This lemma essentially shows that it suffices to determine the existence of arc-disjoint cycles in each of length at most in order to determine if is a yes-instance of ACT. This leads to the following quadratic Erdös-Pósa bound. Recall that for a digraph , denotes the size of a minimum feedback arc set of .

###### Theorem 2.

For every non-negative integer , every tournament either contains arc-disjoint cycles or has a feedback arc set of size .

###### Proof.

Suppose is a maximal set of arc-disjoint cycles in . If , then the claim holds. Otherwise, from Lemma 1, we may assume that each cycle in is of length at most . Let denote the digraph obtained from by deleting the arcs that are in some cycle in . Clearly, is acyclic as is maximal. Then, it follows that . ∎

Next, we strengthen this result to arrive at a linear min-max bound. We will use the following lemma in the process. For a digraph , let denote the number of non-adjacent pairs of vertices in . That is, is the number of pairs of vertices of such that neither nor .

###### Lemma 3.

Let be a triangle-free digraph in which for every pair of distinct vertices, at most one of or is in . Then, .

###### Proof.

We will prove the claim by induction on . The claim trivially holds for . Suppose . First, we apply a simple preprocessing rule on . If has a vertex that either has no in-neighbours or has no out-neighbours, then we delete from to get the digraph . Clearly, there is no cycle in that contains and thus . Therefore, subsequently, we may assume that for every vertex , and .

For a vertex , we define to be the number of induced paths of length 3 with as the first vertex. Similarly, we define to be the number of induced paths of length 3 with as the second vertex. We claim that . Consider an induced path of length 3 in . Then, contributes 1 to and does not contribute to for any . Further, contributes 1 to and does not contribute to for any . Therefore, contributes 1 to and . Hence, . It now follows that there is a vertex such that .

Define the sets , and as , and . That is, is the set of in-neighbours of , is the set of out-neighbours of and is the set of vertices that are not adjacent with . Observe that and . Let and be the subgraphs and , respectively. Then, as and are vertex-disjoint induced subgraphs of , we have . Now, any induced path of length 3 in with as the second vertex satisfies the property that and . Further, due to the facts that is an induced path and is triangle-free. Then, is the number of pairs of non-adjacent vertices with . Therefore, as , and .

Let denote the set of arcs in with and . Let and be feedback arc sets of and , respectively. We claim that is a feedback arc set of . If there is a cycle in the graph obtained from by removing arcs in , then has an arc with and an arc with . However, as has no triangle, any arc with and satisfies and . That is, leading to a contradiction. Therefore, it follows that . Note that as any induced path starting at satisfies , and any arc with corresponds to an induced path starting at . Also, by the choice of , we have . Therefore, . By induction hypothesis, we have and . Hence, . As , we have . ∎

This leads to the following main result of this section.

###### Theorem 4.

For every non-negative integer , every tournament either contains arc-disjoint triangles or has a feedback arc set of size at most that can be obtained in polynomial time.

###### Proof.

Suppose is a maximal set of arc-disjoint triangles in with . Let denote the digraph obtained from by deleting the arcs that are in some triangle in . Clearly, has no triangle and . From Lemma 3, we have . Also, if is a feedback arc set of , then is a feedback arc set of . Therefore, . ∎

We will use this result crucially in showing that ACT can be solved in time and admits a kernel with vertices.

## 4 Fixed-Parameter Tractability and Kernelization Complexity

In this section, we show that ACT is  and admits a polynomial kernel. We show that the first result is a direct consequence of Lemma 1 and the second follows from Theorem 4.

### 4.1 An FPT Algorithm

Consider an instance of ACT. Let denote and denote . Suppose is a yes-instance and is a set of arc-disjoint cycles in . From Lemma 1, we may assume that the total number of arcs that are in cycles in is at most . Using this observation, we proceed as follows. We color the arcs of uniformly at random from the color set where . Let denote this coloring.

###### Proposition 5 ().

If is a subset of of size

, then the probability that the arcs in

are colored with pairwise distinct colors is at least .

Next, we define the notion of a colorful solution for our problem.

###### Definition 4.1.

(Colorful set of cycles) A set of arc-disjoint cycles in that satisfies the property that for any two (not necessarily distinct) cycles and for any two distinct arcs , holds is said to be a colorful set of cycles.

Rephrasing Proposition 5 in the context of our problem, we have the following observation.

###### Observation 6.

If is a solution of with the property that for each , , then is a colorful set of cycles in with probability at least .

Armed with the guarantee that a solution (if one exists) of is colorful with sufficiently high probability, we focus on finding a colorful set of cycles in .

###### Lemma 7.

If has a colorful set of cycles, then such a set can be obtained in time.

###### Proof.

Consider a permutation of . For each , let denote the subgraph of with and . That is, is the set of arcs of that are colored with the first colors, is the set of arcs of that are colored with the next colors and so on. For each , let denote a cycle (if one exists) in . Let denote the set . For each permutation of , we compute the corresponding set . If has a colorful set of cycles, then for some permutation of . Therefore, by computing for every permutation of , we can obtain a colorful set of cycles in (if one exists). ∎

Using the standard technique of derandomization of color coding based algorithms [4, 15, 27], we have the following result by taking .

###### Proposition 8 ([4, 15, 27]).

Given integers , there is a family of coloring functions of size that can be constructed in time satisfying the following property: for every set of size , there is a function such that for any two distinct arcs .

Then, we have the following result.

###### Theorem 9.

ACT can be solved in time.

###### Proof.

Consider an instance of ACT. Let . First, we compute the family of coloring functions using Proposition 8 where is the number of arcs in . Then, for each coloring function in , we determine if has a colorful set of cycles using Lemma 7. Due to the properties of guaranteed by Proposition 8, it follows that is a yes-instance if and only if has a set of cycles that is colorful with respect to at least one of the coloring functions. The overall running time is . ∎

Observe that the running time of the algorithm to find a colorful set of cycles can be improved to by employing a standard dynamic programming scheme. This will result in an time algorithm for ACT. However, we skip the details of the same as we will describe an time algorithm for ACT in Section 5.

### 4.2 A Polynomial Kernel

Now, we show that ACT admits a polynomial kernel. We use Theorem 4 to describe a quadratic vertex kernel.

###### Theorem 10.

ACT admits a kernel with vertices.

###### Proof.

Let denote an instance of ACT. From Theorem 4, we know that has either arc-disjoint triangles or a feedback arc set of size at most . In the former case, we return a trivial yes-instance of constant size as the kernel. In the latter case, is a feedback vertex set of of size at most . Let denote the transitive tournament and denote its unique topological ordering. Observe that for each , the subtournament of induced by is also transitive. If there is a cycle in , then this cycle (which is also a cycle in ) has no arc from leading to a contradiction.

For each , let be the set of first (with respect to ) vertices in . Let be the subtournament of induced by . Clearly, has vertices. We claim that is the required kernel of . We need to show that has arc-disjoint cycles if and only if has arc-disjoint cycles. The reverse direction of the claim holds trivially. Let us now prove the forward direction. Suppose has a set of arc-disjoint cycles. Among all such sets, let be one that minimizes . Suppose there is a cycle in that is not in . Then, there is a vertex that is not in . As argued earlier, any cycle in has at least two vertices from . Let and be two such vertices in where is a path in from to with internal vertices from .

The subtournaments and are transitive with unique topological orderings and , respectively. Observe that for all distinct , if and only if . As is a path in , it follows that for each . Similarly, as is a path in , we have for each . As , it follows that and . Then, there is at least one vertex in such that the arcs and are not in any cycle in . Now, as , and . Thus, we have . As , it follows that as . Then, by replacing the path by , we obtain another set of arc-disjoint cycles such that . However, this leads to a contradiction by the choice of . ∎

## 5 An Improved FPT Algorithm and A Smaller Kernel

Next, we show that ACT can be solved in time and admits a kernel with vertices.

### 5.1 A Linear Vertex Kernel

We show that the linear kernelization described in  for Feedback Arc Set in Tournaments also leads to a linear kernelization for our problem. In order to describe the kernel, we need to state some terminology defined in . Let be a tournament on vertices. First, we apply the following reduction rule.

###### Reduction Rule 5.1.

If a vertex is not in any cycle, then delete from .

This rule is clearly safe as our goal is to find cycles and cannot be in any of them. To describe our next rule, we need to state some terminology and a lemma known from . For an ordering of , let denote the tournament whose vertices are ordered according to . Clearly, and since and denote the same tournament. An arc is called a back arc if and it is called a forward arc otherwise. An interval is a consecutive set of vertices in .

###### Lemma 11 ().
111Lemma 11 is Lemma 3.9 of  that has been rephrased to avoid the use of several definitions and terminology introduced in .

Let be an ordered tournament on which Reduction Rule 5.1 is not applicable. Let denote the set of back arcs in and denote the set of arcs in with endpoints in different intervals. If , then there exists a partition of into intervals with the following properties that can be computed in polynomial time.

• There is at least one arc with .

• There are arc-disjoint cycles using only arcs in .

Our reduction rule that is based on this lemma is as follows.

###### Reduction Rule 5.2.

Let be an ordered tournament on which Reduction Rule 5.1 is not applicable. Let denote the set of back arcs in and denote the set of arcs in with endpoints in different intervals. Let be a partition of into intervals satisfying the properties specified in Lemma 11. Reverse all arcs in and decrease by .

###### Lemma 12.

Reduction Rule 5.2 is safe.

###### Proof.

Let be the tournament obtained from by reversing all arcs in . Suppose has arc-disjoint cycles. Then, it is guaranteed that each such cycle is completely contained in an interval. This is due to the fact that has no back arc with endpoints in different intervals. Indeed, if a cycle in uses a forward (back) arc with endpoints in different intervals, then it also uses a back (forward) arc with endpoints in different intervals. It follows that for each arc , neither nor is used in these cycles. Hence, these cycles in are also cycles in . Then, we can add a set of cycles obtained from the second property of Lemma 11 to these cycles to get cycles in . Conversely, consider a set of cycles in . As argued earlier, we know that the number of cycles that have an arc that is in is at most . The remaining cycles (at least of them) do not contain any arc that is in , in particular, they do not contain any arc from . Therefore, these cycles are also cycles in . ∎

###### Theorem 13.

ACT admits a kernel with vertices.

###### Proof.

Let denote the instance obtained from the input instance by applying Reduction Rule 5.1 exhaustively. From Lemma 4, we know that either has arc-disjoint triangles or has a feedback arc set of size at most that can be obtained in polynomial time. In the first case, we return a trivial yes-instance of constant size as the kernel. In the second case, let be the feedback arc set of size at most of . Let denote a topological ordering of the vertices of the directed acyclic graph . As , is an ordering of such that has at most back arcs. If , then from Lemma 11, there is a partition of into intervals with the specified properties. Therefore, Reduction Rule 5.2 is applicable (and the parameter drops by at least 1). When we obtain an instance where neither of the Reduction Rules 5.1 and 5.2 is applicable, it follows that the tournament in that instance has at most vertices. ∎

### 5.2 A Faster FPT Algorithm

Here, we show that ACT can be solved in time. The idea is to reduce the problem to the following Arc-Disjoint Paths problem in directed acyclic graphs.

Arc-Disjoint Paths Parameter: Input: A digraph on vertices and ordered pairs of vertices of . Question: Do there exist arc-disjoint paths in such that is a path from to for each ?

On directed acyclic graphs, Arc-Disjoint Paths is known to be -complete , -hard  and solvable in time . Despite its fixed-parameter intractability, we will show that we can use the algorithm to describe another (and faster)  algorithm for ACT.

###### Theorem 14.

ACT can be solved in time.

###### Proof.

Consider an instance of ACT. Using Theorem 13, we obtain a kernel such that has vertices. Further, . By definition, is a yes-instance if and only if is a yes-instance. Using Theorem 4, we know that either contains arc-disjoint triangles or has a feedback arc set of size at most that can be obtained in polynomial time. If Theorem 4 returns a set of arc-disjoint triangles in , then we declare that is a yes-instance.

Otherwise, let be the feedback arc set of size at most returned by Theorem 4. Let denote the (acyclic) digraph obtained from by deleting . Observe that has vertices. Suppose has a set of arc-disjoint cycles. For each , we know that as is a feedback arc set of . We can guess that subset of such that . Then, for each cycle , we can guess the arcs from that it contains and also the order in which they appear. This information is captured as a partition of into sets, to and the set of permutations where is a permutation of for each . Any cycle that has contains a -path between every pair , of consecutive arcs of with arcs from . That is, there is a path from and with arcs from for each . The total number of such paths in these cycles is and the arcs of these paths are contained in which is a (simple) directed acyclic graph.

The number of choices for is and the number of choices for a partition of and a set of permutations is . Once such a choice is made, the problem of finding arc-disjoint cycles in reduces to the problem of finding arc-disjoint cycles in such that for each and for each , has a path between and with arcs from . This problem is essentially finding arc-disjoint paths in and can be solved in time using the algorithm in . Therefore, the overall running time of the algorithm is as and . ∎

## 6 Concluding Remarks

We initiated the parameterized complexity study of the Arc-Disjoint Cycle Packing problem on tournaments. We showed that it is  when parameterized by the solution size and admits a linear vertex kernel. However, the classical complexity status of the problem is still open, i.e, we do not know if it is -hard or not. Resolving the same is a natural future research direction. We conjecture that it is indeed -hard. Note that the classical complexity status of the dual problem (Feedback Arc Set in Tournaments) was a long-standing open problem until it was shown to be -hard [2, 12, 14].

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