1 Introduction
A queue layout of a graph consists of a total ordering on its vertices and an assignment of its edges to queues, such that no two edges in a single queue are nested. The minimum number of queues needed in a queue layout of a graph is called its queuenumber and denoted by .
To be more precise, let be a graph and let be a linear order on the vertices of . We say that the edges are nested with respect to if or in . Given a linear order of the vertices of , the edges of form a rainbow of size if in . Given and , the edges of can be partitioned into queues if and only if there is no rainbow of size in , see [10].
The queuenumber was introduced by Heath and Rosenberg in 1992 [10] as an analogy to book embeddings. Queue layouts were implicitly used before and have applications in faulttolerant processing, sorting with parallel queues, matrix computations, scheduling parallel processes, and communication management in distributed algorithm (see [8, 10, 12]).
Perhaps the most intriguing question concerning queuenumbers is whether planar graphs have bounded queuenumber.
Conjecture 1 (Heath and Rosenberg [10])
The queuenumber of planar graphs is bounded by a constant.
In this paper we study queuenumbers of posets. The parameter was introduced in 1997 by Heath and Pemmaraju [9] and the main idea is that given a poset one should lay it out respecting its relation. Two elements of a poset are called comparable if or , and incomparable, denoted by , otherwise. Posets are visualized by their diagrams: Elements are placed as points in the plane and whenever in the poset, and there is no element with , there is a curve from to going upwards (that is monotone). We denote this case as . The diagram represents those relations which are essential in the sense that they are not implied by transitivity, also known as cover relations. The undirected graph implicitly defined by such a diagram is the cover graph of the poset. Given a poset , a linear extension of is a linear order on the elements of such that , whenever . (Throughout the paper we use a subscript on the symbol , if we want to emphasize which order it represents.) Finally, the queuenumber of a poset , denoted by , is the smallest such that there is a linear extension of for which the resulting linear layout of contains no rainbow. Clearly we have , i.e., the queuenumber of a poset is at least the queuenumber of its cover graph. It is shown in [9] that even for planar posets, that is posets admitting crossingfree diagrams, there is no function such that .
Heath and Pemmaraju [9] investigated the maximum queuenumber 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). A set with every two elements being comparable is a chain. A set with every two distinct elements being incomparable is an antichain. They proved that if , then . The lower bound is attained by weak orders, i.e., chains of antichains and is conjectured to be the upper bound as well:
Conjecture 2 (Heath and Pemmaraju [9])
Every poset of width has queuenumber at most .
Furthermore, they made a step towards this conjecture for planar posets: if a planar poset has , then . For the lower bound side they provided planar posets of width and queuenumber .
We improve the bounds for planar posets and get the following:
Theorem 1.1
Every planar poset of width has queuenumber at most . Moreover, there are planar posets of width and queuenumber .
As an ingredient of the proof we show that posets without certain subdivided crowns satisfy Conjecture 2 (c.f. Theorem 2.1). This implies the conjecture for interval orders and planar posets with (unique minimum) 0 and (unique maximum) 1 (c.f. Corollary 2). Moreover, we confirm Conjecture 2 for the first nontrivial case :
Theorem 1.2
Every poset of width has queuenumber at most .
An easy corollary of this is that all posets of width have queuenumber at most (c.f. Corollary 1).
Another conjecture of Heath and Pemmaraju concerns planar posets of bounded height:
Conjecture 3 (Heath and Pemmaraju [9])
Every planar poset of height has queuenumber at most .
We show that Conjecture 3 is false for the first nontrivial case :
Theorem 1.3
There is a planar poset of height with queuenumber at least .
Furthermore, we establish a link between a relaxed version of Conjecture 3 and Conjecture 1, namely we show that the latter is equivalent to planar posets of height having bounded queuenumber (c.f. Theorem 4.1). On the other hand, we show that Conjecture 3 holds for planar posets with 0 and 1:
Theorem 1.4
Every planar poset of height with and has queuenumber at most .
Organization of the paper.
In Section 2 we consider general (not necessarily planar) posets and give upper bounds on their queuenumber in terms of their width, such as Theorem 1.2. In Section 3 we consider planar posets and bound the queuenumber in terms of the width, both from above and below, i.e., we prove Theorem 1.1. In Section 4 we give a counterexample to Conjecture 3 by constructing a planar poset with height and queuenumber at least . Here we also argue that proving any upper bound on the queuenumber of such posets is equivalent to proving Conjecture 1. Finally, we show that Conjecture 3 holds for planar posets with 0 and 1 and that for every there is a planar poset of height and queuenumber (c.f. Proposition 3).
2 General Posets of Bounded Width
By Dilworth’s Theorem [3], the width of a poset coincides with the smallest integer such that can be decomposed into chains of . Let us derive Proposition 1 of Heath and Pemmaraju [9] from such a chain partition.
Proposition 1
For every poset , if then .
Proof
Let be a poset of width and be a chain partition of . Let be any linear extension of and with and . Note that we must have either or . If follows that if , , , and , then . As there are only ordered pairs with , we can conclude that every nesting set of covers has cardinality at most .
Note that in the above proof is any linear extension and that without choosing the linear extension carefully, upper bound is bestpossible. Namely, if with comparabilities for all , then has width and the linear extension creates a rainbow of size .
We continue by showing that every poset of width has queuenumber at most , that is, we prove Theorem 1.2.
Proof (Theorem 1.2)
Let be a poset of width and minimum element and be a chain partition of . Note that the assumption of the minimum causes no loss of generality, since a can be added without increasing the width nor decreasing the queuenumber. Any linear extension of partitions the ground set naturally into inclusionmaximal sets of elements, called blocks, from the same chain in that appear consecutively along , see Figure 2. We denote the blocks by according to their appearance along . We say that is lazy if for each , each element has a relation to some element . A linear extension can be obtained by picking any minimal element , put it into , and recurse on . Lazy linear extensions (with respect to ) can be constructed by the same process where additionally the next element is chosen from the same chain as the element before, if possible. Note that the existence of a is needed in order to ensure the property of laziness with respect to .
Now we shall prove that in a lazy linear extension no three covers are pairwise nesting. So assume that is any cover and that and . As is lazy, is comparable to some element in (if ) and all elements in (if ). With being a cover, it follows from being lazy that . If , then no cover is nested under . If , then no cover is nested above : either and and hence is not a cover, or both endpoints would be inside the same chain, i.e., are the last and first element of and or and , respectively. This implies or , respectively, and cannot nest above . If , then no cover is nested above . Thus, either no cover is nested below , or no cover is nested above , or both. In particular, there is no three nesting covers and .
Corollary 1
Every poset of width has queuenumber at most .
Proof
We take any chain partition of size and pair up chains to obtain a set of disjoint pairs. Each pair from induces a poset of width at most , which by Theorem 1.2 admits a linear order with at most two nesting covers. Let be a linear extension of respecting all these partial linear extensions.
Now, following the proof of Proposition 1 any cover can be labeled by a pair corresponding to the chains containing its endpoint. Thus, in a set of nesting covers any pair appears at most once, but for each such that only two of the four possible pairs can appear simultaneously in a nesting. This yields the upper bound.
For an integer we define a subdivided crown as the poset as follows. The elements of are and the cover relations are given by and for , for , and ; see the left of Figure 3. We refer to the covers of the form as the diagonal covers and we say that a poset has an embedded if contains elements that induce a copy of in with all diagonal covers of that copy being covers of .
Theorem 2.1
If is a poset that for no has an embedded , then the queuenumber of is at most the width of .
Proof
Let be any poset. For this proof we consider the cover graph of as a directed graph with each edge directed from to if in . We call these edges the cover edges. Now we augment to a directed graph by introducing for some incomparable pairs a directed edge. Specifically, we add a directed edge from to if there exists a with in where is a cover relation and is not a cover relation; see the right of Figure 3. We call these edges the gray edges of .
Now we claim that if has a directed cycle, then has an embedded subdivided crown. Clearly, every directed cycle in has at least one gray edge. We consider the directed cycles with the fewest gray edges and among those let be one with the fewest cover edges.s First assume that has a cover edge (hence ), say is a gray edge followed by a cover edge . Consider the element with cover relation and noncover relation in . By we have a noncover relation in . Now if in , then contains the gray edge (see Figure 4(a)) and is a directed cycle with the same number of gray edges as but fewer cover edges, a contradiction. On the other hand, if in (note that is impossible as is a cover), then there is a directed path of cover edges from to (see Figure 4(b)) and contains a directed cycle with fewer gray edges than , again a contradiction.
Hence is a directed cycle consisting solely of gray edges. Note that by the first paragraph is an antichain in . For let be the element of with cover relation and noncover relation , as well as with cover relation and noncover relation . As is an antichain and holds for , we have . Let us assume that in for some . If is a cover relation, then there is a gray edge in (see Figure 4(c)) and the cycle is shorter than , a contradiction. If is a noncover relation, then there is a gray edge in (see Figure 4(d)) and the cycle is shorter than , again a contradiction.
Hence, the only relations between and are cover relations and for and the noncover relations for . Hence are pairwise distinct. Moreover, is an antichain in since the only possible relations among these elements are of the form or , which would contradict that and are cover relations. Finally, we pick for every an element with , which exists as is a noncover relation. Together with the above relations between and we conclude that are pairwise distinct and these elements induce a copy of in with all diagonal covers in that copy being covers of .
Thus, if has no embedded , then the graph we constructed has no directed cycles, and we can pick to be any topological ordering of . As , is a linear extension of . For any two nesting covers we have or or both, since is a cover. However, if in , then there would be a gray edge from to in , contradicting and being a topological ordering of . We conclude that and the left endpoints of any rainbow form an antichain, proving .
Let us remark that several classes of posets have no embedded subdivided crowns, e.g., graded posets, interval orders (since these are 2+2free, see [6]), or (quasi)seriesparallel orders (since these are Nfree, see [7]). Here, 2+2 and N are the fourelement posets defined by and , respectively. Also note that while subdivided crowns are planar posets, no planar poset with 0 and 1 has an embedded crown. Indeed, already looking at the subposet induced by the crown and the 0 and the 1, it is easy to see that there must be a crossing in any diagram. Thus, we obtain:
Corollary 2
For any interval order, seriesparallel order, and planar poset with 0 and 1, we have .
3 Planar Posets of Bounded Width
Heath and Pemmaraju [9] show that the largest queuenumber among planar posets of width lies between and . Here we improve the lower bound to and the upper bound to .
Proposition 2
For each there exists a planar poset with 0 and 1 of width and queuenumber .
Proof
We shall define recursively, starting with being any chain. For , consists of a lower copy and a disjoint upper copy of , three additional elements , and the following cover relations in between:

, where is the 0 of

, where is the 1 of and is the 0 of

, where is the 1 of

It is easily seen that all cover relations of and remain cover relations in , and that is planar, has width , is the 0 of , and is the 1 of . See Figure 5 for an illustration.
To prove that we argue by induction on , with the case being immediate. Let be any linear extension of . Then is the first element in and is the last. Since , all elements in come before all elements of . Now if in the element comes after all elements of , then is nested under cover , and if comes before all elements of , then is nested under cover . We obtain nesting covers by induction on in the former case, and by induction on in the latter case. This concludes the proof.
Next we prove Theorem 1.1, namely that the maximum queuenumber of planar posets of width lies between and .
Proof (Theorem 1.1)
By Proposition 2 some planar posets of width have queuenumber . So it remains to consider an arbitrary planar poset of width and show that has queuenumber at most . To this end, we shall add some relations to , obtaining another planar poset of width that has a and , with the property that . Note that this will conclude the proof, as by Corollary 2 we have .
Given a planar poset of width , there are at most minima and at most maxima. Hence there are at most extrema that are not on the outer face. For each such extremum –say is a minimum– consider the unique face with an obtuse angle at . We introduce a new relation , where is a smallest element at face , see Figure 6. Note that this way we introduce at most new relations, and that these can be drawn ymonotone and crossingfree by carefully choosing the other element in each new relation. Furthermore, every inner face has a unique source and unique sink.
Now consider a cover relation that is not a cover relation in the new poset . For the corresponding edge from to in there is one face with unique source and unique sink . Now either way the other edge in incident to or to must be one of the newly inserted edges, see again Figure 6. This way we assign to one of queues, one for each newly inserted edge. Every such queue contains either at most one edge or two incident edges, i.e., a nesting is impossible, no matter what linear ordering is chosen later.
We create at most queues to deal with the cover relations of that are not cover relations of and spend another queues for dealing with the remaining cover relations of . Thus, .
4 Planar Posets of Bounded Height
Recall Conjecture 3, which states that every planar poset of height has queuenumber at most . In the following, we give a counterexample to this conjecture:
Proof (Theorem 1.3)
Consider the graph that is constructed as follows: Start with with bipartition classes and . For every add four new vertices , each connected to and . The resulting graph has vertices, is planar and bipartite with bipartition classes and . See Figure 7.
Let be the poset arising from by introducing the relation for every edge in with and . Clearly, has height and hence the cover relations of are exactly the edges of . Moreover, by a result of Moore [11] (see also [2]) is planar because is planar, also see the right of Figure 7.
We shall argue that . To this end, let be any linear extension of . Without loss of generality we have . Note that since in one bipartition class of is entirely below the other, any cycle in gives a rainbow. Let be the first two elements of in , be the last two such elements. As there exists such that , i.e., we have , where we use that and are above all elements of in .
Now consider the elements that are above and in . As , there are two elements of that are both below in , or both between and in , or both above in . Consider the rainbow in the cycle . In the first case is nested below the cycle , in the second case the cover is nested below and is nested below the cover , and in the third case cycle is nested below . As each case results in a rainbow, we have .
Even though Conjecture 3 has to be refuted in its strongest meaning, it might hold that planar posets of height have queuenumber , or at least bounded by some function in terms of , or at least that planar posets of height have bounded queuenumber. As it turns out, all these statements are equivalent, and in turn equivalent to Conjecture 1.
Theorem 4.1
The following statements are equivalent:

Planar graphs have queuenumber (Conjecture 1).

Planar posets of height have queuenumber .

Planar posets of height have queuenumber at most for a function .

Planar posets of height have queuenumber .

Planar bipartite graphs have queuenumber .
Proof
 ab

Pemmaraju proves in his thesis [13] (see also [4]) that if is a graph, is a vertex ordering of with no rainbow, are color classes of any proper coloring of , and is the vertex ordering with , where within each the ordering of is inherited, then has no rainbow. So if is any poset of height , its cover graph has by a for some global constant . Splitting into antichains by iteratively removing all minimal elements induces a proper coloring of with color classes . As every vertex ordering of with is a linear extension of , it follows by Pemmaraju’s result that , i.e., .
 bcd

These implications are immediate.
 de
 ea

This is a result of Dujmović and Wood [5].
Finally, we show that Conjecture 3 holds for planar posets with and .
Proof (Theorem 1.4)
Let be a planar poset with and . Then has dimension at most two [1], i.e., it can be written as the intersection of two linear extensions of . A particular consequence of this is, that there is a welldefined dual poset in which two distinct elements are comparable in if and only if they are incomparable in . Poset reflects a “left of”relation for each incomparable pair in in the following sense: Any maximal chain in corresponds to a path in , which splits the elements of into those left of and those right of . Now if and only if is left of the path for every maximal chain containing (equivalently is right of the path for every maximal chain containing ). Due to planarity, if is a cover in and is a maximal chain containing neither nor , then and are on the same side of the path corresponding to . In particular, if for we have and , then and are comparable in , but if we would get a crossing of and . Also see the left of Figure 8. We summarize:

If , for some and for some , then .
Now let be the leftmost linear extension of , i.e., the unique linear extension with the property that for any in we have if and only if in . Assume that is a pair of nesting covers below . Then (hence ) or (hence ) or both. Observe that the latter case is impossible, as for any maximal chain containing we would have with and with , contradicting a. So the nesting of below is either of type A with , or of type B with . See Figure 9.
Now consider the case that cover is nested below another cover , see the right of Figure 8. Then also is nested below and we claim that if both, the nesting of below as well as the nesting of below , are of type A (respectively type B), then also the nesting of below is of type A (respectively type B). Indeed, assuming type B, we would get and , which together with any maximal chain containing contradicts a.
Finally, let be any rainbow and let , i.e., for each the nesting of below is of type A. Then we have just shown that the nesting of below is of type A whenever and of type B whenever . Hence, the set is a chain in of size , and thus . It follows that has queuenumber at most , as desired.
The proof of the following can be found in the appendix.
Proposition 3
For each there exists a planar poset of height and queuenumber .
5 Conclusions
We studied the queuenumber of (planar) posets of bounded height and width. Two main problems remain open: bounding the queuenumber by the width and bounding it by a function of the height in the planar case, where the latter is equivalent to the central conjecture in the area of queuenumbers of graphs. For the first problem the biggest class known to satisfy it are posets without the embedded the subdivided crowns for as defined in Section 2. Note, that proving it for would imply that Conjecture 2 holds for all dimensional posets, which seems to be a natural next step.
Let us close the paper by recalling another interesting conjecture from [9], which we would like to see progress in:
Conjecture 4 (Heath and Pemmaraju [9])
Every planar poset on elements has queuenumber at most .
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6 Appendix
Proof (Proposition 3)
We shall recursively define a planar poset of height and queuenumber , together with a certain set of marked subposets in . Each marked subposet consists of three elements forming a Vsubposet in , i.e., but , with both relations and being cover relations of , and being a minimal element of . We call such a marked subposet in a Vposet. Finally, we ensure that the Vposets are pairwise incomparable, namely that any two elements in distinct Vposets are incomparable in .
For let be the threeelement poset as shown in left of Figure 10, which also forms the only Vposet of . Clearly has height and queuenumber . For assume that we already constructed with a number of Vposets in it. Then is obtained from by replacing each Vposet by the eightelement poset shown in the right of Figure 10, which introduces (for each Vposet) five new elements. Moreover, two new Vposets are identified in as illustrated in Figure 10.
It is easy to check that is planar and has height , since has height and the Vposets in are pairwise incomparable. Moreover, every Vposet in contains a minimal element of and all Vposets in are pairwise incomparable. Finally, observe that, as long as , for every Vposet in there is a unique smallest element that is larger than all elements in , see the right of Figure 10.
In order to show that , we shall show by induction on that for every linear extension of there exists a rainbow in with respect to whose innermost cover is contained in a Vposet of , and, if , whose second innermost cover has the element as its upper end. This clearly holds for . For , consider any linear extension of . This induces a linear extension of as follows: The set of elements in not contained in any Vposet is also a subset of the elements in . The remaining elements of are the minimal elements of the Vposets in . For each minimal element of consider the two corresponding Vposets in with its two corresponding minimal elements . Let be the element that comes first in , i.e., if and only if . Then we define to be the ordering of induced by the ordering of in . Note that is a linear extension of , even though does not necessarily induce a copy of in .
By induction on there exists a rainbow with respect to whose innermost cover is contained in a Vposet and, provided that , its second innermost cover has as its upper end. Consider the elements of with being the minimal element, and the two corresponding Vposets with minimal elements of , where and are covers; see Figure 10. By definition of and , all elements of lie between (included) and (excluded, if ) with respect to .
Assume without loss of generality that . If (), then the Vposet with is nested completely under the cover and replacing in the innermost cover by the cover and any cover with gives a rainbow with the desired properties. If (), then the Vposet with is nested completely under the cover and replacing in the innermost cover by the cover and any cover with gives a rainbow with the desired properties, which concludes the proof.
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