On dually-CPT and strong-CPT posets

04/10/2022
by   Liliana Alcón, et al.
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A poset is a containment of paths in a tree (CPT) if it admits a representation by containment where each element of the poset is represented by a path in a tree and two elements are comparable in the poset if and only if the corresponding paths are related by the inclusion relation. Recently Alcón, Gudiño and Gutierrez introduced proper subclasses of CPT posets, namely dually-CPT, and strongly-CPT. A poset 𝐏 is dually-CPT, if and only if 𝐏 and its dual 𝐏^d both admit a CPT representation. A poset 𝐏 is strongly-CPT, if and only if 𝐏 and all the posets that share the same underlying comparability graph admit a CPT representation. Where as the inclusion between Dually-CPT and CPT was known to be strict. It was raised as an open question by Alcón, Gudiño and Gutierrez whether strongly-CPT was a strict subclass of dually-CPT. We provide a proof that both classes actually coincide.

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1 Introduction

A poset is called a containment order of paths in a tree (CPT for short) if it admits a representation by containment where each element of the poset corresponds to a path in a tree and for two elements and , we have in the poset if and only if the path corresponding to is properly contained in the path corresponding to .

Several classes of posets are known to admit specific containment models, for example, containment orders of circular arcs on a circle [14, 15], containment orders of axis-parallel boxes in [12], or containment orders of disks in the plane [3, 5, 6] to cite just a few. All the aforementioned classes, as well as CPT, generalize the class CI of containment orders of intervals on a line [4]. It is well known that this class coincides with the class of -dimensional posets and are also equivalent to the transitive orientations of permutation graphs [9].

In 1984, Corneil and Golumbic observed that a graph may be the comparability graph of a CPT poset, yet a different transitive orientation of may not necessarily have a CPT representation, (see Golumbic [10]). This stands in contrast to poset dimension, interval orders, unit interval orders, box containment orders, tolerance orders and others which are comparability invariant. Golumbic and Scheinerman [12] called such classes strong containment poset classes.

Recently, interest in CPT posets has been revived and several groups of researchers have considered various aspects of this class [1, 2, 11, 13]. Since the CPT posets are not a strong containment class, Alcón, Gudiño and Gutierrez [1] introduced the study of the subclasses dually-CPT and strongly-CPT posets. A poset is called dually-CPT if and its dual admit a CPT representation. A poset is called strongly-CPT if and all the posets that share the same underlying comparability graph admit CPT representations. From the definition it is clear that the class of strongly-CPT posets is included in the class of dually-CPT posets. Many families of separating examples are now known between the class of dually-CPT and general CPT posets, however, concerning the strongly and dually-CPT, it was left as an open problem for many years to determine whether the inclusion is strict or if the two classes coincide.

We present in this paper a solution to this question with the following main theorem.

Theorem 1.

A poset is strongly-CPT if and only if it is dually-CPT.

To prove our main result we rely on the link between modular decomposition of the underlying comparability graph and its transitive orientations. Our strategy consists of considering a dually-CPT poset and proving that any poset with the same comparability graph also admits a CPT representation. At first we consider the representation and perform some modifications to obtain a representation with particular properties. Once this is done, we rely on the specific structure of modules in dually-CPT posets, and we provide a method to obtain the representation of any poset with the same comparability graph.


The paper is organized as follows: In Section 2, we present the definitions related to posets, CPT and modular decomposition and recall some fundamental results that we will use throughout the paper. In Section 3, we prove that for dually-CPT posets it is possible to obtain a representation where no element of a strong module is represented by a trivial path. Then, in Section 4, we show how to modify a CPT representation of a dually-CPT poset so that either the paths of a strong module do not end on a trivial path or the considered module admits very specific properties. Finally, in Section 5, we show how to use an operation called substitution to prove our main result.

2 Definitions and notations

A partially ordered set or poset is a pair where is a finite non-empty set and is a reflexive, antisymmetric and transitive binary relation on . The elements of are also called vertices of the poset. As usual, we write in for ; and in when and . If or , we say that and are comparable in and write . When there is no relationship between and we say that they are incomparable and write . An element is covered by in P, denoted by in P, when and there is no element for which and . The down-set and the up-set of an element are denoted by and , respectively. We let and . The dual of is the poset where in if and only if in .

A containment representation or model of a poset maps each element of into a set in such a way that in if and only if is a proper subset of . We identify the containment representation with the set family .

A poset is a containment order of paths in a tree, or poset for brevity, if it admits a containment representation where every is a path of a tree , which is called the host tree of the model. When is a path, is said to be a containment order of intervals or poset for short. (We generally consider a path as the set of vertices that induces it.)

The comparability graph of a poset is the simple graph with vertex set and edge set . In what follows, a poset , such that is complete (resp. without edges), is called a total order (resp. an empty order). We say that two posets are associated if their comparability graphs are isomorphic. A graph is a comparability graph if there exists some poset such that .

A transitive orientation of a graph is an assignment of one of the two possible directions, or , to each edge in such a way that if and then . The graphs whose edges can be transitively oriented are exactly the comparability graphs [7, 8, 9]. Furthermore, given a transitive orientation of a graph , we let denote the poset where in if and only if . The comparability graph of is . Thereby, the transitive orientations of are put in one-to-one correspondence with the posets whose comparability graphs are .

Let be a poset. A set is a module (homogeneous set [7]) if for every , either for all , or for all . The whole set and the singleton sets , for any , are modules of . These modules are called trivial modules. A poset is prime or indecomposable if all its modules are trivial. Otherwise is decomposable or degenerate. A module is strong if for all modules either or or .

A module (respectively, strong module) is called maximal if there exists no module (respectively, strong module) such that .

Theorem 2.

(Modular decomposition theorem) [7] Let be a poset with at least two vertices. Then exactly one of the following three conditions is satisfied:

  1. is not connected and the maximal strong modules of are the connected components of .

  2. is not connected and the maximal strong modules of are the connected components of .

  3. and are connected. There is some and a unique partition of such that

    1. ,

    2. is the biggest prime subposet of (in the sense that it is not included in any other prime subposet),

    3. for every part of the partition , is a module of and .

The previous theorem defines a partition of , which is called the canonical partition or maximal modular partition of . In the first case, is said to be parallel or stable and the partition is formed by the vertices of the connected components of . In the second case, is series or clique and the partition is formed by the vertices of each connected component of . And, in the last case, is neighborhood or prime, and the partition is .

The quotient poset of , denoted by , has a vertex for each part of ; and two vertices and of are comparable if and only if for all and for all , in .

The quotient poset is empty (iff is parallel), a total order (iff is series) or indecomposable (iff is neighborhood).

On some occasions, when referring to a module, we will mean the subposet induced by it. For instance, we will say that a module of is or that it is prime, meaning that is. This will be clear from the context and will cause no confusion.

Theorem 3.

[7] Given posets and , if and is indecomposable, then or .

Proposition 4.

[7] Given posets and , if , then and have the same strong modules and, consequently, .

Given a vertex of a poset and a poset , substituting or replacing by in results in the poset such that .

Theorem 5.

Let be the maximal modular partition of a connected poset whose quotient is prime, and call the quotient poset . A poset is associated to if and only if there exist posets for such that is associated to for each , and is obtained by replacing each vertex of by the poset or replacing each vertex of by the poset .

Theorem 6.

[7] A poset is if and only if the quotient poset and all the maximal strong modules of are .

Lemma 7.

[1] If is a vertex of a poset then the subposet induced by the closed down-set of is . In particular, if is dually-, then also the subposet induced by the closed up-set of is .

Remark 8.

[7] Let and be associated posets. Then, is a poset if and only if is a poset. In particular, is a poset if and only if is a poset.

Theorem 9.

Let be a connected dually- poset. Then the quotient poset of is dually- and every maximal strong module of is . In particular, if the quotient poset is , then is .

Proof.

Let be the maximal modular partition of . The quotient poset is a subposet of , so is dually-. We can assume that is not empty, and since is connected we have that is connected, and so every vertex of is in the down-set or in the up-set of some other vertex. Which implies that in the whole module is in the up-set or in the down-set of some other vertex. It follows from Lemma 7 that each is . Therefore, by Theorem 6, if is , then is .


The converse of Theorem 9 is not true in general. For instance, if in the quotient poset there exists a vertex such that in any representation of the corresponding path is reduced to a vertex, then for to be the module has to be a singleton.

In a representation of a CPT poset , a subset of paths of is called one-sided if all the paths that represent arrive at a vertex of the host tree and all paths of , except possibly one trivial path, pass through a vertex of neighbor of . If all the paths of arrive at a vertex and is not one-sided, then it is called two-sided.

Addressing that issue in the proof of the main theorem will requires the following lemmas and properties.

Property 10.

[4, 10] Every poset admits a representation where the intersection of all the intervals used in the representation is a non-trivial interval.

3 Trivial paths into modules

The goal of this section is to prove that for any dually-CPT poset , there exists a representation where all the elements contained in strong modules are represented by non-trivial paths.

At first we prove that if an element of module is represented by a trivial path, it does mean that the module (all its elements) are not greater than any other element not in the module.

Lemma 11.

Let be a poset and let be a strong module of . If there exists a representation where an element of is represented by a trivial path, then all the elements of are not greater than any element of not in .

Proof.

Let us proceed by contradiction and let us assume that there exists an element such that . Then in any representation we have but since is already a trivial path it cannot properly contain some other object.


Hence from the previous lemma, if in a representation an element of a module is represented by a trivial path, the module is a minimal subset of .

Lemma 12.

Let be a strong module of a CPT poset , if in a representation one of its elements is represented as a trivial path, then there exists an element not in such that the path contains all the paths representing the elements of .

Proof.

Since the poset is connected, and by the previous lemma, we know that the module cannot contain any other element, to ensure the connection outside the module, there might be at least one element that is greater than every element of .


Lemma 13.

Let be a strong module of a CPT poset . If in a representation one of its elements is represented as a trivial path, then this path is hosted on some vertex of . If for an element not in its path passes through , then has to contain all the paths corresponding to the elements of .

Proof.

From the definition of a module, every element not in the module is either completely disconnected from or completely connected to . In that case, if for an element , in a representation its path passes through , then it is connected to the element . Hence it has to be connected to every element of . In addition, in a transitive orientation of a graph, the containment relation between and the elements of is the same for every element of . Thus if contains it contains all the paths of the elements of .


Lemma 14.

Let be a strong module of a CPT poset . If in a representation one of its elements is represented as a trivial path and is a clique or prime module, then there exists at least one element of represented as a non-trivial path.

In the case of dually-CPT posets, the next three lemmas consider the presence of trivial paths in a representation of strong modules and show how to obtain an equivalent representation where all the elements of the module are non-trivial paths. For these lemmas, we consider each strong module to be a CI poset and the element of the module represented by a trivial path is denoted by .

Lemma 15.

Let be a strong CI clique module of a dually-CPT poset . If an element of is represented by a trivial path in a representation , then there exists a representation where is represented as non-trivial path.

Proof.

By Lemma 12 we know that there exists an element such that all the paths of are contained in in all CPT representations. Let us consider three cases.


(1) Suppose the trivial path of is not an extremity of any path that represents the elements of . Let be the vertex of that hosts the trivial path of . Since is not an extremity of any path of , admits at least one neighbor in such that all the paths of (except for ) pass through (see Figure 1). Let us subdivide the edge by adding a vertex . Then it suffices to replace the trivial path of by a non-trivial path that goes from to in . The containment relations among are preserved and no new containment relation is added nor deleted with respect to the elements not in .


(2) Suppose now, the trivial path of is a common extremity for all the elements of and is one-sided (see Figure 1). We proceed as in the previous case; we consider a vertex of that is a neighbor of and such that all the paths of except for pass through . Since is a clique, it only admits at most one element represented by a trivial path, such a vertex exists, then we subdivide the edge by adding a vertex and the path of goes from to . Note that the technique still works if some paths of continue after .


(3) Suppose now, the trivial path of is the common extremity for some paths of the module in a 2-sided manner (see Figure 1). Let and be two vertices of that are neighbors of , such that and lie on the path of , being an element not in that contains all elements of . We can partition the elements of into three sets: the elements for which paths arrive at and pass through , defined in a similar way but w.r.t. instead of , and , the paths of that go through and . This time we need to subdivide the edges and of . We add a vertex between and and a vertex between and . Then it suffices to extend the paths of until and the paths of until . The path of now goes from to . By subdividing several times the edges and , we can make sure that all the extremities are distinct.


Figure 1: Representation of cliques modules with trivial paths.
Lemma 16.

Let be a strong CI stable module of a dually-CPT poset . If an element of is represented by a trivial path in a representation , then there exists a representation where is represented as non-trivial path.

Proof.

Let us first remark that in a strong stable module, several elements can be represented as trivial paths in a representation . In addition, if an element of is represented by a trivial path, the trivial path is disjoint from all the other paths representing the elements of . Let be such an element. We will transform such that all the elements of represented by trivial paths in will be represented by non-trivial paths. Let be the vertex of that hosts the path of . Thanks to Lemma 12, we know that there exists an element of such that in the paths of the elements of are contained in the path of . Since is a non-trivial module it contains at least two elements, hence in there exists a vertex of that is adjacent to , and is contained in all the paths of the elements not in that contain , since such a path has to contain and all the other elements of .

Let us denote by the elements of that are represented by trivial paths in . To obtain an equivalent representation , we subdivide times the edge . We then rename as , and we number the newly created vertices (the transformation is presented in Figure 2). In this new representation each element of is replaced by a path that goes from to in .

It remains to prove that this representation is equivalent. First observe that for any element connected to , its path in contains all the elements of . By the choice of vertex to perform the transformation, we can guarantee that any path of such an element will pass through in . Since we subdivided this edge to obtain , this path will still pass through and and all the vertices introduced by the transformation.

Now for any element not connected to , we know by Lemma 13 that no path of such an element will pass through .


Figure 2: Representation of a stable module with elements represented by trivial paths; transformation to eliminate trivial paths from the representation.
Lemma 17.

Let be a strong CI prime module of a dually-CPT poset . If an element of is represented by a trivial path in a representation , then there exists a representation where is represented as non-trivial path.

Proof.

For this proof, we consider three cases: (1) either the trivial path of is properly contained (i.e. is not an extremity of any path of the element of ) in all the paths of the elements of the module , or (2) there exists at least two elements and of such that is the right bound of and the left bound of , or (3) the path is the right (respectively left) bound for some paths representing elements of , and is not the left (respectively right) bound of any elements of . These three cases are illustrated in Figure 3.


(1) Let be the vertex of that hosts , the trivial path representing . By hypothesis, all the paths that represent the elements of properly contain and thus pass through vertex . Since it is a proper containment, no path of elements of (other than ) starts or finishes at . Thus admits at least one neighbor in such that all the paths that represent elements of , except for , pass through . To obtain a new representation we subdivide the edge by a adding a vertex . Then in is replaced by the path . (See Figure 3). Since the representation of is the only modification of the representation, by the previous discussion all the paths that represent the elements of pass through and and as a consequence pass through and since is in between and . By Lemma 13 we know that all the paths of the elements not in that pass through will also contain all the paths of . Hence the modification of preserves the containment relation of .


(2) Let us now consider that there exist at least two elements and of such that in , the vertex is the right bound of the path and the left bound of the path (see Figure 3). Let us denote by the set of elements of for which is the right bound in the representation and similarly let us denote by the set of elements of for which is the left bound in . Let us remark that and some elements of might not be empty. Let be the neighbor of in such that the paths of the elements of pass through . And similarly let be the neighbor of in such that the paths of the elements of pass through . To obtain a new representation we subdivide the edge times, and the edge times. The added vertices are called for the vertices between and and for the vertices between and . Let and be the neighbors of in . The path now goes from to . The left bound of the paths of the elements of are moved on the vertices. The coordinates are chosen to preserve the containment relation. We proceed symmetrically for the paths of the elements in . It remains to prove that the obtained representation still corresponds to . Again we know by Lemma 13 that no path of an element not connected to passes through , by construction it remains valid for and for all the newly introduced vertices. For any other path their relation to and the paths of the elements of and are unchanged. If the path of an element was containing the path of an element of in , it is still the case in . In that case the left bound of is contained in and the right bound of will be at the right of in . This property will be preserved in . Similarly if both paths and were overlapping in , they are still overlapping in .


(3) Since is the right (resp. left) bound of some paths representing some elements of , and is not the left (resp. right) bound of any other elements of , there exists a vertex in adjacent to and such that all the paths representing elements of that end at pass through . To obtain the new representation it suffices to subdivide this edge one time. Let the newly introduced vertex. Then the trivial path in is replaced by a path going from to . By the transformation, we can observe that all the paths that were containing in still contain in . Let be an element of such that in . If was containing it had to pass through and , thus by subdividing we can also conclude that this paths will pass through and , the added vertex, in .


Figure 3: Representation of prime modules with the element represented as trivial path.
Theorem 18.

If is dually-CPT and in a representation some elements of strong modules are represented by trivial paths, then there exists an equivalent representation where all the paths representing elements of strong modules are non-trivial paths.

Proof.

It is a direct consequence of Lemmas 15, 16 and 17 and the fact that each time a trivial path is replaced by a non-trivial one, no trivial path is created in .


From the preceding theorem, we know how to obtain a representation a dually-CPT poset where all the elements contained in non-trivial strong modules are represented by non-trivial paths. Hence, in this representation some elements that do not belong to strong modules might be represented by trivial paths.

4 Ending of modules onto trivial paths

In the previous section we proved that for a dually-CPT poset, one can always obtain a representation where no element of a strong module is represented by a trivial path. It therefore remains to consider how the paths that represent a strong module can connect to an element , not contained in a strong module, where is represented by a trivial path in the representation . Since we need to reconfigure the containment relation inside the module, this operation could be prevented or constrained if the trivial path is misplaced. In the case where the trivial path is in the middle of the paths of the module, it will be easy to reconfigure the containment relation. In the opposite case, if all the paths representing elements of a module arrive at a trivial path, we cannot perform the intended operation as planned. In this section, we will identify the problematic situations, and we will show how to overcome these problems. As in the previous section, we will perform local changes to the representation to suppress problematic cases.

When the paths that represent elements of a module are connected to a trivial path in a representation, several configurations could arise. The most favorable one, is when the trivial path is properly contained in the paths of the module (i.e. the trivial path does not lie on any extremity of the path of the module). Actually this is a configuration we aim at obtaining. The other two configurations is when all the path have their extremities that end at a trivial path, or just some of them end at this trivial path. In most cases we will be able to reconfigure our representation to obtain a representation that is favorable to our purpose.

4.1 Complete ending of a module on a trivial path

Let us assume that all paths in corresponding to elements of have all their extremities end at a vertex of the host tree . In that case, there are several possibilities: either all the paths that represent will arrive at by passing by a vertex of and such that is an edge of , or there is another vertex that is a neighbor of in different from and such that some paths of the module pass through .

In this section, even if it is not explicitly stated, the representation of the module will contain the trivial path of located at the vertex in .

Remark 19.

If a strong module of a dually-CPT poset is two-sided in a representation , then the induced graph is not connected. Hence the strong module is a stable module.

Lemma 20.

Let be a strong module of a dually-CPT poset . If is one-sided in a representation and the poset induced by is connected, then is a clique module.

Proof.

If the graph induced by is connected, is either a clique module or a prime module. If is a clique module, then there is nothing to prove. If is a prime module, then the graph induced by necessarily contains an induced . Let us show that it is not possible to represent a as a CPT representation where all the paths end up at a same vertex of the host tree . Consider the representation of the as presented in Figure 4 with the containment relation represented in Figure 4.

For a contradiction, let us assume that such a representation exists. Since the paths of and have to contain the path of and all these paths have to arrive at vertex of the host tree, we have a configuration similar to the one depicted in Figure 4 and a part of the host tree is depicted in Figure 4(). Since and are not connected, their paths have to diverge in . Call the vertex of where these paths diverge. It remains to represent the path of . Since is connected to but not to , call the vertex of where the path of begins. The vertex has to lie in the proper part of the path of (see Figure 4()), and this path, by hypothesis, has to go all the way to . But in that case it has to contain the path of , hence there is a contradiction.


Figure 4: a , a CI representation of , tentative representation with all the paths arriving at a vertex, the host tree of the tentative representation.

We have proven that if in the representation of a strong module all its paths arrive at a same vertex of the host tree, then the module is either a clique or a stable module. We now consider in which cases can we obtain an alternative representation where all the paths do not arrive at a same vertex of the host tree. When the modification is possible, we will show how by starting from one can obtain an equivalent representation , that is, a containment representation that still corresponds to .

Lemma 21.

Let be a strong module of a dually-CPT poset and a representation of where all the paths of arrive at a same vertex. If there is no element of that contains the elements of , then there exists an alternative representation of where all the paths of will have different endpoints.

Proof.

Let us assume that all the paths of a strong module arrive at a vertex in the representation . If there is no element of that contains all the paths of , then we can add a new branch to the host tree starting at and ending at (see Figure 5). Let us denote by the cardinality of . In order to guarantee that all paths end on a dedicated vertex, the new branch needs to have at least new vertices. It is easy to make sure that the containment relation inside the module is not altered in this new representation. It is simple to notice that the previous containments of are preserved by this modification and no new containment is added since the branch only contains paths of .


Figure 5: Example of modification on a representation of a poset

We now consider the case when there is at least one element not in that is greater than all the elements of . In that case is either one-sided or two-sided. Let us start with this second case.

Lemma 22.

Let be a strong stable module of a dually-CPT poset and let be an element of that contains all the elements of . Let us assume that in a representation all the paths of arrive at a vertex of . Then there exists an equivalent representation where all the endpoints of the paths of near are distinct.

Proof.

By hypothesis, since the elements of are all contained in an element of , it means that in any representation of the union of the paths of is a path. If in a representation of all the paths of arrive at , let and be the immediate neighbors of on along the path that hosts all the paths of . Since the strong module considered is stable and in the representation every element lies under the path of , the module is two-sided at . Since is two-sided in the representation, its elements can be partitioned into two sets and as follows: An element is in if its path in passes by the vertex . Similarly, an element is in if its path in passes by (see Figure 6). To obtain it suffices to subdivide the edges and of . All the paths of the elements of that previously ended at will now end between and . Hence it is necessary to add new vertices between and . In a symmetric manner, the paths of the elements of will be elongated to end on a new vertex between and ; thus it is necessary to add new vertices between and . It is simple to see that the introduced modification does not alter the containment relationship. Any path that contained all the elements of will still contain all the elements of . And any path that crossed the section of tree spanned by the elements of but did not contain them, will still not contain them.


Figure 6: Modification of the representation of a two-sided stable module.
Lemma 23.

Let be a strong clique module of a dually-CPT poset and let be an element of that contains all the elements of . If in a representation all the paths of arrive at a vertex , then does not contain any other strong module.

Proof.

Because of the element , the union of all the paths of the elements of in is included in the path of and hence itself forms a path. Since all these paths are bounded at , then for any pair of elements and of either the path of is contained in the path of or the converse. There is no pair of non-adjacent vertices. As a consequence it does not contain any other module.


Figure 7: Configuration of a clique path. The set of paths represents the paths of a strong clique module

Let be a strong clique module with representation where all the paths of the elements of stop at a vertex . We say that is free in if there is at least one vertex of such that is an edge of , no path of passes through and all the paths that contain the paths of pass through . (See Figure 7 and .)

Lemma 24.

Let be a strong clique module of a dually-CPT poset , and let be an element of that contains all the elements of . If is free in a representation where all the paths of arrive at a vertex , then we can find a representation where all the endpoints of arrive on different vertices of .

Proof.

Since is free in we can re-use the technique used in Lemma 22 by subdividing the edge of .


Thanks to Lemmas 21, 22, and 24, we know how modify a representation in almost all the cases. However, one case is not covered, namely, when the module is a clique and it is blocked. We say that a strong clique module bounded at a vertex in a representation is blocked if it is not free. There are two reasons why may be blocked: (1) It might be because a path that contains the elements of also stops at , or (2) because there are two elements and that contain all the elements of and in their corresponding paths diverge at . (See Figure 7.)

Remark 25.

Let be a strong clique blocked module of a dually-CPT. From Lemma 23, we know that it does not contain any other strong module. Hence, a reconfiguration of this subposet is just a matter of relabelling the elements.

4.2 Partial ending of a module on a trivial path

In the previous section, we proved that whenever a module is connected to an element of represented by a trivial path in a representation and all paths that represent the element of end at this path, we can either alter the representation to ensure that all the paths do not end on that trivial path or, the module is a clique and does not contain any other modules. Hence it is possible to alter the containment relation.

If, in the completely opposite direction, a module is connected to an element represented by a trivial path, but no path that represents an element of ends at this trivial path, it does not create any problem to change the containment relation of the module.

The last case to consider is when is connected to a trivial path, but only some paths of (not all) end at this trivial path. We will prove that in that case an equivalent representation, where no path of ends at this trivial path, can be obtained.

Lemma 26.

Let be a strong module of a dually-CPT poset connected to an element (). If in a representation the element is represented by a trivial path and the paths of some elements of end at the path of and some other paths of elements of properly contain , then there exists an equivalent representation where no element of ends at a trivial path.

Proof.

Let denote the set of elements not in such that the paths of the elements of are contained in the paths of the elements of . In the representation all the paths that represent the element of are all contained in .

Since by hypothesis not all the paths of end at a trivial path, if there are some elements of that end paths represented by trivial paths, there are at most two trivial paths in that situation. Call these trivial paths and .

Let us assume that the part common to all the paths of in is on a horizontal line, and that w.l.o.g. that is the leftmost and is the rightmost of this common part. We assume further, in the representation , that lies on vertex of and lies on vertex of .

We denote by (resp. ) the set of all elements of whose paths in end at (resp. at .) Note that there is at most one element of that belongs to both and , since the containment relation is proper.

There are two cases to consider: (1) either there is no element such that all the paths of are contained in the path of , or (2) such an element exists.


(1) For the first case, let us assume that such an element does not exists. Hence there is no path in that contains any path of the elements of . In that case, to obtain an equivalent representation, in we can add one path with new vertices connected to and another path with new vertices connected to . Since the poset induced by is CI, it suffices to represent this module as a containment of intervals using these new branches for the endpoints. The transformation process is presented in Figure 8.

The containment relation between elements of (resp. ) and remain unchanged. Moreover, for any element not connected to , since the endpoints of the paths of the elements of have been relocated in the two new branches, there is no containment relation between and the paths of the elements of , since does not contain any of the new branches in .


(2) Let us now consider the case when there is an element not in such that in the path of contains all the paths of the elements of . In the host tree we denote by the neighbor of such that no path of passes through but some paths of elements of do (by our initial hypothesis). Let be the neighbor of in such that paths of some elements of pass through but no element of does.

To obtain an alternative representation we subdivide the edge times and subdivide the edge times. (This transformation is presented in Figure 9). Then it is just a matter of extending the paths of the elements in such that they end on a vertex located between and . For each element of , its new ending vertex is determined according to the containment relation in . For the elements of , we proceed in a similar manner.

It remains to prove that the new representation still represents the poset . The only paths that are transformed are the paths that correspond to elements of and . Without loss of generality, let be an element of and let be its path in . Since has been extended, it is clear that all the paths in that were contained in remain contained in . In addition, since the extension occurred between or . Equivalently let be an element of . If in then in . If is an element of , by the transformation we ensure that the containment relation is preserved. If is not an element of , then in , the path passed by vertex of , hence by extending , it will not reach , then it is still contained in in .

Let us now consider an element such that intersects but there is no containment relation in . If is not a path in then it contains a claw pattern and this pattern will be preserved in . Let us now consider the case when forms a path in . If passes through in it has one endpoint contained between the endpoint of . Thus the first endpoint of is at the left of in and the endpoint at the right of (possibly ). Since does not reach in , the overlap relation is preserved in the new representation. If both paths were disjoint, they remain disjoint in .


Figure 8: Illustration of case (1) of Lemma 26. Elements and are parts of the modules. The module is connected to the elements represented by the paths in the box . Elements and belong to and elements and belong to .
Figure 9: The same example as in Figure 8, but this time there is an element not in that contains all the elements of and that prevents performing the modification of case (1).

From Lemmas 21, 22, 24 and Remark 25, we can summarize the results of this section with the following theorem:

Theorem 27.

Let be a dually-CPT poset. Either for each strong module of there exists a representation such that all the paths of do not end on a trivial path, or is a clique blocked module.

We call a representation that fulfills the condition of the previous theorem a normalized representation.

5 Substitution

The last step to obtain our main result is to prove that for any dually-CPT poset all the posets that are associated to admit a CPT representation. Let us consider one particular poset of this set. If is associated to it means by definition that their underlying comparability graphs are identical. We assume that is not CI, otherwise the results already stand from Theorem 6. Thus we deduce that the quotient poset of is not CI, by Theorem 9, and thus is prime. Since and are associated, by Property 4 they admit the same set of strong modules. The quotient of is obtained by keeping one element of each strong maximal module and the quotient of is either equal to or to its dual . Let us consider that is equal to .

To obtain a representation for , we will use the normalized representation obtained for . From it is immediate to obtain a representation for as it suffices to keep one path for each strong module of . In addition, since it is obtained by removing paths from a normalized representation, we can consider that all the paths that correspond to strong modules which are not clique blocked modules, do not end on trivial paths of other elements. Then we will show that for such elements, we can replace this path by an arbitrary CI poset. Finally, to obtain a CPT representation for it suffices to replace each path that is a representative of a strong module, by the corresponding CI poset in . For the clique blocked modules, as they do not contain other strong modules, they correspond to total orders, hence the representation can be preserved, but the labelling has to be changed to suit the total order in .

Let be an element of that is a representative of some maximal strong module of that is not a clique blocked module. Let be its path in . We will assume that is at least . We will show how to replace by a CI poset . Let be a representation of a poset whose vertices are .

Assume that the intervals (subpaths of a path ) are non-trivial, no two of them share an end vertex and there is an edge of contained in the total intersection of the intervals – this assumption is guaranteed by Proposition 10. Name and the end vertices of the interval union of the intervals . Clearly . We also assume that , , and are distinct, and that neither nor are end vertices of an interval .

Replacement process.

The process of replacing in the representation the path by the intervals of the representation consists of:

  • subdividing the edges and of by adding in each one vertices.

  • subdividing the edge of by adding as many vertices as there are in between and .

  • removing from the path and embedding in its place the intervals of in such a way that the vertices , , , and all others between them match with the vertices , , , and all others between them, respectively, as it is shown in Figure 10.

Figure 10: Description of the Replacement process
Lemma 28.

If in the path that represents a module of a dually-CPT poset does not end on trivial paths, then we can obtain the representation by replacing by the intervals of the representation in . If any of the paths contains (resp. is contained in) a path , then all the paths contain (resp. are contained in) .

Moreover, a path of contains (is contained in) if and only if contains (is contained in) every one of the intervals in .

Proof.

This result is a direct consequence of two facts: first, that in no interval of has an end-vertex between and , nor between and , and second, that in , all the intervals contain the interval . See Figure 10.


Lemma 29.

If in the path , that represents a blocked clique module of a dually-CPT poset , ends on a trivial path, then we can obtain the representation by replacing by the a collection of paths that represent a clique.

Proof.

Let us assume that is the trivial path that ends on in . Let us denote by the vertex of the host tree that hosts . Since the containment relation is proper, we can assume that passes through at least two vertices of the host tree. One of the extremities of is . Let us call the other extremity . Since the length of is at least two, we know there exists in the host tree a vertex that is the immediate neighbor of on the path going to . The vertex is possibly equal to . By subdividing an appropiate number of times the edge of the host tree, we can add as many paths as we need to place a clique module. From the transformation, it is easy to see that the containment relation is preserved with respect to the module.


We restate here our main theorem:

Theorem 30.

A poset is strongly-CPT if and only if it is dually-CPT.

Proof.

Let be the quotient poset, where is the maximal modular partition of .

Since is a dually- poset and is a subposet of , then and admit a normalized -representation. If is , by Remark 8 and Theorem 9, is and so strongly-. Thus let us assume that is a prime dually- poset.

Let be an associated poset of and let be its quotient poset. Since and are associated, an immediate consequence is that and are associated; in addition by hypothesis they are both prime, hence by Theorem 3, is either equal to or to . Let us assume, w.l.o.g., that .

We will prove that admits a representation. By Theorem 5 and w.l.o.g, we assume that is obtained by replacing in each vertex of for . By Proposition 4, and possess the same strong modules and by Theorem 9 since is dually-CPT, all the strong modules of and are CI. For each we have a representation.

Let be a representation of , obtained from a normalized representation of . The representation is obtained by only keeping one path for each strong module of .

For each path that corresponds to a module of , if does not end on a trivial path of then it corresponds to a module which is not a blocked clique module, hence by Lemma 28, we can replace by a CI representation of .

The only remaining case is if ends on a trivial path in . In that case, it means that it corresponds to a blocked clique module of in the representation . Hence by Lemma 29, we can replace by a CI representation of the maximal strong clique module .

By proceeding in that way for each maximal strong module, we are able to obtain a CPT representation of .


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