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### Keywords:

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

### Abstract

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 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 , there are at least nonisomorphic minimal non-1-planar graphs of order n. It is also proved that testing 1-planarity is NP-complete.

### 1. INTRODUCTION

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 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 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 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 , where cr(G) denotes the crossing number of G. Let be the graph obtained from G by replacing each edge of G by a path of length t. Then . This implies that is not 1-planar. However, contains an MN-subgraph H. Clearly, H contains at least one subdivided edge of G in its entirety, so . Since t can be arbitrarily large (see, e.g., the well-known lower bound on ), 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 be one of the Kuratowski graphs. For each edge , let be a 5-connected triangulation of the plane and be adjacent vertices of whose degree is at least 6. Let . Now replace each edge of K with 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 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 , is the unique 7-vertex MN-graph and since all 6-vertex graphs are 1-planar, the graph 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 , there are at least 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 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

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 , we say that C lies inside (resp. outside) if every point of C either lies inside (resp. outside) or lies on . 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 of a graph G separate vertices x and y of the graph if x and y belong to different connected components of the graph .

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 under the 1-immersion. Then by a face of a 1-immersion of G we mean any connected component of .

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 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 and vice versa.

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

By a link connecting two vertices x and y we mean any of the graphs shown in Figure 1 where . 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 we mean the graph shown in Figure 2(a). By a B-chain of length we mean the graph shown in Figure 2(c) and, for , every graph obtained from that graph in the following way: for some integers , where and , we replace the link at the left of Figure 2(e) by the link shown at the right, for . 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 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 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 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.

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 and for all for some . We say that the subchain connects the vertices and .

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.

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

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

Proof. If e is an Ω-edge, then 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 . 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 and cross, there is a disk D having 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 are crossed. Similarly, if is a link in a chain, and an edge incident with z crosses an edge incident with , 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 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 , .
2. L is a B-link and , .
3. L is a base link and , .
4. L is a base link and , .
5. L is a base link and , .

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 in this order and that u and v both have precisely one additional neighbor 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 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 either all lie inside or all lie outside C.
2. If all core vertices of lie inside (resp. outside) C, then at most one vertex of lies outside (resp. inside) C.

Proof. First we show (i). If C does not contain the vertex A, then every two core vertices of 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 lie inside and outside C. Then there is a link of such that the vertex z lies inside and the vertex lies outside C. We may assume without loss of generality that Π and are incident to the base vertices Ω(1) and Ω(2), respectively, and (taking (A) into account) that the vertex z (if ) separates the vertices B and in (see Fig. 5(a), where in the dotted line indicates that the link has either edge or ; also if , then the link indicated in Figure 5(a) is a base link).

The 3-cycle C crosses at least two edges of . The vertex z (resp. ) 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 or through the noncore vertices of . 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 is the third edge that crosses C. We conclude that C crosses exactly two edges of and the two edges separate z from in . Thus, the two edges are the core-adjacent edges of the free cycle of . Hence, in φ, the link is 1-immersed as shown in Figure 5(b), where the dotted edges indicate alternative possibilities for the position of z (at top) or (at bottom).

Let 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 . Suppose exactly one of v and is inside the free cycle. If we are in the case of the bottom of Figure 5(b), then the path cannot lie inside C, a contradiction. In the case of the top of Figure 5(b), if the path lies inside C, then x must lie inside a 3-cycle Q of 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, , and the vertices of Q.

If either both v and 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 , there are two edge-disjoint paths of G connecting A and Ω(3) (resp. A and Ω(2)) and not passing through z (resp. ), a contradiction. Reasoning exactly in the same way, we also obtain a contradiction when is a base link.

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

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 are nonadjacent A- and B-chains, respectively, then Π does not cross in φ.

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

1. C lies outside , lies outside C, and exactly one vertex u of 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 lie outside , hence, by Lemma 3, all core vertices of lie outside C and . Since the edge in is crossed by an edge of , also contains precisely one vertex of and has degree 3 and is not a core vertex.

Adjacent trivalent vertices cannot be contained in a base link. Therefore, is not a base link. Let be depicted as shown in Figure 6(a). Because of symmetry, we may assume that and , 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 of are separated by the 3-cycle , whose edges are crossed by three edges different from the edge , a contradiction.

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

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

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.

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 such that every -path in L is crossed by an Ω-edge.

Proof. Suppose that for every link , an -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 be a link in an A- or B-chain Π whose -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 -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, , then L has another cycle that shares with C exactly edges and contains a core vertex not belonging to C. If C lies inside , then we consider the plane as the complex plane and apply the Möbius transformation with the point a taken inside but outside C. This yields a 1-immersion of G such that C does not lie inside and Ω lies inside C. Hence, we may assume that C does not lie inside , that Ω lies inside C and two Ω-edges h and cross two core-adjacent edges of C. Note that any two among the vertices 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 join the vertex Ω with basic vertices and , respectively. If the third basic vertex is Ω(2) or Ω(3), then Ω(2) and Ω(3) lie inside different faces of L, a contradiction, hence .

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.

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

1. In this case, L is a base link and . Consider two edge-disjoint paths in a chain 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 . The 1-planarity implies that the edges and the vertex Ω(1) separate the graph G. Therefore, a and b are core-adjacent edges of a link in the chain . If the edge g (shown in the figure) is crossed by an edge c of , then also and Ω(1) separate the graph. Thus would be core-adjacent edges in a link in 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 lies inside C and Ω(1) is inside C but outside (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 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 . 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 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.

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 , there are at least nonisomorphic MN-graphs of order n.

Proof. The A-chain of length t has vertices and a B-chain of length t has vertices. Consider a chain graph whose three A-chains have length 2, 2, and , respectively, and whose B-chains have length 2, 3, and , respectively. The graph has 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 nonisomorphic chain graphs of order , where and . We claim that for every integer , there are integers and such that . Indeed, if , put , respectively. If , where , then . Hence, there are at least nonisomorphic chain graphs of order . Since every chain graph is an MN-graph, the theorem follows.

### 3. PN-GRAPHS

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 the set of all vertices of the graph not belonging to C but adjacent to C. Two disjoint edges and of a graph G are paired if the four vertices 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 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 are connected by four edge-disjoint paths not passing through the vertices of C.
8. If an edge of a nontriangular peripheral cycle C is paired with an edge of a nontriangular peripheral cycle , then:
1. C and have no vertices in common;
2. any two vertices a and of C and , respectively, such that are nonadjacent and are not connected by a path of length 2, where b does not belong to C and .
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 . 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 cross each other, then there is a vertex of D inside and outside .

If and belong to adjacent 3-cycles and , respectively (see Fig. 9), then, by (C4), there are nontriangular peripheral cycles C and containing e and , respectively. The cycles C and intersect at some point δ different from the intersection point of edges e and . 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 . Property (C8)(ii) implies that D and are nonadjacent 3-cycles. By (a), the proof is complete.

Lemma 6. If 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 .

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

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

By (C1), every vertex of Q has valence at least 4. By (C4), every vertex of is adjacent to at most two vertices of C. We claim that if a vertex is adjacent to two vertices u1 and u2 of C, then v is not adjacent to other vertices of . Suppose, for a contradiction, that v is adjacent to a vertex . Then, by (C4), the vertex w can be adjacent only to u3 and the 4-cycle 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 , , and 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 , (resp. ), 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 .

The claim is obvious for . For , 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 , , and 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 , it suffices to show that no vertex of Q is adjacent to three other vertices in Q. Suppose, for a contradiction, that is adjacent to . If v is adjacent to u1, then the edge belongs to three cycles D1, D2, and D3 such that for , the cycle contains edges and , has length 3 or 4, and if has length 4, then 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 and . Since , h is adjacent to a vertex . By (C2), the edge belongs to a 3-cycle . By (C4), . 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 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 disjoint from C. Let () be all vertices adjacent to v. Suppose there is a 3-cycle , where . If , then the 3-cycle is adjacent to two 3-cycles C and D, contrary to (C4). Hence, . By (C4), at most two vertices of are adjacent to x. Hence, there is a vertex such that a 3-cycle B containing the edge 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 belongs to a 3-cycle . If , then two of the vertices 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 (Fig. 10(c)). Then the vertex x belongs to two adjacent 3-cycles, C and , hence, by (B), . 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 . But then the 3-cycle is adjacent to two 3-cycles, C and , contrary to (C4). If z is adjacent to u and w, then the edge belongs to three peripheral cycles, C, , and , 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 and belongs to two 3-cycles. Here we have two subcases to consider.

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

Now, a is the only vertex inside the 3-cycle (Fig. 12(b)). By (C4), a is nonadjacent to w and b, hence a has valence 4. The edges and (see Fig. 12(a)) belong to a nontriangular peripheral cycle The edge belongs to a 3-cycle and b is the only vertex inside the 3-cycle . The vertex b is not adjacent to c, since the 4-cycle 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 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 belongs to two 3-cycles and (Fig. 13(a)).

Since b has valence at least 4, b lies outside the 3-cycle (Fig. 13(b)). The edges and (resp. and ) belong to a nontriangular peripheral cycle C1 (resp. C2). The cycles C1 and C2 are paired. In φ, the crossing point of edges and 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 of C2 (Fig. 13(c)). The edge is not the edge and belongs to a 3-cycle . If h3 belongs to C2, then in the embedding f the edge is a chord of the embedded peripheral cycle C2 and thus is a separating vertex set of G. But G is 3-connected, a contradiction. Hence, h3 does not belong to C2.

Now suppose that . We have . By (C8)(ii), , 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 (see Fig. 13(c)). By (B), . If , then, since s is the only vertex inside the 3-cycle , a is adjacent to at least one of h2 and h3, a contradiction. Hence, . Since z is the only vertex inside the 3-cycle B, and the 3-cycle is not B (since ), we have . 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, and .

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

By (C4), b is not adjacent to d and h, hence b has valence 4 and belongs to a 3-cycle disjoint from (Fig. 13(f)), where . Now d is the only vertex inside the 3-cycle (Fig. 13(e)). By (C8)(ii), d is not adjacent to t. Since d has valence at least 4, d is adjacent to and has valence 4. The 4-cycle 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 by the edge . 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 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 and, as an example, we shall give two infinite families of graphs in , 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 containing v are opposite peripheral cycles at v if C and have no edges incident with v in common.

Denote by 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 at v, no edge joins a vertex of to a vertex of .

A plane graph G is a medial extension of a graph if G is obtained from H in the following way. The vertex set of G is the set . The edge set of G is defined as follows. For every 3-valent vertex v of H, if are the edges incident with v, then in G the vertices , , and are pairwise adjacent (the three edges of G are said to be associated with v). For every 4-valent vertex w of H, if is the cyclic order of edges incident to w around w in the plane, then G contains the edges of the 4-cycle , and contains either the edge or the edge ; 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 is a 3-connected planar graph satisfying the conditions (C1)–(C6) and (C8) of Theorem 3.

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

There are medial extensions of graphs in 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 satisfies (C7), it is convenient to proceed in the following way. Subdivide every edge e of H by a two-valent vertex of G. We obtain a graph whose vertex set is where the vertices of are all 2-valent vertices of . We will consider paths of G associated with paths of connecting 2-valent vertices.

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

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

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

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 is C-independent if the path does not contain vertices of C. When checking (C7) for a medial extension G of , given a 3-cycle C of G and two 2-valent vertices , 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 are depicted as dashed edges joining 2-valent vertices, the dashed edges are not edges of (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 . If X is a subpath of such that X is associated with a path in , then X is represented by a dashed line passing near the edges of in . If and are subpaths of and (where possibly ), respectively, such that and are edge-disjoint paths associated with a path of , then and are represented by two (parallel) dashed lines passing near the edges of in . Using these conventions, the reader will be able to check that the depicted dashed paths and edges in the figure of represent four C-independent edge-disjoint paths of G connecting x and y.

Now we describe some graphs in . Let us recall that graphs in 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 , let be the Cartesian product of the path P3 of length 2 and the cycle of length n (see 16(a)). Let be the medial extension of shown in Figure 16(b).

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

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

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

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

Case 1. .

Then . If we delete from the vertices 1, 2, …, 14, then the obtained graph has only one connected component U with a vertex in , 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), has four C-independent edge-disjoint paths connecting x and y. Figure 17(b) shows four C-independent edge-disjoint paths in connecting vertices 1 and 4. If we delete from the vertices , then in the resulting nontrivial connected component, for every vertex , there is a path P connecting the vertices 1 and 4, and passing through x; combining two edge-disjoint paths of 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 , 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 ).

Case 2. .

We have . If we delete from 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 that are both in U belong to a cycle of U and then, by (C), has four C-independent edge-disjoint paths connecting x and y. It remains to show that for every vertex x of 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 satisfies (C1)–(C9), hence is a PN-graph for every .

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

### 4. MN-GRAPHS BASED ON PN-GRAPHS

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

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

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

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

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

Proof. If we delete an edge of a central path, then the remaining central paths, each with 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 can be drawn outside B0 with crossing points as shown in the figure and then the remaining 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 can be drawn outside B0 with crossing points (analogously to Fig. 21(a)) and then the remaining central paths can be 1-immersed inside B0. This exhibits all possibilities for the edge e (up to symmetries of ) and henceforth completes the proof.

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

Lemma 10. For every , is a PN-graph.

Proof. The graph contains subgraphs isomorphic to the PN-graph such that for , the graph contains the cycles , , and . Consider an arbitrary 1-immersion φ of . Suppose that in the plane embedding of the PN-graph L1 in φ, the cycle B2 is the boundary cycle of the outer -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 , the subgraph of the PN-graph bounded by and lies outside . As a result, φ is a plane embedding of , hence is a PN-graph.

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

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

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

Consider, for a contradiction, a proper 1-immersion φ of . 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 . We shall show that they all lie in the face of G bounded by the (subdivided) cycle .

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

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

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

Suppose that a vertex v of does not belong to the cycles , , and lies inside a 3- or 4-gonal face F of G. By construction of , the vertex v is adjacent to two vertices w and of some , . By (a), w and 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 . If there is a vertex of outside , then has two adjacent vertices such that one of them is either a vertex of or lies inside , and the other vertex lies outside , a contradiction, since the edge joining the vertices crosses at least five edges of G. This implies that all vertices of lie inside the face of G bounded by . Hence, lies inside and has a proper 1-immersion in φ. If in this 1-immersion of we ignore the 2-valent vertices on the cycle of , then we obtain a proper 1-immersion of the PN-graph , a contradiction.

By the paths of 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 not belonging to the cycles is assigned a type as shown in Figure 22 such that for , the edges of type and 2i are all edges lying between the cycles and , and the edges of type (resp. 2i) are incident to vertices of (resp. ).

Suppose that there is a 1-immersion φ of . By Lemma 11, 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 . 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 edge-disjoint cycles: m cycles and cycles , where the cycle consists of all edges of type i (). The central path can have at most crossing points, hence P cannot enter F0. If P lies between B0 and in φ, then it must cross paths associated either with central vertices or with central vertices (here we interpret all additions modulo ), a contradiction. Hence, in φ any central path either lies inside B0 or crosses some edges of but does not lie entirely between B0 and .

The main goal of this section is to show that 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 has a 1-immersion. Every central path can have at most crossing points, hence, all central paths cannot be 1-immersed inside B0. Then there is a central path that crosses some edges of . Let P be a central path with maximum number of such crossings. Since P can have at most crossing points, some of the other 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 than P does, a contradiction. Before proving Theorem 4, we need some definitions and preliminary Lemmas 12 and 13.

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

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

Proof. Suppose, for a contradiction, that does not cross . 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 (resp. ) becomes a new path (resp. ) connecting the vertices x and (resp. y and ) 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 does not bypass x, the depicted pieces bypassing x belong to .)

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

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

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

where if Π is an end piece and otherwise.

Proof. The piece Π crossing an edge of type t has a point separated from the interior of B0 by edge-disjoint cycles: , and the cycles , where the cycle consists of all edges of type i (). 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 central vertices, hence Π crosses paths associated with those vertices. Hence, we obtain two inequalities,

• (1)

and

• (2)

Now add (1) and (2), apply , 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 and , the graph is not 1-planar.

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

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

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

• (3)

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

The path P has at most crossing points, hence

and, by (3), we obtain

when

• (4)

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

• (5)

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

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

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

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

### 5. TESTING 1-IMMERSIBILITY IS HARD

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 such that 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 , 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 .

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 is the planar graph obtained from G by replacing each pair of crossing edges and by a new vertex of degree 4 joined to , then 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 , 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 only, but one can check that the proof does not depend on whether 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 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.

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 , the h-path connecting the vertices labeled i of the (h)-grid, , is called the basic path of the grid connecting these vertices. Let us denote the ith basic path by . The paths and , , are neighboring basic paths of the grid. For two basic paths and , , denote by the cycle of the graph in Figure 26 consisting of the two paths and of the edges of the two graphs U1 and U2.

By a U-supergraph, we mean every graph obtained in the following way. Consider a plane connected graph H. Now, for every vertex , take a 1-immersed U-graph of order at least and for any two adjacent vertices u and w of the graph, connect and 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, , 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, , 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 and of Γ, , 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 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 of Γ crosses a basic path Q of some grid of M exactly once. If is a basic path of Γ, (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 edges in total, which can be crossed by other edges. If Q is a basic path of a grid different from Γ, then there is a basic path , , of Γ such that is not crossed by and Q. Hence, the cycle is embedded and Q crosses the edges of the cycle exactly once. Then for every other basic path of , the cycle crosses at least twice and crosses or (the edges of different U-subgraphs do not intersect). We have that the seven basic paths of cross and , 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 and of the (2)-grid Γ do not intersect, then the edge e joining the middle vertices of and lies inside the embedded cycle .

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

Suppose that in φ, a basic path of Γ crosses a basic path of a grid twice (i.e., Γ and 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 (resp. ), then there are two neighboring basic paths and of Γ such that one of them, say, , is crossed twice by some basic path Q of Γ (resp. ), and the other basic path is not crossed by Q. Then the cycle is embedded. By (E), the edge e joining the middle vertices of and lies inside . Denote by C1 and C2 the two embedded adjacent 4-cycles each of which consists of e and edges of the 6-cycle . The middle vertex of Q lies outside 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 . 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, , and . 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 and of Γ. The cycle is embedded and, by (E), the edge e joining the middle vertices of and lies inside . It is easy to see that for every edge of G not belonging to U-subgraphs and different from e and the edges of , in the graph the end vertices of 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 . Now, if the edge e is crossed by some other edge , then is not an edge of a U-subgraph, the end vertices of 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.

Now, given a plane graph G every vertex of which has degree 3 or 4, we construct a graph such that G is 3-colorable if and only if is 1-immersible. To obtain , we proceed as follows. First, we construct a subgraph G(1) of 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 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 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 and replace every vertex v by a vertex-block 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 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 Figure 33 shows in more detail. For a k-valent vertex v of G, , the vertex-block has 3k boundary vertices labeled clockwise as ; these vertices do not belong to U-subgraphs of . 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 , respectively.

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

The graph 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 . Every pending path connects either boundary vertices of adjacent vertex blocks or vertices of the same vertex block.

The graph 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 , the pending path connecting vertices labeled h is called the (h)-path connecting the two vertex blocks. In Figure 31 the (h)-path, , is labeled by the letter h.

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

The graph also satisfies the properties stated below. The pending paths connecting vertices of the same vertex-block are divided into three families called, respectively, the -, -, and c-families of . Given the 1-immersion of G(1), for every , the h-family of 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 .
2. The h-family consists of paths such that the graph has exactly two 1-immersions. In the two 1-immersions, every path uses its two embeddings. In one of the 1-immersions, paths of the h-family cross all (h)-paths incident with . In the other 1-immersion, the paths of the h-family do not cross any (h)-path incident with .

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.

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

Figures 33 and 34 show a vertex-block and the h-families of the vertex-block () 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.

In Figures 33 and 34, we use designations of some fragments of ; 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 . The vertex-block 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 , exactly one pending path of the h-family of 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.

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

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

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

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

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

When constructing , 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 is bounded by a constant multiple of k. Similarly, for each , the union of and its three h-families has constant size. Therefore, the whole construction of can be carried over in linear time. This completes the proof of Theorem 5.

### 6. k-PLANARITY TESTING FOR MULTIGRAPHS

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

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

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

First, we show that if edges e1 and e2 of cross in ψ, then each edge of intersects every edge of . Suppose, for a contradiction, that an edge of does not intersect e1 (see Fig. 36(a)). Consider the 2-cell D whose boundary consists of the edges e2 and . Since e2 and 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 intersects every edge of . Delete edges from every k parallel edges. We obtain a plane 1-immersion of G(1), that is, φ. Hence, ψ is .

The graph is obtained from if we add the pending paths of 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 , then each of the edges e and is already crossed by k edges of and , respectively, thus the edges of the pending path incident with the vertex v cannot cross edges e and , a contradiction. Hence, all edges of cross either or . As a result, the h-families and the other pending paths of “behave” in the same way as in . We conclude that has a plane 1-immersion if and only if has a plane k-immersion. Since 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 (), 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

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.

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