A queue layout of a graph consists of a total ordering on its vertices and a partition of its edge set into queues, i.e., no two edges in a single block of the partition are nested. The minimum number of queues needed in a queue layout of a graph is its queue-number and denoted by .
To be more precise, let be a graph and let be a linear order of the vertices. A -rainbow is a set of edges such that in . A pair of edges forming a 2-rainbow is said to be nested. A queue is a set of edges without nesting. Given and , the edges of can be partitioned into queues if and only if there is no rainbow of size in . The queue-number of is the minimum number of queues needed to partition the edges of over all linear orders .
Queue layouts were introduced by Heath and Rosenberg in 1992  as a counterpart of book embeddings. Queue layouts were implicitly used before and have applications in fault-tolerant processing, sorting with parallel queues, matrix computations, scheduling parallel processes, and in communication management in distributed algorithm (see [4, 6, 8]). There is a rich literature exploring bounds on the queue-number of different classes of graphs [4, 6, 9, 2].
Here we study the queue-number of posets. This parameter was introduced in 1997 by Heath and Pemmaraju , inspired by the older concept of the queue-number of directed acyclic graphs. For a queue layout of a directed acyclic graph, it is required that precedes in the total vertex ordering whenever there is a directed edge . I.e., it is a topological ordering of the graph.
A poset is a pair of a finite set of elements, called the ground set, and a transitive (if and , then ) and antisymmetric (if , then ) binary relation on . Two elements are called comparable if either or , and incomparable otherwise. A relation in is a cover if it is not implied by transitivity, i.e., there is no element such that . In the context of drawings, embeddings and layouts for posets , it is natural to work with their directed cover graphs, having vertex set and a directed edge for every cover relation in . For example a diagram of is an upward drawing of the directed cover graph where the direction on edges is usually omitted as each edge is implicitly directed upwards.
Now, a linear extension of is simply a topological ordering of its directed cover graph, and we write in if precedes in (though not necessarily in ). The queue-number of , denoted by , is the smallest such that there is a linear extension of for which the resulting linear layout of the directed cover graph contains no -rainbow. Fig. 1 shows an example.
Clearly, if denotes the undirected cover graph of , then , i.e., the queue-number of a poset is at least as large as the queue-number of its (undirected) cover graph. It was shown by Heath and Pemmaraju  that even for planar posets there is no function such that . They also investigated the maximum queue-number of several classes of posets, in particular with respect to bounded width (the maximum number of pairwise incomparable elements) and height (the maximum number of pairwise comparable elements). In particular they gave a nice argument showing that (see Proposition 1 below). The poset of height 2 and width whose cover graph is the complete bipartite graph attains . Actually, Heath and Pemmaraju conjectured that for every poset .
Knauer, Micek and the second author  showed that the inequality holds for all posets of width 2. Last year, Alam et al.  constructed a non-planar poset of width whose queue-number is ; thus refuting the conjecture of Heath and Pemmaraju. Using a simple lifting argument from , Alam et al. generalized their example and constructed for every a poset with and . Fig. 2 shows their construction. In fact, consider the lifting construction in the middle of Fig. 2 and a fixed linear extension . If in , then the cover edge from the bottommost element to nests above the lower copy of . Symmetrically, if in , the cover edge from to the topmost element nests above the upper copy of . In any case, we extend any rainbow in by one edge. Similarly, in the right of Fig. 2 one of the diagonal cover edges will nest above one of the copies of in any linear extension.
Let us also mention that a second contribution of Alam et al. consists in a slight improvement of the upper bound: They show for all posets of width at most .
Our contribution is the following theorem.
For every there is a poset of width with
These examples (asymptotically) match the upper bound. Besides yielding a strong improvement of the lower bound, we also believe that the analysis of our construction is conceptually simpler than the example provided by Alam et al. to disprove the conjecture of Heath and Pemmaraju. The key difference is that we improve the lifting step rather than the base case. In particular, we show how to lift any poset of width so that the width goes up by only , but the queue-number goes up by at least .
As an open problem we promote the question whether the original conjecture holds for planar posets. In  it was shown that the queue-number of planar posets of width is upper bounded by and that there are such planar posets with .
Before presenting our construction, we like to revisit the nice upper bound argument of Heath and Pemmaraju. Let be a poset of width . Dilworth’s Theorem asserts that can be decomposed into chains of .
Proposition 1 (Heath and Pemmaraju).
For every poset we have .
Let , let be a chain partition, and let be any linear extension of . Partition the cover edges into sets with such that if and . We claim that each is a queue.
Let in support a pair of nesting cover edges and suppose that both edges and belong to . By definition and and from the ordering in we get and in . Now we have and and in and thus the relation is implied by transitivity. This contradicts that is a cover edge. ∎
In fact we have shown a much stronger statement: If and a chain partition are given, then there is a partition of the edges of the cover graph of into parts with such that each is a queue for every(!) linear extension of . Let us remark that for some posets and some linear extensions of , the resulting queue layout indeed has a -rainbow. An example is indicated in Fig. 3.
2.1 Concepts needed for the construction
Let be a poset. The dual of , denoted , is the poset on the same ground set such that: . In terms of its diagram, the dual of is obtained by flipping along a horizontal line.
A poset is 2-dimensional if and only if there are two linear extensions and such that: . Such a pair is called a realizer of .
When drawing 2-dimensional posets, it is common to represent each element by a point with coordinates where is the position of in and is the position of in , see Fig. 4. This is also called a dominance drawing.
3 Proof of Theorem 1.1
We define recursively, focusing on the recursive step. As mentioned in the introduction, the recursive step involves lifting a given poset of width to the desired poset of width such that . Our lifting can be seen as an extension of the situation on the very right of Fig. 2. Specifically, for , the construction of is based on
a copy of ,
a reinforcement poset of width ,
two linear extensions and of , and
the duals of the above.
We invite the reader to take a look at Fig. 5, which shows the construction of using and as a black box. Formally, let denote the number of elements in . Then, contains besides , two additional elements and , and four chains of additional elements , , , and , together with the following additional relations:
is below and above .
is above all elements in and below all elements in .
All elements of are above all elements of , and all elements of are below all elements of .
is above the -th element in the linear extension of , .
is below the -th element in the dual of .
is above the -th element in the linear extension of , .
is below the -th element in the dual of .
All relations that are transitively implied by the above.
First we observe that , as and the additional elements (except , which can be incorporated into an existing chain) can be covered by two chains. Also note that the number of elements of the poset is given by the recursion . (Recall that is the number of elements of .) Further note that and the -th element of in indeed form a cover edge, as is a linear extension of , . Similarly for the edges between and , as well as between and , .
Furthermore, it can be seen that is self-dual; the reflection having two fixed points and . This shows that when analyzing , we can restrict the attention to linear extensions of which have before . With this assumption, a rainbow between and either or nests above every rainbow of . See Fig. 6 for an illustration. If we let be the size of a rainbow between and either or , then we have the recursion:
We think of this use of a self-dual construction as the symmetry trick. Again, let us mention that constructions given in  (proof of Prop. 2) and  (proof of Thm. 4) also use a recursion based on two copies of the poset from the previous level of the recursion, as illustrated in the middle of Fig. 2. However, this only forces one edge to nest over the rainbow from the previous level of the recursion. Our lifting forces a rainbow of edges whose size is linear in the width to nest over the previous level construction and its rainbow; thus giving overall a quadratic lower bound.
It remains to construct the poset together with two linear extensions and such that in any linear extension having entirely before , a large rainbow between and either or appears. Recall that denotes the largest such rainbow and we seek to construct such that is at least linear in .
As the elements in form a chain and thus are ordered in this way in , rainbows between and are in bijection with subsets of elements in that are oppositely ordered in and . Similarly, rainbows between and appear when elements in are oppositely ordered in and . Thus our goal is to construct , and such that for every linear extension of there is a long increasing sequence in which is decreasing in or .
To illustrate this idea, suppose that for each width , we choose the poset to be an antichain of size and the linear extensions and to be a realizer (think of as the identity permutation and of as its reverse). The Lemma of Erdős-Szekeres asserts that in every linear extension of there is an increasing or a decreasing sequence of size at least , i.e., .
This value of together with Inequality (1) yields
For the proof of the theorem we need a better construction for the reinforcement posets . In particular, we seek to have instead of just . A construction of such a is given in Subsection 3.1 and based on the following lemma111The lemma with a different proof was discovered (but not yet published) in October 2020 by the first and the second author together with Francois Dross, Piotr Micek, and Michał Pilipczuk..
For each , there is a 2-dimensional poset of width with a realizer , such that if is a linear extension of and and denote the maximum lengths of an increasing sequence in which is decreasing in and respectively, then .
The lemma says that we can assume the value . With Inequality (1) we get:
The base of our recursive construction is the case or , depending on the parity of . For the validity of Theorem 1.1, it is enough to let with be any poset of width . Of course, it is beneficial to start with a higher queue-number, also given that our bound of is less than the of Alam et al.  for small . The best results are achieved by starting at width or (depending on the parity of the target width ) with the poset of Alam et al.  with queue-number , respectively .
3.1 The construction of for Lemma 1
The construction of is again recursive. Let be a single element. Then clearly . For the construction of for we again use the symmetry trick. We take two copies of and two additional elements and . Then is obtained by a series composition of , and a parallel composition of the result with element . Formally,
is above every element of and below every element of , while
is incomparable to all other elements.
The two linear extensions of the realizer of are obtained as follows.
where is the realizer of the copy of , . We invite the reader to look at Fig. 7 for two illustrations of this recursive construction step for and its realizer .
First, we observe that , as element can be covered by a new chain and element can be incorporated into an existing chain. Also note that the number of elements in is given by the recursion , which with solves for . Further observe that is again self-dual. In particular the two copies and of are isomorphic. The reflection has two fixed points and .
Now let be any linear extension of . First suppose that in . Let be the restriction of to . By induction the lengths and of increasing sequences of which are decreasing in the two linear extensions of the realizer of satisfy . Since precedes in and comes after in , we have . Together with the trivial , we get .
If we have in , then we consider . As before we get the two values and for the restriction of to and know by induction that . This time precedes in but comes after in , which gives . Together with the trivial we again see that . This completes the proof of Lemma 1.
We remark that in both the construction of and , the element is used only for the sake of the exposition. It would suffice to add all relations between and , respectively and . While this gives slightly smaller constructions for and , they would still be exponential in their width. (Recall that and hence .)
We have made substantial progress in the understanding of queue-numbers of partially ordered sets. We take the opportunity to list and comment on open questions in the field.
An obvious question is to ask for improved upper and lower bounds. More precisely, we now know that the growth rate of the maximum queue-number of posets of width is for some constant between and 1. What is the precise value of constant ?
Our reinforcement poset is 2-dimensional for every . However our entire lower bound example is not (already for ), and the same holds for the example of Alam et al. in the left of Fig. 2. We think it is interesting to see whether there exists any 2-dimensional poset with .
What is the maximum queue-number of posets of width with a planar diagram? Knauer, Micek, and the second author  proved the lower bound by observing that the simple lifting operation in the middle of Fig. 2 preserves planarity, while their upper bound is . Clearly, the better lifting operation introduced here necessarily introduces crossing cover edges.
Heath and Pemmaraju  conjectured that planar posets on elements have queue-number at most . Their lower bound construction is an -antichain with realizer together with an -chain matched upward in order of and an -chain matched downward in order of ; see Fig. 8. The Lemma of Erdős-Szekeres implies for this planar poset with elements that . It is open whether there is an asymptotically matching upper bound.
Dujmović and Wood  show that a random vertex ordering for an undirected graph
has with positive probability no rainbow of size, where is the base of the natural logarithm and is the number of edges in . Can a similar result be obtained by considering a random linear extension of a poset ? Note that a positive answer would resolve (up to a constant factor) the previous question of Heath and Pemmaraju about planar posets.
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