Abstract
A permutation graph can be defined as an intersection graph of segments whose endpoints lie on two parallel lines and , one on each. A bipartite permutation graph is a permutation graph which is bipartite. In this paper we study the parameterized complexity of the bipartite permutation vertex deletion problem, which asks, for a given n-vertex graph, whether we can remove at most k vertices to obtain a bipartite permutation graph. This problem is -complete by the classical result of Lewis and Yannakakis [20]. We analyze the structure of the so-called almost bipartite permutation graphs which may contain holes (large induced cycles) in contrast to bipartite permutation graphs. We exploit the structural properties of the shortest hole in a such graph. We use it to obtain an algorithm for the bipartite permutation vertex deletion problem with running time , and also give a polynomial-time 9-approximation algorithm.
Keywords: Permutation graphs, Comparability graphs, Partially ordered set, Graph modification problems
Introduction
Many standard computational problems, including maximum clique, maximum independent set, or minimum coloring, which are -hard in general, have polynomial-time exact or approximation algorithms in restricted classes of graphs. Due to the practical and theoretical applications, some of such graph classes are particularly intensively studied. Among them are:
interval graphs: intersection graphs of intervals on a real line,
proper interval graphs: intersection graphs of intervals none of which is contained in another,
chordal graphs: intersection graphs of subtrees of a tree,
function and permutation graphs: intersection graphs of continuous and linear functions, respectively, defined on the interval [0, 1],
comparability graphs: graphs whose edges correspond to the pairs of vertices comparable in some fixed partial order on the vertex set (such an order is called a transitive orientation of the graph),
co-comparability graphs: the complements of comparability graphs.
It is well known that the class of function graphs corresponds to the class of co-comparability graphs [13], and the class of permutation graphs corresponds to the intersection of comparability and co-comparability graphs [24] (see Fig. 1 for the hierarchy of inclusions). All these classes of graphs are hereditary, which means that they are closed under vertex deletion.
Fig. 1.

Hierarchy of inclusions between graph classes considered in the introduction. An arrow from graph class to graph class indicates that
Being hereditary is a very useful property in algorithmic design as every hereditary class of graphs can also be uniquely characterized in terms of minimal forbidden induced subgraphs: a graph belongs to a class if and only if it does not contain any graph from some family as an induced subgraph. For every graph class introduced above, a characterization by forbidden subgraphs is known, see [8] for perfect graphs, [19] for interval graphs, [12] for comparability and permutation graphs. However, for all of them, the family of forbidden subgraphs is infinite and it may also be quite complex. Moreover, every graph G from any class introduced above is perfect. Grötschel, Lovász, and Schrijver [14] showed that in the class of perfect graphs the maximum clique, the maximum independent set, and the minimum coloring problems can be solved in polynomial time.
Polynomial-time algorithms devised for the above-mentioned graph classes can sometimes be adjusted to also work on graphs that are “close” to graphs from these classes. Usually, the “closeness” of a graph G to a graph class is measured by the number of operations required to transform G into a graph from the class , where a single operation consists either on removing a vertex from G or on adding or removing an edge from G. Such an approach leads us to the following generic problem.
| Problem: | Graph modification problem into a class of graphs |
| Input: | A graph G (typically not from ) and a number k |
| Question: | Can G be transformed into a graph of the class |
| by performing modifications of an appropriate kind? |
Depending on the kind of modifications allowed, we obtain four variants of this problem: vertex deletion problem, edge deletion problem, edge completion problem, and edge edition problem (the latter allowing both deletions and additions of edges). For the class of graphs defined above, all four variants of the modification problem are -hard—see [22] for references to -hardness proofs. In particular, Lewis and Yannakakis [20] showed that the vertex deletion problem into any non-trivial hereditary class of graphs is -hard. This is not surprising, as many classical hard problems can be formulated as vertex deletion problems into particular classes of graphs, for example, Vertex Cover as vertex deletion to edgeless graphs, Feedback Vertex Set as vertex deletion to forests, and Odd Cycle Transversal as vertex deletion to bipartite graphs.
Graph modification problems are a popular research direction in the study of the parameterized complexity of -complete problems. In general, for a problem , an input of a parameterized problem consists of an instance I of and a parameter . Then we say that is fixed parameter tractable () if there exists an algorithm deciding whether I is a yes-instance of in time , where f is some computable function. For a graph modification problem, we often choose the parameter k as a number of allowed modifications, so the instance of such a problem is still a pair (G, k).
It turns out that characterizations by forbidden structures are sometimes useful to design algorithms for graph modification problems. For example, Cai [4] proposed an algorithm for modification problems into classes of graphs characterized by a finite family of forbidden induced subgraphs . His algorithm identifies a forbidden structure in the input graph (which can be done in polynomial time when is finite) and branches over all possible ways of modifying that structure. Since the families of forbidden structures are infinite for graph classes introduced above, modification algorithms for these classes have to be much more sophisticated. For several of them modification problems have satisfactory solutions:
chordal graphs: all four versions of the modification problem are [7, 23];
interval graphs: edge completion and edge deletion are [5, 28], vertex deletion is [7], edge edition remains open;
proper interval graphs: all four versions of the modification problem are [6].
On the other hand, it is known that the vertex deletion to perfect graphs is -hard [15]. It is worth mentioning that for a long time, it was unknown whether there are classes of graphs recognizable in polynomial time for which modification problems are hard. The first such example was given by Lokshtanov [21], who proved that the vertex deletion is -hard for graphs avoiding all wheels (i.e., cycles with an additional vertex adjacent to all other vertices). It is unknown whether comparability graphs, co-comparability graphs, and permutation graphs have modification algorithms. The class of co-comparability graphs, which constitutes the superclass of interval graphs and an important subclass of perfect graphs, seems to be particularly interesting from the parameterized point of view.
Our focus.Like the class of interval graphs, the class of permutation graphs admits polynomial-time algorithms for rich family problems which are -complete in general. Apart from the already mentioned classical hard problems which are polynomial-time solvable for perfect graphs, there also exist polynomial algorithms solving e.g., Hamiltonian Cycle, Feedback Vertex Set or Dominating Set in the class of permutation graphs [3, 9].
In light of the above considerations, since all the modification problems into the class of permutation graphs—and the related classes of comparability and co-comparability graphs—remain open, restricting our attention to the class of bipartite permutation graphs appears to be a natural research direction. Bipartite permutation graphs form an interesting graph class themselves, first investigated by Spinrad, Brandstädt, and Stewart [25], who characterized them by means of appropriately chosen linear orderings of its bipartition classes. One of the most interesting results concerning the bipartite permutation graphs is by Heggernes et al. [16], who showed that the -complete problem of computing the cutwidth of a graph (i.e., finding a linear order of the vertices of a graph that minimizes the maximum number of edges intersected by any line inserted between two consecutive vertices) is polynomial for bipartite permutation graphs.
Our algorithm exploits the absence of some forbidden structures in bipartite permutation graphs. Since these structures cannot, in particular, occur in permutation graphs, we believe that besides being a complete result itself, our research is a step towards understanding the parameterized complexity of modification problems into permutation graphs.
Our results. We focus mainly on the modification by vertex deletion.
Theorem 1
There is an -time algorithm for instances (G, k) of the vertex deletion into bipartite permutation graphs problem.
We prove Theorem 1 in Sect. 4. Our algorithm is based on the characterization of bipartite permutation graphs by forbidden subgraphs. Using the characterization, at first, we get rid of constant-size forbidden subgraphs by branching, which is a standard technique in modification problems on hereditary graph classes [27, 28]. We call graphs without these forbidden subgraphs almost bipartite permutation graphs.
Our main contribution is in the structural analysis of almost bipartite permutation graphs which may contain holes (on more than ten vertices) in contrast to bipartite permutation graphs. This approach is partially inspired by the ideas of van ’t Hof and Villanger [27] who used similar tools in their work on proper interval vertex deletion problem. We use the result of Spinrad, Brandstädt, and Stewart [25], who showed that the vertices of every connected bipartite permutation graph can be embedded into a strip in such a way that the vertices from U are on the bottom edge of the strip, the vertices from W are on the top edge of the strip, the neighbors N(u) of u occur consecutively on the top edge of the strip for every (adjacency property), the vertices from occur consecutively on the top edge of the strip for every (enclosure property), and the analogous properties are satisfied by the vertices in W (see Fig. 2).
Fig. 2.

Embedding of a bipartite permutation graph (U, W, E) into a strip satysfying the adjacency and the enclosure properties
Our structural result (see the discussion after Lemma 1) asserts that, depending on the parity of the length of the shortest hole, a connected almost bipartite permutation graph may be naturally embedded in either a cylinder, or a Möbius strip, locally satisfying adjacency and enclosure properties (see Fig. 3).
Fig. 3.

An embedding of a connected almost bipartite permutation graph in a cylinder or a Möbius strip that locally satisfies the adjacency and enclosure properties
Once we obtain such structure, we show that every minimal vertex cut that destroys all holes lies nearby a few consecutive vertices from the shortest hole. This allows us to check all the possibilities where we can find a minimum cut. Finally, we use a polynomial algorithm for finding maximum flow (and thus a minimum cut).
The approach used to prove Theorem 1 can be slightly modified to obtain a 9-approximation algorithm for the bipartite permutation vertex deletion problem. We show the following.
Theorem 2
There exists a polynomial-time 9-approximation algorithm for vertex deletion into bipartite permutation graphs problem.
Preliminaries
Unless stated otherwise, all graphs considered in this work are simple, i.e., undirected, with no loops and parallel edges. Let be a graph. For a subset , the subgraph of G induced by S is the graph . The neighborhood of a vertex is the set . Similarly, we write for a set . Let . We say that u and v are at distance k (in G) if k is the length of a shortest path between u and v in G. We denote a complete graph and a cycle on n vertices by and , respectively. By hole we mean an induced cycle on at least five vertices. We say that a hole is even (or odd) if it contains even (odd) number of vertices, respectively.
For a graph , a pair is a transitive orientation of G if is a transitive and irreflexive relation on V that satisfies either or iff for every .
A partially ordered set (shortly partial order or poset) is a pair that consists of a set X and a reflexive, transitive, and antisymmetric relation on X. For a poset , let the strict partial order be a binary relation defined on X such that if and only if and . Equivalently, is a strict partial order if is irreflexive and transitive. Two elements are comparable in P if or ; otherwise, x, y are incomparable in P. A linear order is a partial order in which every two vertices are comparable. A strict linear order is a binary relation defined in a way that if and only if and .
Let be a poset. A linear order is called a linear extension of P if . Given a family of posets , we say that P is the intersection of if for every we have if and only if for every . The dimension of a poset P is the minimal number of linear extensions of P that intersect to P. In particular, we say that P is two-dimensional if it is the intersection of two linear extensions of P.
A comparability graph (incomparability graph) of a poset has X as the set of its vertices and the set including every two vertices comparable (incomparable, respectively) in P as the set of its edges. Note the following: if is a poset, then is a transitive orientation of the comparability graph of P. A graph is a comparability graph (co-comparability graph) if G is a comparability (incomparability, respectively) graph of some poset defined on V. So, G is a comparability graph if and only if G admits a transitive orientation. A graph G is a permutation graph if and only if G and the complement of G are comparability graphs [24] (or equivalently, G and the complement of G admit transitive orientations). Baker, Fishburn, and Roberts [1] proved that G is a permutation graph if and only if G is the incomparability graph of a two-dimensional poset.
We say that two sets X and Y are comparable if X and Y are comparable with respect to -relation (that is, or holds). We use the convenient (although non-standard) notation for every . For every such that by [i, j] we mean the set .
The Structure of (Almost) Bipartite Permutation Graphs
The characterization of bipartite permutation graphs presented below was proposed by Spinrad, Brandstädt, and Stewart [25].
Suppose is a connected bipartite graph. A linear order satisfies adjacency property if for each vertex the set N(u) consists of vertices that are consecutive in . A linear order satisfies enclosure property if for every pair of vertices such that N(u) is a subset of , vertices in occur consecutively in . A strong ordering of the vertices of consists of linear orders and such that for every , where , and , it holds that and imply and . Note that, whenever and form a strong ordering of , then and satisfy the adjacency and enclosure properties.
Theorem 3
(Spinrad, Brandstädt, Stewart [25]) The following three statements are equivalent for a connected bipartite graph :
(U, W, E) is a bipartite permutation graph.
There exists a strong ordering of .
There exists a linear order of W satisfying adjacency and enclosure properties.
An example of a bipartite permutation graph with linear order of the vertices of W which satisfies the adjacency and the enclosure properties is shown in Fig. 2.
Another characterization of bipartite permutation graphs can be obtained by listing all minimal forbidden induced subgraphs for this class of graphs. Such a list can be compiled by taking all odd cycles of length (forbidden structures for bipartite graphs) and all bipartite graphs from the list of forbidden structures for permutation graphs obtained by Gallai [12]. The whole list is shown in Fig. 4.
Fig. 4.

Forbidden structures for bipartite permutation graphs
Almost Bipartite Permutation Graphs
The goal of this section is to characterize graphs which do not contain small forbidden subgraphs for the class of bipartite permutation graphs. Following terminology of van ’t Hof and Villanger [27] we call such graphs almost bipartite permutation graphs.
Definition 1
A graph is an almost bipartite permutation graph if G does not contain , , , , for as induced subgraphs.
Suppose is a connected almost bipartite permutation graph.
Proposition 1
Every hole in G is a dominating set.
Proof
Let be a hole in G. Hence, Suppose, for contradiction, that there exists a vertex in the set . As G is connected, there must exist at distance two from C. Let and let be a neighbor of w in C. We now look at the neighborhood of w. As G contains no triangle, and are non-edges. Moreover, as G contains no copy of , vertex w is adjacent to at least one of and , say . Thus, w is nonadjacent to . Therefore, the set induces a copy of in G, which leads to a contradiction.
Let C be a shortest hole in G, m be the size of C, and be the consecutive vertices of C, . In the remaining part of the paper we use the following notation with respect to C. For any integral number i by we denote the unique vertex from the cycle C. For any two different vertices in C, by the set of all vertices between and from C we mean the set , where k is the smallest natural number such that . Note that this notion is not symmetric, i.e., the set of all vertices between and from C contains and all the vertices from C that are not between and .
Proposition 2
For every vertex either:
for some , or
for some .
Proof
Since C is an induced cycle, (2) clearly holds for the vertices from C, so let v be a vertex in . As C is a dominating set, by Proposition 1, vertex v has at least one neighbor in C. If v has exactly one neighbor in C, then (1) holds and we are done. So assume that it has more than one neighbor. We now distinguish two cases. First, suppose that there exist two vertices at distance at least three in C such that v has no neighbor in the set of vertices between and , except and . Then, induces a cycle on at least five vertices in G. As and are at distance at least three in C, is shorter than C. In particular, contradicts either G containing no copy of , for , or C being a shortest hole in G. Therefore, this case never occurs.
Hence, v has either (i) exactly two neighbors in C and those are at distance two as there is no triangle in G, so (2) holds, or (ii) C has an even number of vertices and v is adjacent to every second vertex of C. It remains to show that the latter never occurs. Indeed, if it does, then without loss of generality . But observe that since C has at least ten vertices, the set induces a copy of . This concludes the proof.
Given Proposition 2, for every we can set and . Note that sets form a partition of V. Moreover, for every we have . Following our notation, for any integer i by and we denote the sets and , respectively. Furthermore, for every we set:
We write just A[i, j], B[i, j], and V[i, j], respectively, instead of , , and , when there is no confusion.
We now characterize the neighborhoods of the vertices in sets and , see also Fig. 5.
Fig. 5.

A possible neighborhood of u in and w in
Proposition 3
Let . Then:
and are independent sets.
For every and every we have .
For every we have .
For every we have .
Proof
Statement (1) follows trivially from the fact that G contains no triangle. To show statement (2), assume for a contrary that for some and some . Since , we have , and since we have . Hence, the set induces an in G, which cannot be the case.
To prove statements (3) and (4), consider a graph G induced by the set , where
Since any edge with two endpoints in U (or two endpoints in W) could be extended by some vertices from to an odd cycle of size in G, the graph is bipartite with bipartition classes U and W.
To see that (3) holds, first note that by statement (2). Therefore, suppose that u has a neighbor v in the set .
Consider the case when for some . Since and , u, v and the vertices between and in C as well as u, v and the vertices between and in C induce cycles in G of size . Since , at least one from these cycles has size , and as such it cannot occur in G.
So suppose for some . Since , u, v and the vertices between and in C as well as u, v and the vertices between and in C induce holes in G of size , and as such they cannot occur in G. So, , which completes the proof of statement (3).
Statement (4) is proved by similar arguments.
Proposition 3 asserts that all the neighbors of the vertices from and from are contained in the set and , respectively. The next proposition describes the relations that hold between the neighborhoods of the vertices from restricted to the set and between the neighborhoods of the vertices from restricted to the set .
Proposition 4
Let . For , the following hold:
-
For every the sets and are comparable.
Moreover, if and , then .
-
For every the sets and are comparable.
Moreover, if and , then .
Proof
To prove (1), we consider the case , as the other one follows by symmetry. Suppose that are such that neither nor holds. It means that there are such that , , , and . Since , we have and . Furthermore, and as G contains no triangle. Consequently, the set induces a copy of in G, which cannot be the case. Moreover, if , , then since , the latter statement holds.
To show (2), we again only consider the case . Suppose that are such that neither nor holds. It means that there are such that and . Since , we have and . Furthermore, and as G contains no triangle. Hence, the set induces a copy of in G, which cannot be the case.
To see the second part of the statement, assume that for some , . That is, there is such that and . In particular, it means that . Note that . Consequently, the set induces a copy of in G, which is a contradiction.
Proposition 4 allows us to order vertices of based on two properties. We now define relation which combines them and we show that is a partial order (see Fig. 6 for an illustration). We define for every :
Similarly, we define a relation for every :
Fig. 6.

The neighborhoods of the vertices from restricted to . We have
Proposition 5
The following statements hold for every :
is a strict partial order. Moreover, are incomparable in if and only if .
is a strict partial order. Moreover, are incomparable in if and only if .
Proof
Let be fixed. To prove that is a strict partial order, we need to show that is irreflexive and transitive. The irreflexivity follows from the definition, in aim to show transitivity, we first prove that is antisymmetric. Suppose for a contrary that there are such that , and . Suppose is witnessed by a vertex such that and ; the other case is analogous. By Proposition 4.(1), there is no such that and . Hence, since , there must be a vertex such that and . We have and as . We have also and as and . Moreover, , by Proposition 3. Consequently, the set induces a copy of in G, which cannot be the case.
To show transitivity, suppose for a contrary that there are vertices such that and , but does not hold. Suppose is witnessed by a vertex such that and ; the other case is symmetric. We have as otherwise , by definition. Suppose is witnessed by a vertex . Note that if , then and , which enforces also as and we already proved that is antisymmetric. Thus, and , which shows . Hence, we must have , and so and . Moreover, as otherwise . As , , and , we have , by Proposition 3. Consequently, induces a copy of in G, which is not possible. We conclude that is a strict partial order.
By definition, if , then u and are incomparable in . Hence, for the second statement of (1), it is enough to show that implies that u and are comparable in . Let w be a vertex such that and . By Proposition 3.(2) and (3), . However, if then and if then , by definition. Hence, u and are comparable in .
The proof of (2) is similar. For antisymmetry, suppose that we have such that and . Let and be witnessed by u and from , respectively. Analogously to (1), by Proposition 4.(2), we can assume that and and , . Observe that the set induces a copy of in G, a contradiction.
For transitivity of , suppose that for some , and we have and , but does not hold. By symmetry of the proof of (1), we reach the case and , and . Then, one can easily check that the set induces a copy of in G, a contradiction.
Now, assume that . Without loss of generality assume that there exists such that . Analogously as before, observe that if then and if then . Therefore, w and are comparable, which finishes the proof.
Finally, for every we order arbitrarily the elements inside every antichain of and of , obtaining strict linear orders and . We introduce a binary relation defined on the set V, such that for if one of the following conditions holds for some :
, , and are consecutive in ,
, , and are consecutive in ,
v is the maximum of and is the minimum of ,
v is the maximum of and is the minimum of .
Informally, to get an embedding of G into a cylinder (the shortest hole is even) or into a Möbius strip (the shortest hole is odd) which locally satisfies the adjacency and the enclosure properties, we place the vertices satisfying next to each other, v before assuming that the border of the cylinder or the Möbius strip are oriented as shown in Fig. 3. In what follows we extend relation as follows:
For every by we denote the transitive closure of restricted to ,
For we set if and for some or and for some .
Finally, the following lemma characterizes the global structure of an almost bipartite permutation graph.
Lemma 1
Let i, j be such that , . Let and Then is a bipartite permutation graph with bipartition classes U and W.
Moreover, and are strict linear orders that satisfy the adjacency and enclosure properties in .
Proof
Proposition 3 asserts there is no edge between a vertex in and a vertex in . In particular, is a bipartite graph and and are strict linear orders. Given Theorem 3.(c), to prove the lemma we need to show that satisfies the adjacency and enclosure properties in .
To prove the adjacency property, consider for some suitable k. Recall that by Proposition 3.(3), . To show that N(u) consists of consecutive vertices in W it suffices to note that:
and that analogous statements hold by symmetry for the part of N(u) contained in . If , the case analysis is similar (one needs to swap letters A and B in the reasoning above). Therefore, the adjacency property is proved.
To show that satisfies the enclosure property assume for a contradiction that there are and such that , and , , and .
Claim
There is such that either , or .
Proof of Claim
If , then since , we have , so the claim holds. Therefore, assume that , and suppose that . Then . Assuming (the other case is symmetric), we have that . Due to Proposition 3 and we have . Moreover, as we already proved that is consecutive in (adjacency property), and , we have . Therefore . Note that:
if , then, since , we have that . However, by Proposition 4.(2), we have , so it implies that , a contradiction,
if , then by Proposition 3.(2) we would have , a contradiction.
This concludes the proof of claim.
Suppose . Since and , we have by adjacency property of that . Therefore, we must have and . Observe that witnesses that by definition, however, w witnesses the opposite, that is . We have a contradiction by Proposition 5.
Suppose . An analysis, which is analogous to the one in the previous case (again, it is enough to swap letters A and B in that reasoning above), gives us that we must have and . Again, we obtain a contradiction by the definition of and Proposition 5.
Lemma 1 provides an interesting view on classification of almost bipartite permutation graphs. Specifically, if m is even, then the graph may be drawn on a cylinder, whose boundary consists of two closed curves, one of which traverses the vertices of , and the second one—the vertices of . If in turn m is odd, then the graph can be represented on a Möbius strip, whose boundary traverses the vertices of and then . In both cases we draw the vertices of and of on the opposite side of the strip according to the orders given by and , for (recall Fig. 3).
The following definitions are taken from [27]. A hole cut of G is a vertex set such that is a bipartite permutation graph. Lemma 1 asserts that for every the set is a hole cut in G. A hole cut X of G is minimum if G does not have a hole cut whose size is strictly smaller than the size of X. A hole cut X of G is minimal if any proper subset of X is not a hole cut in G.
The next proposition describes the structure of every hole in G.
Proposition 6
Suppose is a hole of size t in G for some . Then, the consecutive vertices of can be labeled by so as the following conditions hold (the indices are taken modulo k):
for every ,
for every ,
for every .
Proof
Let , , , , be consecutive vertices of denoted in such a way that . We can assume it, since , thus, by Proposition 3.(3) and (4), both belong to or both belong to for some .
Now, we show that if there exists such that , then . Suppose, for contradiction that and . Let be such that . Similarly, as , either if or if . In both cases Lemma 1 implies that restricted to is a strong ordering of . Moreover, are comparable in , by Proposition 3.(3) or (4), and the definition of . Since we assumed that , we must have , and from Theorem 3.(b) we get that , so has a chord—contradiction. Therefore implies for every integer j. Applying the above observation repeatedly for , we get that and
For the last property, suppose for the sake of contradiction that there exists such that . Then, by Lemma 1, due to the adjacency property. But then the edge is a chord in . This completes the proof.
The structure of holes described above asserts that for every the sets and are hole cuts. We use this observation to prove the following statement about minimal hole cuts in G.
Proposition 7
Every minimal hole cut X in G is fully contained in the set for some .
Proof
Let X be a minimal hole cut. Since X is minimal, we can choose elements in V such that the following conditions hold:
we have , the set is non-empty and is contained in X, and the elements are not in X.
Note that either or for some . Otherwise, we have or for some . However, Proposition 6 and Proposition 3.(3) and (4) imply that the sets and are hole cuts for every . Otherwise, the set () is contained in the neighbourhood of some vertex of a hole avoiding (, respectively) in G and the neighborhood of this vertex intersects also and ( and , respectively), which cannot be the case due to Proposition 3.(3) and (4). So, we have either or as X is a minimal hole cut. But then, we have , which completes the proof of our claim.
Therefore, for the rest of the proof we assume ; the other case is proved similarly. Moreover, we may assume that i is picked such that:
, , and .
See Fig. 7 for an illustration.
Fig. 7.

Illustration of the proof: the cycle is marked with a dashed line. The set is shaded
Suppose is the set consisting of all the neighbors of both and ; that is, . Clearly, we have . To complete the proof of the proposition we show that:
every element of is a member of X,
is a hole cut in G.
Then, since , we have by minimality of X and consequently . So, it remains to prove the claims about the set .
Suppose we have such that . We fix some , clearly, since X is a minimal hole cut, there is a hole in . Note that must contain x. Suppose for some are consecutive vertices in chosen such that for every (indices are taken modulo ). Now we pick such that and . Since , we have and . Note that is adjacent to and is adjacent to due to the adjacency property. Next we replace in all the vertices between and (this set includes x) with the vertices and we obtain a cycle containing no elements from X. Clearly, we can easily find a hole among the elements from that avoids all the elements from X. This yields a contradiction as X is a hole cut.
To prove the second claim, suppose there is a hole in . By Proposition 6 there are such that and . However, since and , we have and . So, we have , which is a contradiction.
Proof of Theorem 1
The aim of this section is to provide a complete proof of Theorem 1 using structural results from the previous section. Let us start by showing that the Bipartite Permutation Vertex Deletion problem can be decided in polynomial time on almost bipartite permutation graphs.
Lemma 2
Let (G, k) be an instance of Bipartite Permutation Vertex Deletion where G is an n-vertex almost bipartite permutation graph. Then Bipartite Permutation Vertex Deletion can be decided in time .
Proof
If G is a bipartite permutation graph, (G, k) is a yes-instance, thus, we are done in this case. If G is not connected, we can consider each connected component independently and, at the end, we compare k with the total number of deleted vertices over all components. Let be a connected r-vertex component of G such that is not a bipartite permutation graph (otherwise, clearly, no vertex needs to be deleted). Let be a shortest hole in (it exists as is not a bipartite permutation graph). It can be found in time as follows. We iterate over all possible four-element subsets of . For these S for which is an induced , with consecutive vertices , we construct a graph by removing the vertices from (note that and also get removed). Then we find a shortest --path in in time .
By Proposition 7, every minimal hole cut X in is contained in the set for some . Therefore, we may check all the possibilities where a minimal cut is contained. For every i, we run an algorithm for finding a maximum flow in the following digraph .
Digraph has the vertex set and arc set consisting of:
all arcs of the form , where uv is an edge of ,
if there exists such that uv is an edge of ,
if there exists such that uv is an edge of ,
for all .
Set capacities of arcs of the form to 1 and capacities of all the remaining arcs to (practically ). It is readily seen that minimum (s, t)-cut in the defined network corresponds to minimum hole cut in (arc of unit capacity naturally corresponds to the vertex u of ).
Therefore it remains to apply classical max-flow algorithm to each for and remember the smallest size of minimal (s, t)-cuts. This can be performed in time [11]. Finally, (G, k) is a yes-instance if and only if the sum of remembered sizes over the all considered connected components is at most k. Clearly, the total running time is .
We now propose the algorithm. Given an n-vertex graph and number k, we want to answer the Bipartite Permutation Vertex Deletion problem. We say that (G, k) is the initial instance. We split our algorithm into two parts. The first part consists of a branching algorithm for deletion to almost bipartite permutation graphs. The output of the first part is a set of instances where is an almost bipartite permutation graph and (or no-answer is no such instance exists) such that the initial instance (G, k) is a yes-instance if and only if at least one of these instances is a yes-instance. In the second part, the algorithm runs an -time algorithm for Bipartite Permutation Vertex Deletion for each instance output by the first phase.
Let us start with the first part. We say that is a forbidden set if G[X] is isomorphic to one of the graphs: . We define the following rule.
-
Rule
: Given an instance (G, k), , and a minimal forbidden set X, branch into |X| instances, for each .
Starting with the initial instance, the algorithm applies the rule exhaustively. In other words, the algorithm forms a branching tree with leaves corresponding to instances where or is an almost bipartite permutation graph. Clearly, as at least one vertex from each forbidden set must be removed from G, the initial instance is a yes-instance if and only if at least one of the leaves is a yes-instance.
The algorithm continues to the second part only with such leaves that is an almost bipartite permutation graph (as otherwise, the leaf is no-instance). It runs the algorithm described in Lemma 2 to find if can be transformed into a bipartite permutation graph by using at most vertex deletions. It either finds a yes-instance or concludes after checking all the instances that there is no solution; that is, the initial instance is a no-instance.
We note that such a branching into a bounded number of smaller instances is a standard technique, see e.g., [27] for more details.
We now analyze the running time of the whole algorithm. In the first part, observe that the branching tree has depth at most k and has at most leaves, as k decreases by one whenever the algorithm branches and each of the listed forbidden subgraphs has at most nine vertices. Therefore the total number of nodes in the branching tree is . Moreover, in each node the algorithm works in time as it checks if contains a forbidden set of size at most 9. So, the first part works in time . In the second part, the algorithm does a work in each of at most leaves, by Lemma 2. Thus, the second part works in time . We conclude that the total running time of our algorithm for Bipartite Permutation Vertex Deletion is .
Proof of Theorem 2
In this section, we provide a proof of Theorem 2. The idea of the algorithm is very similar to the algorithm described in Sect. 4.
Let be a graph and let be a subset of vertices of G such that is a bipartite permutation graph. We want to construct a set in polynomial time such that is a bipartite permutation graph and . We construct Z as follows. We start with . Then, as long as contains a set X isomorphic to one of we add all vertices of X to Z. Observe that and .
After this step is an almost bipartite permutation graph. Note that . We find a shortest hole in and find a minimum hole cut X as described in Sect. 4. Since is a hole cut in we have . We add X to Z. Observe that is a bipartite permutation graph.
Since have at most 9 vertices, we have that . This implies that the above algorithm is a 9-approximation algorithm. It runs in polynomial time because finding small forbidden subgraphs can be done in polynomial time and finding a minimum hole cut in an almost bipartite permutation graph can be done in polynomial time.
Conclusion
In this paper we investigate for the first time the modification problems in graph classes related to partial orders. Our main result says that the bipartite permutation vertex deletion problem is fixed parameter tractable. We leave open the following two questions that inspired our research.
Problem 1
What is the parameterized status of the vertex deletion problems to the class of permutation graphs and to the class of co-comparability graphs?
We recall that, due to the result of Lewis and Yannakakis [20], both of these problems are -complete. Problem 1 appears to be difficult; in particular, in contrast to bipartite permutation graphs, in which sufficiently large forbidden structures are just cycles, arbitrarily large forbidden structures for permutation graphs and co-comparability graphs may belong to several different infinite families of graphs. One of the most important results of our work is the description of the structure of almost bipartite permutation graphs, which are defined as graphs which do not induce small graphs from the list of forbidden structures for bipartite permutation graphs. In a similar fashion we can define the class of almost permutation and almost co-comparability graphs. Although the families of forbidden structures for permutation graphs and co-comparability graphs are quite complex, the next two questions seem very natural in order to solve Problem 1.
Problem 2
What is the structure of almost permutation and almost co-comparability graphs?
We are aware that the two problems mentioned above can be quite difficult. Therefore, it is worth considering intermediate problems that may be easier to attack. One of the proposed simplifications relies on the transition from the world of graphs to the world of posets. The following vertex deletion into two-dimensional posets problem seems very natural in the context of our research: we are given in the input a poset P and a number k and we ask whether we can delete at most k points from P so that the remaining points induce a two-dimensional poset in P.
Problem 3
What is the parameterized status of the vertex deletion into two-dimensional poset problem?
Since permutation graphs are co-comparability graphs of two-dimensional posets and since permutation graphs are both comparability and co-comparability graphs, the vertex deletion into two-dimensional poset problem is equivalent to the vertex deletion into co-comparability graph (or into permutation graph) problem if we assume that only comparability graphs can be given in the input. The class of two-dimensional posets is very well understood; in particular, the list of minimal forbidden structures for this class of posets, which is still infinite, is known (obtained independently by Trotter and Moore [26] and by Kelly [18]). Of course, it is natural to ask the following question:
Problem 4
What is the structure of almost two-dimensional posets?
Since the comparability graphs of posets do not contain odd holes of size , we know the structure of almost two-dimensional posets that are bipartite. Indeed, these are the posets whose comparability graphs are almost bipartite permutation graphs embeddable into cylinder stripes. The last problem we want to ask is as follows:
Problem 5
Is there a polynomial kernel for the bipartite permutation vertex deletion problem?
A positive answer to this question obtained by indicating so-called irrelevant vertices may give some hope to solve Problem 1 with the use of irrelevant vertex technique.
Acknowledgements
When this paper was under review, Problem 5 has been solved independently by Kanesh, Madathil, Sahu, Saurabh, and Verma [17] and by Derbisz [10].
The authors are grateful to Bartosz Walczak for valuable comments and help with merging two groups of researchers working on similar projects into one. They also would like to thank the anonymous reviewers for their helpful comments.
Open Access
This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Declarations
Conflict of interest
The authors declare that they have no conflict of interest.
Footnotes
The extended abstract of the work was presented on IPEC 2020 [2]. Ł.B., J.N., and K.O. were supported by the project CUTACOMBS that has received funding from the European Research Council (ERC) under the European Union’s Horizon 2020 research and innovation programme, grant agreement 714704. Their research was partially carried out during the Parameterized Algorithms Retreat of the University of Warsaw, PARUW 2020, held in Krynica-Zdrój in February 2020. T.K. was partially supported by Polish National Science Center (NCN) grant 2015/17/B/ST6/01873. J.N. was supported by the student grant SVV-2020-260578.
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Contributor Information
Łukasz Bożyk, Email: l.bozyk@uw.edu.pl.
Jan Derbisz, Email: jan.derbisz@doctoral.uj.edu.pl.
Tomasz Krawczyk, Email: krawczyk@tcs.uj.edu.pl.
Jana Novotná, Email: janca@kam.mff.cuni.cz.
Karolina Okrasa, Email: k.okrasa@mini.pw.edu.pl.
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