The Queue-Number of Posets of Bounded Width or Height

Abstract

Heath and Pemmaraju [9] conjectured that the queue-number of a poset is bounded by its width and if the poset is planar then also by its height. We show that there are planar posets whose queue-number is larger than their height, refuting the second conjecture. On the other hand, we show that any poset of width 2 has queue-number at most 2, thus confirming the first conjecture in the first non-trivial case. Moreover, we improve the previously best known bounds and show that planar posets of width w have queue-number at most $3w-2$ while any planar poset with 0 and 1 has queue-number at most its width.

K. Knauer was supported by ANR projects GATO: ANR-16-CE40-0009-01 and DISTANCIA: ANR-17-CE40-0015.

P. Micek was partially supported by the National Science Center of Poland under grant no. 2015/18/E/ST6/00299.

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 G is called its queue-number and denoted by ${\text {qn}}(G)$.

To be more precise, let G be a graph and let L be a linear order on the vertices of G. We say that the edges $uv,u'v'\in E(G)$ are nested with respect to L if $u< u'< v'<v$ or $u'<u<v<v'$ in L. Given a linear order L of the vertices of G, the edges $u_1v_1,\ldots , u_kv_k$ of G form a rainbow of size k if $u_1<\cdots<u_k<v_k<\cdots <v_1$ in L. Given G and L, the edges of G can be partitioned into k queues if and only if there is no rainbow of size $k+1$ in L, see [10].

The queue-number 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 fault-tolerant processing, sorting with parallel queues, matrix computations, scheduling parallel processes, and communication management in distributed algorithm (see [8, 10, 13]).

Perhaps the most intriguing question concerning queue-numbers is whether planar graphs have bounded queue-number.

Conjecture 1

(Heath and Rosenberg [10]).

The queue-number of planar graphs is bounded by a constant.

In this paper we study queue-numbers 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 a, b of a poset are called comparable if $a < b$ or $b < a$, and incomparable, denoted by $a \parallel b$, otherwise. Posets are visualized by their diagrams: Elements are placed as points in the plane and whenever $a<b$ in the poset, and there is no element c with $a<c<b$, there is a curve from a to b going upwards (that is y-monotone). We denote this case as $a\prec b$. 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 P, a linear extension L of P is a linear order on the elements of P such that $x <_L y$, whenever $x <_P y$. (Throughout the paper we use a subscript on the symbol <, if we want to emphasize which order it represents.) Finally, the queue-number of a poset P, denoted by ${\text {qn}}(P)$, is the smallest k such that there is a linear extension L of P for which the resulting linear layout of $G_P$ contains no $(k+1)$-rainbow. Clearly we have ${\text {qn}}(G_P)\le {\text {qn}}(P)$, i.e., the queue-number of a poset is at least the queue-number of its cover graph. It is shown in [9] that even for planar posets, that is posets admitting crossing-free diagrams, there is no function f such that ${\text {qn}}(P)\le f({\text {qn}}(G_P))$ (Fig. 1).

Fig. 1.

A poset and a layout with two queues (gray and black). Note that the order of the elements on the spine is a linear extension of the poset.

Heath and Pemmaraju [9] 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). 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 ${{\mathrm{width}}}(P)\le w$, then ${\text {qn}}(P)\le w^2$. 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 w has queue-number at most w.

Furthermore, they made a step towards this conjecture for planar posets: if a planar poset P has ${{\mathrm{width}}}(P)\le w$, then ${\text {qn}}(P)\le 4w-1$. For the lower bound side they provided planar posets of width w and queue-number $\lceil \sqrt{w}\rceil$.

We improve the bounds for planar posets and get the following:

Theorem 1

Every planar poset of width w has queue-number at most $3w-2$. Moreover, there are planar posets of width w and queue-number w.

As an ingredient of the proof we show that posets without certain subdivided crowns satisfy Conjecture 2 (c.f. Theorem 5). 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 non-trivial case $w=2$:

Theorem 2

Every poset of width 2 has queue-number at most 2.

An easy corollary of this is that all posets of width w have queue-number at most $w^2-w+1$ (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 h has queue-number at most h.

We show that Conjecture 3 is false for the first non-trivial case $h=2$:

Theorem 3

There is a planar poset of height 2 with queue-number at least 4.

Furthermore, we establish a link between a relaxed version of Conjectures 3 and 1, namely we show that the latter is equivalent to planar posets of height 2 having bounded queue-number (c.f. Theorem 6). On the other hand, we show that Conjecture 3 holds for planar posets with 0 and 1:

Theorem 4

Every planar poset of height h with 0 and 1 has queue-number at most $h-1$.

Organization of the paper. In Sect. 2 we consider general (not necessarily planar) posets and give upper bounds on their queue-number in terms of their width, such as Theorem 2. In Sect. 3 we consider planar posets and bound the queue-number in terms of the width, both from above and below, i.e., we prove Theorem 1. In Sect. 4 we give a counterexample to Conjecture 3 by constructing a planar poset with height 2 and queue-number at least 4. Here we also argue that proving any upper bound on the queue-number 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 h there is a planar poset of height h and queue-number $h-1$ (c.f. Proposition 3).

2 General Posets of Bounded Width

By Dilworth’s Theorem [3], the width of a poset P coincides with the smallest integer w such that P can be decomposed into w chains of P. Let us derive Proposition 1 of Heath and Pemmaraju [9] from such a chain partition.

Proposition 1

For every poset P, if ${{\mathrm{width}}}(P)\le w$ then ${\text {qn}}(P)\le w^2$.

Proof

Let P be a poset of width w and $C_1,\ldots ,C_w$ be a chain partition of P. Let L be any linear extension of P and $a<_L b<_L c <_L d$ with $a\prec d$ and $b\prec c$. Note that we must have either $a \parallel b$ or $c \parallel d$. If follows that if $a \in C_i$, $b \in C_j$, $c \in C_k$, and $d \in C_\ell$, then $(i,\ell ) \ne (j,k)$. As there are only $w^2$ ordered pairs (x, y) with $x,y \in [w]$, we can conclude that every nesting set of covers has cardinality at most $w^2$.    $\square$

Note that in the above proof L is any linear extension and that without choosing the linear extension L carefully, upper bound $w^2$ is best-possible. Namely, if $P=\{a_1, \ldots , a_k,b_1, \ldots , b_k\}$ with comparabilities $a_i<b_j$ for all $1\le i,j\le k$, then P has width k and the linear extension $a_1< \ldots<a_k<b_k< \ldots < b_1$ creates a rainbow of size $k^2$.

We continue by showing that every poset of width 2 has queue-number at most 2, that is, we prove Theorem 2.

Proof

(Theorem 2). Let P be a poset of width 2 and minimum element 0 and $C_1,C_2$ be a chain partition of P. Note that the assumption of the minimum causes no loss of generality, since a 0 can be added without increasing the width nor decreasing the queue-number. Any linear extension L of P partitions the ground set X naturally into inclusion-maximal sets of elements, called blocks, from the same chain in $\{C_1,C_2\}$ that appear consecutively along L, see Fig. 2. We denote the blocks by $B_1,\ldots ,B_k$ according to their appearance along L. We say that L is lazy if for each $i = 2,\ldots ,k$, each element $x \in B_i$ has a relation to some element $y \in B_{i-1}$. A linear extension L can be obtained by picking any minimal element $m\in P$, put it into L, and recurse on $P\setminus \{m\}$. Lazy linear extensions (with respect to $C_1,C_2$) 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 0 is needed in order to ensure the property of laziness with respect to $B_2$.

Fig. 2.

A poset of width 2 with a 0 and a chain partition $C_1,C_2$ and the blocks $B_1,\ldots ,B_5$ induced by a lazy linear extension with respect to $C_1,C_2$.

Now we shall prove that in a lazy linear extension no three covers are pairwise nesting. So assume that $a\prec b$ is any cover and that $a \in B_i$ and $b \in B_j$. As L is lazy, b is comparable to some element in $B_{j-1}$ (if $j \ge 2$) and all elements in $B_1,\ldots ,B_{j-2}$ (if $j \ge 3$). With $a\prec b$ being a cover, it follows from L being lazy that $i \in \{j-2,j-1,j\}$. If $i=j$, then no cover is nested under $a\prec b$. If $i=j-1$, then no cover $c\prec d$ is nested above $a\prec b$: either $c\in B_i$ and $d\in B_j$ and hence $c\prec d$ is not a cover, or both endpoints would be inside the same chain, i.e., c, d are the last and first element of $B_{j-2}$ and $B_{j}$ or $B_{i}$ and $B_{i+2}$, respectively. This implies $c<_L a<_L d<_L b$ or $a<_L c<_L b<_L r$, respectively, and $c\prec d$ cannot nest above $a\prec b$. If $i=j-2$, then no cover is nested above $a\prec b$. Thus, either no cover is nested below $a\prec b$, or no cover is nested above $a\prec b$, or both. In particular, there is no three nesting covers and ${\text {qn}}(P) \le 2$.    $\square$

Corollary 1

Every poset of width w has queue-number at most $w^2-2\lfloor w/2 \rfloor$.

Proof

We take any chain partition of size w and pair up chains to obtain a set S of $\lfloor w/2 \rfloor$ disjoint pairs. Each pair from S induces a poset of width at most 2, which by Theorem 2 admits a linear order with at most two nesting covers. Let L be a linear extension of P respecting all these partial linear extensions.

Now, following the proof of Proposition 1 any cover can be labeled by a pair (i, j) corresponding to the chains containing its endpoint. Thus, in a set of nesting covers any pair appears at most once, but for each i, j such that $(i,j)\in S$ only two of the four possible pairs can appear simultaneously in a nesting. This yields the upper bound.    $\square$

For an integer $k \ge 2$ we define a subdivided k-crown as the poset $P_k$ as follows. The elements of $P_k$ are $\{a_1,\ldots ,a_k,b_1,\ldots ,b_k,c_1,\ldots ,c_k\}$ and the cover relations are given by $a_i \prec b_i$ and $b_i \prec c_i$ for $i = 2,\ldots ,k$, $a_i \prec c_{i-1}$ for $i = 1,\ldots ,k-1$, and $a_1 \prec c_k$; see the left of Fig. 3. We refer to the covers of the form $a_i \prec c_j$ as the diagonal covers and we say that a poset P has an embedded $P_k$ if P contains 3k elements that induce a copy of $P_k$ in P with all diagonal covers of that copy being covers of P.

Fig. 3.

Left: The posets $P_2$, $P_3$, and $P_4$. Right: The existence of an element z with cover relation $z \prec x$ and non-cover relation $z < y$ gives rise to a gray edge from x to y.

Theorem 5

If P is a poset that for no $k \ge 2$ has an embedded $P_k$, then the queue-number of P is at most the width of P.

Proof

Let P be any poset. For this proof we consider the cover graph $G_P$ of P as a directed graph with each edge xy directed from x to y if $x \prec y$ in P. We call these edges the cover edges. Now we augment $G_P$ to a directed graph G by introducing for some incomparable pairs $x \parallel y$ a directed edge. Specifically, we add a directed edge from x to y if there exists a z with $z < x,y$ in P where $z \prec x$ is a cover relation and $z < y$ is not a cover relation; see the right of Fig. 3. We call these edges the gray edges of G.

Now we claim that if G has a directed cycle, then P has an embedded subdivided crown. Clearly, every directed cycle in G has at least one gray edge. We consider the directed cycles with the fewest gray edges and among those let $C = [c_1,\ldots ,c_{\ell }]$ be one with the fewest cover edges. First assume that C has a cover edge (hence ${\ell } \ge 3$), say $c_1c_2$ is a gray edge followed by a cover edge $c_2c_3$. Consider the element z with cover relation $z \prec c_1$ and non-cover relation $z < c_2$ in P. By $z < c_2 \prec c_3$ we have a non-cover relation $z < c_3$ in P. Now if $c_1 \parallel c_3$ in P, then G contains the gray edge $c_1c_3$ (see Fig. 4(a)) and $[c_1,c_3,\ldots ,c_{\ell }]$ is a directed cycle with the same number of gray edges as C but fewer cover edges, a contradiction. On the other hand, if $c_1 < c_3$ in P (note that $c_3 < c_1$ is impossible as $z \prec c_1$ is a cover), then there is a directed path Q of cover edges from $c_1$ to $c_3$ (see Fig. 4(b)) and $C + Q - \{c_1c_2, c_2c_3\}$ contains a directed cycle with fewer gray edges than C, again a contradiction.

Fig. 4.

Illustrations for the proof of Theorem 5.

Hence $C = [c_1,\ldots ,c_{\ell }]$ is a directed cycle consisting solely of gray edges. Note that by the first paragraph $\{c_1,\ldots ,c_{\ell }\}$ is an antichain in P. For $i=2,\ldots ,{\ell }$ let $a_i$ be the element of P with cover relation $a_i \prec c_{i-1}$ and non-cover relation $a_i < c_i$, as well as $a_1$ with cover relation $a_1 \prec c_{\ell }$ and non-cover relation $a_1 < c_1$. As $\{c_1,\ldots ,c_{\ell }\}$ is an antichain and $a_i < c_i$ holds for $i=1, \ldots , {\ell }$, we have $\{c_1,\ldots ,c_{\ell }\} \cap \{a_1,\ldots ,a_{\ell }\} = \emptyset$. Let us assume that $a_1 < c_j$ in P for some $j \ne 1,{\ell }$. If $a_1 \prec c_j$ is a cover relation, then there is a gray edge $c_jc_1$ in G (see Fig. 4(c)) and the cycle $[c_1,\ldots ,c_j]$ is shorter than C, a contradiction. If $a_1 < c_j$ is a non-cover relation, then there is a gray edge $c_{\ell }c_j$ in G (see Fig. 4(d)) and the cycle $[c_j,\ldots ,c_{\ell }]$ is shorter than C, again a contradiction.

Hence, the only relations between $a_1,\ldots ,a_{\ell }$ and $c_1,\ldots ,c_{\ell }$ are cover relations $a_1 \prec c_{\ell }$ and $a_i \prec c_{i-1}$ for $i=2,\ldots ,{\ell }$ and the non-cover relations $a_i < c_i$ for $i = 1,\ldots ,\ell$. Hence $a_1,\ldots ,a_{\ell }$ are pairwise distinct. Moreover, $\{a_1,\ldots ,a_{\ell }\}$ is an antichain in P since the only possible relations among these elements are of the form $a_1 < a_{\ell }$ or $a_i < a_{i-1}$, which would contradict that $a_1 \prec c_{\ell }$ and $a_i \prec c_{i-1}$ are cover relations. Finally, we pick for every $i=1,\ldots ,{\ell }$ an element $b_i$ with $a_i< b_i < c_i$, which exists as $a_i < c_i$ is a non-cover relation. Together with the above relations between $a_1,\ldots ,a_{\ell }$ and $c_1,\ldots ,c_{\ell }$ we conclude that $b_1,\ldots ,b_{\ell }$ are pairwise distinct and these $3{\ell }$ elements induce a copy of $P_{\ell }$ in P with all diagonal covers in that copy being covers of P.

Thus, if P has no embedded $P_k$, then the graph G we constructed has no directed cycles, and we can pick L to be any topological ordering of G. As $G_P \subseteq G$, L is a linear extension of P. For any two nesting covers $x_2<_L x_1<_L y_1 <_L y_2$ we have $x_1 \parallel x_2$ or $y_1 \parallel y_2$ or both, since $x_2 \prec y_2$ is a cover. However, if $x_2 < x_1$ in P, then there would be a gray edge from $y_2$ to $y_1$ in G, contradicting $y_1 <_L y_2$ and L being a topological ordering of G. We conclude that $x_1 \parallel x_2$ and the left endpoints of any rainbow form an antichain, proving ${\text {qn}}(P)\le {{\mathrm{width}}}(P)$.    $\square$

Let us remark that several classes of posets have no embedded subdivided crowns, e.g., graded posets, interval orders (since these are 2 + 2-free, see [6]), or (quasi-)series-parallel orders (since these are N-free, see [7]). Here, 2 + 2 and N are the four-element posets defined by $a<b, c<d$ and $a<b, c<d, c<b$, respectively. Also note that while subdivided crowns are planar posets, no planar poset with 0 and 1 has an embedded k-crown. Indeed, already looking at the subposet induced by the k-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, series-parallel order, and planar poset with 0 and 1, P we have ${\text {qn}}(P) \le {{\mathrm{width}}}(P)$.

3 Planar Posets of Bounded Width

Heath and Pemmaraju [9] show that the largest queue-number among planar posets of width w lies between $\lceil \sqrt{w} \rceil$ and $4w-1$. Here we improve the lower bound to w and the upper bound to $3w-2$.

Proposition 2

For each w there exists a planar poset $Q_w$ with 0 and 1 of width w and queue-number w.

Proof

We shall define $Q_w$ recursively, starting with $Q_1$ being any chain. For $w \ge 2$, $Q_w$ consists of a lower copy P and a disjoint upper copy $P'$ of $Q_{w-1}$, three additional elements a, b, c, and the following cover relations in between:

• $a \prec x$, where x is the 0 of P

• $y \prec x'$, where y is the 1 of P and $x'$ is the 0 of $P'$

• $y' \prec c$, where $y'$ is the 1 of $P'$

• $a \prec b \prec c$

It is easily seen that all cover relations of P and $P'$ remain cover relations in $Q_w$, and that $Q_w$ is planar, has width w, a is the 0 of $Q_w$, and c is the 1 of $Q_w$. See Fig. 5 for an illustration.

Fig. 5.

Recursively constructing planar posets $Q_w$ of width w and queue-number w. Left: $Q_1$ is a two-element chain. Middle: $Q_w$ is defined from two copies $P,P'$ of $Q_{w-1}$. Right: The general situation for a linear extension of $Q_w$.

To prove that ${\text {qn}}(Q_w) = w$ we argue by induction on w, with the case $w=1$ being immediate. Let L be any linear extension of $Q_w$. Then a is the first element in L and c is the last. Since $y \prec x'$, all elements in P come before all elements of $P'$. Now if in L the element b comes after all elements of P, then P is nested under cover $a\prec b$, and if b comes before all elements of $P'$, then $P'$ is nested under cover $b\prec c$. We obtain w nesting covers by induction on P in the former case, and by induction on $P'$ in the latter case. This concludes the proof.    $\square$

Next we prove Theorem 1, namely that the maximum queue-number of planar posets of width w lies between w and $3w-2$.

Proof

(Theorem 1). By Proposition 2 some planar posets of width w have queue-number w. So it remains to consider an arbitrary planar poset P of width w and show that P has queue-number at most $3w-2$. To this end, we shall add some relations to P, obtaining another planar poset Q of width w that has a 0 and 1, with the property that ${\text {qn}}(P) \le {\text {qn}}(Q) + 2w-2$. Note that this will conclude the proof, as by Corollary 2 we have ${\text {qn}}(Q) \le w$.

Given a planar poset P of width w, there are at most w minima and at most w maxima. Hence there are at most $2w-2$ extrema that are not on the outer face. For each such extremum x – say x is a minimum – consider the unique face f with an obtuse angle at x. We introduce a new relation $y < x$, where y is a smallest element at face f, see Fig. 6. Note that this way we introduce at most $2w-2$ new relations, and that these can be drawn y-monotone and crossing-free by carefully choosing the other element in each new relation. Furthermore, every inner face has a unique source and unique sink.

Fig. 6.

Inserting new relations (dashed) into a face of a plane diagram. Note that relation $a < b$ is a cover relation in P but not in Q.

Now consider a cover relation $a \prec _P b$ that is not a cover relation in the new poset Q. For the corresponding edge e from a to b in Q there is one face f with unique source a and unique sink b. Now either way the other edge in f incident to a or to b must be one of the $2w-2$ newly inserted edges, see again Fig. 6. This way we assign $a \prec b$ to one of $2w-2$ 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 $2w-2$ queues to deal with the cover relations of P that are not cover relations of Q and spend another w queues for Q dealing with the remaining cover relations of P. Thus, ${\text {qn}}(P) \le {\text {qn}}(Q) + 2w - 2 \le 3w-2$.    $\square$

4 Planar Posets of Bounded Height

Recall Conjecture 3, which states that every planar poset of height h has queue-number at most h. In the following, we give a counterexample to this conjecture:

Proof

(Theorem 3). Consider the graph G that is constructed as follows: Start with $K_{2,10}$ with bipartition classes $\{a_1,a_2\}$ and $\{b_1,\ldots ,b_{10}\}$. For every $i=1,\ldots ,9$ add four new vertices $c_{i,1},\ldots ,c_{i,4}$, each connected to $b_i$ and $b_{i+1}$. The resulting graph G has 46 vertices, is planar and bipartite with bipartition classes $X = \{b_1,\ldots ,b_{10}\}$ and $Y = \{a_1,a_2\} \cup \{c_{i,j} \mid 1\le i \le 9, 1\le j \le 4\}$. See Fig. 7.

Fig. 7.

A planar poset P of height 2 and queue-number at least 4. Left: The cover graph $G_P$ of P. Right: A part of a planar diagram of P.

Let P be the poset arising from G by introducing the relation $x < y$ for every edge xy in G with $x \in X$ and $y \in Y$. Clearly, P has height 2 and hence the cover relations of P are exactly the edges of G. Moreover, by a result of Moore [12] (see also [2]) P is planar because G is planar, also see the right of Fig. 7.

We shall argue that ${\text {qn}}(P) \ge 4$. To this end, let L be any linear extension of P. Without loss of generality we have $a_1 <_L a_2$. Note that since in P one bipartition class of G is entirely below the other, any 4-cycle in G gives a 2-rainbow. Let $b_{i_1},b_{i_2}$ be the first two elements of X in L, $b_{j_1},b_{j_2}$ be the last two such elements. As $|X| = 10$ there exists $1\le i \le 9$ such that $\{i,i+1\} \cap \{i_1,i_2,j_1,j_2\} = \emptyset$, i.e., we have $b_{i_1},b_{i_2}<_L b_i,b_{i+1}<_L b_{j_1},b_{j_2}<_L a_1 <_L a_2$, where we use that $a_1$ and $a_2$ are above all elements of X in P.

Now consider the elements $C = \{c_{i,1},\ldots ,c_{i,4}\}$ that are above $b_i$ and $b_{i+1}$ in P. As $|C| \ge 4$, there are two elements $c_1,c_2$ of C that are both below $a_1,a_2$ in L, or both between $a_1$ and $a_2$ in L, or both above $a_1,a_2$ in L. Consider the 2-rainbow R in the 4-cycle $[c_1,b_i,c_2,b_{i+1}]$. In the first case R is nested below the 4-cycle $[a_1,b_{i_1},a_2,b_{i_2}]$, in the second case the cover $b_{j_1}\prec a_1$ is nested below R and R is nested below the cover $b_{i_1}\prec a_2$, and in the third case 4-cycle $[a_1,b_{j_1},a_2,b_{j_2}]$ is nested below R. As each case results in a 4-rainbow, we have ${\text {qn}}(P) \ge 4$.    $\square$

Even though Conjecture 3 has to be refuted in its strongest meaning, it might hold that planar posets of height h have queue-number O(h), or at least bounded by some function f(h) in terms of h, or at least that planar posets of height 2 have bounded queue-number. As it turns out, all these statements are equivalent, and in turn equivalent to Conjecture 1.

Theorem 6

The following statements are equivalent:

• (i) Planar graphs have queue-number O(1) (Conjecture 1).

• (ii) Planar posets of height h have queue-number O(h).

• (iii) Planar posets of height h have queue-number at most f(h) for a function f.

• (iv) Planar posets of height 2 have queue-number O(1).

• (v) Planar bipartite graphs have queue-number O(1).

Proof

• (i)$\Rightarrow$(ii) Pemmaraju proves in his thesis [14] (see also [4]) that if G is a graph, $\pi$ is a vertex ordering of G with no $(k+1)$-rainbow, $V_1,\ldots ,V_m$ are color classes of any proper m-coloring of G, and $\pi '$ is the vertex ordering with $V_1<_{\pi '} \cdots <_{\pi '} V_m$, where within each $V_i$ the ordering of $\pi$ is inherited, then $\pi '$ has no $(2(m-1)k+1)$-rainbow. So if P is any poset of height h, its cover graph $G_P$ has ${\text {qn}}(G_P) \le c$ by (i) for some global constant $c >0$. Splitting P into h antichains $A_1,\ldots ,A_h$ by iteratively removing all minimal elements induces a proper h-coloring of $G_P$ with color classes $A_1,\ldots ,A_h$. As every vertex ordering $\pi '$ of G with $A_1<_{\pi '} \cdots <_{\pi '} A_h$ is a linear extension of P, it follows by Pemmaraju’s result that ${\text {qn}}(P) \le 2(h-1){\text {qn}}(G_P) \le 2ch$, i.e., ${\text {qn}}(P) \in O(h)$.

(ii) $\Rightarrow$(iii)$\Rightarrow$(iv) These implications are immediate.

(iv)$\Rightarrow$(v) Moore proves in his thesis [12] (see also [2]) that if G is a planar and bipartite graph with bipartition classes A and B, and $P_G$ is the poset on element set $A \cup B = V(G)$ where $x<y$ if and only if $x \in A, y \in B, xy \in E(G)$, then $P_G$ is a planar poset of height 2. As G is the cover graph of $P_G$, we have ${\text {qn}}(G) \le {\text {qn}}(P_G) \le c$ for some constant $c >0$ by (iv), i.e., ${\text {qn}}(G) \in O(1)$.

(v)$\Rightarrow$(i) This is a result of Dujmović and Wood [5].    $\square$

Finally, we show that Conjecture 3 holds for planar posets with 0 and 1.

Proof

(Theorem 4). Let P be a planar poset with 0 and 1. Then P has dimension at most two [1], i.e., it can be written as the intersection of two linear extensions of P. A particular consequence of this is, that there is a well-defined dual poset $P^\star$ in which two distinct elements x, y are comparable in P if and only if they are incomparable in $P^\star$. Poset $P^\star$ reflects a “left of”-relation for each incomparable pair $x \parallel y$ in P in the following sense: Any maximal chain C in P corresponds to a 0-1-path Q in $G_P$, which splits the elements of $P\setminus C$ into those left of Q and those right of Q. Now $x <_{P^\star } y$ if and only if x is left of the path for every maximal chain containing y (equivalently y is right of the path for every maximal chain containing x). Due to planarity, if $a\prec b$ is a cover in P and C is a maximal chain containing neither a nor b, then a and b are on the same side of the path Q corresponding to C. In particular, if for $x,y\in C$ we have $a <_{P^\star } x$ and $b \parallel y$, then b and y are comparable in $P^\star$, but if $y <_{P^\star } b$ we would get a crossing of C and $a\prec b$. Also see the left of Fig. 8. We summarize:

($\star$):

If $a\prec b$, $a <_{P^\star } x$ for some $x \in C$ and $b \parallel y$ for some $y \in C$, then $b <_{P^\star } y$.

Fig. 8.

Left: Illustration of ($\star$): If $a <_{P^\star } x$, $b \parallel y$, $x < y$, and $a\prec b$ is a cover, then $b <_{P^\star } y$ due to planarity. Right: If $a_3<_L a_2<_L a_1<_L b_1<_L b_2 <_L b_3$ is a 3-rainbow with $a_2,a_3 < a_1$, then $a_3 < a_2$.

Now let L be the leftmost linear extension of P, i.e., the unique linear extension L with the property that for any $x \parallel y$ in P we have $x <_L y$ if and only if $x < y$ in $P^\star$. Assume that $a_2<_L a_1<_L b_1 <_L b_2$ is a pair of nesting covers $a_1\prec b_1$ below $a_2\prec b_2$. Then $a_1 \parallel a_2$ (hence $a_2 <_{P^\star } a_1$) or $b_1 \parallel b_2$ (hence $b_1 <_{P^\star } b_2$) or both. Observe that the latter case is impossible, as for any maximal chain C containing $a_1\prec b_1$ we would have $a_2 <_{P^\star } a_1$ with $a_1 \in C$ and $b_1 <_{P^\star } b_2$ with $b_1 \in C$, contradicting ($\star$). So the nesting of $a_1\prec b_1$ below $a_2\prec b_2$ is either of type A with $a_2 < a_1$, or of type B with $b_1 < b_2$. See Fig. 9.

Fig. 9.

A nesting of $a_1\prec b_1$ below $a_2\prec b_2$ of type A (left) and type B (right).

Now consider the case that cover $a_2\prec b_2$ is nested below another cover $a_3\prec b_3$, see the right of Fig. 8. Then also $a_1\prec b_1$ is nested below $a_3\prec b_3$ and we claim that if both, the nesting of $a_1\prec b_1$ below $a_2\prec b_2$ as well as the nesting of $a_1\prec b_1$ below $a_3\prec b_3$, are of type A (respectively type B), then also the nesting of $a_2\prec b_2$ below $a_3\prec b_3$ is of type A (respectively type B). Indeed, assuming type B, we would get $a_3 <_{P^\star } a_2$ and $b_1 <_{P^\star } b_3$, which together with any maximal chain C containing $a_2< a_1 <b_1$ contradicts ($\star$).

Finally, let $a_k<_L \cdots<_L a_1<_L b_1<_L \cdots <_L b_k$ be any k-rainbow and let $I = \{ i \in [k] \mid a_i < a_1 \}$, i.e., for each $i \in I$ the nesting of $a_1\prec b_1$ below $a_i\prec b_i$ is of type A. Then we have just shown that the nesting of $a_j\prec b_j$ below $a_i\prec b_i$ is of type A whenever $i,j \in I$ and of type B whenever $i,j \notin I$. Hence, the set $\{a_i \mid i \in I\} \cup \{a_1,b_1\} \cup \{b_i \mid i \notin I\}$ is a chain in P of size $k+1$, and thus $k \le h-1$. It follows that P has queue-number at most $h-1$, as desired.    $\square$

The proof of the following can be found in the arXiv version of the present paper, [11].

Proposition 3

For each h there exists a planar poset $Q_h$ of height h and queue-number $h-1$.

5 Conclusions

We studied the queue-number of (planar) posets of bounded height and width. Two main problems remain open: bounding the queue-number 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 queue-numbers of graphs. For the first problem the biggest class known to satisfy it are posets without the embedded the subdivided k-crowns for $k\ge 2$ as defined in Sect. 2. Note, that proving it for $k\ge 3$ would imply that Conjecture 2 holds for all 2-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 n elements has queue-number at most $\lceil \sqrt{n} \rceil$.