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Keywords:

  • topological graph;
  • crossing edges;
  • 1-planar graph;
  • 1-immersion

Abstract

  1. Top of page
  2. Abstract
  3. 1. INTRODUCTION
  4. 2. CHAIN GRAPHS BASED ON K3, 3
  5. 3. PN-GRAPHS
  6. 4. MN-GRAPHS BASED ON PN-GRAPHS
  7. 5. TESTING 1-IMMERSIBILITY IS HARD
  8. 6. k-PLANARITY TESTING FOR MULTIGRAPHS
  9. ACKNOWLEDGMENTS
  10. REFERENCES

A graph is 1-planar if it can be drawn on the plane so that each edge is crossed by no more than one other edge (and any pair of crossing edges cross only once). A non-1-planar graph G is minimal if the graph inline image is 1-planar for every edge e of G. We construct two infinite families of minimal non-1-planar graphs and show that for every integer inline image, there are at least inline image nonisomorphic minimal non-1-planar graphs of order n. It is also proved that testing 1-planarity is NP-complete.

1. INTRODUCTION

  1. Top of page
  2. Abstract
  3. 1. INTRODUCTION
  4. 2. CHAIN GRAPHS BASED ON K3, 3
  5. 3. PN-GRAPHS
  6. 4. MN-GRAPHS BASED ON PN-GRAPHS
  7. 5. TESTING 1-IMMERSIBILITY IS HARD
  8. 6. k-PLANARITY TESTING FOR MULTIGRAPHS
  9. ACKNOWLEDGMENTS
  10. REFERENCES

A graph drawn in the plane is 1-immersed in the plane if any edge is crossed by at most one other edge (and any pair of crossing edges cross only once). A graph is 1-planar if it can be 1-immersed into the plane. It is easy to see that if a graph has 1-immersion in which two edges inline image with a common end vertex cross, then the drawing of e and f can be changed so that these two edges no longer cross. Consequently, we may assume that adjacent edges are never crossing each other and that no edge is crossing itself. We take this assumption as a part of the definition of 1-immersions since this limits the number of possible cases when discussing 1-immersions.

The notion of 1-immersion of a graph was introduced by Ringel [14] when trying to color the vertices and faces of a plane graph so that adjacent or incident elements receive distinct colors. In the last two decades this class of graphs received additional attention because of its relationship to the family of map graphs, see [7, 8] for further details.

Little is known about 1-planar graphs. Borodin [1, 2] proved that every 1-planar graph is 6-colorable. Some properties of maximal 1-planar graphs are considered in [15]. It was shown in [3] that every 1-planar graph is acyclically 20-colorable. The existence of subgraphs of bounded vertex degrees in 1-planar graphs is investigated in [10]. It is known (see [4–6]) that a 1-planar graph with n vertices has at most inline image edges and that this upper bound is tight. In the paper [9], it was observed that the class of 1-planar graphs is not closed under the operation of edge-contraction.

Much less is known about non-1-planar graphs. The basic question is how to recognize 1-planar graphs. This problem is clearly in NP, but it is not clear at all if there is a polynomial time recognition algorithm. We shall answer this question by proving that 1-planarity testing problem is NP-complete.

The recognition problem is closely related to the study of minimal obstructions for 1-planarity. A graph G is said to be a minimal non-1-planar graph (MN-graph, for short) if G is not 1-planar, but inline image is 1-planar for every edge e of G. An obvious question is:

How many MN-graphs are there? Is their number finite? If not, can they be characterized?

The answer to the first question is not hard: there are infinitely many. This was first proved in [12]. Here, we present two additional simple arguments implying the same conclusion.

Example 1. Let G be a graph such that inline image, where cr(G) denotes the crossing number of G. Let inline image be the graph obtained from G by replacing each edge of G by a path of length t. Then inline image. This implies that inline image is not 1-planar. However, inline image contains an MN-subgraph H. Clearly, H contains at least one subdivided edge of G in its entirety, so inline image. Since t can be arbitrarily large (see, e.g., the well-known lower bound on inline image), this shows that there are infinitely many MN-graphs.

Before giving the next example, it is worth noticing that 3-cycles must be embedded in a planar way in every 1-immersion of a graph in the plane.

Example 2. Let inline image be one of the Kuratowski graphs. For each edge inline image, let inline image be a 5-connected triangulation of the plane and inline image be adjacent vertices of inline image whose degree is at least 6. Let inline image. Now replace each edge inline image of K with inline image by identifying x with u and y with v. It is not hard to see that the resulting graph G is not 1-planar (since two of graphs inline image must “cross each other,” but that is not possible since they come from 5-connected triangulations). Again, one can argue that they contain large MN-graphs.

The paper [12] and the above examples prove the existence of infinitely many MN-graphs but do not give any concrete examples. They provide no information on properties of MN-graphs. Even the most basic question, if there are infinitely many MN-graphs whose minimum degree is at least 3, cannot be answered by considering these constructions. In [12], two specific MN-graphs of orders 7 and 8, respectively, are given. One of them, the graph inline image, is the unique 7-vertex MN-graph and since all 6-vertex graphs are 1-planar, the graph inline image is the MN-graph with the minimum number of vertices. Surprisingly enough, the two MN-graphs in [12] are the only explicit MN-graphs known in the literature.

The main problem when trying to construct 1-planar graphs is that we have no characterization of 1-planar graphs. The set of 1-planar graphs is not closed under taking minors, so 1-planarity cannot be characterized by forbidding some minors.

In the present paper, we construct two explicit infinite families of MN-graphs whose minimum degree is at least 3 and, correspondingly, we give two different approaches how to prove that a graph has no plane 1-immersion.

In Section 'CHAIN GRAPHS BASED ON K3, 3', we construct MN-graphs based on the Kuratowski graph K3, 3. To obtain them, we replace six edges of K3, 3 by some special subgraphs. The minimality of these examples is easy to verify, but their non-1-planarity needs long and delicate arguments. Using these MN-graphs, we show that for every integer inline image, there are at least inline image nonisomorphic minimal non-1-planar graphs of order n. In Section 'PN-GRAPHS', we describe a class of 3-connected planar graphs that have no plane 1-immersions with at least one crossing point (PN-graphs, for short). Every PN-graph has a unique plane 1-immersion, namely, its unique plane embedding. Hence, if a 1-planar graph G contains a PN-graph H as a subgraph, then in every plane 1-immersion of G the subgraph H is 1-immersed in the plane in the same way.

Having constructions of PN-graphs, we can construct 1-planar and non-1-planar graphs with some desired properties: 1-planar graphs that have exactly inline image different plane 1-immersions; MN-graphs, etc.

In Section 'MN-GRAPHS BASED ON PN-GRAPHS', we construct MN-graphs based on PN-graphs. Each of these MN-graphs G has as a subgraph a PN-graph H and the unique plane 1-immersion of H prevents 1-immersion of the remaining part of G in the plane.

Despite the fact that minimal obstructions for 1-planarity (i.e., the MN-graphs) have diverse structure, and despite the fact that discovering 1-immersions of specific graphs can be very tricky, it turned out to be a hard problem to establish hardness of 1-planarity testing. In Section 'TESTING 1-IMMERSIBILITY IS HARD' we show that 1-planarity testing is NP-complete. The proof is geometric in the sense that the reduction is from 3-colorability of planar graphs (or similarly, from planar 3-satisfiability).

In Section 'k-PLANARITY TESTING FOR MULTIGRAPHS', we show how the proof of Theorem 5 can be modified to obtain a proof that k-planarity testing for multigraphs is NP-complete. An extended abstract of this paper was published in Graph Drawing 2008 [13].

2. CHAIN GRAPHS BASED ON K3, 3

  1. Top of page
  2. Abstract
  3. 1. INTRODUCTION
  4. 2. CHAIN GRAPHS BASED ON K3, 3
  5. 3. PN-GRAPHS
  6. 4. MN-GRAPHS BASED ON PN-GRAPHS
  7. 5. TESTING 1-IMMERSIBILITY IS HARD
  8. 6. k-PLANARITY TESTING FOR MULTIGRAPHS
  9. ACKNOWLEDGMENTS
  10. REFERENCES

Two cycles of a graph are adjacent if they share a common edge. If a graph G is drawn in the plane, then we say that a vertex x lies inside (resp. outside) an embedded (i.e., nonself-intersecting) cycle C, if x lies in the interior (resp. exterior) of C, and does not lie on C. Having two embedded adjacent cycles C and inline image, we say that C lies inside (resp. outside) inline image if every point of C either lies inside (resp. outside) inline image or lies on inline image. From this point on, by a 1-immersion of a graph we mean a plane 1-immersion. We assume that in 1-immersions, adjacent edges do not cross each other and no edge crosses itself. Thus, every 3-cycle of a 1-immersed graph is embedded in the plane. Hence, given a 3-cycle of a 1-immersed graph, we can speak about its interior and exterior. We say that an embedded cycle separates two vertices x and y on the plane, if one of the vertices lies inside and the other one lies outside the cycle. Two edges e and inline image of a graph G separate vertices x and y of the graph if x and y belong to different connected components of the graph inline image.

Throughout the paper, we will deal with 1-immersed graphs. When an immersion of a graph G is clear from the context, we shall identify vertices, edges, cycles, and subgraphs of G with their image in inline image under the 1-immersion. Then by a face of a 1-immersion of G we mean any connected component of inline image.

By using Möbius transformations combined with homeomorphisms of the plane, it is always possible to exchange the interior and exterior of any embedded cycle and change any face of a given 1-immersion into the outer face of a 1-immersion. Formally, we have the following observation (which we will use without referring to it every time):

  1. Let C be a cycle of a graph G. If G has a 1-immersion φ in which C is embedded, then G has a 1-immersion inline image with the same number of crossings as φ, in which C is embedded and all vertices of G, which lie inside C in φ, lie outside C in inline image and vice versa.

Now we begin describing a family of MN-graphs based on the graph K3, 3.

By a link inline image connecting two vertices x and y we mean any of the graphs shown in Figure 1 where inline image. We say that the vertices x and y are incident with the link. The links in Figure 1(A) and (B) are called A-link and B-link, respectively, and the one in Figure 1(C) is called a base link. Every link has a free cycle: both 3-cycles in an A-link are its free cycles, while every B-link or base link has exactly one free cycle (the cycle indicated by thick lines in Fig. 1).

By an A-chain of length inline image we mean the graph shown in Figure 2(a). By a B-chain of length inline image we mean the graph shown in Figure 2(c) and, for inline image, every graph obtained from that graph in the following way: for some integers inline image, where inline image and inline image, we replace the link at the left of Figure 2(e) by the link shown at the right, for inline image. Note that, by definition, A- and B-chains have length at least 2. We say that the chains in Figure 2(a) and (c) connect the vertices v(0) and inline image which are called the end vertices of the chain. Two chains are adjacent if they share a common end vertex. A- and B-chains will be represented as shown in Figure 2(b) and (d), respectively, where the arrow points to the end vertex incident with the base link. The vertices inline image are the core vertices of the chains. Every free cycle of a link contains exactly one core vertex. The two edges of a free cycle C incident to the core vertex are the core-adjacent edges of C. It is easy to see that two edges e and inline image of a chain separate the end vertices of the chain if and only if the edges are the core-adjacent edges of a free cycle of a link of the chain.

image

Figure 2. A- and B-chains.

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By a subchain of a chain shown in Figure 2(a) and (c), we mean a subgraph of the chain consisting of links incident with inline image and inline image for all inline image for some inline image. We say that the subchain connects the vertices inline image and inline image.

A chain graph is any graph obtained from K3, 3 by replacing three of its edges incident with the same vertex by A-chains and three edges incident with another vertex in the same chromatic class by B-chains, where the chains can have arbitrary lengths ⩾2. These changes are to be made as shown in Figure 3(a). The vertices Ω(1), Ω(2), and Ω(3) are the base vertices of the chain graph. The edges joining the vertex Ω to the base vertices are called the Ω-edges.

image

Figure 3. A chain graph G and 1-planarity of the graph inline image.

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image

Figure 4. Eliminating the crossing between the edges e and f.

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We will show that every chain graph is an MN-graph.

Lemma 1. Let G be a chain graph and inline image. Then inline image is 1-planar.

Proof. If e is an Ω-edge, then inline image is planar and hence 1-planar. Suppose now that e is not an Ω-edge. By symmetry, we may assume that e belongs to an A- or B-chain incident to Ω(2). If e is the “middle” edge of a B-link, then Figure 3(b) shows that the corresponding B-chain can be crossed by an A-chain, and it is easy to see that this can be made into a 1-immersion of inline image. In all other cases, 1-immersions are made by crossing the link L whose edge e is deleted with the edges incident with the vertex Ω. The upper row in Figure 3(c) shows the cases when L is a base link. The lower row covers the cases when L is an A-link or a B-link. The edge e is shown in all cases as the dotted edge.

Our next goal is to show that chain graphs are not 1-planar. In what follows, we let G be a chain graph and φ a (hypothetical) 1-immersion of G.

Lemma 2. Let φ be a 1-immersion of a chain graph G such that the number of crossings in φ is minimal among all 1-immersions of G. If L is a link in an A- or B-chain of G, then no two edges of L cross each other in φ.

Proof. The first thing to observe is that whenever edges inline image and inline image cross, there is a disk D having inline image on its boundary, and D contains these two edges but no other points of G. In 1-immersions with minimum number of crossings, this implies that no other edges between the vertices inline image are crossed. Similarly, if inline image is a link in a chain, and an edge incident with z crosses an edge incident with inline image, the whole link L can be drawn in D without making any crossings. This shows that the only possible cases for a crossing of two edges inline image in L are the following ones, where we take the notation from Figure 1 and let u be the vertex of L that is not labeled in the figure:

  1. L is a B-link and inline image, inline image.
  2. L is a B-link and inline image, inline image.
  3. L is a base link and inline image, inline image.
  4. L is a base link and inline image, inline image.
  5. L is a base link and inline image, inline image.

Let D be a disk as discussed above corresponding to the crossing of e and f. In cases (b), (d), and (e), the 3-valent vertex u has all neighbors on the boundary of D, so the crossing between e and f can be eliminated by moving u inside D onto the other side of the edge e (see Fig. 4(a)).

It remains to consider cases (a) and (c). Observe that the boundary of D contains vertices inline image in this order and that u and v both have precisely one additional neighbor inline image outside of D. Therefore, we can turn φ into another 1-immersion of G by swapping u and v and only redraw the edges inside D (see Fig. 4(b)). However, this eliminates the crossing in D and yields a 1-immersion with fewer crossings, a contradiction.

Lemma 3. Let φ be a 1-immersion of a chain graph Gsuch that the number of crossings in φ is minimal among all 1-immersions of G. If Π and inline image are nonadjacent A- and B-chains, respectively, then in φ the following holds for every 3-cycle C of Π:

  1. The core vertices of inline image either all lie inside or all lie outside C.
  2. If all core vertices of inline image lie inside (resp. outside) C, then at most one vertex of inline image lies outside (resp. inside) C.

Proof. First we show (i). If C does not contain the vertex A, then every two core vertices of inline image are connected by four edge-disjoint paths not passing through the vertices of C, hence (i) holds for C.

Suppose now that C contains the vertex A and that core vertices of inline image lie inside and outside C. Then there is a link inline image of inline image such that the vertex z lies inside and the vertex inline image lies outside C. We may assume without loss of generality that Π and inline image are incident to the base vertices Ω(1) and Ω(2), respectively, and (taking (A) into account) that the vertex z (if inline image) separates the vertices B and inline image in inline image (see Fig. 5(a), where in inline image the dotted line indicates that the link has either edge inline image or inline image; also if inline image, then the link indicated in Figure 5(a) is a base link).

The 3-cycle C crosses at least two edges of inline image. The vertex z (resp. inline image) is connected to each of the vertices Ω(1) and Ω(3) (resp. to the vertex Ω(2)) by two edge-disjoint paths not passing through inline image or through the noncore vertices of inline image. Hence, Ω(1) and Ω(3) lie inside C (resp. Ω(2) lies outside C). It follows that the vertex Ω lies inside C and the edge inline image is the third edge that crosses C. We conclude that C crosses exactly two edges of inline image and the two edges separate z from inline image in inline image. Thus, the two edges are the core-adjacent edges of the free cycle of inline image. Hence, in φ, the link inline image is 1-immersed as shown in Figure 5(b), where the dotted edges indicate alternative possibilities for the position of z (at top) or inline image (at bottom).

Let inline image be the vertices of C different from A and let x be the fourth vertex of the link containing C. The vertex x is connected to Ω(1) by two edge-disjoint paths not passing through the vertices of C, hence x lies inside C. At most two vertices of C lie inside the free cycle of inline image. Suppose exactly one of v and inline image is inside the free cycle. If we are in the case of the bottom of Figure 5(b), then the path inline image cannot lie inside C, a contradiction. In the case of the top of Figure 5(b), if the path inline image lies inside C, then x must lie inside a 3-cycle Q of inline image incident to z, whereas A lies outside Q, a contradiction, since Q is not incident to B and in G there are two edge-disjoint paths connecting x to A and not passing through the vertices v, inline image, and the vertices of Q.

If either both v and inline image or none of them lie inside the free 4-cycle, then in the case of Figure 5(c) (resp. (d)), where we depict the two possible placements of the nonbase link inline image, there are two edge-disjoint paths of G connecting A and Ω(3) (resp. A and Ω(2)) and not passing through z (resp. inline image), a contradiction. Reasoning exactly in the same way, we also obtain a contradiction when inline image is a base link.

Now we prove (ii). By (A), we may assume that all core vertices of inline image lie inside C. By inspecting Figure 1, it is easy to check that for every link L, for every set W of noncore vertices of L such that inline image, there are at least four edges joining W with of inline image. Hence, if at least two noncore vertices belonging to the same link of inline image lie outside C, then at least four edges join them with the vertices of inline image lying inside C, a contradiction. Every noncore vertex of inline image has valence at least 3. Hence, if exactly n (inline image) noncore vertices of inline image lie outside C and if they all belong to different links, then at least inline image edges join them with the vertices of inline image lying inside C, a contradiction.

image

Figure 5. Cases in the proof of Lemma 3.

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Two chains cross if an edge of one crosses an edge of the other.

Lemma 4. Let φ be a 1-immersion of a chain graph G such that the number of crossings in φ is minimal among all 1-immersions of G. If Π and inline image are nonadjacent A- and B-chains, respectively, then Π does not cross inline image in φ.

Proof. Suppose, for a contradiction, that Π crosses inline image. Then an edge of a link inline image of inline image crosses a 3-cycle inline image of a link L of Π. Let inline image be a 3-cycle that is adjacent to C in L. (If L is not a base link, then inline image and inline image are the core vertices of L.) By Lemma 3, we may assume that all core vertices of inline image lie outside C and that exactly one vertex u of inline image lies inside C. The vertex u is 3-valent and is not a core vertex. The three edges incident with u cross all three edges of C, hence inline image does not cross C. If inline image lies inside C, then one of the three edges incident with u crosses C and inline image, a contradiction. If C lies inside inline image, then we consider the plane as the complex plane and apply the Möbius transformation inline image with the point a taken inside inline image but outside C. This yields a 1-immersion of G such that

  1. C lies outside inline image, inline image lies outside C, and exactly one vertex u of inline image lies inside C.

Therefore, we may assume that in φ we have (a). Since the three edges incident with u cross all three edges of C, at least two vertices of inline image lie outside inline image, hence, by Lemma 3, all core vertices of inline image lie outside C and inline image. Since the edge inline image in inline image is crossed by an edge of inline image, also inline image contains precisely one vertex inline image of inline image and inline image has degree 3 and is not a core vertex.

Adjacent trivalent vertices inline image cannot be contained in a base link. Therefore, inline image is not a base link. Let inline image be depicted as shown in Figure 6(a). Because of symmetry, we may assume that inline image and inline image, and that the crossings are as shown in Figure 6(b) and (c).

In the case of Figure 6(b), the adjacent vertices z and inline image of inline image are separated by the 3-cycle inline image, whose edges are crossed by three edges different from the edge inline image, a contradiction.

Consider the case in Figure 6(c). If x and inline image are core vertices of Π, then they are separated by the 3-cycle inline image of inline image, a contradiction, since there are four edge-disjoint paths between x and inline image that avoid this 3-cycle (the 3-cycle does not contain the vertex B).

Suppose that x and inline image are not two core vertices. This is possible only when L is a base link. The 3-cycle inline image of inline image crosses the three edges joining the vertex inline image of L with three vertices x, v, and inline image of L. The fifth vertex of L is adjacent to at least one vertex from x, v, and inline image, hence it lies outside inline image and is not adjacent to inline image. Thus, inline image has valence 3 in L. If inline image is a core vertex, then, since the 3-cycle inline image does not contain the vertex B, inline image is connected to one of the vertices x, v, and inline image by a path passing through B and not passing through the vertices of inline image, a contradiction. Hence, inline image is not a core vertex. The link L has exactly one noncore vertex inline image of valence 3 and the vertices x and inline image are adjacent. Hence, the 3-cycle inline image separates two core vertices z and inline image of inline image, a contradiction, since z and inline image are connected by a path not passing through the vertices of the 3-cycle inline image.

Theorem 1. Every chain graph is an MN-graph.

Proof. Let G be a chain graph. By Lemma 1, it suffices to prove that G is not 1-planar. Consider, for a contradiction, a 1-immersion φ of G and suppose that φ has minimum number of crossings among all 1-immersions of G.

image

Figure 6. Cases in the proof of Lemma 4.

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We know by Lemma 4 that nonadjacent chains do not cross each other. In the sequel, we will consider possible ways that the Ω-edges cross with one of the chains. Let us first show that such a crossing is inevitable.

Claim 1. At least one of the chains contains a link inline image such that every inline image-path in L is crossed by an Ω-edge.

Proof. Suppose that for every link inline image, an inline image-path in L is not crossed by any Ω-edge. Then every chain contains a path joining its end vertices that is not crossed by the Ω-edges. All six such paths plus the Ω-edges form a subgraph of G that is homeomorphic to K3, 3. By Lemma 4, the only crossings between subdivided edges of this K3, 3-subgraph are among adjacent paths. However, it is easy to eliminate crossings between adjacent paths and obtain an embedding of K3, 3 in the plane. This contradiction completes the proof of the claim.

Let inline image be a link in an A- or B-chain Π whose inline image-paths are all crossed by the Ω-edges. We may assume that L is contained in a chain connecting the vertex Ω(1) with A or B and that x separates y and Ω(1) in Π. By Lemma 2, the induced 1-immersion of L is an embedding. The vertex Ω lies inside a face of L and all Ω-edges that cross L cross the edges of the boundary of the face. Considering the possible embeddings of L, it is easy to see that all inline image-paths are crossed by Ω-edges only in the case when Ω lies inside a face of L whose boundary contains two core-adjacent edges of a free cycle C of L, and two Ω-edges cross the two core-adjacent edges. By (A), we may assume that Ω lies inside C.

If C is a k-cycle, inline image, then L has another cycle inline image that shares with C exactly inline image edges and contains a core vertex not belonging to C. If C lies inside inline image, then we consider the plane as the complex plane and apply the Möbius transformation inline image with the point a taken inside inline image but outside C. This yields a 1-immersion of G such that C does not lie inside inline image and Ω lies inside C. Hence, we may assume that C does not lie inside inline image, that Ω lies inside C and two Ω-edges h and inline image cross two core-adjacent edges of C. Note that any two among the vertices inline image are joined by four edge-disjoint paths not using any edges in the chain Π containing L. Therefore, these four vertices of G are all immersed in the same face of L.

Let the Ω-edges h and inline image join the vertex Ω with basic vertices inline image and inline image, respectively. If the third basic vertex inline image is Ω(2) or Ω(3), then Ω(2) and Ω(3) lie inside different faces of L, a contradiction, hence inline image.

The vertex Ω(1) is connected to one of the vertices A and B by two edge-disjoint paths, not passing the vertices of C. Hence, if C is a 3-cycle, then Ω(1) is not inside C.

Now the embeddings of possible links L (so that we can join the vertices Ω and Ω(1) by an edge not violating the 1-planarity) are shown in Figure 7.

image

Figure 7. The Ω-edges crossing a link.

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Let us now consider particular cases (a)–(f) of Figure 7.

  1. In this case, L is a base link and inline image. Consider two edge-disjoint paths in a chain inline image joining Ω(1) with the vertex A or B, which is not incident with Π. These paths must cross the edges e and f indicated in the figure. Let a be the edge crossing e and b be the edge crossing f. It is easy to see that a and b cannot be both incident with Ω(1) since Ω(1) is incident with three edges of the base link in the chain inline image. The 1-planarity implies that the edges inline image and the vertex Ω(1) separate the graph G. Therefore, a and b are core-adjacent edges of a link in the chain inline image. If the edge g (shown in the figure) is crossed by an edge c of inline image, then also inline image and Ω(1) separate the graph. Thus inline image would be core-adjacent edges in a link in inline image as well, a contradiction. But if g is not crossed and a is not incident with Ω(1), then the edge a and the vertex Ω(1) separate G, a contradiction. If a is incident with Ω(1), then b is not (as proved above). Now we get a contradiction by considering the separation of G by the edge b and the vertex Ω(1).
  2. In this case, L is a B-link, the cycle inline image lies inside C and Ω(1) is inside C but outside inline image (see Fig. 7(b)). In the figure, the vertex labeled x is actually x and not y because the vertex is not B and is connected to Ω(1) by two edge-disjoint paths not passing through the vertices of C, and y is connected to the vertex B (lying outside C) by two edge-disjoint paths not passing through the vertices of C. Again, consider two edge-disjoint paths in the A-chain inline image joining Ω(1) with the vertex A. Their edges a and b (say) must cross the edges e and f, respectively. As in the proof of case (a), we see that the edges a, b and the vertex Ω(1) separate G, thus a and b are core-adjacent edges of a free cycle of a link of inline image. This free cycle has length 3 and separates x from Ω(1), a contradiction, since there is a B-subchain from x to Ω(1) disjoint from the free 3-cycle and not containing the edges of L incident with x.
  3. In this case, L is a B-link, the cycle inline image lies outside C and Ω(1) is inside C. The vertex Ω(1) is connected by four edge-disjoint paths with vertices x and A such that the paths do not pass through noncore vertices of L, a contradiction.
  4. In these cases, L is a B-link and Ω(1) lies inside a 3-cycle of L. Two edge-disjoint paths from Ω(1) to A cross the edges e and f of L. One of the paths also crosses the edge g. Then the edges crossing e and g separate G, a contradiction, since G is 3-edge-connected.
  5. In this case, L is an A-link and the chain Π is joining Ω(1) with the vertex A. Again, consider two edge-disjoint paths in the B-chain inline image joining Ω(1) with B. Their edges a and b (say) must cross the edges e and f, respectively. As in the proof of case (a), we see that the edges inline image and the vertex Ω(1) separate G, thus inline image are core-adjacent edges in a link inline image in inline image. Let z be the core vertex of inline image incident with the edges inline image and inline image. Note that in inline image, vertices inline image have two common neighbors, a vertex r of degree 3 in G and the core vertex inline image, and that inline image is adjacent to r. Inside L, we have the subchain inline image of the A-chain Π connecting x with Ω(1). Hence, the free cycle of inline image containing the edges a and b must have length at least 4 (i.e., inline image is not a base link) and z is immersed outside L, while inline image are inside. The subchain inline image has two edge-disjoint paths that are crossed by the paths inline image and inline image. Each of the paths inline image and inline image crosses core-adjacent edges in Π since the crossed edges and Ω(1) separate G. Thus, they enter faces of these links that are bounded by free cycles of the links. However, the edge inline image would need to cross one edge of each of these two free cycles, hence the two free cycles are adjacent, that is, they are the two free cycles of an A-link of Π. Then the vertex inline image lies inside one of the two free cycles and the free cycle separates inline image from Ω(1), a contradiction, since inline image is connected with Ω(1) by a B-subchain.

The following theorem shows how chain graphs can be used to construct exponentially many nonisomorphic MN-graphs of order n.

Theorem 2. For every integer inline image, there are at least inline image nonisomorphic MN-graphs of order n.

Proof. The A-chain of length t has inline image vertices and a B-chain of length t has inline image vertices. Consider a chain graph whose three A-chains have length 2, 2, and inline image, respectively, and whose B-chains have length 2, 3, and inline image, respectively. The graph has inline image vertices. Applying the modification shown in Figure 2(e) to links of the two B-chains of the graph, which have length at least 3, we obtain inline image nonisomorphic chain graphs of order inline image, where inline image and inline image. We claim that for every integer inline image, there are integers inline image and inline image such that inline image. Indeed, if inline image, put inline image, respectively. If inline image, where inline image, then inline image. Hence, there are at least inline image nonisomorphic chain graphs of order inline image. Since every chain graph is an MN-graph, the theorem follows.

3. PN-GRAPHS

  1. Top of page
  2. Abstract
  3. 1. INTRODUCTION
  4. 2. CHAIN GRAPHS BASED ON K3, 3
  5. 3. PN-GRAPHS
  6. 4. MN-GRAPHS BASED ON PN-GRAPHS
  7. 5. TESTING 1-IMMERSIBILITY IS HARD
  8. 6. k-PLANARITY TESTING FOR MULTIGRAPHS
  9. ACKNOWLEDGMENTS
  10. REFERENCES

By a proper 1-immersion of a graph, we mean a 1-immersion with at least one crossing point. Let us recall that a PN-graph is a planar graph that does not have proper 1-immersions. In this section, we describe a class of PN-graphs and construct some graphs of the class. They will be used in Section 'MN-GRAPHS BASED ON PN-GRAPHS' to construct MN-graphs.

For every cycle C of G, denote by inline image the set of all vertices of the graph not belonging to C but adjacent to C. Two disjoint edges inline image and inline image of a graph G are paired if the four vertices inline image are all four vertices of two adjacent 3-cycles (two cycles are adjacent if they share an edge).

Following Tutte, we call a cycle C of a graph G peripheral if it is an induced cycle in G and inline image is connected. If G is 3-connected and planar, then the face boundaries in its (combinatorially unique) embedding in the plane are precisely its peripheral cycles.

Theorem 3. Suppose that a 3-connected planar graph G satisfies the following conditions:

  1. Every vertex has degree at least 4 and at most 6.
  2. Every edge belongs to at least one 3-cycle.
  3. Every 3-cycle is peripheral (in other words, there are no separating 3-cycles).
  4. Every 3-cycle is adjacent to at most one other 3-cycle.
  5. No vertex belongs to three mutually edge-disjoint 3-cycles.
  6. Every 4-cycle is either peripheral or is the boundary of two adjacent triangular faces (this means that there are no separating 4-cycles).
  7. For every 3-cycle C, any two vertices of inline image are connected by four edge-disjoint paths not passing through the vertices of C.
  8. If an edge inline image of a nontriangular peripheral cycle C is paired with an edge inline image of a nontriangular peripheral cycle inline image, then:
    1. C and inline image have no vertices in common;
    2. any two vertices a and inline image of C and inline image, respectively, such that inline image are nonadjacent and are not connected by a path inline image of length 2, where b does not belong to C and inline image.
  9. G does not contain the subgraphs shown in Figure 8 (in this figure, 4-valent (resp. 5-valent) vertices of G are encircled (resp. encircled twice) and the two starred vertices can be the same vertex).

Then G has no proper 1-immersion.

Proof. Denote by f, the unique plane embedding of G. Suppose, for a contradiction, that there is a proper 1-immersion φ of G. Below we consider the 1-immersion and show that then G has a subgraph that is excluded by (C8) and (C9), thereby obtaining a contradiction. In the figures below, the encircled letter f (resp. φ) at the top left of a figure means that the figure shows a fragment of the plane embedding f (resp. 1-immersion φ).

Lemma 5. In φ, there is a 3-cycle such that there is a vertex inside and a vertex outside the cycle.

Proof. The 1-immersion φ has crossing edges e and inline image. By (C2), the crossing edges belong to different 3-cycles. If the 3-cycles are nonadjacent, then we apply the following obvious observation:

  1. If two nonadjacent 3-cycles D and inline image cross each other, then there is a vertex of D inside and outside inline image.

If inline image and inline image belong to adjacent 3-cycles inline image and inline image, respectively (see Fig. 9), then, by (C4), there are nontriangular peripheral cycles C and inline image containing e and inline image, respectively. The cycles C and inline image intersect at some point δ different from the intersection point of edges e and inline image. By (C8)(i), the two cycles do not have a common vertex, hence δ is the intersection point of two edges. By (C2), these two edges belong to some 3-cycles, D and inline image. Property (C8)(ii) implies that D and inline image are nonadjacent 3-cycles. By (a), the proof is complete.

Lemma 6. If inline image is a 3-cycle such that there is a vertex inside and a vertex outside C, then there is only one vertex inside C or only one vertex outside C, and this vertex belongs to inline image.

Proof. By (C7), we may assume that all vertices of inline image lie outside C. Then there can be only vertices of inline image inside C. To prove the lemma, it suffices to show the following:

  1. For every inline image, inline image, at least four edges join vertices of Q to vertices in inline image.

    By (C1), every vertex of Q has valence at least 4. By (C4), every vertex of inline image is adjacent to at most two vertices of C. We claim that if a vertex inline image is adjacent to two vertices u1 and u2 of C, then v is not adjacent to other vertices of inline image. Suppose, for a contradiction, that v is adjacent to a vertex inline image. Then, by (C4), the vertex w can be adjacent only to u3 and the 4-cycle inline image is not the boundary of two adjacent 3-cycles, hence, by (C6), the 4-cycle is peripheral. Then, by (C3), any two of the three edges inline image, inline image, and inline image are two edges of a peripheral cycle, hence u2 has valence 3, contrary to (C1).

    Now to prove (a), it suffices to prove the following claim:

  2. For every inline image, inline image (resp. inline image), such that every vertex of Q is adjacent to exactly one vertex of C, at least 2 (resp. 4) edges join vertices of Q to vertices of inline image.

The claim is obvious for inline image. For inline image, it suffices to show that the three vertices of Q are not all pairwise adjacent. Suppose, for a contradiction, that the vertices v1, v2, and v3 of Q are pairwise adjacent. Then, by (C4), the three vertices of Q are not adjacent to the same vertex of C. Let v1 and v2 be adjacent to the vertices u1 and u2 of C, respectively. Then any two of the edges inline image, inline image, and inline image are two edges of a 3-cycle (peripheral cycle) or a 4-cycle that is not the boundary of two adjacent 3-cycles, so by (C6), that 4-cycle is also peripheral. Hence, v1 has valence 3, contrary to (C1).

For inline image, it suffices to show that no vertex of Q is adjacent to three other vertices in Q. Suppose, for a contradiction, that inline image is adjacent to inline image. If v is adjacent to u1, then the edge inline image belongs to three cycles D1, D2, and D3 such that for inline image, the cycle inline image contains edges inline image and inline image, has length 3 or 4, and if inline image has length 4, then inline image is not the boundary of two adjacent 3-cycles. By (C3) and (C6), these three cycles are peripheral. This contradiction completes the proof of (b).

Suppose that a vertex h belongs to two adjacent 3-cycles inline image and inline image. Since inline image, h is adjacent to a vertex inline image. By (C2), the edge inline image belongs to a 3-cycle inline image. By (C4), inline image. Hence, we have the following:

  1. If an edge e is contained in two 3-cycles of G, then both end vertices of e have valence at least 5.

In the remainder of the proof of Theorem 3, we will show that any two crossing edges of the proper 1-immersion φ belong to a subgraph that is excluded by (C8) and (C9).

By Lemma 5, there is a 3-cycle inline image such that there is a vertex inside and a vertex outside C. By Lemma 6, there is only one vertex v inside C and v is adjacent to x.

Now we show that there is a 3-cycle inline image disjoint from C. Let inline image (inline image) be all vertices adjacent to v. Suppose there is a 3-cycle inline image, where inline image. If inline image, then the 3-cycle inline image is adjacent to two 3-cycles C and D, contrary to (C4). Hence, inline image. By (C4), at most two vertices of inline image are adjacent to x. Hence, there is a vertex inline image such that a 3-cycle B containing the edge inline image is disjoint from C.

Since there is only one vertex v inside C, exactly two edges of B cross edges of C. First, suppose that B separates x from y and z (Fig. 10(a)). By (C2), the edge inline image belongs to a 3-cycle inline image. If inline image, then two of the vertices inline image lie inside R and the other two vertices lie outside R (see Fig. 10(b)), contrary to Lemma 6. So, we may assume, without loss of generality, that inline image (Fig. 10(c)). Then the vertex x belongs to two adjacent 3-cycles, C and inline image, hence, by (B), inline image. By (C4), x is not adjacent to u or w. Since x is the only vertex inside B, x has valence at most 4, a contradiction. Hence, B cannot separate x from y and z.

Now suppose that B separates the vertices y and z, and, without loss of generality, let z lie inside B (Fig. 11(a)). If z is adjacent to v, then, by (B), z has valence at least 5, hence z is adjacent to a vertex inline image. But then the 3-cycle inline image is adjacent to two 3-cycles, C and inline image, contrary to (C4). If z is adjacent to u and w, then the edge inline image belongs to three peripheral cycles, C, inline image, and inline image, a contradiction, since every edge belongs to at most two peripheral cycles. Hence, since z is the only vertex inside B, z has valence 4, z is not adjacent to v and is adjacent to exactly one of the vertices u and w. The vertices u and w are not symmetric in Figure 11(a), so we have to consider two cases.

Case 1. The vertex z is adjacent to u (Fig. 11(b)).

Consider the vertex v. If v is adjacent to y, then (see Fig. 11(b)) x is incident with exactly three peripheral cycles and has valence 3, a contradiction. Hence, v is not adjacent to y and since v is the only vertex inside C, v has valence 4 (see Fig. 11(c)). By (C2), we obtain a subgraph shown in Figure 11(d). By (C9), at least one of the vertices u and x, say u, is not 4-valent. Then, by (C5), at least one of the edges inline image and inline image belongs to two 3-cycles. Here we have two subcases to consider.

Subcase 1.1. The edge inline image belongs to two 3-cycles inline image and inline image (Fig. 12(a)).

Now, a is the only vertex inside the 3-cycle inline image (Fig. 12(b)). By (C4), a is nonadjacent to w and b, hence a has valence 4. The edges inline image and inline image (see Fig. 12(a)) belong to a nontriangular peripheral cycle inline image The edge inline image belongs to a 3-cycle inline image and b is the only vertex inside the 3-cycle inline image. The vertex b is not adjacent to c, since the 4-cycle inline image cannot be peripheral (see Fig. 12(a)). Since b has valence at least 4, b is adjacent to d and has valence 4. The 4-cycle inline image is peripheral, so we obtain a subgraph of G shown in Figure 12(c). Note that, by (C3)–(C6), the vertex at the top of Figure 12(c) is different from all other vertices shown in the figure. This contradicts (C9).

Subcase 1.2. The edge inline image belongs to two 3-cycles inline image and inline image (Fig. 13(a)).

Since b has valence at least 4, b lies outside the 3-cycle inline image (Fig. 13(b)). The edges inline image and inline image (resp. inline image and inline image) belong to a nontriangular peripheral cycle C1 (resp. C2). The cycles C1 and C2 are paired. In φ, the crossing point of edges inline image and inline image is an intersection point of C1 and C2. The cycles C1 and C2 have at least one other crossing point, denote this intersection point by δ. By (C8)(i), C1 and C2 are vertex disjoints, hence, δ is the crossing point of C1 and an edge inline image of C2 (Fig. 13(c)). The edge inline image is not the edge inline image and belongs to a 3-cycle inline image. If h3 belongs to C2, then in the embedding f the edge inline image is a chord of the embedded peripheral cycle C2 and thus inline image is a separating vertex set of G. But G is 3-connected, a contradiction. Hence, h3 does not belong to C2.

Now suppose that inline image. We have inline image. By (C8)(ii), inline image, a is not adjacent to h2 and h3, and C1 does not pass through h3 (i.e., h3 does not belong to C1 and C2). Hence, a vertex s of C1 lies inside the 3-cycle inline image (see Fig. 13(c)). By (B), inline image. If inline image, then, since s is the only vertex inside the 3-cycle inline image, a is adjacent to at least one of h2 and h3, a contradiction. Hence, inline image. Since z is the only vertex inside the 3-cycle B, and the 3-cycle inline image is not B (since inline image), we have inline image. Now, since s has valence at least 4, s is adjacent to at least one of the vertices h1, h2, and h3, contrary to (C8)(ii). Hence, inline image and inline image.

We have inline image and the edge inline image belongs to a 3-cycle inline image (see Fig. 13(d) and (e)). Considering Figure 13(e), if there is a vertex inside the 3-cycle inline image, the vertex has valence at most 3, a contradiction. Hence, no edge crosses the edge inline image. If h lies inside the 3-cycle inline image, then, by (C4), h is not adjacent to b and u, h has valence at most 3, a contradiction. Hence, h lies outside the 3-cycle inline image and b is the only vertex inside the 3-cycle inline image (see Fig. 13(e)). Note that the edge inline image belongs to C1, hence C1 does not cross both inline image and inline image.

By (C4), b is not adjacent to d and h, hence b has valence 4 and belongs to a 3-cycle inline image disjoint from inline image (Fig. 13(f)), where inline image. Now d is the only vertex inside the 3-cycle inline image (Fig. 13(e)). By (C8)(ii), d is not adjacent to t. Since d has valence at least 4, d is adjacent to inline image and has valence 4. The 4-cycle inline image is peripheral. We obtain a subgraph of G shown in Figure 13(g), contrary to (C9). The obtained contradiction completes the proof in the Case 1.

Case 2. The vertex z is adjacent to w.

This case is dealt in much the same way as Case 1. Here, we describe only what figures will be in Case 2 instead of the Figures 1113 in Case 1. We hope that the reader is familiar enough with the proof of Case 1 to supply the missing details himself.

In Figures 11(a) and (d) and 12(a) and (c), interchange the letters u and w. In Figures 11(c) and 12(b), replace the edge inline image by the edge inline image. In Figures 13(a), (d), (f), and (g) interchange the letters u and w. Figures 13(b), (c), and (e) are replaced by the Figures 14(a), (b), and (c), respectively.

Denote by inline image the class of all 3-connected plane graphs G satisfying the conditions (C1)–(C9) of Theorem 3. In what follows, we show how to construct some graphs in inline image and, as an example, we shall give two infinite families of graphs in inline image, one of which will be used in Section 'MN-GRAPHS BASED ON PN-GRAPHS' to construct MN-graphs.

First, we describe a large family of 3-connected plane graphs satisfying the conditions (C1)–(C6) and (C8) of Theorem 3.

Given a 4-valent vertex v of a 3-connected plane graph, two peripheral cycles C and inline image containing v are opposite peripheral cycles at v if C and inline image have no edges incident with v in common.

Denote by inline image the class of all 3-connected (simple) planar graphs H satisfying the following conditions (H1)–(H4):

  • (H1) Every vertex has valence 3 or 4.
  • (H2) H has no 3-cycles.
  • (H3) Every 4-cycle is peripheral.
  • (H4) For every 4-valent vertex v and for any two opposite peripheral cycles C and inline image at v, no edge joins a vertex of inline image to a vertex of inline image.

A plane graph G is a medial extension of a graph inline image if G is obtained from H in the following way. The vertex set of G is the set inline image. The edge set of G is defined as follows. For every 3-valent vertex v of H, if inline image are the edges incident with v, then in G the vertices inline image, inline image, and inline image are pairwise adjacent (the three edges of G are said to be associated with v). For every 4-valent vertex w of H, if inline image is the cyclic order of edges incident to w around w in the plane, then G contains the edges of the 4-cycle inline image, and contains either the edge inline image or the edge inline image; these five edges of G are said to be associated with the 4-valent vertex of H. Note that G can be obtained from the medial graph of H by adding a diagonal to every 4-cycle associated with a 4-valent vertex of H.

Lemma 7. Every medial extension G of any graph inline image is a 3-connected planar graph satisfying the conditions (C1)–(C6) and (C8) of Theorem 3.

Proof. By the construction of G, if inline image is a separating vertex set of G, then the graph inline image is disconnected, a contradiction, since H is 3-connected. Hence, G is 3-connected. Every peripheral cycle of H induces a peripheral cycle of G. It is easy to see that all peripheral cycles of G that are not induced by the peripheral cycles of H are the 3-cycles formed by the edges associated with the vertices of H.

It is easy to see that G satisfies (C1)–(C5). To show that G satisfies (C6), let inline image be a 4-cycle of G. By the construction of G, if vertices inline image and inline image of G are adjacent, then the edges e and inline image of H are adjacent, too. Since in H no three edges among e1, e2, e3, e4 form a cycle (by (H2)), these four edges either form a 4-cycle (in this case J is peripheral) or are the edges incident to a 4-valent vertex of H (in this case, by the construction of G, J is the boundary of two adjacent faces of G). Hence, G satisfies (C6).

It remains to show that G satisfies (C8). Let C and inline image be nontriangular peripheral cycles of G such that an edge a of C is paired with an edge inline image of inline image. Then, by the construction of G, the peripheral cycles C and inline image are induced by peripheral cycles inline image and inline image of H, respectively, which are opposite at some 4-valent vertex u. If C and inline image have a common vertex inline image, then inline image and inline image have a common edge e, hence H has a separating vertex set inline image, where w is a vertex incident to e, a contradiction, since H is 3-connected. Now suppose that G has an edge joining a vertex inline image of C to a vertex inline image of inline image such that at least one of the vertices inline image and inline image is not incident to a and inline image. Then the edges e and inline image are incident to the same vertex w of H and the cycles inline image and inline image pass through w. It follows that inline image is a separating vertex set of H, a contradiction. Next suppose that G has a path inline image connecting a vertex inline image of C to a vertex inline image of inline image such that inline image does not belong to C and inline image, and at least one of the vertices inline image and inline image is not incident to a or inline image. If in H the edges e, b, and inline image are incident to the same vertex w, then inline image is a separating vertex set of H, a contradiction. If in H the edges a and b (resp. b and inline image) are incident to a vertex w (resp. inline image) such that inline image, then the edge b joins the vertex w of inline image with the vertex inline image of inline image, contrary to (H4). Hence, G satisfies (C8). The proof is complete.

image

Figure 8. Forbidden subgraphs.

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image

Figure 9. Crossing two adjacent 3-cycles.

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image

Figure 10. The 3-cycle B separates x from y and z.

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image

Figure 11. The 3-cycle B separates y and z.

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image

Figure 12. The edge inline image belongs to two 3-cycles.

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image

Figure 13. The edge inline image belongs to two 3-cycles.

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There are medial extensions of graphs in inline image that do not satisfy conditions (C7) and (C9). In the sequel we shall describe a way to verify the conditions (C7), and henceforth give examples of graphs satisfying (C1)–(C9). To show that a medial extension G of inline image satisfies (C7), it is convenient to proceed in the following way. Subdivide every edge e of H by a two-valent vertex inline image of G. We obtain a graph inline image whose vertex set is inline image where the vertices of inline image are all 2-valent vertices of inline image. We will consider paths of G associated with paths of inline image connecting 2-valent vertices.

Two paths P and inline image of inline image are H-disjoint if inline image, that is, inline image.

Consider a path inline image in inline image where inline image are the H-vertices on P. It is easy to see that the edges of G associated with the vertices inline image of H contain two edge-disjoint paths connecting in G the vertices inline image and inline image (see Fig. 15); any two such paths in G are said to be associated with the path P of inline image. Since H has no multiple edges, every edge of G is associated with exactly one vertex of H. Hence, if P and inline image are H-disjoint paths in inline image, each of which is connecting 2-valent vertices, then every path in G associated with P is edge-disjoint from every path in G associated with inline image. As a consequence, we have the following conclusion:

  1. If inline image has a cycle containing 2-valent vertices inline image and inline image, then G has four edge-disjoint paths connecting inline image and inline image.

The fact that a path in H gives rise to two edge-disjoint paths in G (paths associated with the path of H) can be used to check the property (C7) of G.

For a 3-cycle C of G, a path of inline image is C-independent if the path does not contain vertices of C. When checking (C7) for a medial extension G of inline image, given a 3-cycle C of G and two 2-valent vertices inline image, four C-independent edge-disjoint paths P1, P2, P3, and P4 of G connecting x and y in G will be represented in some subsequent figures (see, e.g., Fig. 17) in the following way. The edges of the paths incident to vertices of inline image are depicted as dashed edges joining 2-valent vertices, the dashed edges are not edges of inline image (see, e.g., Figure 17(b), where the edges of H are given as solid lines). All other edges of the paths are represented by paths in inline image. If X is a subpath of inline image such that X is associated with a path inline image in inline image, then X is represented by a dashed line passing near the edges of inline image in inline image. If inline image and inline image are subpaths of inline image and inline image (where possibly inline image), respectively, such that inline image and inline image are edge-disjoint paths associated with a path inline image of inline image, then inline image and inline image are represented by two (parallel) dashed lines passing near the edges of inline image in inline image. Using these conventions, the reader will be able to check that the depicted dashed paths and edges in the figure of inline image represent four C-independent edge-disjoint paths of G connecting x and y.

Now we describe some graphs in inline image. Let us recall that graphs in inline image are precisely those 3-connected planar graphs that satisfy conditions (C1)–(C9). To simplify the arguments, we construct graphs with lots of symmetries so that, for example, to check the condition (C7) we will have to consider only two 3-cycles of a graph.

For inline image, let inline image be the Cartesian product of the path P3 of length 2 and the cycle inline image of length n (see 16(a)). Let inline image be the medial extension of inline image shown in Figure 16(b).

Lemma 8. Each graph inline image, inline image, is a PN-graph.

Proof. We show that inline image satisfies (C1)–(C9) for every inline image.

By Lemma 7, inline image satisfies (C1)–(C6) and (C8). Every 4-gonal face of inline image is incident with a 6-valent vertex and inline image has no 5-valent vertices, hence inline image satisfies (C9).

Now we show that inline image satisfies (C7). Consider a fragment of inline image shown in Figure 17(a) (in the figure we introduce notation for some vertices and also depict in dashed lines some edges of inline image). Because of the symmetries of inline image, it suffices to consider the following two cases for a 3-cycle C, when checking (C7):

Case 1. inline image.

Then inline image. If we delete from inline image the vertices 1, 2, …, 14, then the obtained graph has only one connected component U with a vertex in inline image, and U is 2-connected. Hence, any two 2-valent vertices x and y of U belong to a cycle in U and then, by (C), inline image has four C-independent edge-disjoint paths connecting x and y. Figure 17(b) shows four C-independent edge-disjoint paths in inline image connecting vertices 1 and 4. If we delete from inline image the vertices inline image, then in the resulting nontrivial connected component, for every vertex inline image, there is a path P connecting the vertices 1 and 4, and passing through x; combining two edge-disjoint paths of inline image associated with P and the two edge-disjoint paths connecting 1 and 4, shown in Figure 17(b), we obtain four C-independent edge-disjoint paths connecting the vertices 4 and x (and, analogously, for the vertices 1 and x). Now, because of the symmetries of inline image, it remains to show that there are four C-independent edge-disjoint paths connecting the vertex 4 with each of the vertices 15, 16, 17, 18; Figure 17(c) shows the paths (since inline image).

Case 2. inline image.

We have inline image. If we delete from inline image the vertices 1, 2, …, 7 and 9, 10, 13, then the obtained graph has only one connected component U with at least two vertices and U is 2-connected. Hence any two vertices x and y in inline image that are both in U belong to a cycle of U and then, by (C), inline image has four C-independent edge-disjoint paths connecting x and y. It remains to show that for every vertex x of inline image belonging to U, there are four C-independent edge-disjoint paths connecting x and the vertex 13. These four paths are shown in Figures 17(d)–(f), depending on the choice of x. We conclude that inline image satisfies (C1)–(C9), hence inline image is a PN-graph for every inline image.

image

Figure 14. The edge inline image belongs to two 3-cycles.

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image

Figure 15. Two paths of G associated with a path of inline image.

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image

Figure 16. The graph inline image and its extension inline image.

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image

Figure 17. Verifying (C7) for the graph inline image.

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Figure 18 gives another example of an extension inline image of a graph inline image. By Lemma 7, inline image satisfies (C1)–(C6) and (C8). Since inline image has no 4-cycles, it satisfies (C9). Using the symmetry of inline image, one can easily check that for every 3-cycle C of inline image, if we delete from inline image the vertices inline image of inline image, then the obtained graph has only one connected component U with at least two vertices and U is 2-connected. Then, by (C), any two vertices of inline image in U are connected by four C-independent edge-disjoint paths. Hence, inline image satisfies (C7) and is a PN-graph.

image

Figure 18. The graph inline image and its extension inline image.

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4. MN-GRAPHS BASED ON PN-GRAPHS

  1. Top of page
  2. Abstract
  3. 1. INTRODUCTION
  4. 2. CHAIN GRAPHS BASED ON K3, 3
  5. 3. PN-GRAPHS
  6. 4. MN-GRAPHS BASED ON PN-GRAPHS
  7. 5. TESTING 1-IMMERSIBILITY IS HARD
  8. 6. k-PLANARITY TESTING FOR MULTIGRAPHS
  9. ACKNOWLEDGMENTS
  10. REFERENCES

In this section, we construct MN-graphs based on the PN-graphs inline image described in Section 'PN-GRAPHS'.

For inline image, denote by inline image, the graph shown in Figure 19. The graph has inline image disjoint cycles of length inline image labeled by inline image as shown in the figure. The vertices of B0 are called the central vertices of inline image and are labeled by inline image (see Fig. 19). For every central vertex inline image, denote by inline image its “opposite” vertex inline image if inline image and the vertex inline image if inline image. In inline image, any pair inline image of central vertices is connected by a central path inline image of length inline image with inline image two-valent vertices. There are exactly inline image central paths.

image

Figure 19. The graph inline image.

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For any integers inline image and inline image, denote by inline image the set of all inline image-tuples inline image of nonnegative integers such that inline image. For every inline image, denote by inline image the graph obtained from inline image by replacing, for every central vertex inline image, the eight edges marked by short crossings in Figure 20(a) by inline image new edges marked by crossings in Figure 20(b) (the value inline image in that figure is to be considered modulo inline image). The graph inline image has inline image   inline image-cycles inline image and three inline image-cycles inline image.

image

Figure 20. Obtaining the graph inline image.

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We want to show that for every inline image and for every inline image, inline image, the graph inline image is an MN-graph.

Lemma 9. For every inline image, inline image and inline image, the graph inline image is 1-planar for every edge e.

Proof. If we delete an edge of a central path, then the remaining inline image central paths, each with inline image edges, can be 1-immersed inside B0 in Figure 19. If we delete one of the edges shown in Figure 21(a) by a thick line, then the central path inline image can be drawn outside B0 with inline image crossing points as shown in the figure and then the remaining inline image central paths can be 1-immersed inside B0. If we delete one of the two edges depicted in Figure 21(a) by a dotted line, then Figure 21(b) shows how to place the central vertex x so that the path inline image can be drawn outside B0 with inline image crossing points (analogously to Fig. 21(a)) and then the remaining inline image central paths can be 1-immersed inside B0. This exhibits all possibilities for the edge e (up to symmetries of inline image) and henceforth completes the proof.

image

Figure 21. The central path inline image immersed outside B0.

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Given a 1-immersion of a graph G and an embedded cycle C, we say that G lies inside (resp. outside) C, if the exterior (resp. interior) of C does not contain vertices and edges of G.

Denote by inline image, the graph obtained from the graph inline image in Figure 19 by deleting the 2-valent vertices of all central paths and all vertices lying outside the cycle inline image.

Lemma 10. For every inline image, inline image is a PN-graph.

Proof. The graph inline image contains inline image subgraphs inline image isomorphic to the PN-graph inline image such that for inline image, the graph inline image contains the cycles inline image, inline image, and inline image. Consider an arbitrary 1-immersion φ of inline image. Suppose that in the plane embedding of the PN-graph L1 in φ, the cycle B2 is the boundary cycle of the outer inline image-gonal face of the embedding. Then the embedding of L1 determines an embedding of the subgraph of L2 bounded by the cycles B1 and B2. Since L2 is a PN-graph, the subgraph of L2 bounded by B2 and B3 lies outside the cycle B2. Reasoning similarly, we obtain that for inline image, the subgraph of the PN-graph inline image bounded by inline image and inline image lies outside inline image. As a result, φ is a plane embedding of inline image, hence inline image is a PN-graph.

Denote by inline image the graph obtained from inline image, where inline image and inline image, by deleting the 2-valent vertices of all central paths.

Lemma 11. For every inline image, inline image and inline image, inline image is a PN-graph.

Proof. The graph inline image contains a subgraph G isomorphic to the PN-graph inline image and contains a subgraph inline image homeomorphic to the PN-graph inline image. The graph G contains the cycles inline image, inline image, and inline image of inline image, and the graph inline image contains the cycles inline image of inline image and is obtained from inline image by subdividing the edges of the cycle inline image (by using n 2-valent vertices in total).

Consider, for a contradiction, a proper 1-immersion φ of inline image. In φ, the graph G has a plane embedding and we shall investigate in which faces of the embedding of G lie the vertices of inline image. We shall show that they all lie in the face of G bounded by the (subdivided) cycle inline image.

In the graph inline image the cycles inline image and inline image, inline image are connected by inline image edge-disjoint paths. This implies that no 3- or 4-gonal face of G contains all vertices of inline image in its interior.

Any two vertices of inline image are connected by six edge-disjoint paths in inline image. Therefore

  1. No 3- or 4-gonal face of G contains any vertex of the cycles inline image, inline image, in its interior.

Suppose that a vertex v of inline image does not belong to the cycles inline image, inline image, and lies inside a 3- or 4-gonal face F of G. By construction of inline image, the vertex v is adjacent to two vertices w and inline image of some inline image, inline image. By (a), w and inline image do not lie inside F, hence they lie, respectively, in faces F1 and F2 of G adjacent to F. However, at least one of F1 and F2 is 3- or 4-gonal, contrary to (a). Therefore, no 3- or 4-gonal face of G contains any vertex of inline image. If there is a vertex of inline image outside inline image, then inline image has two adjacent vertices such that one of them is either a vertex of inline image or lies inside inline image, and the other vertex lies outside inline image, a contradiction, since the edge joining the vertices crosses at least five edges of G. This implies that all vertices of inline image lie inside the face of G bounded by inline image. Hence, inline image lies inside inline image and has a proper 1-immersion in φ. If in this 1-immersion of inline image we ignore the 2-valent vertices on the cycle inline image of inline image, then we obtain a proper 1-immersion of the PN-graph inline image, a contradiction.

By the paths of inline image associated with any central vertex x, we mean the two paths shown in Figure 22; one of them is depicted in thick line and the other in dashed line. Every edge of inline image not belonging to the cycles inline image is assigned a type inline image as shown in Figure 22 such that for inline image, the edges of type inline image and 2i are all edges lying between the cycles inline image and inline image, and the edges of type inline image (resp. 2i) are incident to vertices of inline image (resp. inline image).

image

Figure 22. The paths associated with a central vertex and the types of edges.

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Suppose that there is a 1-immersion φ of inline image. By Lemma 11, inline image is a PN-graph. Thus, φ induces an embedding of this graph. We shall assume that the outer face F0 of this embedding is bounded by the cycle inline image. We shall first show that F0 is also a face of φ. To prove this, it suffices to see that no central path can enter F0.

Any central vertex x is separated from F0 by inline image edge-disjoint cycles: m cycles inline image and inline image cycles inline image, where the cycle inline image consists of all edges of type i (inline image). The central path inline image can have at most inline image crossing points, hence P cannot enter F0. If P lies between B0 and inline image in φ, then it must cross inline image paths associated either with inline image central vertices inline image or with inline image central vertices inline image (here we interpret all additions modulo inline image), a contradiction. Hence, in φ any central path either lies inside B0 or crosses some edges of inline image but does not lie entirely between B0 and inline image.

The main goal of this section is to show that inline image has no 1-immersions (see Theorem 4 in the sequel). Roughly speaking, the main idea of the proof is as follows. Suppose, for a contradiction, that inline image has a 1-immersion. Every central path can have at most inline image crossing points, hence, all inline image central paths cannot be 1-immersed inside B0. Then there is a central path that crosses some edges of inline image. Let P be a central path with maximum number of such crossings. Since P can have at most inline image crossing points, some of the other inline image central paths do not cross P and have to “go around” P and, in doing so, one of the paths has to cross more edges of inline image than P does, a contradiction. Before proving Theorem 4, we need some definitions and preliminary Lemmas 12 and 13.

Consider a 1-immersion of inline image (if it exists). If a central path inline image does not lie inside B0, consider the sequence inline image (inline image), where inline image and inline image, obtained by listing the intersection points of the path and B0 when traversing the path from the vertex x to the vertex inline image (here inline image are crossing points). By a piece of P, we mean the segment of P from inline image to inline image for some inline image; denote the piece by inline image. A piece of P with an end point x or inline image is called an end piece of P at the vertex x or inline image, respectively. An outer piece of P is every piece of P that is immersed outside B0. Clearly, either inline image or inline image are all outer pieces of P. The end points δ and inline image of an outer piece Π of P partition B0 into two curves A and inline image such that the curve A lies inside the closed curve consisting of Π and inline image (see Fig. 23). The central vertices belonging to A and different from δ and inline image are said to be bypassed by Π and P (cf. Fig. 23).

Lemma 12. If inline image bypasses neither a central vertex y nor its opposite vertex inline image, and inline image bypasses neither x nor inline image, then inline image crosses inline image.

Proof. Suppose, for a contradiction, that inline image does not cross inline image. For every outer piece of the two paths, we can replace a curve of a path containing the piece by a new curve lying inside B0 so that the path inline image (resp. inline image) becomes a new path inline image (resp. inline image) connecting the vertices x and inline image (resp. y and inline image) such that the two new paths lie inside B0 and do not cross each other, a contradiction. How the replacements can be done is shown in Figure 24, where the new curves are depicted in thick line. (Note that in Fig. 24(b), since inline image does not bypass x, the depicted pieces bypassing x belong to inline image.)

image

Figure 23. The central vertices bypassed by an outer piece Π.

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image

Figure 24. Transforming paths inline image into paths inline image.

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By a type of an outer piece of a central path, we mean the maximal type of an edge of inline image crossed by the path.

For an outer piece Π of a central path inline image, denote by inline image the number of central vertices bypassed by Π, and by inline image the number of intersection points of Π and inline image, including the crossings at the end points of Π (except if an end point is x or inline image).

Lemma 13. If Π is an outer piece of type t of a central path P, then

  • display math

where inline image if Π is an end piece and inline image otherwise.

Proof. The piece Π crossing an edge of type t has a point separated from the interior of B0 by inline image edge-disjoint cycles: inline image, and the cycles inline image, where the cycle inline image consists of all edges of type i (inline image). Thus, Π crosses each of these cycles twice, except that for an end piece, we may miss one crossing with C1. The piece Π bypasses inline image central vertices, hence Π crosses inline image paths associated with those inline image vertices. Hence, we obtain two inequalities,

  • display math(1)

and

  • display math(2)

Now add (1) and (2), apply inline image, then divide by 2 and rearrange to get the result.

By the type of a central path not lying (entirely) inside B0, we mean the maximal type of the outer pieces of the path. If t different central paths bypass a central vertex x, then all the paths cross edges of the same path T associated with x and since the edges of T have pairwise different types, we obtain that one of the central paths crosses an edge of type at least t. Hence we have

  1. If t different central paths bypass the same central vertex, then one of the paths has type at least t.

Theorem 4. For every inline image and inline image, the graph inline image is not 1-planar.

Proof. Consider, for a contradiction, a 1-immersion φ of inline image and a path inline image of maximal type inline image.

As above, let inline image be the number of crossing points of P and inline image, and let inline image be the number of distinct central vertices bypassed by P and different from x and inline image.

There are at least inline image different pairs inline image (inline image) of central vertices such that P does not bypass y and inline image; denote by inline image the set of the corresponding (at least inline image) paths inline image. If inline image does not bypass y and inline image, then inline image either does not bypass x and inline image (in this case inline image crosses inline image by Lemma 12) or bypasses at least one of the vertices x and inline image. Hence, we have

  • display math(3)

where β is the number of paths of inline image that cross P and do not bypass x or inline image; γ is the number of paths of inline image that bypass x or inline image and do not cross P; ε is the number of paths of inline image that cross P and bypass x or inline image. We are interested in the number inline image of paths of inline image that bypass x or inline image.

The path P has at most inline image crossing points, hence

  • display math

and, by (3), we obtain

  • display math

when

  • display math(4)

Let inline image (inline image) be all outer pieces of P, and Π1 be of the maximal type t. We have inline image and inline image (the vertices x and inline image can be bypassed by P, and some central vertices can be bypassed by P more than once). By Lemma 13, inline image for every inline image. Hence, by (4) and Lemma 13, we obtain

  • display math(5)

where inline image if Π1 is an end piece and inline image otherwise. If Π1 is not an end piece or inline image, then, by (5), inline image, hence one of the vertices x and inline image is bypassed by at least inline image paths of inline image. Now, by (D), one of the inline image paths has type at least inline image, a contradiction. Now suppose that Π1 is an end piece at the vertex x and inline image. Then every path of inline image either crosses P or bypasses x or inline image, and at least 2t paths of inline image bypass x or inline image. If no one of the 2t paths bypasses x, then all the inline image paths bypass inline image and, by (D), one of the paths has type at least inline image. If one of the 2t paths, say, inline image, bypasses x, then inline image has an outer piece inline image that bypasses x and does not cross Π1 (since inline image, inline image does not cross P). The piece Π1 has type t and is an end piece at x, hence inline image has type at least inline image, a contradiction.

We have shown that every graph inline image, where inline image and inline image, is an MN-graph. These graphs have order inline image. Clearly, graphs inline image and inline image, where inline image and inline image, are nonisomorphic for inline image and for inline image and inline image.

Claim 2. For any integers inline image and inline image, there are at least inline image nonisomorphic MN-graphs inline image, where inline image.

Proof. The automorphism group of inline image is the dihedral group inline image of order inline image. Now the claim follows by recalling a well-known fact that inline image.

5. TESTING 1-IMMERSIBILITY IS HARD

  1. Top of page
  2. Abstract
  3. 1. INTRODUCTION
  4. 2. CHAIN GRAPHS BASED ON K3, 3
  5. 3. PN-GRAPHS
  6. 4. MN-GRAPHS BASED ON PN-GRAPHS
  7. 5. TESTING 1-IMMERSIBILITY IS HARD
  8. 6. k-PLANARITY TESTING FOR MULTIGRAPHS
  9. ACKNOWLEDGMENTS
  10. REFERENCES

In this section, we prove that testing 1-immersibility is NP-hard. This shows that it is extremely unlikely that there exists a nice classification of MN-graphs.

Theorem 5. It is NP-complete to decide if a given input graph is 1-immersible.

Since 1-immersions can be represented combinatorially, it is clear that 1-immersibility is in NP. To prove its completeness, we shall make a reduction from a known NP-complete problem, that of 3-colorability of planar graphs of maximum degree at most 4 [11].

The rest of this section is devoted to the proof of Theorem 5.

Let G be a given plane graph of maximum degree 4 whose 3-colorability is to be tested. We shall show how to construct, in polynomial time, a related graph inline image such that inline image is 1-immersible if and only if G is 3-colorable. We may assume that G has no vertices of degree less than 3 (since degree 1 and 2 vertices may be deleted without affecting 3-colorability).

To construct inline image, we will use as building blocks graphs that have a unique 1-immersion. These building blocks are connected with each other by edges to form a graph that also has a unique 1-immersion. Then we add some additional paths to obtain inline image.

We say that a 1-planar graph G has a unique 1-immersion if, whenever two edges e and f cross each other in some 1-immersion, then they cross each other in every 1-immersion of G, and second, if inline image is the planar graph obtained from G by replacing each pair of crossing edges inline image and inline image by a new vertex of degree 4 joined to inline image, then inline image is 3-connected (and thus has combinatorially unique embedding in the plane – the one obtained from 1-immersions of G).

It was proved in [12] that for every inline image, the graph with 4n vertices and 13n edges shown in Figure 25(a) has a unique 1-immersion. (To be precise, the paper [12] considers the graph for even values of inline image only, but one can check that the proof does not depend on whether inline image is even or odd.) We call the graph a U-graph. Figure 25(b) shows a designation of the U-graph used in what follows. In the 1-immersion of the U-graph shown in Figure 25, the vertices inline image which lie on the boundary of the outer face of the spanning embedding (the boundary is called the outer boundary cycle of the 1-immersed U-graph), are called the boundary vertices of the U-graph in the 1-immersion. If a graph has a U-graph as a subgraph, then the U-graph is called the U-subgraph of the graph.

image

Figure 25. The U-graph.

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Take two 1-immersed U-graphs U1 and U2 such that each of them has the outer boundary cycle of length at least 7, and construct the 1-immersed graphs shown in Figure 26(a) and (b), respectively, where by 1, 2, …, 7 we denote seven consecutive vertices on the outer boundary cycle of each of the 1-immersed graphs. We say that in Figure 26(a) (resp. (b)) the U-graphs U1 and U2 are connected by a (1)-grid (resp. (2)-grid). The vertices labeled 1, 2, …, 7 are the basic vertices of the grid and for inline image, the h-path connecting the vertices labeled i of the (h)-grid, inline image, is called the basic path of the grid connecting these vertices. Let us denote the ith basic path by inline image. The paths inline image and inline image, inline image, are neighboring basic paths of the grid. For two basic paths inline image and inline image, inline image, denote by inline image the cycle of the graph in Figure 26 consisting of the two paths and of the edges inline image of the two graphs U1 and U2.

image

Figure 26. Two U-graphs connected by a grid.

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By a U-supergraph, we mean every graph obtained in the following way. Consider a plane connected graph H. Now, for every vertex inline image, take a 1-immersed U-graph inline image of order at least inline image and for any two adjacent vertices u and w of the graph, connect inline image and inline image by a (1)- or (2)-grid as shown in Figure 27 such that any two distinct grids have no basic vertices in common. We obtain a 1-immersed U-supergraph.

Theorem 6. Every U-supergraph M has a unique 1-immersion.

Proof. It suffices to show the following:

  1. The graph consisting of two U-graphs connected by an (h)-grid, inline image, has a unique 1-immersion.
  2. In every 1-immersion φ of M, the edges of distinct grids do not intersect.

Note that M contains no subgraph that can be 1-immersed inside the boundary cycle of a 1-immersed U-subgraph of M in a 1-immersion of M as shown in Figure 28 in dashed line. Hence, in every 1-immersion of M, the boundary edges of the U-subgraphs of M are not crossed.

We prove (a) and (b) in the following way. We consider a 1-immersed subgraph W of M (cf. Fig. 26) consisting of two U-graphs U1 and U2 connected by an (h)-grid Γ, inline image, and we show that in every 1-immersion φ of M, the graph W has the same 1-immersion and the edges of Γ are not crossed by edges of other grids.

Suppose, for a contradiction, that U1 and U2 are 1-immersed under φ as shown in Figure 29(a) (the situation described in the figure arises when U1 is drawn clockwise and U2 counter-clockwise (or vice versa)). Clearly, there are two basic paths inline image and inline image of Γ, inline image, which do not intersect. Then the cycle shown in Figure 29(a) in thick line is embedded in the plane, a contradiction, since the cycle is crossed by five other basic paths of Γ, but the cycle has only inline image edges that can be crossed by other edges. Hence, U1 and U2 are 1-immersed as shown in Figure 26.

Suppose that in φ, a basic path inline image of Γ crosses a basic path Q of some grid of M exactly once. If inline image is a basic path of Γ, inline image (see Fig. 29(b)), then the closed curves C1 and C2 shown in Figure 29(b) by dashed cycles, are embedded in the plane and each of the other five basic paths of Γ crosses an edge of C1 or C2, a contradiction, since C1 and C2 have inline image edges in total, which can be crossed by other edges. If Q is a basic path of a grid inline image different from Γ, then there is a basic path inline image, inline image, of Γ such that inline image is not crossed by inline image and Q. Hence, the cycle inline image is embedded and Q crosses the edges of the cycle exactly once. Then for every other basic path inline image of inline image, the cycle inline image crosses inline image at least twice and inline image crosses inline image or inline image (the edges of different U-subgraphs do not intersect). We have that the seven basic paths of inline image cross inline image and inline image, a contradiction. Hence, in φ, if two basic paths intersect, then they intersect twice. In particular, only basic paths of (2)-grids can intersect.

Now we claim the following:

  1. If in φ two neighboring basic paths inline image and inline image of the (2)-grid Γ do not intersect, then the edge e joining the middle vertices of inline image and inline image lies inside the embedded cycle inline image.

Indeed, if e lies outside inline image, then the 4-cycle shown in Figure 29(c) in thick line is crossed by five basic paths inline image, inline image, a contradiction.

Suppose that in φ, a basic path of Γ crosses a basic path of a grid inline image twice (i.e., Γ and inline image are (2)-grids). Since the number of basic paths of Γ is odd (namely, seven), it cannot be that every basic path of Γ crosses some other basic path of Γ twice. Then there is a basic path of Γ, which is not crossed by other basic paths of Γ. Hence, if inline image (resp. inline image), then there are two neighboring basic paths inline image and inline image of Γ such that one of them, say, inline image, is crossed twice by some basic path Q of Γ (resp. inline image), and the other basic path inline image is not crossed by Q. Then the cycle inline image is embedded. By (E), the edge e joining the middle vertices of inline image and inline image lies inside inline image. Denote by C1 and C2 the two embedded adjacent 4-cycles each of which consists of e and edges of the 6-cycle inline image. The middle vertex of Q lies outside inline image and the two end vertices of Q lie inside C1 and C2, respectively. The end vertices of Q belong to two U-subgraphs connected by inline image. Since the edges of the two U-subgraphs do not cross the edges of C1 and C2, we obtain that one of the U-subgraphs lies inside C1 and the other lies inside C2, a contradiction, since the two U-subgraphs are connected by at least four basic paths different from Q, inline image, and inline image. Hence, no basic path of Γ crosses some other basic path twice.

We conclude that the basic paths of the grids connecting U-subgraphs do not intersect.

Now it remains to show that if Γ is a (2)-grid, then the edges joining the middle vertices of the basic paths of Γ are not crossed. Consider any two neighboring basic paths inline image and inline image of Γ. The cycle inline image is embedded and, by (E), the edge e joining the middle vertices of inline image and inline image lies inside inline image. It is easy to see that for every edge inline image of G not belonging to U-subgraphs and different from e and the edges of inline image, in the graph inline image the end vertices of inline image are connected by a path that consists of edges of U-subgraphs and basic paths of grids and which does not pass through the vertices of inline image. Now, if the edge e is crossed by some other edge inline image, then inline image is not an edge of a U-subgraph, the end vertices of inline image lie inside the cycles C1 and C2, respectively (where C1 and C2 are defined as in the preceding paragraph), whose edges are not crossed by edges of U-subgraphs and basic paths, a contradiction.

Therefore, the edges of M do not intersect and that the graph W has a unique 1-immersion. This completes the proof of the theorem.

image

Figure 27. Constructing a U-supergraph.

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image

Figure 28. A 1-immersion of a subgraph.

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image

Figure 29. Cycles of two adjacent 1-immersed U-subgraphs.

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Now, given a plane graph G every vertex of which has degree 3 or 4, we construct a graph inline image such that G is 3-colorable if and only if inline image is 1-immersible. To obtain inline image, we proceed as follows. First, we construct a subgraph G(1) of inline image such that G(1) has a unique 1-immersion. The graph G(1) is obtained from a U-supergraph W by adding some additional vertices and edges. By inspection of the subsequent figures that illustrate the construction of G(1) and its 1-immersion, the reader will easily identify the additional vertices and edges: they do not belong to U-subgraphs and grids. Then one can easily check that given the 1-immersion of W, the additional vertices and edges can be placed in the plane in a unique way to obtain a 1-immersion of G(1), hence G(1) has a unique 1-immersion also. Now, given the unique 1-immersion of G(1), to construct inline image we place some new additional paths “between” 1-immersed U-subgraphs of G(1). Notice that due to circle inversion, one may assume that in a 1-planar drawing of inline image each U-subgraph is drawn in such a way that the outer boundary cycle is containing all other vertices of the U-subgraph inside the region it bounds.

The graph G(1) is obtained from the plane graph G if we replace every face F of the embedding of G by a U-graph inline image and replace every vertex v by a vertex-block inline image as shown at the top of Figure 30. At the bottom of Figure 30, we show the designation of a (1)-grid used at the top of the figure and at what follows. The vertex-block inline image has a unique 1-immersion and is obtained from a U-supergraph by adding some additional vertices and edges. Figure 30 shows schematically the boundary of inline image Figure 33 shows inline image in more detail. For a k-valent vertex v of G, inline image, the vertex-block inline image has 3k boundary vertices labeled clockwise as inline image; these vertices do not belong to U-subgraphs of inline image. In Figures 30 and 33, we only show the case of a 3-valent vertex v; for a 4-valent vertex, the construction is analogous—there are three more boundary vertices labeled inline image, respectively.

We say that vertex-blocks inline image and inline image are adjacent if v and w are adjacent vertices of G.

image

Figure 30. Constructing the graph G(1).

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The graph inline image is obtained from G(1) if we take a collection of additional disjoint paths of length ⩾1 (they are called the pending paths) and identify the end vertices of every path with two vertices, respectively, of G(1). The graph G(1) has a unique 1-immersion and the edges of the U-subgraphs of G(1) cannot be crossed by the pending paths, hence the 1-immersed G(1) restricts the ways in which the pending paths can be placed in the plane to obtain a 1-immersion of inline image. Every pending path connects either boundary vertices of adjacent vertex blocks or vertices of the same vertex block.

The graph inline image is such that for any two adjacent vertex blocks, there are exactly three pending paths connecting the vertices of the vertex blocks. The paths have length 3 and are shown in Figure 31; we say that these pending paths are incident with the two vertex blocks. Each of the three pending paths connects the boundary vertices labeled by the same letter: a, b, or c. For inline image, the pending path connecting vertices labeled h is called the (h)-path connecting the two vertex blocks. In Figure 31 the (h)-path, inline image, is labeled by the letter h.

image

Figure 31. The pending paths connecting adjacent vertex blocks.

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Denote by G(2) the graph obtained from G(1) if we add all triples of pending paths connecting vertex-blocks inline image, for all edges inline image.

The graph inline image also satisfies the properties stated below. The pending paths connecting vertices of the same vertex-block inline image are divided into three families called, respectively, the inline image-, inline image-, and c-families of inline image. Given the 1-immersion of G(1), for every inline image, the h-family of inline image has the following properties:

  1. Every path P of the h-family admits exactly two embeddings in the plane such that we obtain a 1-immersion of inline image.
  2. The h-family consists of paths inline image such that the graph inline image has exactly two 1-immersions. In the two 1-immersions, every path inline image uses its two embeddings. In one of the 1-immersions, paths of the h-family cross all (h)-paths incident with inline image. In the other 1-immersion, the paths of the h-family do not cross any (h)-path incident with inline image.

Figure 32 shows fragments of the two 1-immersions of the union of G(2) and the pending paths of an h-family. In the figure, each of the depicted (in thick line) six edges of the family, they are labeled by 1, 2, …, 6, respectively, uses its two embeddings in one of the two 1-immersions.

image

Figure 32. Pending paths of an h-family.

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If in a 1-immersion of inline image, paths of an h-family of inline image, inline image, cross (h)-paths incident with inline image, then we say that the h-family of inline image is activated in the 1-immersion of inline image.

Figures 33 and 34 show a vertex-block inline image and the h-families of the vertex-block (inline image) in the case where v is 3-valent (the generalization for a 4-valent vertex v is straightforward). The pending paths of the three h-families are shown by thick lines and the three families are activated. To avoid cluttering a figure, Figure 33 contains a fragment denoted by R, which is given in more detail in Figure 34. Recall that the gray areas in Figures 33 and 34 represent U-graphs.

image

Figure 33. A vertex block and the activated h-families.

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image

Figure 34. The fragment R of the vertex block in Figure 33.

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In Figures 33 and 34, we use designations of some fragments of inline image; the designations are given at the left of Figure 35 and the corresponding fragments are given at the right of Figure 35 (i.e., the connections between the gray areas in Figures 33 and 34 consist of seven basic paths). The reader can easily check that for every pending path P of the three families, there are exactly two ways to embed the path so that we obtain a 1-immersion of inline image. The vertex-block inline image contains a 2-path connecting the vertices labeled 0 in Figure 34 and a 1-path connecting the vertices labeled 1 in Figure 34; we call the paths the (0)- and (1)-blocking paths, respectively). For every inline image, exactly one pending path of the h-family of inline image crosses a blocking path: the pending path has length 33, crosses the (1)-blocking (resp. (0)-blocking) path when the h-family is activated (resp. not activated), and the pending path in each of its two embeddings crosses exactly one pending path of each of the other two families. Figure 34 shows the two embeddings of the pending 33-path of the b-family (one of them is in thick line, the other, when the family is not activated, is in dashed line). Note that Figure 34 shows something that is not a 1-immersion, since all three families of paths are activated, and the (1)-blocking path is crossed three times.

image

Figure 35. The designations of fragments of the vertex-block inline image.

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image

Figure 36. Different plane 1-immersions of G(1).

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Figure 37. Edges of an h-family in inline image.

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Denote by inline image the union of G(2) and the paths of all three h-families of inline image. Now the reader can check that inline image and the h-families of inline image are constructed in such a way that the following holds:

  1. In every 1-immersion of inline image (and, hence, of inline image) exactly one h-family of inline image is activated, and for each inline image, there is a 1-immersion of inline image in which the h-family of inline image is activated.

By construction of inline image, if inline image has a 1-immersion, then in the 1-immersion for every vertex v of G, exactly one h-family (inline image) of inline image is activated and taking Figure 31 into account we obtain that in the 1-immersion the h-families of the vertex blocks adjacent to inline image are not activated.

Now take a 1-immersion of inline image (if it exists) and assign every vertex v of G a color inline image such that the h-family of inline image is activated in the 1-immersion of inline image. We obtain a proper 3-coloring of G with colors inline image.

Take a proper 3-coloring of G (if it exists) with colors inline image and for every vertex v of G, if inline image is the color of v, take the inline image-family of inline image to be activated and the other two families not to be activated. By the construction of inline image, and the mentioned properties of 1-immersions of its subgraphs inline image, it follows that we obtain a 1-immersion of inline image.

When constructing inline image, we choose the order of every U-subgraph such that every boundary vertex of the U-subgraph is incident with an edge not belonging to the U-subgraph. This implies that for every face F of size k of the plane embedding of G, the number of edges in the U-graph inline image is bounded by a constant multiple of k. Similarly, for each inline image, the union of inline image and its three h-families has constant size. Therefore, the whole construction of inline image can be carried over in linear time. This completes the proof of Theorem 5.

6. k-PLANARITY TESTING FOR MULTIGRAPHS

  1. Top of page
  2. Abstract
  3. 1. INTRODUCTION
  4. 2. CHAIN GRAPHS BASED ON K3, 3
  5. 3. PN-GRAPHS
  6. 4. MN-GRAPHS BASED ON PN-GRAPHS
  7. 5. TESTING 1-IMMERSIBILITY IS HARD
  8. 6. k-PLANARITY TESTING FOR MULTIGRAPHS
  9. ACKNOWLEDGMENTS
  10. REFERENCES

A graph drawn in the plane is k-immersed in the plane (inline image) if any edge is crossed by at most k other edges (and any pair of crossing edges cross only once). A graph is k-planar if it can be k-immersed into the plane.

It appears that we can slightly modify the proof of Theorem 5 so as to obtain a proof that k-planarity testing (inline image) for multigraphs is NP-complete. Below we give only a sketch of the proof, the reader can easily fill in the missing details.

Denote by inline image, inline image, and inline image, respectively, the multigraphs obtained from the graphs G, inline image, and G(1) if we replace every edge by k parallel edges. For an edge e of the multigraphs denote by inline image the set consisting of e and all other inline image edges parallel to e. Denote by φ the unique plane 1-immersion of G(1), and by inline image the plane k-immersion of inline image obtained from φ if we replace every edge of G(1) by k parallel edges.

Lemma 14. The multigraph inline image, inline image, has a unique plane k-immersion.

Proof. We consider an arbitrary plane k-immersion ψ of inline image and show that ψ is inline image.

First, we show that if edges e1 and e2 of inline image cross in ψ, then each edge of inline image intersects every edge of inline image. Suppose, for a contradiction, that an edge inline image of inline image does not intersect e1 (see Fig. 36(a)). Consider the 2-cell D whose boundary consists of the edges e2 and inline image. Since e2 and inline image can have at most 2k crossings in total, there are at most two vertices lying outside D that are adjacent to vertices inside D. This means (see Fig. 36(b)) that G(1) has two different plane 1-immersions (the edge of G(1) joining u and w has different positions in the two different plane 1-immersions), a contradiction. Hence, each edge of inline image intersects every edge of inline image. Delete inline image edges from every k parallel edges. We obtain a plane 1-immersion of G(1), that is, φ. Hence, ψ is inline image.

The graph inline image is obtained from inline image if we add the pending paths of inline image where every edge is replaced by k parallel edges. Now, considering a pending path of an h-family, we have (see Fig. 37, where each thick edge represents k parallel edges) that if inline image, then each of the edges e and inline image is already crossed by k edges of inline image and inline image, respectively, thus the edges of the pending path incident with the vertex v cannot cross edges e and inline image, a contradiction. Hence, all edges of inline image cross either inline image or inline image. As a result, the h-families and the other pending paths of inline image “behave” in the same way as in inline image. We conclude that inline image has a plane 1-immersion if and only if inline image has a plane k-immersion. Since inline image has a plane 1-immersion if and only if G has a proper 3-coloring, we get that k-planarity testing for multigraphs is NP-complete.

If we restrict ourselves to simple graphs only, then to have a proof analogous to the proof of Theorem 5 we need simple graphs that have a unique plane k-immersion (inline image), but the construction of such graphs seems to be nontrivial and does not readily follow from the construction of U-graphs in Section 'TESTING 1-IMMERSIBILITY IS HARD'.

ACKNOWLEDGMENTS

  1. Top of page
  2. Abstract
  3. 1. INTRODUCTION
  4. 2. CHAIN GRAPHS BASED ON K3, 3
  5. 3. PN-GRAPHS
  6. 4. MN-GRAPHS BASED ON PN-GRAPHS
  7. 5. TESTING 1-IMMERSIBILITY IS HARD
  8. 6. k-PLANARITY TESTING FOR MULTIGRAPHS
  9. ACKNOWLEDGMENTS
  10. REFERENCES

The authors are grateful to anonymous referee for pointing out that our proof in Section 'TESTING 1-IMMERSIBILITY IS HARD' might be used to derive a corresponding result for k-immersions.

REFERENCES

  1. Top of page
  2. Abstract
  3. 1. INTRODUCTION
  4. 2. CHAIN GRAPHS BASED ON K3, 3
  5. 3. PN-GRAPHS
  6. 4. MN-GRAPHS BASED ON PN-GRAPHS
  7. 5. TESTING 1-IMMERSIBILITY IS HARD
  8. 6. k-PLANARITY TESTING FOR MULTIGRAPHS
  9. ACKNOWLEDGMENTS
  10. REFERENCES
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