# Planar Graphs that Need Four Pages

We show that there are planar graphs that require four pages in any book embedding.

## Authors

• 7 publications
• ### Four Pages Are Indeed Necessary for Planar Graphs

An embedding of a graph in a book consists of a linear order of its vert...
04/16/2020 ∙ by Michael A. Bekos, et al. ∙ 0

• ### Identifying KDM Model of JSP Pages

In this report, we propose our approach that identifies a KDM model of J...
03/14/2018 ∙ by Anas Shatnawi, et al. ∙ 0

• ### Modeling pages left blank in university examination: A resolution in higher education process

Trees are the main sources of paper production, in most of the cases, as...
10/11/2019 ∙ by Suman K. Ghosh, et al. ∙ 0

• ### Book Embeddings of k-Map Graphs

A map is a partition of the sphere into regions that are labeled as coun...
12/12/2020 ∙ by Franz J. Brandenburg, et al. ∙ 0

• ### Artificial chemistry experiments with chemlambda, lambda calculus, interaction combinators

Given a graph rewrite system, a graph G is a quine graph if it has a non...
03/31/2020 ∙ by Marius Buliga, et al. ∙ 0

• ### Efficiently Reclaiming Space in a Log Structured Store

A log structured store uses a single write I/O for a number of diverse a...
04/30/2020 ∙ by David Lomet, et al. ∙ 0

• ### DRAW: Deep networks for Recognizing styles of Artists Who illustrate children's books

This paper is motivated from a young boy's capability to recognize an il...
04/10/2017 ∙ by Samet Hicsonmez, et al. ∙ 0

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## 1 Introduction

A book embedding of a graph consists of a linear order of its nodes and a partitioning of its edges, so that the nodes can be placed in order on a straight line (the “spine” of the book) and the edges in each part can be drawn on a separate half-plane bordered by the line (a “page” of the book) so that the edges on the same page do not intersect. The objective is to find a book embedding that uses the minimum number of pages. This minimum number is called the pagenumber (or book thickness) of the graph.

Book embeddings were introduced in [12] and [3]. They were studied in connection with an approach to fault-tolerant VLSI design [13, 5], and have applications also in sorting using stacks, in graph drawing, complexity theory, and other areas. Computationally, the problem of computing the pagenumber of a graph is hard: it is NP-complete to decide whether a planar graph has pagenumber 2 [14, 5]. Note that in the case of two pages, the crux of the problem is the node-embedding part (the linear ordering of the nodes): once the node ordering is fixed, it is easy to test whether two pages suffice. In general, the subproblem of minimizing the number of pages for a given fixed node ordering is itself also NP-hard [8].

Graphs with pagenumber one are exactly the outerplanar graphs. Graphs with pagenumber two are exactly the planar subhamiltonian graphs, i.e. the subgraphs of planar Hamiltonian graphs [3]. In [15, 16] we showed that all planar graphs can be embedded in four pages, and gave a linear time algorithm for this purpose (references [4] and [9] gave earlier algorithms that embed planar graphs in nine and seven pages respectively). In the conference paper [15] we stated also that there are planar graphs that require four pages, and outlined briefly the approach and the structure of the construction. The present paper gives the full details of the construction and the proof that the constructed planar graph has pagenumber at least four.

Besides planar graphs, there has been extensive work on book embeddings for various other classes of graphs, for example graphs of bounded genus [10, 11], bounded treewidth [7], 1-planar graphs [1].

Before getting into the technical details of the construction and the proof, we make a few remarks regarding the issues involved and our approach to address them. To show a lower bound of four on the pagenumber of planar graphs means finding a planar graph and showing that no matter how we order its nodes and how we partition its edges into three pages, there will be a violation (two conflicting edges on the same page). A major obstacle in this regard stems from the computational complexity of the problem: The problem is NP-complete, which means that if , as is widely believed to be the case, there is in general no short way to prove the nonexistence of a suitable node ordering and edge partitioning for a given graph; that is, any proof has to be in general at least superpolynomially long in the size of the graph, and will likely amount to examining essentially all (or at least a large number of) possibilities to ensure that they do not work. An important difference here is that we do not have to deal with an arbitrary general graph, but with a specific graph of our own design that is suitable for our purpose. If there was a very small suitable graph , then the complexity would not be an issue. This is the case for example in the graph coloring problem, where there is an extremely small planar graph (namely, ) that needs 4 colors, so the lower bound is trivial. Unfortunately, this does not seem to be the case in the book embedding problem. For example, [2] studied experimentally the book embedding problem by formulating it in terms of the SAT (Boolean formula satisfiability) problem and using a SAT solver; the software handles graphs with up to 500-700 nodes. In their experiments searching for planar graphs that require four pages (they tried both random graphs and crafted graphs in certain families) none was encountered, which led the authors to hypothesize that perhaps three pages are enough for all planar graphs.

The way we deal with the complexity of the problem in our construction is by building up the graph (and the proof) in stages, to control the exponential explosion in case analysis. As in NP-completeness reductions, we design and use gadgets (‘small’ graphs that have useful properties) as building blocks. We start with a small graph (10 nodes) which we analyze in some detail to characterize its book embeddings under some strong restrictions. Then we use this gadget to analyze a somewhat larger graph , and this is used in turn for a larger gadget under weaker restrictions. The gadgets are used in the construction of the final graph that cannot be embedded in three pages. The properties we showed for the gadgets restrict significantly the possible book embeddings of and make the analysis tractable.

The rest of the paper is organized as follows. Section 2 provides basic definitions and some simple observations. In Section 3 we design the gadgets and prove their properties. Section 4 gives the definition of the graph and proves that requires four pages.

## 2 Preliminaries

Let be a (undirected) graph, and a linear ordering (a permutation) of its nodes. We say that two edges conflict in the ordering if or ; that is, if we place the nodes on a line ordered according to and draw the edges (as curves) on a half-plane bordered by the line, two edges conflict iff they intersect. A book embedding of in pages consists of (1) a linear ordering of its nodes, and (2) a coloring of its edges with colors (the “pages”) so that conflicting edges in receive different colors. The pagenumber of is the minimum number of pages such that has a book embedding in pages.

We will usually show in figures a (partial) book embedding in 3 pages by showing the positions of the nodes on a line and showing edges of color 1 as red dashed curves, color 2 as solid blue, and color 3 as dotted green. To simplify notation, we will identify the positions of the nodes on the line with the nodes, and refer for example to node on the line (instead of point ), to interval of the line (instead of interval ), and so forth.

Alternatively, a book embedding of a graph can be defined by a mapping of its nodes to (distinct) points on a circle, drawing the edges as straight line segments (chords) inside the circle and coloring them with colors so that intersecting edges receive different colors. From such a circle embedding one can obtain a linear embedding by cutting the circle at any point and then ordering the nodes on the line in either of the two directions, clockwise or counterclockwise (the edges retain their colors). Conversely, given a linear embedding, one can obtain a circle embedding by connecting the two ends of the line to form a circle that encloses all the edges. Note that a circle embedding corresponds to linear embeddings that are essentially equivalent.

Given a book embedding of a graph , we will say that an edge exits an interval if one node of the edge is in the open interval and the other node is outside the closed interval . Similarly, for a circle embedding, an edge exits an arc if the two nodes of the edge are on the two different open arcs of the circle between and .

The following proposition gives some basic simple observations that are used throughout the paper, usually without making explicit reference to the proposition.

###### Proposition 1.

(1) If in a book embedding of a graph there is a path between two nodes all of whose edges have the same color , then there is no edge of color that exits the interval and connects two nodes that are not on the path .
(2) If in a 3-page embedding of there are paths of all 3 colors respectively between two nodes , then there is no edge that exits the interval and connects two nodes that are not in any of these paths.
(3) If in a 3-page embedding of there are paths of all 3 colors between two nodes , then for every connected component of the graph obtained by deleting the nodes of from , either all the nodes of are in the interval or they are all outside .

###### Proof.

(1) Let be the path , and let be a color- edge whose nodes are not on . If a node of the path is in the interval , then the next node must be also in because the edges have the same color, hence they do not conflict. Thus, by induction, if is in the interval then all the nodes of the path are in , hence also . Similarly, if is not in the interval then is not in the interval either. In either case, does not exit the interval .

(2) Follows from (1).

(3) Any two nodes of are connected by a path that does not contain any node of the paths . By (2), is in the interval iff is also in . Hence by induction, is in iff every node of the path, and in particular , is in . ∎

We will say that a node reaches an interval (or any subset of the line) if there is an edge from to a node in the interval (resp., in the subset). Thus, for example by the above Proposition (part 2), if in a 3-page embedding there are paths of all three colors between nodes , then the nodes embedded in the interval cannot reach any nodes outside except possibly the nodes of the paths .

Figure 1 shows our first “gadget” . We will refer to nodes 1, 2 as the outer terminals, to nodes as the inner terminals, and to nodes as the centers. We will use to denote the graph with the additional edge connecting the outer terminals, and we will use to denote the graph with the additional edge connecting the inner terminals.

###### Lemma 1.

There is no 3-page embedding of or of such that (1) the inner terminals lie in the interval between the outer terminals, (2) the centers of lie in the interval , and (3) all edges from node 1 to the closed interval use the same color, say color 1, and all edges from node 2 to use color 2.

###### Proof.

The proof is by contradiction. Consider any 3-page embedding of that satisfies conditions 1-3 of the lemma. Suppose without loss of generality that the nodes are laid out in the order shown in Figure 2; the other possible embeddings with the order of and/or reversed, are symmetric. The outer terminals 1,2 are not shown explicitly in the figure: one of them, node is at the left end of the line and the other outer terminal is to the right of , but we do not specify which is which so that we do not have to distinguish cases; it is irrelevant for the following arguments which one of 1,2 is at the left end and which is on the right. The edges to 1, 2 are shown in the figure as ‘dangling’ segments with only one node. Thus, the red dashed dangling edges have color 1 and go to node 1, and the solid blue dangling edges have color 2 and go to terminal 2. We shall prove that edge has color 2 and has color 1, as shown in the figure.

Edge conflicts with the edge which is colored 1 by the hypothesis, hence is colored 2 or 3. Similarly edge conflicts with , hence is colored 1 or 3. The two edges , conflict with each other, so they cannot both be colored 3. Suppose without loss of generality that is not colored 3, hence it is colored 2.

Suppose that edge has color 3. We shall argue that it is impossible then to embed legally node . Observe that node cannot reach any node, other than 1,2, outside the interval because of the color-3 edge , the color-1 path and the color-2 path . Hence must lie in the interval . Since the edges from node 2 to this interval are colored 2 and is also colored 2, node must lie in the interval . Then there is no legal color available for the edge because it conflicts with the color-1 edge , the color-2 edge and the color-3 edge . We conclude that the edge does not have color 3, hence it has color 1.

Consider now the possible positions of nodes and . We show first that they must be outside the interval . Suppose that is inside the interval , to derive a contradiction. Since is adjacent to node 1 and edge is colored 1, node cannot be in the interval , hence it must be in . The edge must be colored 3 (since it conflicts with the color-1 edge and the color-2 edge ). Hence, node cannot reach any node, other than 1, 2, outside the interval because of the color-1 path , the color-2 path , and the color-3 edge . Therefore, node must then also be inside the interval . However, node 2 cannot reach the subinterval because of the color-2 edge , and node cannot reach the subinterval because of the color-1 edge , the color-2 path and the color-3 edge Thus, there is no legal position for node , contradiction. We conclude that must be outside the interval . By a symmetric argument, also lies outside the interval .

The edges and must be colored 3, because they exit the interval and there are color-1 and -2 paths and connecting and .

We can prove the lemma now for : The edge connecting the inner terminals cannot be colored legally because it intersects the color-1 edge , the color-2 edge and the color-3 edge . This proves the claim for the graph .

It remains to prove the claim for . Since the edges and have the same color, either is to the left of and hence of , or is to the right of and hence of (or both). Suppose without loss of generality that is to the right of . If it is left of , i.e., within the same arc in the corresponding embedding on a circle, then the edge must be also colored 3 (because it conflicts with the edges ), but then either or conflicts with the edge , a contradiction. We conclude that is right of , i.e., lies in the opposite arc . By a symmetric argument, the same holds for the node .

Since and have the same color (3), hence do not intersect, nodes appear in this order. Consider the three edges . They all intersect a color-3 edge, namely respectively, hence they can only use the colors 1, 2. However, the three edges conflict with each other, hence they cannot all be colored with two colors. This proves the lemma for . ∎

###### Lemma 2.

It is not possible to embed in three pages, such that (1) the inner terminals lie in a subinterval between the outer terminals 1,2, (2) both centers lie outside the interval , (3) all edges from node 1 to the interval use the same color, say color 1, and all edges from node 2 to use color 2, and (4) all other edges exiting the interval (not connecting to nodes 1, 2) are colored 3.

###### Proof.

The proof is by contradiction. Assume an embedding as in the lemma. If is outside then both edges , must have color 3 (by condition (4)), and similarly for . We cannot have both outside the interval because then (at least) one of the edges , would intersect one of the edges , . Therefore, at least one of must be in the interval .

Assume without loss of generality that is in the interval , and, since it is not in the interval , assume wlog that it is left of , i.e., that it is in the subinterval . The edge must be colored 3 because of the edges . Node cannot have any edge exiting the interval to a node other than 1,2 because of condition (4), the color-3 edge and the color-1 and -2 edges . Therefore, lies in the interval , and since it is not in , it is in . Assume without of loss of generality that is closer to than ; see Figure 3.

The edge is colored 3, because of the edges . This implies that the edge must be colored 1 because of the color-2 edge . Consider the possible position of node . Node cannot exit the interval because of the color-1 path , the color-2 path , and the color-3 edge . On the other hand, node 1 cannot reach the interval because of the color-1 edge . Furthermore, cannot reach the interval because of the color-1 path , the color-2 path and the color-3 edge (see Figure 3). Therefore, there is no legal position for node , a contradiction. ∎

Let be the graph shown in Figure 4 where each internal face (triangle) is stellated twice more. That is, inside each triangle (e.g. etc.) we insert a center node with edges to the nodes of the triangle, and repeat this once more for each resulting triangle; these additional nodes are not shown in the Figure so that it will not become too cluttered. Note that contains many copies of . For example there is one copy with outer terminals 1,2 and inner terminals . Another copy of has outer terminals and inner terminals ; another has outer terminals and inner terminals . And so forth. We will use again to denote the graph with the additional edge .

###### Lemma 3.

Suppose that in a 3-page embedding of , (1) the inner terminals lie in a subinterval between the outer terminals 1,2, (2) all edges from node 1 to the interval use the same color, say color 1, and all edges from node 2 to use color 2, and (3) all other edges exiting the interval (not connecting to nodes 1, 2) are colored 3. Then one of the center nodes is inside the interval and the other one is outside but inside . Furthermore, the embedding is as shown in Figure 5 (up to reversing the order, switching and/or switching the indices ). That is, if the center node is outside and it is in the interval , then is in the interval and the edges all have color 3.

###### Proof.

By Lemmas 1, 2, one of the centers must lie outside the interval and one inside. Assume without loss of generality that is outside and is inside. The edge must have color 3 because of the color-1 path and the color-2 path .

We will show that the embedding conforms to the more detailed Figure 6 (or a symmetric one obtained by reversing the order, switching and/or switching ). We will do this in several steps. First we will show that is in the interval . Second we will show that and have the indicated positions. Third, we will identify the positions of and . Finally, we will identify the position of to complete the proof.

###### Claim 1.

Node is in the interval .

###### Proof.

Suppose that is outside . Then both edges have color 3 by condition (3). Since and are connected by paths , , of all three colors, no edge can exit the interval to any node other than 1, 2, . Since is inside the interval , it follows that all the nodes of in the interior of the graph bounded by the cycle (which is connected) must be inside the interval . Note that the subgraph bounded by the cycle contains with as the outer terminals and as the inner terminals, which are adjacent. The inner terminals lie in the same arc (in a cyclic ordering), all edges from to the interval must be colored 3 and all edges from 2 to must be colored 2. By Lemma 1, at least one of the two centers must be outside the interval . Hence it must be in the interval , and the edge connecting it to node must be colored 1 because it conflicts with the edges and . Similarly, the subgraph of bounded by the cycle contains with as the outer terminals and as the inner terminals, which are adjacent. By a symmetric argument, at least one of the two centers must be in the interval , and the edge that connects it to is also colored 1. Thus, there are two color-1 edges that intersect, a contradiction. We conclude that is in the interval . ∎

Since is outside the interval (see the beginning of the proof) and is inside the interval (by Claim 1), it follows that is either in the interval or in . Assume without loss of generality that is in the interval as in Fig. 5.

###### Claim 2.

Node is in the interval and node is in the interval . Edge has color 1 and edge has color 3.

###### Proof.

Nodes , are both adjacent to nodes and . Node cannot reach outside the interval because of the color-1 path , color-2 path and the color-3 edge . Node cannot reach the interval because of the edges and . Hence and must be in the interval .

Suppose that is in , to derive a contradiction. Then the edge must have color 1 and color 3, hence the edge must have color 2. Then there is no legal position for node : cannot reach outside the interval (because of the color-1 path , the color-2 path and the color-3 path ), node 2 cannot reach the interval with color 2, and node cannot reach the interval (because of the color-1 edge , the color-2 edge and the color-3 edge ). We conclude that is in interval .

The subgraph of bounded by the cycle contains a copy of with as the outer terminals and as the inner terminals. The inner terminals are adjacent, they lie in the same () arc (in the cyclic order), node can reach the interval only with color 3 and node can reach only with color 2. The conditions of Lemma 1 are satisfied, therefore the interval cannot contain both centers . Since is in the interval , the other center is not, thus it is in the interval . The edge must have color 1 (because of the conflicting edges and ), and the edge has color 2, therefore the edge must have color 3. ∎

###### Claim 3.

Node is in the interval . Node is in interval . Edge has color 2.

###### Proof.

Node is adjacent to nodes . Node can only reach the interval . Node cannot reach the subinterval (because of the color-1 edge , the color-2 edge and the color-3 edge ) and node cannot reach the subinterval (because of the edges ). Therefore, is in the interval .

The subgraph of bounded by the cycle contains a copy of with outer terminals , and inner terminals . The inner terminals are adjacent and are in the same arc. Furthermore, all edges from to the interval are colored 1, and all edges from to must be colored 3. The conditions of Lemma 1 are satisfied, therefore we cannot have both centers in the interval . Since is in the interval, it follows that must be outside the interval . Node cannot reach outside the interval (because of the color-1 path , the color-2 edge and the the color-3 edge ), and node cannot reach the interval . Therefore, must be in the interval . It follows that the edge must have color 2, since it conflicts with edges and . ∎

###### Claim 4.

Node is in interval . Edge is colored 3.

###### Proof.

Node is adjacent to . Node cannot reach the interval , node cannot reach the interval and 1 cannot reach (because of the edge ). Therefore lies outside the interval . First, we claim that it must be in the interval . Suppose to the contrary that it lies outside the interval (possibly even outside the interval ). Then it is easy to see that both edges and must have color 3: If is outside the interval then this holds because of condition (3), and if is in this holds because of the color-1 edge and the color-2 edges and . Then there is no legal position to place the center of the triangle : If is outside then must both have color 3 and one of them intersects one of , . Similarly, if is in , the edge must have color 3 and intersects . If is in the interval then must be colored 3 (by condition (3) or because of the edges ) and it intersects the edge . If is in the interval then must have color 3 (because of the edges ) and intersects the edge . We conclude that is in the interval . Hence it is either in or in .

Suppose that is in the interval , to derive a contradiction. The edge is colored 1 and the edge must be colored 3, hence the edge must be colored 2. Consider the subgraph of bounded by the cycle and recall that all the triangles in Figure 4 are stellated twice. The subgraph contains a copy of with as the outer terminals, as the inner terminals, which are adjacent, and are embedded inside the interval . Node can reach the interval only with color 2 (because of the color-1 edge and the color-3 edge ), and can reach only with color 3. The center of the triangle is inside the interval . Hence, by Lemma 1, the center of the other triangle cannot be in the interval . This leaves no possible position for the center of : Node cannot reach outside or left of (because of the edges and condition 3), node cannot reach inside the interval or , and cannot reach the interval .

We conclude that is not in the interval , hence it must be in the interval . The edge must have color 3 because it conflicts with the edges . ∎

This completes the proof of the lemma. ∎

Let be the graph formed by taking 15 copies of , identifying their outer terminals 1,2, and identifying terminal of the -th copy with terminal of the -th copy; i.e., is formed by glueing together back-to-back 15 copies of with the same outer terminals, see Figure 7. We call a quad, nodes 1,2 the outer terminals of quad and call the inner terminals of the copies of the inner terminals of . We use to denote the th copy of . Let denote the graph consisting of and the edge .

###### Lemma 4.

There is no embedding of in three pages such that all the inner terminals are embedded in a subinterval of the interval , and all edges from nodes 1 and 2 to this subinterval use only two (the same two) colors (i.e., one of the three colors is not used by any edge connecting nodes 1 and 2 to this subinterval).

###### Proof.

Consider a book embedding of the nodes of the quad with all the inner terminals embedded in a subinterval of the interval and edges from 1, 2 to this subinterval using two colors. Let be the inner terminals closest to 1 and the inner terminals closest to 2; thus, if we assume wlog that the embedding on the line starts with terminal 1, then are the first two inner terminals and the last two. The order of these nodes is . Let 1 be the color of the edge and let 2 be the color of the edge . By the hypothesis, all edges from nodes 1 and 2 to the interval are colored 1 or 2. Since the edge is colored 2, all edges from node 1 to the interval (including nodes ) must be colored 1. Similarly, since the edge is colored 1, all edges from node 2 to the interval (including and ) must be colored 2. Because of the color-1 path and the color-2 path , all other edges that exit the interval (and are not going to 1,2) must be colored 3. Thus, the conditions of Lemma 3 are satisfied for every copy of whose inner terminals are in the interval . Since there are 16 inner terminals, there are at least three consecutive terminals that are not in the set , i.e. that are embedded in the interval . Thus, there are two consecutive copies of , for which Lemma 3 holds.

Consider any copy of with inner terminals in the interval . Observe from the conclusion of Lemma 3 (see Fig. 5) that the inner terminals are connected by paths of all 3 colors, , and thus no edge can exit the interval unless it connects to , or . Second, observe that there are nodes of outside the interval on both sides, e.g. the nodes . Applying these observations to the -th and -th copy of , tells us that cannot lie inside the interval , and similarly cannot lie inside the interval . For, suppose to the contrary that is in . Since no edge of the -th copy can exit the interval , all nodes of must lie inside the interval , contradicting the fact that some nodes must lie on both sides outside . Therefore, is outside the interval . Similarly is outside the interval .

Thus, we may assume without loss of generality that the nodes appear in this order. Observe from Lemma 3 for each of copy of that there is a color-3 edge (edge in Fig. 5) that connects two nodes of that lie outside, and on both sides, of the interval and that furthermore, one of these two nodes of the edge (node in Fig. 5) has a color-3 edge to an inner terminal ( in Fig. 5). Let be the edge for , and for . The left endpoint of is left of and the right endpoint is right of , and since it cannot be in the interval (no edge that does not belong to can exit this interval unless it goes to 1 or 2), it must be right of . Similarly, the left endpoint of must be left of and the right endpoint right of . Since the edges both have color 3, they are nested. Suppose without loss of generality that is nested inside . Then neither endpoint of can have a color-3 edge to or , because it would conflict with . This contradicts Lemma 3. ∎

We remark that the properties of Lemmas 3 and 4 depend crucially on the fact that in we included the edge between the two centers rather than the edge between the two inner terminals. It can be shown that with the edge instead, even after stellating the faces an arbitrary number of times, and gluing back-to-back an arbitrary number of copies of the resulting graph, yields a graph that does not have the property of Lemma 4: the graph can be embedded between the two outer terminals so that all edges to each outer terminal have the same color.

Attaching quads to the edges of a graph restricts the possible embeddings into 3 pages. The following lemma illustrates how can be used. The lemma will be used often in the sequel.

###### Lemma 5.

Consider a graph formed by taking a triangle on the plane, attaching (adding) a quad to each edge of the triangle, by identifying the outer terminals of the quad with the nodes of the edge. There is no embedding of in three pages such that all inner terminals of the quads are embedded in the arc that does not contain and all edges from to the inner terminals have the same color.

###### Proof.

Suppose that there is such a 3-page embedding, consider its linearization with lying outside the interval . Assume first that among all the inner terminals of the quads, the one embedded closest to does not belong to the quad, i.e., it belongs to the or the quad. Let be this terminal and assume wlog that the edge has color 1. All the inner terminals of the quad are in the interval ; this interval can be reached from and only with colors 2 and 3. By Lemma 4 this is impossible.

Therefore, the inner terminal closest to belongs to the quad. By a symmetric argument, the inner terminal closest to belongs to the quad. The edges and have the same color, say color 1. Then all inner terminals of the quad are in the interval and and can reach this interval only with colors 2 and 3. This is impossible by Lemma 4. ∎

## 4 The Graph

Our ‘hard’ graph is constructed as follows. Take a long path , of nodes, where is sufficiently large, say . Take two other nodes 1, 2 and connect them to all the nodes of the path, as in Fig. 8. This forms 2 triangles, which we call the big triangles. Subdivide each big triangle to three small triangles by inserting a center node and connecting it to the 3 nodes of the triangle. Inside each small triangle, attach a copy of the quad to each edge of the triangle, and add a central node connecting it to the three nodes of the triangle and the innermost inner terminals of the three quads. The construction is shown in Figure 8 (except for the quads attached to the edges, which we omitted for clarity). Note that all small triangles as well as all big triangles satisfy the conditions of Lemma 5. Note also the copies of with outer terminals 1,2 and inner terminals . We call the nodes 1, 2 the terminals of , we call the nodes the vertical nodes of , and call the edges the vertical edges.

We will show that cannot be embedded in three pages. For this purpose, fix any 3-page embedding of . We will show that the embedding has to satisfy a sequence of properties, and derive eventually a contradiction.

Given a 3-page circle embedding of , we designate one of the two arcs between the two terminals 1,2 as the major (1,2) arc, and the other as the minor (1,2) arc as follows: the major arc is an arc (1,2) that contains at least half of the vertical nodes, and the other (1,2) arc is the minor arc (if both (1,2) arcs contain exactly half of the vertical nodes, we arbitrarily designate one as the major and the other as the minor arc). Let be the node in the major arc that is closest to node 1 and adjacent to 2, and let be the node in the major arc that is closest to node 2 and adjacent to 1 - see Fig. 9. We show first that there are not many vertical edges with nodes on both (1,2) arcs.

###### Lemma 6.

There are at most 4 vertical nodes in the arc , whose successor on the path is in the minor arc (1,2).

###### Proof.

By contradiction. Suppose that there are 5 such vertical nodes in whose successor is in the minor arc (1,2). Denote these 5 nodes as in the order that they appear in , and let their successors be respectively. Assume wlog that edge has color 1 and edge has color 2 (they intersect so they must have different colors). The 5 vertical edges intersect both edges , so they must all have color 3, hence they do not intersect each other; see Fig. 9. The edges must have color 1 because they intersect and . Similarly the edges must have color 2 because they intersect and . Note that a node inside the arc can reach nodes other than nodes 1,2 only in the same arc or the arc , because of the color-1 path , the color-2 path and the color-3 edges . Similarly, a node inside the arc can reach nodes other than nodes 1,2 only in the same arc or the arc . Node can only reach nodes in the arcs and .

Similar observations hold for the edges connecting nodes 1, 2 to the vertical nodes on the minor arc (1,2). Edges , must have color 1 or 2 because they intersect . If has color 1 then must have color 2, hence all edges must have color 1 and all the edges must have color 2. If has color 2 then must have color 1, all edges have color 2 and the edges have color 1. (Figure 9 depicts the latter case.) In either case, note again that a node in the arc can reach nodes other than 1,2 only in the same arc or the arc . Similarly, a node inside the arc can reach nodes other than 1, 2 only in the same arc or the arc . Node can only reach nodes in the arcs and .

Consider the big triangle of . From the observations in the previous two paragraphs it follows that either all the internal nodes of this triangle are in the strip or they are all in the strip . Assume without loss of generality that they are in the strip (the argument is the same in the other case). Any edge from to the arc must be colored 3 and likewise any edge from to the arc must be colored 3, therefore there cannot exist both kinds of edges. Hence, either all the inner terminals of the quads attached to edges are in arc or all inner terminals of the quads attached to the edges are in arc . Assume wlog that the former holds, i.e. all the inner terminals of the quads attached to edges are in arc .

If the inner terminal closest to belongs to the quad, then the edge connecting it to 1 must be colored 1, hence all edges from the inner terminals of the quad to must be colored 2 or 3 and all edges from these inner terminals to must be colored 3, contradicting Lemma 4. On the other hand, if the inner terminal closest to belongs to the quad, then the edge to is colored 3, hence all edges from the inner terminals of the quad to must be colored 1 or 2 and all edges from these inner terminals to must be colored 1, contradicting again Lemma 4. The lemma follows. ∎

We view the 3-page embedding as an embedding on the line, starting with one of the terminals, followed by the major (1,2) arc (which is now an interval on the line), then the other terminal, followed by the minor arc (1,2). We will focus on the interval corresponding to the major arc. Let (resp. ) be again the node in the interval that is closest to node 1 (resp. 2) and is adjacent to 2 (resp. 1). Define the stretch between two points , of the linear embedding, denoted , to be the number of vertical nodes in the interval .

###### Lemma 7.

(1) If are two nodes in the interval connected by an edge then .
(2) If are two nodes in the interval that have incident edges that exit the interval then .

###### Proof.

The proof is essentially the same for both parts. Suppose that are two nodes as in part (1) or (2) of the lemma such that . We will derive a contradiction. If there is an edge , say of color 3, then all edges from the terminals 1, 2 to all the nodes in the interval