## 1 Introduction

Graph connectivity is a crucial graph property studied in the context of network robustness. Well-studied notions of connectivity consider for example hamiltonicity, edge-disjoint spanning trees, edge cuts, vertex cuts, etc. In this paper, we study the notion of a *disconnected cut*, which is a vertex set of a connected graph such that is disconnected and the subgraph induced by is disconnected as well. Alternatively, we may say that can be partitioned into sets such that no vertex of is adjacent to a vertex of (that is, is anti-complete to ) and is anti-complete to ; then both and form a disconnected cut. See Figure 1 for an example. The Disconnected Cut problem asks whether a given
connected
graph has a disconnected cut.

The Disconnected Cut problem is intimately connected to at least five other problems studied in the literature. We give a brief overview here, and refer to the related work section for more details. The name Disconnected Cut originates from Fleischner et al. [15], who determined the complexity of partitioning the vertices of a graph into exactly bicliques (complete bipartite graphs with at least one edge), except for the case . For , this problem is polynomially equivalent to Disconnected Cut (by taking the complement of the input graph). The Disconnected Cut problem can also be seen as an -Partition problem for appropriately defined 4-vertex graphs . Dantas et al. [8] proved that -Partition is polynomial-time solvable for each 4-vertex graph except for the two cases equivalent to Disconnected Cut. If the input graph has diameter , then Disconnected Cut is equivalent to -Compaction [15], which asks for a homomorphism from a graph to the graph (the 4-vertex-cycle with a self-loop in each vertex) such that for every with there is an edge with and . The diameter- case is also equivalent to testing if a graph can be modified to a biclique by a series of edge contractions [23]. The restriction to graphs of diameter is natural, as Disconnected Cut is trivial otherwise [15]. Finally, Disconnected Cut fits in the broad study of vertex cut problems with extra properties on the cut set, such as -cuts, -cuts, (strict) -clique cuts, stable cuts, matching cuts, etc.; see e.g. [24] for an overview.

The above demonstrates that Disconnected Cut is of central importance to understanding many different types of problems, ranging from cut problems to homomorphism and graph contractibility problems. Therefore, there has been broad interest to determine its computational complexity. Indeed, numerous papers [6, 8, 9, 10, 15, 22, 23, 31] asked about its complexity on general graphs. NP-completeness was proven independently in [28] and by Vikas, as announced in [33]. The strong interest in Disconnected Cut also led to a study on graph classes. We know polynomial-time algorithms for many classes, including graphs of bounded maximum degree, graphs not locally connected, graphs with a dominating edge (which include cobipartite graphs and -free graphs) [15], -free graphs, co-spiders, co--sparse graphs, co-circular arc graphs [6], apex-minor-free graphs (which generalize planar graphs), chordal graphs [23], -free graphs (graphs with independence number at most 4), graphs of bounded treewidth, -free graphs (co-diamond-free graphs), -free graphs, co-planar graphs, co--graphs (for every fixed integer ), and -free graphs (which contains the class of triangle-free graphs) [9]. The latter is the complement of the well-known class of claw-free graphs (graphs with no induced claw ).

Our interest in Disconnected Cut on claw-free graphs is heightened by the close relation of this problem to -Contractibility, which is to decide if a graph contains the -vertex cycle as a contraction. This problem is NP-complete if [3] and stays NP-complete for claw-free graphs as long as [14]. Given that the case is polynomial-time solvable even for general graphs [3], this leaves open on claw-free graphs the cases where . Ito et al. [23] showed that -Contractibility on claw-free graphs of diameter is equivalent to Disconnected Cut. As Disconnected Cut is trivial if the input graph does not have diameter , this led Ito et al. [23] to explicitly ask the following:

What is the computational complexity of Disconnected Cut on claw-free graphs?

### 1.1 Our Contribution

We answer the open question of Ito et al. [23] by giving a polynomial-time algorithm for Disconnected Cut on claw-free graphs. This immediately implies that besides -Compaction, also -Contractibility is polynomial-solvable on claw-free graphs of diameter , thus improving our understanding of these problems too. As claw-free graphs are not closed under edge contraction, the latter is certainly not expected beforehand.

Our result is grounded in a new graph-theoretic theorem that proves that claw-free graphs of diameter belong to one of four basic graph classes after performing two types of elementary operations. We believe this novel structural theorem to be of independent interest. The theorem builds on one of the algorithmic decomposition theorems for claw-free graphs developed by Hermelin et al. [20, 21], and relies on the pioneering works of Chudnovsky and Seymour (see [5]). Several other algorithmic decomposition theorems for claw-free graphs have been built on the ideas of Chudnovsky and Seymour, see e.g. [11, 25], which jointly have had a broad impact on our algorithmic understanding of claw-free graphs (see [20] for an overview). Our structural theorem and resulting algorithm for Disconnected Cut expand this line of research, and we hope it will prove similarly useful for future work.

The crux of the proof of our structural theorem is to exploit the extra structure offered by claw-free graphs of diameter to show that the so-called strip-structures, which are central to the aforementioned decomposition theorems, only contain trivial strips. An important ingredient in the proof is to exclude not only twins (vertices for which ), but also vertices with nested neighbourhoods (vertices for which there exists a vertex such that ). Using this operation, one can simplify the decomposition theorem of Hermelin et al. [20], and thus we think this observation has an impact beyond this work. Indeed, our final decomposition for claw-free graphs of diameter is much cleaner to state and easier to understand than the one for general claw-free graphs.

Using the structural theorem, Disconnected Cut on claw-free graphs reduces to understanding its behavior under the elementary operations and on the basic graph classes. The crucial elementary operation is to remove certain cobipartite structures called W-joins. Intuitively, a W-join is a cobipartite induced subgraph such that each vertex of the rest of the graph is complete to one or two sides of the cobipartition of the W-join, or wholly anti-complete to the W-join. We develop the notion of unshatterable proper W-joins, which are essentially W-joins that cannot be broken into smaller W-joins, and exhibit how unshatterable proper W-joins interact with disconnected cuts. We then show that unshatterable proper W-joins can be removed from the graph by a simple operation. We complete our arguments by proving that all W-joins in the graph must be in fact be unshatterable proper W-joins, and that we can find unshatterable proper W-joins in polynomial time.

The main basic graph classes in the structural theorem are line graphs and proper circular-arc graphs. Prior to our work, the complexity of Disconnected Cut was unknown for these classes as well. We present a polynomial-time algorithm for line graphs and even for general circular-arcs graphs (not only proper-circular arcs). Both algorithms rely on the existence of a small induced cycle passing through a disconnected cut in a highly structured matter. In addition, for line graphs, we prove that the pre-image of the line graph is -free, and thus has diameter at most . The hardest part of the proof is then to prove that if the pre-image has diameter exactly , then the line graph has in fact no disconnected cut.

### 1.2 Related Work

As mentioned, the name Disconnected Cut stems from Fleischner et al. [15], who studied how to partition the vertices of a graph into exactly bicliques, where Disconnected Cut is equivalent to the case . However, Disconnected Cut originates from -partitions, introduced in [8]. A model graph on vertices has solid and dotted edges. An -partition of a graph is a partition of into nonempty sets such that for every pair of vertices and : if is a solid edge of , then ; and if is a dotted edge of , then (if , then or are both allowed). The corresponding decision problem is called -Partition. Dantas et al. [8] proved -Partition is polynomial-time solvable for every 4-vertex model graph except , which has solid edges , and no dotted edges, and , which has dotted edges and no solid edges. As a graph has a disconnected cut if and only if it has a -partition if and only if its complement has a -partition, these two cases are polynomial-time equivalent to Disconnected Cut. Hence, we now know that, as a matter of exception, -Partition is NP-complete if [28].

We can encode a model graph as a matrix in which every entry is either 0 (dotted edge), 1 (solid edge) or (no restriction). If we allow sets in a solution for -Partition to be empty, then we obtain the -Partition problem, introduced by Feder et al. [13].
This well-known problem generalizes many classical problems involving vertex cuts and partitions, including -Colouring and -Colouring;
see also [18].
An even more general variant is to give every vertex a list and to search for a solution, in which each vertex may only belong to a set with . This yields the List -Partition problem, which includes well-known cases, such as the Stubborn problem,
which turned out to be polynomial-time solvable [7], in contrast to Disconnected Cut.
A homomorphism from to is a *retraction* if contains as an induced subgraph and for every . The corresponding decision version is called -Retraction. Let be the 4-cycle with a self-loop in each vertex. Then -Retraction is a special case of List -Partition where the input graph contains a cycle on four specified vertices with for and for .
This problem is a generalization of Disconnected Cut.
Feder and Hell [12] proved that -Retraction is NP-complete. Hence, List -Partition and List -Partition are NP-complete. Note that this result is also implied by the NP-completeness of -Partition [28].

Vikas [32] solved an open problem of Winkler (see [13, 32]) by proving NP-completeness of -Compaction, the variant of the -Partition problem with the extra constraint that there must be at least one edge with and for (where ). Generally, a homomorphism from a graph to a graph is a compaction if is edge-surjective, i.e., for every with there is an edge with and . The corresponding decision problem is called -Compaction. If , then the problem is equivalent to Disconnected Cut when restricted to graphs of diameter [15]. Hence, -Compaction is NP-complete for graphs of diameter [28] (the result of [32] holds for graphs of diameter at least 3). Similarly, a homomorphism from a graph to a graph is (vertex-)surjective if for every there is a vertex such that . The decision problem is called Surjective -Colouring (or -Vertex Compaction, or Surjective -Homomorphism) and is equivalent to Disconnected Cut if . The complexity classifications of -Compaction and Surjective -Colouring are wide open despite many partial results; see [2] for a survey and [16] for a more recent overview focussing on Surjective -Colouring .

### 1.3 Overview

In Section 2 we state several underlying structural observations for graphs of diameter . In Sections 3 and 4, respectively, we prove that Disconnected Cut can be solved in polynomial time for circular-arc graphs and line graphs, respectively. In Section 5 we prove our main result, and in particular, our new structural theorem for claw-free graphs of diameter . In Section 6 we show that Disconnected Cut is polynomial-time solvable on paw-free graphs, co-paw-free graphs and on diamond-free graphs. By combining these results with our result for claw-free graphs we prove that Disconnected Cut is polynomial-time solvable for -free graphs whenever is a graph on at most four vertices not isomorphic to the complete graph . We pose the case where as an open problem in Section 7, together with some other relevant open problems.

## 2 Preliminaries and Basic Results

In the remainder of our paper, graphs are finite, undirected, and have neither multiple edges nor self-loops unless explicitly stated otherwise.

Let be a graph. For a set , is the subgraph of induced by .
We say that is connected if is connected.
We write , and if , we write instead.
For a vertex , let be the neighbourhood of
and .
The complement of has vertex set and edge set
.
The *contraction* of an edge is the operation that removes the vertices and from , and replaces and by a new vertex that is made adjacent to precisely those vertices that were adjacent to or in (without introducing self-loops nor multiple edges).
The distance between two vertices and of is the number of edges in a shortest path between them.
If and are in different connected components of , then .
The diameter of is equal to .

The following lemma was observed by Fleischner et al.

###### Lemma 1 ([15])

Let be a graph. If has diameter , then has no disconnected cut. If has diameter at least , then has a disconnected cut.

A subset is a dominating set of a graph if every vertex of is adjacent to at least one vertex of . If , then is a dominating vertex of . An edge of a graph is dominating if is dominating. A vertex has a disconnected neighbourhood if induces a disconnected graph.

We need the following two lemmas, the first one of which is a straightforward observation.

###### Lemma 2

If a graph contains a dominating vertex, then has no disconnected cut.

###### Lemma 3

If a graph contains a non-dominating vertex with a disconnected neighbourhood, then has a disconnected cut.

###### Proof

Let be the connected components of for some . As is not dominating, is nonempty. We define , , and and find that (or ) is a disconnected cut of . ∎

Two disjoint vertex sets and in a graph are complete if there is an edge between every vertex of and every vertex of , and and are anticomplete if there is no edge between a vertex of and a vertex of . Recall that has a disconnected cut if can be partitioned into four (nonempty) sets , , , , such that is anticomplete to and is anticomplete to . We say that form a disconnected partition of .

###### Lemma 4

Let , , , be a disconnected partition of a graph of diameter . Then has an induced cycle with such that for .

###### Proof

Let and . As has diameter 2, there exists a vertex in or , say , such that is adjacent to and to . Let . As has diameter 2, there exists a vertex in or , say , such that is adjacent to and . If and are adjacent, then we can take as the cycle on vertices , , , in that order. Otherwise, as has diameter 2, there exists a vertex , such that is adjacent to and to . In that case we can take as the cycle on vertices , , , , . ∎

Two adjacent vertices and of graph have a nested neighbourhood if or .
We say that has *distinct neighbourhoods* if has no two vertices that have nested neighbourhoods.
In our proof we will apply the following lemma exhaustively.

###### Lemma 5

Let be a graph of diameter that contains two vertices and such that . Then has a disconnected cut if and only if has a disconnected cut. Moreover, has diameter at most .

###### Proof

As has diameter 2 and , we find that has diameter at most 2.

First suppose that has a disconnected cut. Then has a disconnected partition . We may assume without loss of generality that belongs to . First assume that . Hence, and thus, form a disconnected partition of . Hence, has a disconnected cut. Now assume that . As is adjacent to , we may assume without loss of generality that belongs to . As belongs to and is anticomplete to , we find that has no neighbours in . As , this means that has no neighbours in . As has diameter 2, this means that contains a vertex that has a neighbour in , thus is nonempty. As a consequence, form a disconnected partition of . Hence, has a disconnected cut.

Now suppose that has a disconnected cut. Then has a disconnected partition . We may assume without loss of generality that belongs to . Then, as , we find that form a disconnected partition of . Hence, has a disconnected cut. ∎

A pair of vertices and of a graph is a universal pair if is a dominating set and there exist distinct vertices and in , such that and ; note that this implies that and have at least one neighbour in . Let be a graph. Then is -free if contains no induced subgraph isomorphic to . The disjoint union of two vertex-disjoint graphs and is the graph . The disjoint union of copies of a graph is denoted by . The graphs and denote the cycle and path on vertices, respectively. The graph denotes the complete graph on vertices. The independence number of a graph is the largest such that contains an induced subgraph isomorphic to .

###### Lemma 6 ([6])

A -free graph has a disconnected cut if and only if its complement has a universal pair.

###### Lemma 7 ([9])

Disconnected Cut is -time solvable for -free graphs.

A graph is bipartite if can be partitioned into two classes and such that every edge of has an endpoint in and an endpoint in . If is complete to , then is a complete bipartite graph. The graph denotes the complete bipartite graph wit partition classes of size and , respectively. The graph is the claw . A cobipartite graph is the complement of a bipartite graph.

The line graph of a graph with edges is the graph with vertices such that there is an edge between any two vertices and if and only if and have a common endpoint in . Note that every line graph is claw-free. We call the preimage of . Every connected line graph except has a unique preimage [17].

A circular-arc graph is a graph that
has a representation in which each vertex
corresponds to an arc of a circle, such that two vertices
are adjacent if and only if their corresponding arcs intersect.
An interval graph is a graph that has representation in which each vertex corresponds to an interval of the line, such that two vertices are adjacent if and only if their corresponding intervals intersect. Note that circular-arc graphs generalize interval graphs.
A circular-arc or interval graph is *proper* if it has a representation where the arcs respectively intervals are such that no one is contained in another.
A chordal graph is a graph in which every induced cycle is a triangle; note that every interval graph is chordal, but that there exist circular-arc graphs, such as cycles, that are not chordal.

## 3 Circular-Arc Graphs

In this section we prove that Disconnected Cut is polynomial-time solvable for circular-arc graphs. This result is known already for interval graphs, as it follows from the result that Disconnected Cut is polynomial-time solvable for the class of chordal graphs [23], which contains the class of interval graphs. In fact, we have an -time algorithm for interval graphs. Due to Lemma 4 and the fact that interval graphs are chordal, no interval graph of diameter 2 has a disconnected cut. Consequently, an interval graph has a disconnected cut if and only if its diameter is at least 3 due to Lemma 1. To show that Disconnected Cut is polynomial-time solvable for circular-arc graphs requires significant additional work.

Let be a circular-arc graph. For each vertex we can associate an arc where we say that is the clockwise left endpoint of and is the clockwise right endpoint of . The following result of McConnell shows that we may assume that all left and right endpoints of the vertices of are unique.

###### Lemma 8 ([29])

A circular-arc graph on vertices and edges can be recognized in time. In the same time, a representation of can be constructed with distinct arc endpoints that are clockwise enumerated as .

We need the following lemma.

###### Lemma 9

Let be a circular-arc graph of diameter with a disconnected cut. Then has a disconnected partition , , , such that each is connected.

###### Proof

By Lemma 8 we may assume that has a representation with distinct arc endpoints clockwise enumerated as . Let , , , be a disconnected partition of . By Lemma 4, contains a cycle with vertices for (with ) and , such that for . We may assume without loss of generality that if , then .

Let be the connected component of that contains for . Note that if , then . We let denote the arc corresponding to the arc covered by the vertices of (or equivalently, the arc associated with the vertex obtained after contracting the connected graph into a single vertex). Note that the arcs , …, cover the whole circle, as they contain the arcs corresponding to the vertices of . Moreover, intersects with and for (where ).

We set . If for , then , , , is our desired partition of . Assume that this is not the case. Then . Consider an arbitrary vertex that is not in any . Let be the number of arcs that intersect with . We claim that .

First suppose that . Then would be adjacent to a vertex of every . This is not possible, because , , , is a disconnected partition of . Now suppose that , say intersects with , and . Then must be in , that is, is in the same set as the vertices of . As intersects with we would have put in , a contradiction. Now suppose that . By construction, intersects with two consecutive arcs, say with arcs and . Then or . If then we would have put in , and if then we would have put . Hence, this situation is not possible either.

From the above, we conclude that , that is, for each , we have that intersects with at most one . As the arcs cover the whole circle, this means that is contained in some , and we can safely put in without destroying the connectivity or the anticompleteness of , or of . Afterwards, the four sets , , , form a disconnected partition of , such that each is connected. ∎

We are now ready to prove the main result of this section.

###### Theorem 3.1

Disconnected Cut is -time solvable for circular-arc graphs.

###### Proof

Let be a circular-arc graph on vertices. We compute the diameter of in time using the
(more general) -time
algorithm of [4] (or of [1] or the linear-time algorithm of [27]).
Lemma 1 tells us that if has diameter 1, then has no disconnected cut, and if has diameter at least 3, then has a disconnected cut.
Assume that has diameter . Lemma 9 tells us that if has a disconnected cut, then has a disconnected partition , , , such that is connected for .
We say that the *arc of a set * is the union of all the arcs of the vertices in . As has diameter 2, the union of the arcs of the sets cover the whole circle.
Moreover, the arcs of and are disjoint and the arcs of and are disjoint.

We now compute, in linear time, a representation of with distinct arc endpoints clockwise enumerated as via Lemma 8. After sorting the arcs in time, we apply the following procedure for each . We take a neighbour with right-most right endpoint amongst all neighbours of . We then take a neighbour with right-most right endpoint amongst all neighbours of . We check if is also a neighbour of . If so, then form a triangle if and an edge if such that the corresponding arcs cover the whole circle. In that case has no disconnected partition , , , such that is connected for (as the arcs of the three vertices of the triangle must be placed in the corresponding arcs of the sets ), and then, by Lemma 9, we find that has no disconnected cut. Note that this procedure takes time in total.

Suppose has no pair or triple of vertices whose arcs cover the whole circle. We perform the following procedure exhaustively. We pick a vertex of and take a neighbour with right-most right endpoint amongst all neighbours of . We then do the same for , and so on. If this procedure ends without yielding an induced cycle, then has no induced cycle on four or more vertices. Hence, has no disconnected cut due to Lemma 4. Otherwise, we have found in time, an induced cycle with vertices , in that order, for some . As has diameter 2, we find that . Moreover, by construction and because is circular-arc, the arcs corresponding to the vertices must cover the whole circle.

If has a disconnected partition , , , such that is connected for , then the above implies the following. If , we may assume without loss of generality that for . If , two vertices belong to the same set , whereas the other sets with each contain a single vertex from . If , then we guess which two vertices will be put in the same set, say ; this does not influence the asymptotic running time. Now we build up the sets from scratch by putting in the vertices from .

We will always maintain that each induces a connected graph, and thus, the union of the arcs of the vertices in indeed always form an arc.
We say that a vertex *intersects* a set if the arc of intersects the arc of . Note that, since the arcs corresponding to cover the entire circle, so do the arcs of the sets that we are constructing.
If there is a vertex that intersects each of the sets constructed so far, then there is no disconnected cut with each connected. If , this mean that has no disconnected cut due to Lemma 9. If , our guess of vertices to belong to may have been incorrect, and we need to put two other consecutive vertices of in the same set before concluding that has no disconnected cut.

Otherwise, we do as follows. Note that any vertex that intersects two sets and for some (say, without loss of generality), also intersects or (where ). We now put a vertex that intersects two sets and for some into set if intersects as well; otherwise, intersects and we put in .

Let be the set of vertices of that we have not placed in some set yet. We claim that each vertex of must intersect with exactly two sets and such that, in addition, holds. For contradiction, assume this does not hold for . Then intersects exactly one set , say . There must exist a path from to of length 2, as has diameter 2. Hence, there exists a vertex that is adjacent to both and . This means that the arc corresponding to must intersect and . Hence, has already been placed in or and thus the arc of intersects two sets, namely and , or and , a contradiction. The claim follows.

As every vertex in intersects two sets and , we can model the remaining instance as an instance of 2-Satisfiability as follows. Let , where must be placed in, say, or . We introduce two variables and with clauses and . For each edge where must be placed in or and must be placed in or , we introduce the clause . For each edge where must be placed in or and must be placed in or , we introduce the clauses and .

The correctness of our algorithm follows from the above description. Assigning the vertices to the sets takes time, whereas solving the corresponding instance of 2-Satisfiability takes time as well. As computing the diameter takes time and all other steps take time as well, the total running time is . ∎

## 4 Line Graphs

In this section we prove that Disconnected Cut is polynomial-time solvable for line graphs. We start with the following lemma due to Ito et al. [23].

###### Lemma 10 ([23])

Let be a graph with diameter whose line graph also has diameter . Then has a disconnected cut if and only if has a disconnected cut.

We need a lemma on graphs whose line graph has diameter 2.

###### Lemma 11

Let be a graph that is neither a triangle nor a star. Then has diameter if and only if is -free.

###### Proof

Let be a graph that is neither a triangle nor a star. First suppose that has diameter 2. In order to obtain a contradiction, assume that is not -free. Then contains an induced subgraph with vertices and edges and . As is an induced subgraph of , we find that and are non-adjacent vertices in . Then, because has diameter 2, contains a vertex that is adjacent to and . However, then is an edge with one endvertex in and the other one in . This means that is not induced, a contradiction.

Now suppose that is -free. In order to obtain a contradiction, assume that has diameter not equal to 2. If the diameter of is 1, then is a complete graph implying that is a triangle or star, which is not what we assume. If the diameter of is at least 3, then contains two vertices and that are of distance at least 3. This means that and form an induced in , a contradiction. ∎

We are now ready to prove the main result of this section.

###### Theorem 4.1

Disconnected Cut is -time solvable for line graphs of -vertex graphs.

###### Proof

Let be a graph on vertices and edges. We will show how to decide in time if has a disconnected cut. We first check in time if is a triangle or star. If so, then is a complete graph and thus has no disconnected cut. From now on suppose that is neither a triangle nor a star. By Lemma 11 we find that has diameter 2 if and only if is -free. Hence we can check in time, via checking if has an induced by brute force, if has diameter 2.

First assume that does not have diameter 2. As is not a triangle or a star, has diameter at least 3. By Lemma 1 we find that has a disconnected cut. Now assume that has diameter 2. We check in time if has an edge such that every vertex of is adjacent to at least one of . If so, then is a dominating vertex of , and has no disconnected cut due to Lemma 2. If not, then has no dominating vertices, and we proceed as follows. First we check if has a vertex with a disconnected neighbourhood, or equivalently, if contains an edge such that and have degree at least 2 and no common neighbours. This takes time. If has a vertex with a disconnected neighbourhood, then has a disconnected cut by Lemma 3. From now on assume that has no vertex with a disconnected neighbourhood. As is neither a triangle nor a star, is -free by Lemma 11. Hence, has diameter at most 3. We can determine in time the diameter of and consider each case separately.

Case 1. has diameter 1.

We claim that has no disconnected cut. For contradiction, assume that has a disconnected cut. Let , , , be a disconnected partition of .
By Lemma 4, contains a cycle
with vertices for (with ) and ,
such that for .
Then we may assume without loss of generality that for and .
As has diameter 1, is an edge of and thus a vertex of . In , is adjacent
to every vertex in , and thus to a vertex in for , a contradiction.

Case 2. has diameter 2.

Then has a disconnected cut if and only if has a disconnected cut due to Lemma 10.
By Lemma 6 it suffices to check if has a universal pair. This takes time.

Case 3. has diameter 3.

We will prove that has no disconnected cut.
As has diameter 3, does have a disconnected cut by Lemma 1. We need the following claim.

Claim. Let be a disconnected partition of . Then every cycle of with contains vertices of at most three distinct sets from .

We prove the Claim as follows. For contradiction, assume that has a cycle with vertices for , such that for . We may assume without loss of generality that for and . As is -free, we may assume without loss of generality that is in a singleton connected component of . If , then we may also assume without loss of generality that is in a singleton connected component of . If , then must be in a singleton connected component of due to the edge , which is contained in . This means that the sets and are disjoint. As and are edges of , both and are nonempty. Hence, the vertex has a disconnected neighbourhood in , a contradiction. This proves the Claim.

Now, for contradiction, assume that has a disconnected cut. Let , , , be a disconnected partition of . By Lemma 4, contains a cycle with vertices for (with ) and , such that for . Then we may assume without loss of generality that for and .

We define the following partition , , , of . Let . If is incident to only edges from one set , then we put in . Suppose is incident to edges from more than one set . As , , , is a disconnected partition of , we find that is incident to edges from and for some (where ) and to no other sets . In that case we put into .

We now prove that is anticomplete to . For contradiction, suppose that contains a vertex and contains a vertex such that . As , we find that is incident to edges only in and . Hence, . As , we find that is incident to edges only in and . Hence , a contradiction. By the same argument we can show that is anticomplete to . Let be the cycle with vertices in . Then , and thus , for . Hence, , , , is a disconnected partition of , and is a cycle in with for every . This is not possible due to the Claim. We conclude that has no disconnected cut.

The correctness of our algorithm follows from the above description. If has diameter 1 (Case 1) or diameter 3 (Case 3), no additional running time is required, as we showed that has no disconnected cut in both these cases. Hence, only executing Case 2 takes additional time, namely time . Hence the total running time of our algorithm is ), as desired. ∎

## 5 Claw-Free Graphs

In this section, we prove that Disconnected Cut is polynomial-time solvable on claw-free graphs. The proof consists of two parts. In Section 5.1 we show how to get rid of certain cobipartite structures in the graph, called W-joins. We remark that Disconnected Cut can be solved in polynomial time on cobipartite graphs [15]. Although this is a necessary condition for Disconnected Cut to be solvable in polynomial time on claw-free graphs, the algorithm for cobipartite graphs is not sufficient to deal with W-joins. In Section 5.2 we present our new decomposition theorem for claw-free graphs of diameter 2 and combine this theorem with the results from the previous sections and Section 5.1 to show our main result.

### 5.1 Cobipartite Structures versus Disconnected Cuts

We consider the following cobipartite structures that might be present in claw-free graphs [5, 20, 21].
A pair of disjoint non-empty sets of vertices is a *W-join* in graph if , and are cliques, is neither complete nor anticomplete to , and every vertex of is either complete or anticomplete to and either complete or anticomplete to . A W-join is a *proper W-join* if each vertex in is neither complete nor anticomplete to and each vertex in is neither complete nor anticomplete to . Observe that for a proper W-join , it must hold that . For any W-join , it holds that is a cobipartite induced subgraph in .

We assume that an input graph of Disconnected Cut has diameter and that has distinct neighbourhoods, by Lemmas 1 and 5 respectively. We show how to use these assumptions to remove all W-joins in a claw-free graph and obtain an equivalent instance of Disconnected Cut. As a first step, we show that we can focus on proper W-joins.

###### Lemma 12

Let be a graph with distinct neighbourhoods. If admits a W-join , then is a proper W-join.

###### Proof

We need to show that no vertex of (respectively ) is complete or anticomplete to (respectively ). Suppose there exists a vertex such that is anticomplete to . Since is not anticomplete to , there exists a vertex such that is adjacent to some vertex of . This implies that , which is a contradiction to the fact that has distinct neighbourhoods. Hence, no vertex of is anticomplete to

Comments

There are no comments yet.