1 Introduction
Queue layouts of graphs form a wellknown type of linear layouts and play an important role in various fields, e.g., in sorting [2], scheduling [20], VLSI circuit design [15], and graph drawing [13, 17]. A queue layout of a graph consists of a vertex ordering and a partition of its edges into queues, so that no two independent edges of the same queue are nested [14]; see Fig. 0(a). The queue number of a graph is the smallest number of queues required by any of its queue layouts. Note that queue layouts form the “dual” concept of stack layouts [16, 24] (widely known as book embeddings), in which two edges of the same stack are allowed to nest but not to cross.
It is known that there exist nonplanar graphs on vertices with queue number, for example, the queue number of the complete graph is [14]. Moreover, there exist graphs of bounded degree that may require arbitrarily many queues [23]. Among the graphs having sublinear queue number are those with a subquadratic number of edges [13], and those that belong to any minorclosed graph family [7]. Constant queue number is achieved by all graphs of bounded treewidth. In particular, a graph with treewidth has queue number [9, 21]. Improved bounds (linear in the parameter) are known for graphs of bounded pathwidth [7], bounded track number [9], or bounded bandwidth [13]; for a survey we refer the reader to [9].
A rich body of literature focuses on planar graphs. In fact, it is known that the graphs that admit queue layouts with only one queue are the archedlevel planar graphs [14], which are planar graphs with at most edges over vertices (note that testing whether a graph is archedlevel planar is NPcomplete [13]). Trees are archedlevel planar and therefore have queue number one [14]. Outerplanar graphs have queue number at most two [13], Halin graphs and seriesparallel graphs have queue number at most three [12, 19], and planar trees have queue number at most five [1]. However, it is still unknown whether the queue number of planar graphs is constant. In particular, back in 1992, Heath, Leighton and Rosenberg [13] conjectured that every planar graph has constant queue number. Notably, this conjecture has not been settled after almost three decades. The bestknown upper bound is due to Dujmović [5], who showed that the queue number of planar graphs on vertices is improving upon a previous polylogarithmic bound by Di Battista et al. [4]. On the other hand, the bestknown lower bound is due to a family of planar trees that require four queues [1].
It is worth noting that a positive answer to the conjecture by Heath, Leighton and Rosenberg [13] would have several remarkable implications. Here we name few of them. A first implication is that every planar graph would admit a threedimensional grid drawing in linear volume [22], which is a major open problem in graph drawing first posed by Felsner et al. [10] back in 2003. A second one is that every planar graph would have constant track number [8]. Moreover, every bipartite planar graph would have a layer drawing and a corresponding edgecoloring of constant size, in which no two edges of the same color cross [9]. Finally, it is known that if the queue number of planar graphs is constant, then the same holds for the broader family of planar graphs, i.e., of those graphs that can be drawn in the plane such that each edge is crossed at most times (for fixed values of ) [9].
Our contribution.
In this paper, we make an important step towards settling the conjecture by Heath, Leighton and Rosenberg [13] that planar graphs have constant queue number. Namely, we prove that every planar graph of maximum degree has queue number , where is a small constant. This implies that the conjecture holds for planar graphs of bounded degree. More precisely, the main contribution of this paper is the following theorem.
Theorem 1.
Every planar graph of maximum degree has queue number at most .
The proof of Theorem 1 is constructive and yields an algorithm that computes a queue layout of the input graph with at most queues in polynomial time. In particular, the algorithm works in time, where is the number of vertices of , under the unitcost RAM model of computation with word size linear in . We remark that this model of computation makes use of reasonable assumptions, which fit the reality of common programming languages [11]; for example, Chan [3] used it to efficiently solve fundamental problems in computational geometry.
Wood [22] proves that a graph on vertices that belongs to a proper minorclosed family of graphs, such as planar graphs, has a straightline drawing on a threedimensional grid if and only if its queue number is constant. This result combined with Theorem 1 yields the following corollary. We remark that the best upper bound currently known for the volume of threedimensional grid drawings of planar graphs is [5].
Corollary 1.
Every planar graph of bounded degree admits a straightline drawing on a threedimensional grid.
Closely related to the queue layouts are the socalled track layouts. In a track layout, the vertices of a graph are partitioned into sequences, called tracks, such that the vertices in each track form an independent set and the edges between each pair of tracks do not cross. The track number of a graph is the minimum number of tracks required by any of its track layouts. Whether planar graphs have constant track number is still an open question, with the current best upper bound being [5]. Dujmović, Pór and Wood [8] prove that a graph with queue number at most has track number , which together with Theorem 1 implies that planar graphs of bounded degree have constant track number. More precisely, we have the following.
Corollary 2.
Every planar graph of maximum degree has track number .
Another related problem is the track thickness problem, in which one seeks for a layer drawing of a bipartite graph and a corresponding coloring of its edges with as few colors as possible, such that no two edges of the same color cross; the minimum number of colors required in any such drawing is referred to as track thickness. Again, it is not known whether the track thickness of bipartite planar graphs is constant, with the best upper bound being [9], which is due to a result by Dujmović and Wood [9] that relates the queue number of a bipartite planar graph and its track thickness. Using the same result, together with Theorem 1, we obtain that every bipartite planar graph of bounded degree has constant track thickness.
Corollary 3.
Every bipartite planar graph of maximum degree has track thickness .
A last corollary of Theorem 1 stems from another result by Dujmović and Wood [9], who prove that if the queue number of planar graphs is constant, then the same holds for planar graphs when is a fixed constant. Their proof relies on a planarization technique, in which each crossing is replaced by a degree vertex. If the input is a planar graph of bounded degree, the obtained planarization is also of bounded degree, and we obtain the following corollary. Note that, Dujmović and Frati [6] recently proved that planar graphs have queue number .
Corollary 4.
For fixed , every planar graph of bounded degree has constant queue number.
Proof strategy.
Our approach starts with a layering of the vertices of the input graph obtained from a breadthfirst search (BFS) traversal of it. This layering only serves as a basis for computing the linear order of the vertices in the queue layout. In particular, the vertices that belong to earlier layers in the BFStree will precede those belonging to subsequent layers of . As a consequence, all edges that belong to do not nest in such a linear order. The main challenge with this approach is to deal with edges that are not part of . Such edges either connect vertices on the same layer or vertices lying on consecutive layers. While the second type of edges can be eliminated by subdividing them a constant number of times, the first type of edges may still result in arbitrarily large groups of pairwise nesting edges, commonly called rainbows. To cope with this issue, we change the order of the vertices on each layer so as to eliminate all nestings between edges connecting vertices of the same layer. On the other hand, this will unavoidably introduce rainbows formed by edges of . However, we reorder the vertices such that the maximum number of edges of in the same rainbow is bounded by a polynomial in . For ease of description, we first discuss our approach in a special case, namely when the edges that do not belong to form a perfect matching on its leaves. Then we show how to use the solution of the special case for the general case.
Paper organization.
In Section 2, we introduce notation and definitions that are used throughout this paper. We also present results from the literature that we exploit in our algorithm. In Section 3, we start by giving an overview of the proof technique and then continue with its detailed description. We conclude in Section 4 by posing a question which might shed some further light towards solving the conjecture by Heath, Leighton and Rosenberg.
2 Preliminaries
Queue Layouts.
Let be an vertex simple connected undirected graph, that is, a graph with neither selfloops nor multiedges. We denote an edge between vertices and by . Let be a linear order of the vertices of . Consider two independent edges and , that is, edges and do not share a common endvertex. Up to a renaming of their endvertices, we may assume that and . We say that nests with respect to if and only if ; see Fig. 0(b). Consider now edges that are pairwise independent, such that for each . If , then we say that edges form a necklace; see Fig. 0(c). On the other hand, if , then we say that edges form a rainbow; see Fig. 0(d).
A preliminary result by Heath and Rosenberg [14] shows that a graph admits a queue layout with queues if and only if there exist a linear order of its vertices in which no rainbow is formed. Another useful tool for analyzing the queue number of graphs is the fact that the queue number of a graph is bounded by the queue number of any of its subdivisions. More precisely:
Lemma 1 (Dujmović and Wood [9]).
Let be a subdivision of a graph obtained by subdividing each of the edges of at most times. If , then .
Drawings and Representations.
A drawing of a graph is a mapping of the vertices of to distinct points of the plane, and of the edges of to Jordan arcs connecting their corresponding endvertices but not passing through any other vertex. A drawing is planar if no two edges intersect, except possibly at a common endvertex. A graph is planar if it admits a planar drawing. A planar drawing subdivides the plane into topologically connected regions, called faces. The infinite region is called the outer face.
Central in our approach are also the socalled ordered concentric representations, which were recently studied by Pupyrev [18]. An ordered concentric representation of a planar graph is a planar drawing of where the vertices are located at concentric circles with decreasing radii centered at a point of the plane, except for a single vertex , called the center of the representation, which is located at point . All vertices on circle have graphtheoretic distance from , for . It follows that each edge of either has both its endvertices at the same circle (level edge) or at two consecutive circles (binding edge). Moreover, in an ordered concentric representation, the following three properties hold: (R.1) each level edge is drawn outside the circle on which its endvertices are located, (R.2) each binding edge consists of at most two segments, one required segment which is drawn between the two circles on which its endvertices are located and one optional segment which is drawn outside both of these circles, and (R.3) vertex is incident to the outer face; for an illustration refer to Fig. 2. To compute an ordered concentric representation of a planar graph , Pupyrev [18] first computes a BFStree of . The root of this tree is the center of the representation, while its edges are binding edges drawn between the two corresponding circles (see R.2). His result is summarized as follows.
Lemma 2 (Pupyrev [18]).
Given an vertex planar graph , it is possible to compute in time an ordered concentric representation of with center , such that is the height of a breadthfirst search tree of rooted at .
Given a spanning tree of a graph rooted at a vertex , and a vertex of , we denote by the graphtheoretic distance of from in . Clearly, , while for each leaf of , it holds that , where is the height of . For a vertex of , we refer to the value as the layer of in , which we denote by . Note that given an ordered concentric representation of a graph computed using a BFStree of , for each vertex that belongs to circle in , its layer in is , for each .
3 Description of the Algorithm
In this section, we prove that every planar graph of bounded degree has constant queue number. To obtain this result, we employ a constructive approach. Namely, given a planar graph of maximum degree , we describe an efficient algorithm that computes a queue layout of this graph with at most queues. This section is organized as follows: First, we introduce in Section 3.1 a special subfamily of plane graphs with maximum degree , for which it is possible to compute a queue layout with at most queues. Then, in Section 3.2, we reduce the problem of finding a queue layout of a general planar graph of degree to the already discussed special case. This reduction increases the number of queues used to . We conclude in Section 3.3 by discussing the time complexity of our algorithm.
3.1 The Special Case: matched Graphs
In this section we consider a special subfamily of planar graphs of degree , which we call matched graphs. Formally, a matched graph consists of a ary tree with vertices, rooted at a vertex , and a perfect matching on the leaves of , such that there exists a planar drawing of with the following properties (refer to Fig. 3 for an illustration):

all leaves of lie on a horizontal line ,

all edges of (of ) are drawn above (below, respectively)

vertex lies on the outer face of , and

the vertices of layer in are drawn on a horizontal line .
Note that the root of has degree at most , while the leaves of have degree in . All other vertices of have degree at most . Hence, graph has maximum degree at most . Our goal is to compute a queue layout of with at most queues. Observe that drawing can be easily converted into an ordered concentric representation of with center in which all binding edges are part of .
First, we assign an integer value, called nestingvalue, to all edges of . Recall that each edge of connects two vertices of that are leaves in , and by Property P.1 are along the horizontal line in the drawing of . We denote the order in which the endvertices of the edges of appear along by . Consider now an edge of . Edge has nestingvalue zero if is not nested by any other edge in . Edge has nestingvalue if the maximum nestingvalue of all edges nesting is equal to ; refer to Fig. 3 for an illustration.
Based on the nestingvalues of the edges of , we compute in a bottomup traversal of an integer value for each vertex of , the socalled matchingvalue of , . For a vertex on layer of , the matchingvalue of is the nestingvalue of the unique edge of incident to it. For a vertex on layer of , the matchingvalue of equals the minimum matchingvalue of its neighbors in layer in ; refer to Fig. 3 for an illustration. In other words, the matchingvalue of vertex equals the minimum nestingvalue of the edges of incident to the leaves of the subtree of rooted at . In addition, the matchingvalues of any two consecutive leaves of along differ by at most one. Hence, if the leaves of a subtree rooted at a vertex have minimum matchingvalue and maximum matchingvalue , then for every value in there exists at least one leaf of this subtree with matchingvalue .
Finally, based on the matchingvalues, we partition the vertices of that belong to a certain layer of to layergroups. Formally, the layergroup of a vertex of is defined as follows:
(1) 
Observe that if belongs to layer . We further denote by the set of vertices of that belong to layer in , and that are contained in layergroup . Remark that is a partition of the vertices of . We are now ready to present the main result of this section.
Lemma 3.
Every matched graph has queue number at most .
Proof.
Let be a matched graph, and let be a partition of the vertexset of as described above. Recall that is a planar drawing of satisfying Properties P.1–P.4. We construct a linear order of the vertices of as follows. For every two distinct vertices and of , we have that if and only if one of the following conditions holds:

, or

and , or

, and is to the left of along in .
Observe that all edges of form a necklace. Indeed, if is an edge of , then the endvertices of belong to the same layergroup and they are consecutive in . Therefore, we can assign all edges of in one queue, say .
The remaining edges of belong to . As a result, their endvertices belong to consecutive layers of . Let be a vertex of that belongs to layer in , and assume that is contained in layergroup , i.e., . Let also be the neighbors of in layer (not necessarily in this lefttoright order), where . Without loss of generality, we assume that . This implies that . Since is contained in , it follows by Eq. (1) that
(2) 
Hence
(3) 
It follows that for vertex of layer we have
Alternatively,
Now consider vertex , such that . We claim that
(4) 
Recall that is the minimum nestingvalue of the edges of incident to the leaves of the subtree of rooted at . Since , for every integer value such that , there exists a leaf in the subtree of rooted at with matchingvalue . Since vertex belongs to layer in , the subtree of rooted at has at most leaves. Therefore
holds, which applied times, gives
(5) 
Since , we conclude that
holds, which implies our initial claim in Eq. (4), as desired. Using Eq. (4) and the fact that , we get for
(6) 
Recall that we have already assigned all edges of that belong to in a single queue denoted by . We are now ready to describe how to assign each edge of that is in to the remaining queues. For each layer and for each layergroup of layer in , we assign the edges between and to queue , for ; for an illustration refer to Fig. 4. By Eq. (6), all edges of have been assigned to one of the queues .
To complete the proof of the lemma, it remains to show that no two independent edges of the same queue are nested. As already mentioned, all edges of form a necklace and therefore do not nest. For some , consider two independent edges and assigned to the same queue , such that and . Without loss of generality, we may further assume that . We will prove that , which implies that edges and are not nested. Since , by Conditions C.1C.3, we have that . Since the endvertices of and belong to consecutive layers in , it follows that and . For the sake of contradiction, assume that , which implies that . By combining the above inequalities, we may conclude that and , for some .
Let and . Since both and are assigned to queue , it follows that and . Since , by Conditions C.2 and C.3, we conclude that , which implies that . On the other hand, since , by Conditions C.2 and C.3 we similarly get . Therefore, and consequently .
Now since , by Condition C.3 it follows that is to the left of along in the drawing of . Similarly, since and , it follows by Condition C.3 again that is to the left of along in the drawing of . However, this implies that edges and cross in , which is a contradiction to the fact that is a planar drawing of . Therefore, , and edges and are not nested, as desired, which concludes the proof. ∎
3.2 The Generalization: Planar Graphs of Maximum Degree
In this section, we use the approach presented in the previous section to general planar graphs of maximum degree . In a highlevel description, our approach consists of three main steps. Given a planar graph of maximum degree , we first compute an auxiliary planar graph of maximum degree by subdividing some of the edges of a constant number of times. Then, we exploit structural properties of graph to obtain a matched graph of maximum degree by replacing some of the vertices of with appropriatelydefined matched instances. It follows, by Lemma 3, that the queue number of is at most . In a third step, we show that a queue layout of can be obtained from a queue layout of by introducing a number of additional queues that is polynomial in , thus confirming Theorem 1.
First, let us argue that we may assume without loss of generality that has minimum degree at least two. Note that Theorem 1 clearly holds for , as all graphs of maximum degree at most two have queue number one [14]. So, assume for the remainder that . Suppose that is a vertex of degree one in . We introduce two new vertices and , and three new edges , and in . It follows that vertex has degree three and each of the two introduced vertices and has degree two. By applying this procedure to all degree vertices in , we obtain a planar supergraph of with minimum degree at least two and maximum degree . It is not difficult to see that the queue number of is at most the queue number of this supergraph. So, in the following we will assume that every vertex in has degree at least two.
We start by describing how to construct graph from graph . If has no degree vertex, then we subdivide an edge of once, in order to introduce such a degree vertex. As a consequence, in the following we will assume that contains at least one degree vertex, which we denote by . Let be an ordered concentric representation of centered at . Let be the BFStree of that was used in order to compute ; refer to Section 2. Recall that an edge of that is not in , is either a level edge (if it connects two vertices of the same level in ) or a binding edge (if it connects two vertices on consecutive levels in ).
We proceed by subdividing each binding edge of that does not belong to (if any) once. By this simple operation, each binding edge of is split into a level edge and an edge that can be assigned to tree . For an illustration of this operation refer to the binding edge in Fig. 4(a), which is subdivided once by introducing vertex in Fig. 4(b); as a result, the edge is binding and part of , while the edge is level. Additionally, we subdivide each level edge twice, if or has degree greater than two. This yields three edges , and , the first and last of which we assign to the tree , while the middle edge becomes a level edge. This guarantees that the endvertices of all level edges have degree two. For an illustration refer to the level edge in Fig. 4(b) whose endvertices have degree three; by subdividing this edge twice with vertices and in Fig. 4(c), we guarantee that two of the newly formed edges, that is, and , become binding and part of , while the middle edge becomes a level edge whose endvertices have degree two. The resulting subdivision of is graph . Note that each edge of is subdivided at most three times in order to obtain .
Based on graph , we next describe how to construct graph . In this step, we need to guarantee one additional property of matched graphs, that is, all level edges of a matched graph belong to layer . Suppose that in the concentric representation of there exists a level edge at layer . We create two complete ary trees of height , say and , and we identify their roots with vertices and , respectively. Since the height of each of and is , it follows that all leaves of and all leaves of can be placed along consecutively while preserving planarity. We replace edge by edges forming a matching , such that the th leaf of from the left along is connected to the th leaf of from the right along . Observe that the edges of form a rainbow. For an illustration of this operation, refer to edges and in Fig. 4(c), which are two level edges that do not have their endvertices along the outermost circle . Edge is replaced by the two graycolored binary trees and of height two in Fig. 4(d), while the edge is replaced by the two blackcolored binary trees and of height one. Also, the four edges of and the two edges of are drawn dotted with mostly rectilinear segments in Fig. 4(d). Once we apply the aforementioned procedure to all level edges of , we obtain graph together with an ordered concentric representation , in which all level edges have their endvertices along the outermost circle . Since we assumed that has minimum degree at least two, all leaves of lie on .
We now claim that can be converted into a drawing of that satisfies Properties P.1–P.4. Since belongs to the outer face of , there is a curve that starts at , cuts all circles of once, and crosses no edge of . Hence, we can use to cut all the circles of and stretch so that each circle becomes a line segment by preserving the planarity of the drawing. In other words, we can obtain the drawing of through a suitable homeomorphic transformation of . Properties P.1 and P.4 hold by construction in , while Properties P.2 and P.3 follow from Properties R.1 and R.3 of the ordered concentric representation , respectively. It follows that graph is a matched graph. Thus, by Lemma 3, graph admits a queue layout with queues . Also, graph contains a subdivision of as an induced subgraph.
Next we derive a queue layout for from the queue layout for . Recall that was obtained from by replacing each level edge of by two complete ary trees and , and by a set of matching edges. For the desired queue layout of we order the vertices of according to their ordering in . For every edge of we assign the same queue as in , provided that this edge is also an edge of . Otherwise, such an edge is a level edge of and we assign it to the queue . Thus, in queue layout , all level edges are assigned to queue , while queues contain all edges of the BFStree .
It remains to show that no two (level) edges in nest. Recall that the ordering of vertices is inherited from queue layout and hence satisfies Conditions C.1–C.3. For any level in we have . Moreover, the set of matching edges in incident to the leaves of is given by , and the same holds for . This implies that and consequently by Eq. (1). Now suppose for the sake of contradiction that level edge nests level edge with . By Condition C.1 it follows that and as , all four vertices have the same level . Secondly, in all edges of nest all edges of . Hence, , i.e., the difference between the minimum nestingvalue of edges in and the minimum nestingvalue of edges in is . This however, by Eq. (1) gives , which contradicts the fact that according to Condition C.2.
3.3 Time Complexity
In this section, we analyze the runtime of our algorithm to construct a queue layout with at most queues for a given planar graph of maximum degree . First assume that the input graph is a matched graph. In this case, the runtime of our algorithm is dominated by the computation of the nestingvalues of the edges of and of the matchingvalues of the vertices of . The former can be easily accomplished in time by the drawing of , while the latter in by a bottomup traversal of . Having computed these values, the calculation of the layergroup of each vertex can be done in time in total. The linear order can be determined in time by a single iteration through the vertices of each layer of in the order that they appear along ; recall Conditions C.1–C.3. Finally, the assignment of the edges of into queues can be performed in time, since for each edge the corresponding queue is determined based on the layergroups of its endvertices. Hence, the algorithm supporting Lemma 3 runs in time, which is in , since is planar and hence .
For the case of a general planar graph of maximum degree , however, the construction of the graph may increase the runtime dramatically. In particular, note that graph may have size exponential in , when is sparse. To see this, consider a graph obtained from a path on vertices by adding an edge connecting to . If is chosen as the center of the ordered concentric representation , then edge becomes a level edge; see Fig. 5(a). Since the endvertex of has degree greater than two, the edge will be subdivided twice. Let and be the subdivision vertices; refer to Fig. 5(b) for an illustration. Observe that the edge is a new level edge, whose endvertices are not in the outermost circle of the representation. Then each of the trees and replacing has vertices; see Fig. 5(c). So, in order to keep the runtime of our algorithm polynomial in the size of , we avoid introducing trees and explicitly for each level edge . In fact, the introduction of these trees was convenient for proving the correctness of our approach. But, as we will argue below, to determine the correct queue layout we only need to know the size of the set .
By Lemma 2, the ordered concentric representation of can be computed in linear time. Based on , the computation of graph needs an additional time. To avoid introducing trees and for each level edge of (as required for the computation of graph ), we first reroute each level edge in , so that one part of it lies outside the outermost circle of . This can be done without introducing crossings in time in total, by a single traversal over all level edges in starting from those level edges of the outermost circle and moving inwards in the representation . This edge rerouting guarantees that for any two level edges and , all edges of nest all edges of if and only if the part of that lies outside of nests the corresponding part of in . Therefore, instead of introducing two trees and for each level edge of , we assign a weight to the edge equal to and compute the nestingvalues of the edges of and the matchingvalues of the vertices of based on the weights of these edges as follows; e.g., the weight of the level edge of Fig. 5(b) is four. Consider an edge such that either belongs to or is a level edge of some layer ; note that in the former case we assume that . Observe that the order in which the endvertices of the edges of and the edge segments of the level edges lying outside appear along defines a linear order on their endvertices. Edge has nestingvalue zero if is not nested by any other edge in this order. Otherwise, let be the edge with maximum nestingvalue that nests edge . Then, edge has nestingvalue equal to the nestingvalue of plus . Note that once all nestingvalues of edges are computed, the computation of the matchingvalues of vertices can be done as in the unweighted case.
Up to this point, the time complexity of the algorithm is in , assuming a RAM model of computation that supports standard operations on bit words with unit cost such that , so that each value (in particular, each weight, nestingvalue, and matchingvalue) associated with a vertex or with an edge fits in a word. Since each of these values does not exceed the maximum number of leaves of a ary tree of height at most , that is, or equivalently , words of size linear in suffice. Thus, the running time of our algorithm is , as the computation of the weights of all level edges can be done in time based on the drawing of .
4 Open Problems
Clearly, the major question that remains open is to settle the conjecture by Heath, Leighton and Rosenberg for planar graphs without any restriction in the maximum degree (either in the positive or in the negative). As a further intermediate question, we ask the following: Given a planar graph, is it possible to compute a linear order of its vertices and a partition of its edges into one stack and a constant number of queues? These layouts are known as mixed layouts and have been introduced in the paper by Heath, Leighton and Rosenberg, who conjectured that every planar graph admits a mixed layout with one stack and one queue. Pupyrev [18] recently disproved this conjecture by demonstrating a planar graph for which one stack and one queue do not suffice, and conjectured that for bipartite planar graphs one stack and one queue are always sufficient.
Acknowledgement.
We thank David R. Wood for pointing out an issue in an earlier version.
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