On the dimension of posets with cover graphs of treewidth 2

On the dimension of posets with cover graphs of treewidth 2

Gwenaël Joret Computer Science Department
Université Libre de Bruxelles
Brussels
Belgium
Piotr Micek Theoretical Computer Science Department
Faculty of Mathematics and Computer Science
Jagiellonian University
Kraków
Poland
William T. Trotter Ruidong Wang School of Mathematics
Georgia Institute of Technology
Atlanta, Georgia 30332
U.S.A.
and  Veit Wiechert Institut für Mathematik
Technische Universität Berlin
Berlin
Germany
July 12, 2019
Abstract.

In 1977, Trotter and Moore proved that a poset has dimension at most whenever its cover graph is a forest, or equivalently, has treewidth at most . On the other hand, a well-known construction of Kelly shows that there are posets of arbitrarily large dimension whose cover graphs have treewidth . In this paper we focus on the boundary case of treewidth . It was recently shown that the dimension is bounded if the cover graph is outerplanar (Felsner, Trotter, and Wiechert) or if it has pathwidth (Biró, Keller, and Young). This can be interpreted as evidence that the dimension should be bounded more generally when the cover graph has treewidth . We show that it is indeed the case: Every such poset has dimension at most .

Key words and phrases:
Poset, dimension, treewidth
2010 Mathematics Subject Classification:
06A07, 05C35
G. Joret was supported by a DECRA Fellowship from the Australian Research Council.
P. Micek is supported by the Mobility Plus program from The Polish Ministry of Science and higher Education.
V. Wiechert is supported by the Deutsche Forschungsgemeinschaft within the research training group ‘Methods for Discrete Structures’ (GRK 1408).

1. Introduction

The purpose of this paper is to show the following:

Theorem 1.

Every poset whose cover graph has treewidth at most has dimension at most .

Let us provide some context for our theorem. Already in 1977, Trotter and Moore [11] showed that if the cover graph of a poset is a forest then and this is best possible, where denotes the dimension of . Recalling that forests are exactly the graphs of treewidth at most , it is natural to ask how big can the dimension be for larger treewidths. Motivated by this question, we proceed with a brief survey of relevant results about the dimension of posets and properties of their cover graphs.

One such result, due to Felsner, Trotter and Wiechert [3], states that if the cover graph of a poset is outerplanar then . Again, the bound is best possible. Note that outerplanar graphs have treewidth at most . Note also that one cannot hope for a similar bound on the dimension of posets with a planar cover graph. Indeed, already in 1981 Kelly [6] presented a family of posets with planar cover graphs and (see Figure 1). One interesting feature of Kelly’s construction for our purposes is that the cover graphs also have treewidth at most (with equality for ), as is easily verified. In fact, they even have pathwidth at most (with equality for ).

Very recently, Biró, Keller and Young [1] showed that if the cover graph of a poset has pathwidth at most , then its dimension is bounded: it is at most . Furthermore, they proved that the treewidth of the cover graph of any poset containing the standard example with is at least , thus showing in particular that Kelly’s construction cannot be modified to have treewidth .

To summarize, while the dimension of posets with cover graphs of treewidth is unbounded, no such property is known to hold for the case of treewidth , and we cannot hope to obtain it by constructing posets containing large standard examples. Moreover, as mentioned above, the dimension is bounded for two important classes of graphs of treewidth at most , outerplanar graphs and graphs of pathwidth at most . All this can be interpreted as strong evidence that the dimension should be bounded more generally when the cover graph has treewidth at most , which is exactly what we prove in this paper.

We note that the bound on the dimension we obtain is large (), and is most likely far from the truth. Furthermore, while we strove to make our arguments as simple as possible—and as a result did not try to optimize the bound—the proofs are lengthy and technical. We believe that there is still room for improvements, and it could very well be that a different approach would give a better bound and/or more insight into these problems.

We conclude this introduction by briefly mentioning a related line of research. Recently, new bounds for the dimension were found for certain posets of bounded height. Streib and Trotter [8] proved that for every positive integer , there is a constant such that if a poset has height at most and its cover graph is planar, then . Joret, Micek, Milans, Trotter, Walczak, and Wang [4] showed that for every positive integers and , there is a constant so that if has height at most and the treewidth of its cover graph is at most , then .

These two results are closely related. In particular, one can deduce the result for planar cover graphs from the result for bounded treewidth cover graphs using a ‘trick’ introduced in [8] that reduces the problem to the special case where there is a special minimal element in the poset that is smaller than all the maximal elements. This implies that the diameter of the cover graph is bounded from above by a function of the height of the poset, and it is well-known that planar graphs with bounded diameter have bounded treewidth (see for instance [2]). This trick of having a special minimal element below all maximal elements turned out to be very useful in the context of this paper as well (though for different reasons), see Observation 6 in Section 2.

Finally, we mention that several new results on bounding the dimension of certain posets in terms of their height have recently been obtained [12, 7, 5], the interested reader is referred to [5] for a detailed overview of that area.

The paper is organized as follows. In Section 2 we give the necessary definitions and present a number of reductions, culminating in a more technical version of our theorem, Theorem 7. Then, in Section 3, we prove the result.

2. Definitions and Preliminaries

Let be a finite poset. The cover graph of , denoted , is the graph on the elements of where two distinct elements , are adjacent if and only if they are in a cover relation in ; that is, either or in , and this relation cannot be deduced from transitivity. Informally, the cover graph of can be thought of as its order diagram seen as an undirected graph. The dimension of , denoted , is the least positive integer for which there are linear extensions of so that in if and only if in for each . We mention that an introduction to the theory of posets and their dimension can be found in the monograph [9] and in the survey article [10].

When and are distinct elements in , we write to denote that and are incomparable. Also, we let denote the set of ordered pairs of incomparable elements in . We denote by the set of minimal elements in and by the set of maximal elements in . The downset of a set of elements is defined as , and similarly we define the upset of to be .

A set of incomparable pairs is reversible if there is a linear extension of with in for every . It is easily seen that if is not a chain, then is the least positive integer for which there exists a partition of into reversible sets.

A subset of with is said to be an alternating cycle if in for each , where indices are taken cyclically (thus in is required). For example, in the poset of Figure 1 the pairs , form an alternating cycle of length for all such that . An alternating cycle is strict if, for each , we have in if and only if (cyclically). Note that in that case are all distinct, and are all distinct. Notice also that every non-strict alternating cycle can be made strict by discarding some of its incomparable pairs.

Observe that if is an alternating cycle in then cannot be reversed by a linear extension of . Indeed, otherwise we would have in for each , which cannot hold cyclically. Hence, alternating cycles are not reversible. It is easily checked—and this was originally observed by Trotter and Moore [11]—that every non-reversible subset contains an alternating cycle, and thus a strict alternating cycle:

Observation 2.

A set of incomparable pairs of a poset is reversible if and only if contains no strict alternating cycle.

An incomparable pair of a poset is said to be a min-max pair if is minimal in and is maximal in . The set of all min-max pairs in is denoted by . Define as the least positive integer such that can be partitioned into reversible subsets if , and as being equal to otherwise. For our purposes, when bounding the dimension we will be able to focus on reversing only those incomparable pairs that are min-max pairs. This is the content of Observation 3 below. In order to state this observation formally we first need to recall some standard definitions from graph theory.

By ‘graph’ we will always mean an undirected finite simple graph in this paper. The treewidth of a graph is the least positive integer such that there exist a tree and non-empty subtrees of for each such that

1. for each edge , and

2. for each node of the tree .

The pathwidth of is defined as treewidth, except that the tree is required to be a path. A graph is a minor of a graph if can be obtained from a subgraph of by contracting edges. (We note that since we only consider simple graphs, loops and parallel edges resulting from edge contractions are deleted.) Recall that the class of graphs of treewidth at most () is closed under taking minors, thus for every graph and minor of .

Given a class of graphs, we let denote the class of graphs that can be obtained from a graph by adding independently for each vertex of zero, one, or two new pendant vertices adjacent to . We will use the easy observation that when is the class of graphs of treewidth at most (provided ). We note that holds for other classes of interest, such as planar graphs.

The next elementary observation is due to Streib and Trotter [8], who were interested in the case of planar cover graphs. We provide a proof for the sake of completeness.

Observation 3.

Let be a class of graphs. If is a poset with then there exists a poset such that

1. , and

2. .

Proof.

If is a chain then we set and the statement can be easily verified.

Otherwise, let be the poset constructed from as follows: For each non-minimal element of , add a new element below (and its upset) such that is the only cover relation involving in . Also, for each non-maximal element of , add a new element above (and its downset) such that is the only cover relation involving in . Now, the cover graph of is the same as the cover graph of except that we attached up to two new pendant vertices to each vertex.

For convenience, we also define an element for each minimal element of , simply by setting . Similarly, we let , for each maximal element of .

Observe that if a set of linear extensions of reverses all min-max pairs of then it must reverse all incomparable pairs of . Indeed, for each pair consider the min-max pair in . There is some linear extension reversing . Given that and in , it follows that in . Hence, restricting the linear orders in to the elements of we deduce that reverses all pairs in so (as is not a chain). Therefore, . ∎

As a corollary, for treewidth we obtain:

Observation 4.

For every poset there exists a poset such that

1. , and

2. .

Proof.

This follows from Observation 3 if . If, on the other hand, , then is an antichain and we can simply take . ∎

In the next observation we consider posets with disconnected cover graphs. As expected, we define the components of a poset as the subposets of induced by the components of its cover graph.

Observation 5.

If is a poset with components then either

1. is a disjoint union of chains and we have , or

2. and
.

Proof.

If for each the subposet of is a chain then it is easy to see that . Thus we may assume that this is not the case, that is, for some .

For each let be a family of linear extensions of witnessing the dimension of . We construct a family of linear extensions of in the following way. First, let . Then, as long as there is a set which is not empty,

1. choose a linear extension for each such that is not empty;

2. choose any linear extension of for each such that is empty;

3. add to the linear extension of defined by , and

4. remove from for each .

Clearly, . Now consider one arbitrarily chosen linear extension ; say we had when it was defined above, and replace by in . It is easy to verify that the resulting family reverses all incomparable pairs in . In particular, all incomparable pairs of with elements from distinct components are reversed by and any other linear extension in (note there is at least one more as for some ). This shows that .

The proof for goes along the same lines and is thus omitted. ∎

To prove the next observation we partition the minimal and maximal elements of a poset by ‘unfolding’ the poset from an arbitrary minimal element, and contract some part of the poset into a single element. This proof idea is due to Streib and Trotter [8], and is very useful for our purposes. In [8] it was used in the context of planar cover graphs but it works equally well for any minor-closed class of graphs.

Observation 6.

For every poset there exists a poset such that

1. is a minor of (and thus in particular );

2. there is an element with in for all , and

3. .

Proof.

First of all, we note that it is enough to prove the statement in the case where is connected. Indeed, if is disconnected then by Observation 5 either is a disjoint union of chains and , in which case the observation is trivial, or and we can simply consider a component of with .

From now on we suppose that is connected. We are going to build a small set of linear extensions of reversing all min-max pairs of . Partition the minimal and maximal elements of as follows. Choose an arbitrary element , let , and for let

 Bi ={b∈max(P)−⋃1⩽j

Let be the least index such that is empty. See Figure 2 for an illustration. The fact that each minimal and maximal element of is included in one of the sets defined above follows from the connectivity of . If then is below all maximal elements of and hence itself satisfies conditions (i)-(iii). So we may assume from now on.

Let be the poset on the set of elements with order relation inherited from . Figure 3 illustrates this definition. Let

 t=max{dim∗(Qi+1i)∣i=0,…,k−1}.

For each consider linear extensions of that reverse all pairs from the set . Combining these we define linear extensions of . For let be a linear extension of that contains the linear order

 Lk−1j<⋯

Then, reverse all pairs with and where . In a similar way we are able to reverse the pairs where .

Let be the poset on the set of elements being ordered as in . We set and for each we fix linear extensions of reversing all pairs from . Again, we combine these to obtain linear extensions of . For let be a linear extension of that contains the linear order

 L1j

Clearly, reverse all pairs with and where . It follows that reverse the set and hence .

Now suppose first , so in particular . Then let such that . Note that we must have since . We define to be the poset that is obtained from by adding an extra element which is such that for all , and incomparable to all other elements of (here we need so that exists). In particular, for all . Observe that the cover graph of is an induced subgraph of with an extra vertex linked to some of the other vertices. Here, can be seen as the result of the contraction of the connected set plus the deletion of some of the edges incident to the contracted vertex (see Figure 3 with , dashed edges indicate deletions). The deletion step is necessary, as after the contraction it might be that some edges incident to do not correspond to cover relations anymore. It follows that is a minor of . Furthermore, it holds that

 dim∗(P)⩽2t=2dim∗(Qℓ+1ℓ)⩽2dim∗(Q).

Therefore, the dual of satisfies conditions (i)-(iii).

The case goes along similar lines as in the first case (with the slight difference that we do not need to exclude the subcase ). We leave the details to the reader. ∎

Applying these observations we move from Theorem 1 to a more technical statement.

Theorem 7.

Let be a poset with

1. a cover graph of treewidth at most , and

2. a minimal element such that for all .

Then the set can be partitioned into reversible sets.

In order to deduce Theorem 1 from Theorem 7 consider any poset with cover graph of treewidth at most . By Observation 4 there is a poset with and . Now by Observation 6 and applying Theorem 7 there is a poset with , a minimal element such that for all , and

 dim(P)⩽dim∗(Q)⩽2dim∗(R)⩽2⋅638=1276,

as desired.

From now on we focus on the proof of Theorem 7. Let be a poset fulfilling the conditions of Theorem 7. Consider a tree decomposition of width at most of , consisting of a tree and subtrees for each . We may assume that the width of the decomposition is exactly , since otherwise by the result of Trotter and Moore [11], and the theorem follows trivially.

For each node of let denote its bag, namely, the set . Since the tree decomposition has width , every bag has size at most , and at least one bag has size exactly . Modifying the tree decomposition if necessary, we may suppose that every bag has size . Indeed, say is an edge of with and . Then choose arbitrarily elements from and add them to . Repeating this process as many times as necessary, we eventually ensure that every bag has size . Note that the subtrees () of the tree decomposition are uniquely determined by the bags, and vice versa; thus, it is enough to specify how and the bags are modified. The above modification repeatedly adds leaves to some of the subtrees (), which clearly keeps the fact that and the subtrees () form a tree decomposition of .

Recall that, by the assumptions of Theorem 7, the poset has a minimal element with for all . This implies that the cover graph of is connected. Using this, we may suppose without loss of generality that for each edge of . For if this does not hold, then the bags of one of the two components of are all empty (as is easily checked), and thus the nodes of that component can be removed from without affecting the tree decomposition.

In fact, we may even assume that holds for every edge of . To see this, consider the following iterative modification of the tree decomposition: Suppose that is an edge of such that . If then simply identify and , and contract the edge in . If then subdivide the edge in with a new node , and let the bag of be the set , where and are arbitrarily chosen elements in and , respectively. These modifications are valid, in the sense that the bags still define a tree decomposition of of width , and in order to ensure the desired property it suffices to apply them iteratively until there is no problematic edge left.

To summarize, in the tree decomposition we have for every node of , and for every edge of . We will need to further refine our tree decomposition so as to ensure a few extra properties. These changes will be explained one by one below. Let us mention that we will keep the fact that for every edge of , and that for every internal node of . However, we will add new leaves to having bags of size only.

Choose an arbitrary node with . Add a new node to and make it adjacent to . The bag of is defined as the union of and one arbitrarily chosen element from . (Observe that the size of is only ; on the other hand, we do have .) We call the root of , and thus see as being rooted at . (For a technical reason we need the root to be a leaf of , which explains why we set it up this way.) Every non-root node in has a parent in , namely, the neighbor of on the path from to in . Now we have an order relation on the nodes of , namely in if is on the path from to in . The following observation will be useful later.

Observation 8.

If is a sequence of nodes of such that consecutive nodes are comparable in (that is or in for each ), then there is an index such that in for each .

Proof.

We prove this by induction on . For it is immediate. So suppose that . Then we can apply the induction hypothesis on the sequence and get such that for each . As and are comparable in , we have or in . In the first case we conclude in and we are done. In the second case we have in , which makes and comparable in . But clearly, from this it follows that in for each or in for each . ∎

Fix a planar drawing of the tree with the root at the bottom. Suppose that and are two nodes of that are incomparable in . Take the maximum node (with respect to the order in ) such that and in . We denote this node by . Observe that has degree at least in , and hence is distinct from the root . (Ensuring this is the reason why we made sure that the root is a leaf.) Consider the edge from to , the edge from towards and the edge from towards . All these edges are distinct. If the clockwise order around in the drawing is for these three edges, then we say that is to the left of in , otherwise the clockwise order around is and we say that is to the right of in . Observe that the relations “is left of in ” and “is right of in ” both induce a linear order on any set of nodes which are pairwise incomparable in .

Observation 9.

Let and be incomparable nodes in with left of in , and let . If and are the neighbors of on the paths towards and in , respectively, then for each node in we have that

1. is left of in if in , and

2. is left of in if in .

Proof.

If in , then we also have , and the first edge on the path from to in is the same as that of the path from to in . Since is left of in , it follows that is left of as well. The proof for the second item is analogous. ∎

Next we modify once more the tree decomposition. For each element such that for some , choose arbitrarily a node of such that . Similarly, for each element such that for some , choose arbitrarily a node of such that . (Note that the same node of could possibly be chosen more than once.) Now that all these choices are made, for each minimal element of considered above, add a new leaf to adjacent to with bag , where is an arbitrarily chosen element from . Similarly, for each maximal element of considered above, add a new leaf to adjacent to with bag , where is an arbitrarily chosen element from .

This concludes our modifications of the tree decomposition. Notice that we made sure that for every internal node of , and that for every edge of . Observe also that for every pair , the two nodes and are incomparable in , and thus one is to the left of the other in . Figure 4 provides an illustration. (We also note that while the tree has been modified since stating Observations 8 and 9, they obviously still apply to the new tree .)

Let be the intersection graph of the subtrees () of . Thus two distinct elements are adjacent in if and only if . The graph is chordal and the maximum clique size in is . Hence the vertices of can be (properly) colored with three colors. We fix a -coloring of which is such that receive distinct colors whenever . In particular, if and are two distinct elements of such that for some then and receive different colors.

We end this section with a fundamental observation which is going to be used repeatedly in a number of forthcoming arguments. We say that a relation in hits a set if there exists with in .

Observation 10.

Let in and let be such that , .

1. If lies on the path from to in then hits .

2. If lies on the path from to in then hits .

3. If are nodes on the path from to in appearing in this order, then there exist for each such that in .

Proof.

Suppose that lies on a path from to in . Since in there is a path in such that is a cover relation in for each . This means that is a (connected) subtree of containing and . Thus, contains and therefore there exists with . The proof of 2 is analogous.

We prove 3 by induction on . For this corresponds to 1, so let us assume and consider the inductive case. By induction there exist for each such that in . Applying 1 with relation and the path, we obtain that in for some . Combining, we obtain in , as desired. ∎

3. The Proof

We aim to partition into a constant number of sets, each of which is reversible. This will be realized with the help of a signature tree, which is depicted on Figure 5. This plane tree , rooted at node , assigns to each pair a corresponding leaf of according to properties of the pair .

The nodes of are enumerated by depth-first and left-to-right search. Each node which is distinct from the root and not a leaf has a corresponding function of the form , where and is a finite set, whose size does not depend on . We put and the other domains will be defined one by one in this section. To give an example, let us look forward to upcoming subsections where we define and as follows.

• encodes whether is to the left or to the right of in ;

• is the answer to the question “Is there an element with in ?”.

Furthermore, for each internal node with children in , the edges of are respectively labeled by subsets of such that , that is, so that the sets form a partition of . For example,

 Σ(ν1,ν2)={left,right}=Σ1; Σ(ν2,ν3)={no}; Σ(ν2,ν4)={yes}.

Observe that each internal node of has either one or two children; in particular, if has only one child then the corresponding edge is labeled with the full set .

The reader may wonder why we do not refine the tree and have an edge out of for every possible value in . This is because sometimes several values in will correspond to analogous cases in our proofs which can be treated all at once. To give a concrete example, consider : When proving that a set of min-max pairs is reversible, the case that is left of for every is analogous to the case that is right of for every , as one is obtained from the other by exchanging the notion of left and right in (that is, by replacing the plane tree by its mirror image). Hence it will be enough to only consider, say, the case where is to the left of for every .

Now for an internal node of distinct from the root (), let be the path from the root to in . Define the signature of as the set

 Σ(νi)=Σ(νi1,νi2)×…×Σ(νil−1,νil) (1)

and let

 MM(P,νi) ={(a,b)∈MM(P)∣(αi1(a,b),…,αil−1(a,b))∈Σ(νi)}; MM(P,νi,Σ) ={(a,b)∈MM(P)∣(αi1(a,b),…,αil−1(a,b))=Σ}for Σ∈Σ(νi).

Observe that by this definition, for each internal node of with children we get the partition

 MM(P,νi) =⋃1⩽j⩽lMM(P,νij).

Therefore, by construction the sets with a leaf of (so for ) form a partition of . With a further refinement it follows that

 MM(P)=⋃νi leaf of Ψ⋃Σ∈Σ(νi)MM(P,νi,Σ)

and the proof below boils down to showing that is reversible for each leaf of and each .

Once this is established we get an upper bound on just by counting the number of sets in our partition of , namely

 dim∗(P)⩽∑νi leaf of% Ψ|Σ(νi)| =|Σ(ν3)|+|Σ(ν7)|+|Σ(ν10)|+|Σ(ν15)| =2+2⋅6+2⋅6⋅2⋅2+2⋅6⋅6⋅4⋅2=638.

Note that the calculation for the particular summands follows from (1) and the edge labelings in Figure 5. Our proof will follow a depth-first, left-to-right search of the signature tree , defining the functions one by one in that order, and showing that for each the set is reversible when encountering a leaf . Hence, the tree also serves as a road map of the proof. Moreover, for the reader’s convenience we included a table collecting all functions and their meanings, see Table 1. It is not necessary to read this table now, but it might be helpful while going through the main proof.