Minimal obstructions for 1-immersions and hardness of 1-planarity testing
MMinimal obstructions for 1-immersions andhardness of 1-planarity testing
Vladimir P. Korzhik ∗ National University of Chernivtsi, ChernivtsiandInstitute of Applied Problems of Mechanics and Mathematicsof National Academy of Science of Ukraine, LvivUkraineBojan Mohar †‡ Department of MathematicsSimon Fraser UniversityBurnaby, B.C. V5A 1S6Canada ∗ This paper was done while the author visited Simon Fraser University. † Supported in part by the Research Grant P1–0297 of ARRS (Slovenia), by an NSERCDiscovery Grant (Canada) and by the Canada Research Chair program. ‡ On leave from: IMFM & FMF, Department of Mathematics, University of Ljubljana,Ljubljana, Slovenia. a r X i v : . [ m a t h . C O ] O c t bstract A graph is 1 -planar if it can be drawn on the plane so that eachedge is crossed by no more than one other edge (and any pair ofcrossing edges cross only once). A non-1-planar graph G is minimal if the graph G − e is 1-planar for every edge e of G . We constructtwo infinite families of minimal non-1-planar graphs and show thatfor every integer n ≥
63, there are at least 2 ( n − / nonisomorphicminimal non-1-planar graphs of order n . It is also proved that testing1-planarity is NP-complete. Running head:
Obstructions for 1-immersions
Corresponding author:
Bojan Mohar
AMS classifications:
Keywords:
Topological graph, crossing edges, 1-planar graph, 1-immersion.2
Introduction
A graph drawn in the plane is 1 -immersed in the plane if any edge is crossedby 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 tosee that if a graph has 1-immersion in which two edges e, f with a commonendvertex cross, then the drawing of e and f can be changed so that thesetwo edges no longer cross. Consequently, we may assume that adjacent edgesare never crossing each other and that no edge is crossing itself. We take thisassumption as a part of the definition of 1-immersions since this limits thenumber of possible cases when discussing 1-immersions.The notion of 1-immersion of a graph was introduced by Ringel [15] whentrying to color the vertices and faces of a plane graph so that adjacent orincident elements receive distinct colors. In the last two decades this class ofgraphs received additional attention because of its relationship to the familyof 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 areconsidered in [16]. It was shown in [3] that every 1-planar graph is acyclically20-colorable. The existence of subgraphs of bounded vertex degrees in 1-planar graphs is investigated in [10]. It is known (see [4, 5, 6]) that a 1-planargraph with n vertices has at most 4 n − G is said to be a minimal non-1-planargraph (MN -graph , for short) if G is not 1-planar, but G − e is 1-planar forevery 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. Thiswas first proved in [13]. Here we present two additional simple argumentsimplying the same conclusion. 3 xample 1. Let G be a graph such that t = (cid:100) cr( G ) / | E ( G ) |(cid:101) − ≥ G ) denotes the crossing number of G . Let G t be the graph obtainedfrom G by replacing each edge of G by a path of length t . Then | E ( G t ) | = t | E ( G ) | < cr( G ) = cr( G t ). This implies that G t is not 1-planar. However, G t contains an MN-subgraph H . Clearly, H contains at least one subdividededge of G in its entirety, so | V ( H ) | > t . Since t can be arbitrarily large (see,for example, the well-known lower bound on cr( K n )), this shows that thereare infinitely many MN-graphs.Before giving the next example, it is worth noticing that 3-cycles mustbe embedded in a planar way in every 1-immersion of a graph in the plane. Example 2.
Let K ∈ { K , K , } be one of the Kuratowski graphs. Foreach edge xy ∈ E ( K ), let L xy be a 5-connected triangulation of the plane and u, v be adjacent vertices of L xy whose degree is at least 6. Let L (cid:48) xy = L xy − uv .Now replace each edge xy of K with L (cid:48) xy by identifying x with u and y with v . It is not hard to see that the resulting graph G is not 1-planar (sincetwo of graphs L (cid:48) xy must “cross each other”, but that is not possible sincethey come from 5-connected triangulations). Again, one can argue that theycontain large MN-graphs.The paper [13] and the above examples prove the existence of infinitelymany MN-graphs but do not give any concrete examples. They provide noinformation on properties of MN-graphs. Even the most basic question ifthere are infinitely many MN-graphs whose minimum degree is at least threecannot be answered by considering these constructions. In [13], two specificMN-graphs of order 7 and 8, respectively, are given. One of them, the graph K − E ( K ), is the unique 7-vertex MN-graph and since all 6-vertex graphsare 1-planar, the graph K − E ( K ) is the MN-graph with the minimumnumber of vertices. Surprisingly enough, the two MN-graphs in [13] are theonly explicit MN-graphs known in the literature.The main problem when trying to construct 1-planar graphs is that wehave no characterization of 1-planar graphs. The set of 1-planar graphs isnot closed under taking minors, so 1-planarity can not be characterized byforbidding some minors.In the present paper we construct two explicit infinite families of MN-graphs whose minimum degree is at least three and, correspondingly, we givetwo different approaches how to prove that a graph has no plane 1-immersion.In Sect. 2 we construct MN-graphs based on the Kuratowski graph K , .To obtain them, we replace six edges of K , by some special subgraphs. The4inimality of these examples is easy to verify, but their non-1-planarity needslong and delicate arguments. Using these MN-graphs, we show that for everyinteger n ≥
63, there are at least 2 ( n − / nonisomorphic minimal non-1-planar graphs of order n . In Sect. 3 we describe a class of 3-connected planargraphs 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 haveexactly k > 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 theplane.Despite the fact that minimal obstructions for 1-planarity (i.e., the MN-graphs) have diverse structure, and despite the fact that discovering 1-immer-sions of specific graphs can be very tricky, it turned out to be a hard problemto establish hardness of 1-planarity testing. A solution is given in Sect. 5,where we show that 1-planarity testing is NP-complete, see . The proof isgeometric in the sense that the reduction is from 3-colorability of planargraphs (or similarly, from planar 3-satisfiability).In Sect. 6 we show how the proof of Theorem 5 can be modified to obtaina proof that k -planarity testing for multigraphs is NP-complete.An extended abstract of this paper was published in Graph Drawing 2008[14]. K , 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 (that is, non-self-intersecting) cycle C , if x lies in the interior(resp. exterior) of C , and does not lie on C . Having two embedded adjacentcycles C and C (cid:48) , we say that C lies inside (resp. outside) C (cid:48) if every pointof C either lies inside (resp. outside) C (cid:48) or lies on C (cid:48) . From this point on,by a 1-immersion of a graph we mean a plane 1-immersion. We assume that5n 1-immersions, adjacent edges do not cross each other and no edge crossesitself. 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 interiorand 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 outsidethe cycle. Two edges e and e (cid:48) of a graph G separate vertices x and y ofthe graph if x and y belong to different connected components of the graph G − e − e (cid:48) .Throughout the paper we will deal with 1-immersed graphs. When animmersion of a graph G is clear from the context, we shall identify ver-tices, edges, cycles and subgraphs of G with their image in R under the1-immersion. Then by a face of a 1-immersion of G we mean any connectedcomponent of R \ G .By using M¨obius transformations combined with homeomorphisms of theplane it is always possible to exchange the interior and exterior of any em-bedded cycle and it is possible to change any face of a given 1-immersion intothe outer face of a 1-immersion. Formally, we have the following observation(which we will use without referring to it every time):(A) 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 ϕ (cid:48) with the same number of crossingsas ϕ , in which C is embedded and all vertices of G , which lie inside C in ϕ ,lie outside C in ϕ (cid:48) and vice versa.Now we begin describing a family of MN-graphs based on the graph K , .By a link L ( x, y ) connecting two vertices x and y we mean any of thegraphs shown in Fig. 1 where { z, z } = { x, y } . We say that the vertices x and y are incident with the link. The links in Figs. 1(A) and (B) are called A-link and
B-link , respectively, and the one in Fig. 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 indicatedby thick lines in Fig. 1).By an A- chain of length n ≥ chain of length n ≥ n ≥
3, every graph obtained from that graph in the following way: for someintegers h , h , . . . , h t , where t ≥ ≤ h < h < · · · < h t ≤ n −
2, wereplace the link at the left of Fig. 2(e) by the link shown at the right, for i = 1 , , . . . , t . Note that, by definition, A- and B-chains have length at least2. We say that the chains in Figs. 2(a) and (c) connect the vertices v (0) and6igure 1: Links.Figure 2: A- and B-chains. v ( n ) which are called the end vertices of the chain. Two chains are adjacent ifthey share a common end vertex. A-chains and B-chains will be representedas shown in Figs. 2(b) and (d), respectively, where the arrow points to theend vertex incident with the base link. The vertices v (0) , v (1) , v (2) , . . . , v ( n )are the core vertices of the chains. Every free cycle of a link contains exactlyone core vertex. The two edges of a free cycle C incident to the core vertexare the core-adjacent edges of C . It is easy to see that two edges e and e (cid:48) ofa chain separate the end vertices of the chain if and only if the edges are thecore-adjacent edges of a free cycle of a link of the chain.By a subchain of a chain shown in Figs. 2(a) and (c) we mean a subgraphof the chain consisting of links incident with v ( i ) and v ( i + 1) for all i = m, m + 1 , . . . , m (cid:48) − ≤ m < m (cid:48) ≤ n . We say that the subchain7igure 3: A chain graph G and 1-planarity of the graph G − e . connects the vertices v ( m ) and v ( m (cid:48) ).A chain graph is any graph obtained from K , by replacing three of itsedges incident with the same vertex by A-chains and three edges incidentwith another vertex in the same chromatic class by B-chains, where thechains can have arbitrary lengths ≥
2. These changes are to be made asshown in Fig. 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 arecalled the Ω- edges .We will show that every chain graph is an MN-graph.
Lemma 1
Let G be a chain graph and e ∈ E ( G ) . Then G − e is -planar.Proof. If e is an Ω-edge, then G − e is planar and hence 1-planar. Supposenow that e is not an Ω-edge. By symmetry, we may assume that e belongsto an A- or B-chain incident to Ω(2). If e is the “middle” edge of a B-link,8hen Fig. 3(b) shows that the corresponding B-chain can be crossed by anA-chain, and it is easy to see that this can be made into a 1-immersion of G − e . In all other cases, 1-immersions are made by crossing the link L whoseedge e is deleted with the edges incident with the vertex Ω. The upper rowin Fig. 3(c) shows the cases when L is a base link. The lower row covers thecases when L is an A-link or a B-link. The edge e is shown in all cases asthe dotted edge. (cid:3) Our next goal is to show that chain graphs are not 1-planar. In whatfollows we let G be a chain graph and ϕ a (hypothetical) 1-immersion of G . Lemma 2
Let ϕ be a -immersion of a chain graph G such that the numberof crossings in ϕ is minimal among all -immersions of G . If L is a link inan 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 ab and cd cross,there is a disk D having a, c, b, d on its boundary, and D contains these twoedges but no other points of G . In 1-immersions with minimum number ofcrossings this implies that no other edges between the vertices a, c, b, d arecrossed. Similarly, if L = L ( z, ¯ z ) is a link in a chain, and an edge incidentwith z crosses an edge incident with ¯ z , the whole link L can be drawn in D without making any crossings. This shows that the only possible cases fora crossing of two edges e, f in L are the following ones, where we take thenotation from Figure 1 and we let u be the vertex of L that is not labeled inthe figure:(a) L is a B-link and e = zv , f = uw .(b) L is a B-link and e = ¯ zv , f = uw .(c) L is a base link and e = zv , f = uw .(d) L is a base link and e = vw , f = ¯ zu .(e) L is a base link and e = ¯ zw , f = vu .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 neigbors on theboundary 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 boundaryof D contains vertices z, u, v, w in this order and that u and v both haveprecisely one additional neighbor ¯ z outside of D . Therefore, we can turn ϕ into another 1-immersion of G by swapping u and v and only redraw the9igure 4: Eliminating the crossing between the edges e and f .edges inside D (see Fig. 4(b)). However, this eliminates the crossing in D and yields a 1-immersion with fewer crossings, a contradiction. (cid:3) Lemma 3
Let ϕ be a -immersion of a chain graph G such that the numberof crossings in ϕ is minimal among all -immersions of G . If Π and Π (cid:48) arenonadjacent A- and B-chains, respectively, then in ϕ the following holds forevery 3-cycle C of Π : (i) The core vertices of Π (cid:48) either all lie inside or all lie outside C . (ii) If all core vertices of Π (cid:48) lie inside ( resp. outside ) C , then at most onevertex of Π (cid:48) lies outside ( resp. inside ) C .Proof. First we show (i). If C does not contain the vertex A , then everytwo core vertices of Π (cid:48) are connected by four edge-disjoint paths not passingthrough the vertices of C , hence (i) holds for C .Suppose now that C contains the vertex A and that core vertices of Π (cid:48) lie inside and outside C . Then there is a link L ( z, z ) of Π (cid:48) such that thevertex z lies inside and the vertex z lies outside C . We may assume withoutloss of generality that Π and Π (cid:48) are incident to the base vertices Ω(1) andΩ(2), respectively, and (taking (A) into account) that the vertex z (if z (cid:54) = B )separates the vertices B and z in Π (cid:48) (see Fig. 5(a), where in L ( z, z ) the dottedline indicates that the link has either edge εz or εz ; also if ¯ z = Ω(2), thenthe link indicated in Fig. 5(a) is a base link).The 3-cycle C crosses at least two edges of L ( z, z ). The vertex z (resp. z )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 V ( C ) or through the noncorevertices of L ( z, z ). 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 (Ω , Ω(2)) is the10igure 5: Cases in the proof of Lemma 3.third edge that crosses C . We conclude that C crosses exactly two edges of L ( z, z ) and the two edges separate z from z in L ( z, z ). Thus, the two edgesare the core-adjacent edges of the free cycle of L ( z, z ). Hence, in ϕ , the link L ( z, z ) is 1-immersed as shown in Fig. 5(b), where the dotted edges indicatealternative possibilities for the position of z (at top) or z (at bottom).Let v, ¯ v be the vertices of C different from A and let x be the fourthvertex of the link containing C . The vertex x is connected to Ω(1) by twoedge-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 L ( z, z ). Suppose11xactly one of v and v is inside the free cycle. If we are in the case of thebottom of Fig. 5(b), then the path vxv can not lie inside C , a contradiction.In the case of the top of Fig. 5(b), if the path vxv lies inside C , then x must lie inside a 3-cycle Q of L ( z, z ) 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 , v ,and the vertices of Q .If either both v and v or none of them lie inside the free 4-cycle, then inthe case of Fig. 5(c) (resp. (d)), where we depict the two possible placementsof the nonbase link L ( z, z ), there are two edge-disjoint paths of G connecting A and Ω(3) (resp. A and Ω(2)) and not passing through z (resp. z ), a contra-diction. Reasoning exactly in the same way, we also obtain a contradictionwhen L ( z, z ) is a base link.Now we prove (ii). By (A), we may assume that all core vertices of Π (cid:48) lie inside C . By inspecting Fig. 1, it is easy to check that for every link L ,for every set W of noncore vertices of L such that | W | ≥
2, there are atleast four edges joining W with of V ( L ) \ W . Hence, if at least two noncorevertices belonging to the same link of Π (cid:48) lie outside C , then at least fouredges join them with the vertices of Π (cid:48) lying inside C , a contradiction. Everynoncore vertex of Π (cid:48) has valence at least 3. Hence if exactly n ( n ≥ (cid:48) lie outside C and if they all belong to different links,then at least 3 n ≥ (cid:48) lying inside C ,a contradiction. (cid:3) Two chains cross if an edge of one crosses an edge of the other.
Lemma 4
Let ϕ be a -immersion of a chain graph G such that the numberof crossings in ϕ is minimal among all -immersions of G . If Π and Π (cid:48) arenonadjacent A- and B-chains, respectively, then Π does not cross Π (cid:48) in ϕ .Proof. Suppose, for a contradiction, that Π crosses Π (cid:48) . Then an edge of alink L ( z, z ) of Π (cid:48) crosses a 3-cycle C = xv ¯ v of a link L of Π. Let C = ¯ xv ¯ v bea 3-cycle that is adjacent to C in L . (If L is not a base link, then L = C ∪ C and x, ¯ x are the core vertices of L .) By Lemma 3, we may assume that allcore vertices of Π (cid:48) lie outside C and that exactly one vertex u of Π (cid:48) liesinside C . The vertex u is 3-valent and is not a core vertex. The three edgesincident with u cross all three edges of C , hence C does not cross C . If C lies inside C , then one of the three edges incident with u crosses C and C , a12igure 6: Cases in the proof of Lemma 4.contradiction. If C lies inside C , then we consider the plane as the complexplane and apply the M¨obius transformation f ( z ) = 1 / ( z − a ) with the point a taken inside C but outside C . This yields a 1-immersion of G such that(a) C lies outside C , C lies outside C , and exactly one vertex u of Π (cid:48) liesinside C .Therefore, we may assume that in ϕ we have (a). Since the three edgesincident with u cross all three edges of C , at least two vertices of Π (cid:48) lieoutside C , hence, by Lemma 3, all core vertices of Π (cid:48) lie outside C and C .Since the edge v ¯ v in C ∩ C is crossed by an edge of L ( z, ¯ z ), also C containsprecisely one vertex u (cid:48) of L ( z, ¯ z ) and u (cid:48) has degree 3 and is not a core vertex.Adjacent trivalent vertices u, u (cid:48) cannot be contained in a base link. There-fore, L ( z, z ) is not a base link. Let L ( z, z ) be depicted as shown in Fig. 6(a).Because of symmetry, we may assume that u = w and u (cid:48) = ε , and that thecrossings are as shown in Fig. 6(b) and (c).In the case of Fig. 6(b) the adjacent vertices z and w of L ( z, z ) areseparated by the 3-cycle zwε , whose edges are crossed by three edges differentfrom the edge zw , a contradiction.Consider the case in Fig. 6(c). If x and x are core vertices of Π, thenthey are separated by the 3-cycle zwε of Π (cid:48) , a contradiction, since there are4 edge-disjoint paths between x and x that avoid this 3-cycle (the 3-cycledoes not contain the vertex B ).Suppose that x and x are not two core vertices. This is possible only when L is a base link. The 3-cycle zwε of L ( z, z ) crosses the three edges joiningthe vertex v of L with three vertices x , v , and x of L . The fifth vertex of L is adjacent to at least one vertex from x , v , and x , hence it lies outside zwε and is not adjacent to v . Thus, v has valence 3 in L . If v is a core vertex,13hen, since the 3-cycle zwε does not contain the vertex B , v is connected toone of the vertices x , v , and x by a path passing through B and not passingthrough the vertices of zwε , a contradiction. Hence v is not a core vertex.The link L has exactly one noncore vertex v of valence 3 and the vertices x and x are adjacent. Hence the 3-cycle xvx separates two core vertices z and z of Π (cid:48) , a contradiction, since z and z are connected by a path not passingthrough the vertices of the 3-cycle xvx . (cid:3) 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 not1-planar. Consider, for a contradiction, a 1-immersion ϕ of G and supposethat ϕ has minimum number of crossings among all 1-immersions of G .We know by Lemma 4 that non-adjacent chains do not cross each other.In the sequel we will consider possible ways that the Ω-edges cross with oneof the chains. Let us first show that such a crossing is inevitable. Claim 1
At least one of the chains contains a link L = L ( x, y ) such thatevery ( x, y ) -path in L is crossed by an Ω -edge.Proof. Suppose that for every link L = L ( x, y ), an ( x, y )-path in L is notcrossed by any Ω-edge. Then every chain contains a path joining its endvertices 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 K , . By Lemma 4,the only crossings between subdivided edges of this K , -subgraph are amongadjacent paths. However, it is easy to eliminate crossings between adjacentpaths and obtain an embedding of K , in the plane. This contradictioncompletes the proof of the claim. (cid:3) Let L = L ( x, y ) be a link in an A- or B-chain Π whose ( x, y )-paths areall crossed by the Ω-edges. We may assume that L is contained in a chainconnecting the vertex Ω(1) with A or B and that x separates y and Ω(1) inΠ. By Lemma 2, the induced 1-immersion of L is an embedding. The vertexΩ lies inside a face of L and all Ω-edges that cross L cross the edges of theboundary of the face. Considering the possible embeddings of L , it is easyto see that all ( x, y )-paths are crossed by Ω-edges only in the case when Ωlies inside a face of L whose boundary contains two core-adjacent edges of14 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, k ∈ { , } , then L has another cycle C (cid:48) that shares with C exactly k − C . If C lies inside C (cid:48) , then we consider the plane as the complex plane and apply theM¨obius transformation f ( z ) = 1 / ( z − a ) with the point a taken inside C (cid:48) butoutside C . This yields a 1-immersion of G such that C does not lie inside C (cid:48) and Ω lies inside C . Hence, we may assume that C does not lie inside C (cid:48) , that Ω lies inside C and two Ω-edges h and h (cid:48) cross two core-adjacentedges of C . Note that any two among the vertices A, B,
Ω(2) , Ω(3) are joinedby 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 h (cid:48) join the vertex Ω with basic vertices Ω( i ) andΩ( j ), respectively. If the third basic vertex Ω( (cid:96) ) is Ω(2) or Ω(3), then Ω(2)and Ω(3) lie inside different faces of L , a contradiction, hence Ω( (cid:96) ) = Ω(1).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, thenΩ(1) is not inside C .Now the embeddings of possible links L (so that we can join the verticesΩ 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.(a): In this case, L is a base link and x = Ω(1). Consider two edge-disjoint paths in a chain Π joining Ω(1) with the vertex A or B which is notincident with Π. These paths must cross the edges e and f indicated in thefigure. Let a be the edge crossing e and b be the edge crossing f . It is easyto see that a and b cannot be both incident with Ω(1) since Ω(1) is incidentwith three edges of the base link in the chain Π. The 1-planarity impliesthat the edges a, b and the vertex Ω(1) separate the graph G . Therefore, a and b are core-adjacent edges of a link in the chain Π. If the edge g (shownin the figure) is crossed by an edge c of Π, then also a, c and Ω(1) separatethe graph. Thus a, c would be core-adjacent edges in a link in Π as well, acontradiction. But if g is not crossed and a is not incident with Ω(1), thenthe edge a and the vertex Ω(1) separate G , a contradiction. If a is incidentwith Ω(1), then b is not (as proved above). Now we get a contradiction byconsidering the separation of G by the edge b and the vertex Ω(1).(b): In this case, L is a B-link, the cycle C (cid:48) lies inside C and Ω(1) isinside C but outside C (cid:48) (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)15igure 7: The Ω-edges crossing a linkby two edge-disjoint paths not passing through the vertices of C , and y isconnected to the vertex B (lying outside C ) by two edge-disjoint paths notpassing through the vertices of C . Again, consider two edge-disjoint pathsin the A-chain Π 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), wesee that the edges a , b and the vertex Ω(1) separate G , thus a and b arecore-adjacent edges of a free cycle of a link of Π. This free cycle has length 3and 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 .(c): In this case, L is a B-link, the cycle C (cid:48) lies outside C and Ω(1) isinside C . The vertex Ω(1) is connected by 4 edge-disjoint paths with vertices x and A such that the paths do not pass through noncore vertices of L , acontradiction.(d) and (e): In these cases, L is a B-link and Ω(1) lies inside a 3-cycleof 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.16f): In this case, L is an A-link and the chain Π is joining Ω(1) withthe vertex A . Again, consider two edge-disjoint paths in the B-chain Πjoining 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 a, b and the vertex Ω(1) separate G , thus a, b are core-adjacent edges in a link L (cid:48) = L ( z, z (cid:48) ) in Π. Let z be the core vertex of L (cid:48) incident with the edges a = zp and b = zq . Note that in L (cid:48) , vertices p, q have two common neighbors,a vertex r of degree 3 in G and the core vertex z (cid:48) , and that z (cid:48) is adjacentto r . Inside L we have the subchain Π (cid:48) of the A-chain Π connecting x withΩ(1). Hence, the free cycle of L (cid:48) containing the edges a and b must havelength at least 4 (that is, L (cid:48) is not a base link) and z is immersed outside L ,while p, q, r, z (cid:48) are inside. The subchain Π (cid:48) has two edge-disjoint paths thatare crossed by the paths prq and pz (cid:48) q . Each of the paths prq and pz (cid:48) q crossescore-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 rz (cid:48) would need to cross one edge of each of these two freecycles, hence the two free cycles are adjacent, that is, they are the two freecycles of an A-link of Π. Then the vertex z (cid:48) lies inside one of the two freecycles and the free cycle separates z (cid:48) from Ω(1), a contradiction, since z (cid:48) isconnected with Ω(1) by a B-subchain. (cid:3) The following theorem shows how chain graphs can be used to constructexponentially many nonisomorphic MN-graphs of order n . Theorem 2
For every integer n ≥ , there are at least ( n − / noniso-morphic MN-graphs of order n .Proof. The A-chain of length t has 3 t + 2 vertices and a B-chain of length t has 4 t + 1 vertices. Consider a chain graph whose three A-chains have length2, 2, and (cid:96) ≥
2, respectively, and whose B-chains have length 2, 3, and t ≥ (cid:96) + 4 t vertices. Applying the modificationshown in Fig. 2(e) to links of the two B-chains of the graph which have lengthat least 3, we obtain 2 t − nonisomorphic chain graphs of order 35 + 3 (cid:96) + 4 t ,where (cid:96) ≥ t ≥
4. We claim that for every integer n ≥
63, thereare integers 2 ≤ (cid:96) ≤ t ≥ n = 35 + 3 (cid:96) + 4 t . Indeed, if n ≡ , , , (cid:96) = 3 , , ,
4, respectively. If n = 35 + 3 (cid:96) + 4 t ,where 2 ≤ (cid:96) ≤
5, then t ≥ n/ − /
4. Hence, there are at least 2 n − nonisomorphic chain graphs of order n ≥
63. Since every chain graph is aMN-graph, the theorem follows. (cid:3) PN-graphs
By a proper graph is a planar graph that does nothave proper 1-immersions. In this section we describe a class of PN-graphsand construct some graphs of the class. They will be used in Section 4 toconstruct MN-graphs.For every cycle C of G , denote by N ( C ) the set of all vertices of the graphnot belonging to C but adjacent to C . Two disjoint edges vw and v (cid:48) w (cid:48) of agraph G are paired if the four vertices v, w, v (cid:48) , w (cid:48) are all four vertices of twoadjacent 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 aninduced cycle in G and G − V ( C ) is connected. If G is 3-connected andplanar, then the face boundaries in its (combinatorially unique) embeddingin the plane are precisely its peripheral cycles. Theorem 3
Suppose that a 3-connected planar graph G satisfies the follow-ing conditions: (C1) Every vertex has degree at least and at most . (C2) Every edge belongs to at least one -cycle. (C3) Every -cycle is peripheral ( in other words, there are no separating 3-cycles ) . (C4) Every -cycle is adjacent to at most one other -cycle. (C5) No vertex belongs to three mutually edge-disjoint -cycles. (C6) Every -cycle is either peripheral or is the boundary of two adjacenttriangular faces ( this means that there are no separating 4-cycles ) . (C7) For every -cycle C , any two vertices of V ( G ) \ ( V ( C ) ∪ N ( C )) areconnected by four edge-disjoint paths not passing through the verticesof C . (C8) If an edge vw of a nontriangular peripheral cycle C is paired with anedge v (cid:48) w (cid:48) of a nontriangular peripheral cycle C (cid:48) , then: (i) C and C (cid:48) have no vertices in common; (ii) any two vertices a and a (cid:48) of C and C (cid:48) , respectively, such that { a, a (cid:48) } (cid:54)⊆ { v, w, v (cid:48) , w (cid:48) } are non-adjacent and are not connected bya path aba (cid:48) of length , where b does not belong to C and C (cid:48) . G does not contain the subgraphs shown in Fig. 8 ( in this figure, -valent ( resp. -valent ) vertices of G are encircled ( resp. encircled twice ) andthe two starred vertices can be the same vertex ) .Then G has no proper -immersion.Proof. Denote by f the unique plane embedding of G . Suppose, for acontradiction, that there is a proper 1-immersion ϕ of G . Below we considerthe 1-immersion and show that then G has a subgraph which is excluded by(C8) and (C9), thereby obtaining a contradiction. In the figures below, theencircled letter f (resp. ϕ ) at the top left of a figure means that the figureshows a fragment of the plane embedding f (resp. 1-immersion ϕ ). Lemma 5 In ϕ , there is a -cycle such that there is a vertex inside and avertex outside the cycle.Proof. The 1-immersion ϕ has crossing edges e and e (cid:48) . By (C2), the crossingedges belong to different 3-cycles. If the 3-cycles are nonadjacent, then weapply the following obvious observation:(a) If two nonadjacent 3-cycles D and D (cid:48) cross each other, then there isa vertex of D inside and outside D (cid:48) .If e = xy and e (cid:48) = x (cid:48) y (cid:48) belong to adjacent 3-cycles xyy (cid:48) and x (cid:48) yy (cid:48) , respec-tively (see Fig. 9), then, by (C4), there are nontriangular peripheral cycles C and C (cid:48) containing e and e (cid:48) , respectively. The cycles C and C (cid:48) intersectat some point δ different from the intersection point of edges e and e (cid:48) . By(C8)(i), the two cycles do not have a common vertex, hence δ is the intersec-tion point of two edges. By (C2), these two edges belong to some 3-cycles, D and D (cid:48) . Property (C8)(ii) implies that D and D (cid:48) are nonadjacent 3-cycles.By (a), the proof is complete. (cid:3) Lemma 6 If C = u u u is a -cycle such that there is a vertex inside anda vertex outside C , then there is only one vertex inside C or only one vertexoutside C , and this vertex belongs to N ( C ) .Proof. By (C7), we may assume that all vertices of V ( G ) \ ( V ( C ) ∪ N ( C ))lie outside C . Then there can be only vertices of N ( C ) inside C . To provethe lemma, it suffices to show the following:(a) For every Q ⊆ N ( C ), | Q | ≥
2, at least four edges join vertices of Q to vertices in V ( G ) \ ( V ( C ) ∪ Q ).By (C1), every vertex of Q has valence at least 4. By (C4), every vertexof N ( C ) is adjacent to at most two vertices of C . We claim that if a vertex v ∈ N ( C ) is adjacent to two vertices u and u of C , then v is not adjacentto other vertices of N ( C ). Suppose, for a contradiction, that v is adjacent toa vertex w ∈ N ( C ). Then, by (C4), the vertex w can be adjacent only to u and the 4-cycle vwu u 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 u v , u u , and u u are two edges of a peripheral cycle, hence u has valence3, contrary to (C1).Now to prove (a) it suffices to prove the following claim:(b) For every Q ⊆ N ( C ), | Q | = 1 (resp. | Q | ≥ Q is adjacent to exactly one vertex of C , at least 2 (resp. 4) edges joinvertices of Q to vertices of V ( G ) \ ( V ( C ) ∪ Q ).The claim is obvious for | Q | ∈ { , } . For | Q | = 3, it suffices to showthat the three vertices of Q are not all pairwise adjacent. Suppose, for acontradiction, that the vertices v , v , and v of Q are pairwise adjacent.Then, by (C4), the three vertices of Q are not adjacent to the same vertexof C . Let v and v be adjacent to the vertices u and u of C , respectively.Then any two of the edges v u , v v , and v v are two edges of a 3-cycle(peripheral cycle) or a 4-cycle which is not the boundary of two adjacent3-cycles, so by (C6), that 4-cycle is also peripheral. Hence, v has valence 3,contrary to (C1). 20or | Q | ≥
4, it suffices to show that no vertex of Q is adjacent to threeother vertices in Q . Suppose, for a contradiction, that v ∈ Q is adjacent to w , w , w ∈ Q . If v is adjacent to u , then the edge vu belongs to threecycles D , D , and D such that for i = 1 , ,
3, the cycle D i contains edges vu and vw i , has length 3 or 4, and if D i has length 4, then D i is not theboundary of two adjacent 3-cycles. By (C3) and (C6), these three cycles areperipheral. This contradiction completes the proof of (b). (cid:3) Suppose that a vertex h belongs to two adjacent 3-cycles hvw and hvw (cid:48) .Since deg( h ) ≥ h is adjacent to a vertex u (cid:54)∈ { v, w, w (cid:48) } . By (C2), the edge hu belongs to a 3-cycle huu (cid:48) . By (C4), u (cid:48) (cid:54)∈ { v, w, w (cid:48) , u } . Hence, we havethe following:(B) If an edge e is contained in two 3-cycles of G , then both endverticesof e have valence at least 5.In the remainder of the proof of Theorem 3, we will show that any twocrossing edges of the proper 1-immersion ϕ belong to a subgraph that isexcluded by (C8) and (C9).By Lemma 5, there is a 3-cycle C = xyz such that there is a vertex insideand 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 B = vuw disjoint from C . Let x, a , a , . . . , a t ( t ≥
3) be all vertices adjacent to v . Suppose there is a 3-cycle D = va i b , where b ∈ { x, y, z } . If b ∈ { y, z } , then the 3-cycle xvb isadjacent to two 3-cycles C and D , contrary to (C4). Hence D = va i x . By(C4), at most two vertices of { a , a , . . . , a t } are adjacent to x . Hence, thereis a vertex a j such that a 3-cycle B containing the edge va j is disjoint from C . Since there is only one vertex v inside C , exactly two edges of B crossedges of C . First, suppose that B separates x from y and z (Fig. 10(a)). By(C2), the edge xv belongs to a 3-cycle R = xva . If a (cid:54)∈ { y, z, u, w } , then twoof the vertices y, z, u, w 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 ofgenerality, that a = z (Fig. 10(c)). Then the vertex x belongs to two adjacent3-cycles, C and vxz , hence, by (B), deg( x ) ≥
5. By (C4), x is not adjacentto u or w . Since x is the only vertex inside B , x has valence at most 4, acontradiction. Hence, B can not separate x from y and z .21igure 10: The 3-cycle B separates x from y and z .Figure 11: The 3-cycle B separates y and z .Now suppose that B separates the vertices y and z , and, without loss ofgenerality, 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 b ∈ { u, w } . But thenthe 3-cycle xzv is adjacent to two 3-cycles, C and vbz , contrary to (C4). If z is adjacent to u and w , then the edge xz belongs to three peripheral cycles, C , xvuz , and xvwz , a contradiction, since every edge belongs to at most twoperipheral 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 Fig. 11(a), so we have to considertwo 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 isincident with exactly three peripheral cycles and has valence 3, a contra-diction. 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 shownin Fig. 11(d). By (C9), at least one of the vertices u and x , say u , is not4-valent. Then, by (C5), at least one of the edges au and uw belongs to two3-cycles. Here we have two subcases to consider. Subcase 1.1. The edge uw belongs to two 3-cycles uwv and uwb ( Fig. 12 ( a )) . Now, a is the only vertex inside the 3-cycle uwb (Fig. 12(b)). By (C4), a uw belongs to two 3-cycles.is non-adjacent to w and b , hence a has valence 4. The edges yz and za (seeFig. 12(a)) belong to a nontriangular peripheral cycle yzac . . . . The edge ac belongs to a 3-cycle acd and b is the only vertex inside the 3-cycle acd . Thevertex b is not adjacent to c , since the 4-cycle bcau can not be peripheral (seeFig. 12(a)). Since b has valence at least 4, b is adjacent to d and has valence4. The 4-cycle dbua is peripheral, so we obtain a subgraph of G shown inFig. 12(c). Note that, by (C3)–(C6), the vertex at the top of Figure 12(c) isdifferent from all other vertices shown in the figure. This contradicts (C9). Subcase 1.2. The edge au belongs to two 3-cycles auz and aub ( Fig. 13 ( a )) . Since b has valence at least 4, b lies outside the 3-cycle auz (Fig. 13(b)).The edges yz and za (resp. wu and ub ) belong to a nontriangular peripheralcycle C (resp. C ). The cycles C and C are paired. In ϕ , the crossingpoint of edges uw and az is an intersection point of C and C . The cycles C and C have at least one other crossing point, denote this intersectionpoint by δ . By (C8)(i), C and C are vertex-disjoint, hence, δ is the crossingpoint of C and an edge h h of C (Fig. 13(c)). The edge h h is not theedge uw and belongs to a 3-cycle h h h . If h belongs to C , then in theembedding f the edge h h is a chord of the embedded peripheral cycle C and thus { h , h } is a separating vertex set of G . But G is 3-connected, acontradiction. Hence h does not belong to C .Now suppose that h h (cid:54) = bu . We have h (cid:54)∈ { a, z, b, u } . By (C8)(ii), h (cid:54) = a , a is not adjacent to h and h , and C does not pass through h (that is, h does not belong to C and C ). Hence, a vertex s of C lies insidethe 3-cycle h h h (see Fig. 13(c)). By (B), deg( a ) ≥
5. If s = a , then, since s is the only vertex inside the 3-cycle h h h , a is adjacent to at least one of h and h , a contradiction. Hence, s (cid:54) = a . Since z is the only vertex insidethe 3-cycle B , and the 3-cycle h h h is not B (since h h (cid:54) = uw ), we have s (cid:54) = z . Now, since s has valence at least 4, s is adjacent to at least one of the23igure 13: The edge au belongs to two 3-cycles.vertices h , h , and h , contrary to (C8)(ii). Hence h h = bu and h = a .We have C = yzah . . . and the edge ah belongs to a 3-cycle ahd (seeFigs. 13(d) and (e)). Considering Fig. 13(e), if there is a vertex inside the3-cycle auz , the vertex has valence at most 3, a contradiction. Hence no edgecrosses the edge au . If h lies inside the 3-cycle aub , then, by (C4), h is notadjacent to b and u , h has valence at most 3, a contradiction. Hence, h liesoutside the 3-cycle aub and b is the only vertex inside the 3-cycle ahd (seeFig. 13(e)). Note that the edge ah belongs to C , hence C does not crossboth bu and ab .By (C4), b is not adjacent to d and h , hence b has valence 4 and belongsto a 3-cycle btt (cid:48) disjoint from aub (Fig. 13(f)), where C = wubt . . . . Now d is the only vertex inside the 3-cycle btt (cid:48) (Fig. 13(e)). By (C8)(ii), d is notadjacent to t . Since d has valence at least 4, d is adjacent to t (cid:48) and has24igure 14: The edge aw belongs to two 3-cycles.valence 4. The 4-cycle dt (cid:48) ba is peripheral. We obtain a subgraph of G shownin Fig. 13(g), contrary to (C9). The obtained contradiction completes theproof in the Case 1. Case 2. The vertex z is adjacent to w . This case is dealt in much the same way as the Case 1. Here we describeonly what figures will be in the Case 2 instead of the Figs. 11-13 in the Case1. We hope that the reader is familiar enough with the proof of the Case 1to supply the missing details himself.In Figs. 11(a), 11(d), 12(a), and 12(c) interchange the letters u and w .In Figs. 11(c) and 12(b) replace the edge uz by the edge zw . In Figs. 13(a),(d), (f), and (g) interchange the letters u and w . Figs. 14(b), (c), and (e) arereplaced by the Figs.(a), (b), and (c), respectively. (cid:3) Denote by A the class of all 3-connected plane graphs G satisfying theconditions (C1)–(C9) of Theorem 3. In what follows we show how to con-struct some graphs in A and, as an example, we shall give two infinite familiesof graphs in A , one of which will be used in Section 4 to construct MN-graphs.First we describe a large family of 3-connected plane graphs satisfyingthe conditions (C1)–(C6) and (C8) of Theorem 3.Given a 4-valent vertex v of a 3-connected plane graph, two peripheralcycles C and C (cid:48) containing v are opposite peripheral cycles at v if C and C (cid:48) have no edges incident with v in common.Denote by H the class of all 3-connected (simple) planar graphs H satis-fying the following conditions (H1)–(H4):(H1) Every vertex has valence 3 or 4.(H2) H has no 3-cycles. 25H3) Every 4-cycle is peripheral.(H4) For every 4-valent vertex v and for any two opposite peripheral cycles C and C (cid:48) at v , no edge joins a vertex of C − v to a vertex of C (cid:48) − v .A plane graph G is a medial extension of a graph H ∈ H if G is obtainedfrom H in the following way. The vertex set of G is the set { v ( e ) : e ∈ E ( H ) } .The edge set of G is defined as follows. For every 3-valent vertex v of H , if e , e , e are the edges incident with v , then in G the vertices v ( e ), v ( e ), and v ( e ) are pairwise adjacent (the three edges of G are said to be associatedwith v ). For every 4-valent vertex w of H , if ( e , e , e , e ) is the cyclic orderof edges incident to w around w in the plane, then G contains the edges ofthe 4-cycle v ( e ) v ( e ) v ( e ) v ( e ), and contains either the edge v ( e ) v ( e ) orthe edge v ( e ) v ( e ); these five edges of G are said to be associated with the4-valent vertex of H . Note that G can be obtained from the medial graphof H by adding a diagonal to every 4-cycle associated with a 4-valent vertexof H . Lemma 7
Every medial extension G of any graph H ∈ H is a 3-connectedplanar graph satisfying the conditions (C1)–(C6) and (C8) of Theorem 3.Proof. By the construction of G , if { v ( e ) , v ( e (cid:48) ) } is a separating vertex setof G , then the graph H − e − e (cid:48) is disconnected, a contradiction, since H is3-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 thatare not induced by the peripheral cycles of H are the 3-cycles formed by theedges associated with the vertices of H .It is easy to see that G satisfies (C1)–(C5). To show that G satisfies(C6), let J = v ( e ) v ( e ) v ( e ) v ( e ) be a 4-cycle of G . By the constructionof G , if vertices v ( e ) and v ( e (cid:48) ) of G are adjacent, then the edges e and e (cid:48) of H are adjacent, too. Since in H no three edges among e , e , e , e forma cycle (by (H2)), these four edges either form a 4-cycle (in this case J isperipheral) 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 C (cid:48) be nontriangularperipheral cycles of G such that an edge a of C is paired with an edge a (cid:48) of C (cid:48) .Then, by the construction of G , the peripheral cycles C and C (cid:48) are inducedby peripheral cycles C and C (cid:48) of H , respectively, that are opposite at some26igure 15: Two paths of G associated with a path of H .4-valent vertex u . If C and C (cid:48) have a common vertex v ( e ), then C and C (cid:48) have a common edge e , hence H has a separating vertex set { u, w } , where w is a vertex incident to e , a contradiction, since H is 3-connected. Nowsuppose that G has an edge joining a vertex v ( e ) of C to a vertex v ( e (cid:48) ) of C (cid:48) such that at least one of the vertices v ( e ) and v ( e (cid:48) ) is not incident to a and a (cid:48) .Then the edges e and e (cid:48) are incident to the same vertex w of H and the cycles C and C (cid:48) pass through w . It follows that { u, w } is a separating vertex set of H , a contradiction. Next suppose that G has a path v ( e ) v ( b ) v ( e (cid:48) ) connectinga vertex v ( e ) of C to a vertex v ( e (cid:48) ) of C (cid:48) such that v ( b ) does not belong to C and C (cid:48) , and at least one of the vertices v ( e ) and v ( e (cid:48) ) is not incident to a or a (cid:48) . If in H the edges e , b , and e (cid:48) are incident to the same vertex w , then { u, w } is a separating vertex set of H , a contradiction. If in H the edges a and b (resp. b and e (cid:48) ) are incident to a vertex w (resp. w (cid:48) ) such that w (cid:54) = w (cid:48) ,then the edge b joins the vertex w of C with the vertex w (cid:48) of C (cid:48) , contrary to(H4). Hence G satisfies (C8). The proof is complete. (cid:3) There are medial extensions of graphs in H 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 showthat a medial extension G of H ∈ H satisfies (C7) it is convenient to proceedin the following way. Subdivide every edge e of H by a two-valent vertex v ( e ) of G . We obtain a graph H whose vertex set is V ( H ) ∪ V ( G ) where thevertices of V ( G ) are all 2-valent vertices of H . We will consider paths of G associated with paths of H connecting 2-valent vertices.Two paths P and P (cid:48) of H are H-disjoint if P ∩ P (cid:48) ∩ V ( H ) = ∅ , i.e., P ∩ P (cid:48) ⊆ V ( G ).Consider a path P = v ( e ) w v ( e ) w v ( e ) . . . w n − v ( e n ) in H where w ,w , . . . , w n − are the H -vertices on P . It is easy to see that the edges of G associated with the vertices w , w , . . . , w n − of H contain two edge-disjointpaths connecting in G the vertices v ( e ) and v ( e n ) (see Fig. 15); any two suchpaths in G are said to be associated with the path P of H . Since H has no27ultiple edges, every edge of G is associated with exactly one vertex of H .Hence, if P and P (cid:48) are H -disjoint paths in H , each of which is connecting2-valent vertices, then every path in G associated with P is edge-disjointfrom every path in G associated with P (cid:48) . As a consequence, we have thefollowing conclusion:(C) If H has a cycle containing 2-valent vertices v ( e ) and v ( e (cid:48) ), then G has four edge-disjoint paths connecting v ( e ) and v ( e (cid:48) ).The fact that a path in H gives rise to two edge-disjoint paths in G (pathsassociated 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 H is C -independent if the path doesnot contain vertices of C . When checking (C7) for a medial extension G of H ∈ H , given a 3-cycle C of G and two 2-valent vertices x, y ∈ V ( G ) \ ( V ( C ) ∪ N ( C )) ⊂ V ( H ), four C -independent edge-disjoint paths P , P , P ,and P of G connecting x and y in G will be represented in some subsequentfigures (see, e.g., Figure 17) in the following way. The edges of the pathsincident to vertices of N ( C ) are depicted as dashed edges joining 2-valentvertices, the dashed edges are not edges of H (see, for example, Fig. 17(b),where the edges of H are given as solid lines). All other edges of the paths arerepresented by paths in H . If X is a subpath of P i such that X is associatedwith a path X in H , then X is represented by a dashed line passing near theedges of X in H . If X i and X j are subpaths of P i and P j (where possibly i = j ), respectively, such that X i and X j are edge-disjoint paths associatedwith a path X of H , then X i and X j are represented by two (parallel) dashedlines passing near the edges of X in H . Using these conventions, the readerwill be able to check that the depicted dashed paths and edges in the figureof H represent four C -independent edge-disjoint paths of G connecting x and y . Now we describe some graphs in A . Let us recall that graphs in A areprecisely those 3-connected planar graphs that satisfy conditions (C1)–(C9).To simplify the arguments, we construct graphs with lots of symmetries sothat, for example, to check the condition (C7) we will have to consider onlytwo 3-cycles of a graph.For n ≥
6, let H n ∈ H be the Cartesian product of the path P of length2 and the cycle C n of length n (see 16(a)). Let G n be the medial extensionof H n shown in Fig. 16(b). Lemma 8
Each graph G n , n ≥ , is a PN-graph. H n ∈ H and its extension G n . Proof.
We show that G n satisfies (C1)–(C9) for every n ≥ G n satisfies (C1)–(C6) and (C8). Every 4-gonal face of G n is incident with a 6-valent vertex and G n has no 5-valent vertices, hence G n satisfies (C9).Now we show that G n satisfies (C7). Consider a fragment of H n shownin Fig. 17(a) (in the figure we introduce notation for some vertices and alsodepict in dashed lines some edges of G n ). Because of the symmetries of G n ,it suffices to consider the following two cases for a 3-cycle C , when checking(C7): Case 1. V ( C ) = { , , } . Then N ( C ) = { , , , , , , , , } . If we delete from H n the ver-tices 1 , , . . . ,
14, then the obtained graph has only one connected component U with a vertex in H n , and U is 2-connected. Hence, any two 2-valent ver-tices x and y of U belong to a cycle in U and then, by (C), G n has four C -independent edge-disjoint paths connecting x and y . Fig. 17(b) showsfour C -independent edge-disjoint paths in G n connecting vertices 1 and 4. Ifwe delete from H n the vertices { , , . . . , } \ { , } , then in the resultingnon-trivial connected component, for every vertex x ∈ V ( G n ) \ { , , . . . , } ,there is a path P connecting the vertices 1 and 4, and passing through x ;combining two edge-disjoint paths of G n associated with P and the twoedge-disjoint paths connecting 1 and 4, shown in Fig. 17(b), we obtain four C -independent edge-disjoint paths connecting the vertices 4 and x (and, anal-ogously, for the vertices 1 and x ). Now, because of the symmetries of H n , itremains to show that there are four C -independent edge-disjoint paths con-necting the vertex 4 with each of the vertices 15 , , ,
18; Fig. 17(c) shows29igure 17: Verifying (C7) for the graph G n .the paths (since n ≥ Case 2: V ( C ) = { , , } . We have N ( C ) = { , , , , , } . If we delete from H n the vertices1 , , . . . , , ,
13, then the obtained graph has only one connectedcomponent U with at least two vertices and U is 2-connected. Hence anytwo vertices x and y in H n that are both in U belong to a cycle of U andthen, by (C), G n has four C -independent edge-disjoint paths connecting x and y . It remains to show that for every vertex x of H n belonging to U , thereare four C -independent edge-disjoint paths connecting x and the vertex 13.These four paths are shown in Figs. 17(d)–(f), depending on the choice of x .We conclude that G n satisfies (C1)–(C9), hence G n is a PN-graph for every n ≥ (cid:3) Fig. 18 gives another example of an extension G (cid:48) n of a graph H (cid:48) n ∈ H .By Lemma 7, G (cid:48) n satisfies (C1)–(C6) and (C8). Since G (cid:48) n has no 4-cycles,it satisfies (C9). Using the symmetry of H (cid:48) n , one can easily check that forevery 3-cycle C of G (cid:48) n , if we delete from H (cid:48) n the vertices V ( C ) ∪ N ( C ) of G (cid:48) n , then the obtained graph has only one connected component U with atleast two vertices and U is 2-connected. Then, by (C), any two vertices of G (cid:48) n in U are connected by four C -independent edge-disjoint paths. Hence,30igure 18: The graph H (cid:48) n ∈ H and its extension G (cid:48) n . G (cid:48) n satisfies (C7) and is a PN-graph. In this section we construct MN-graphs based on the PN-graphs G n describedin Section 3.For m ≥
2, denote by S m , the graph shown in Fig. 19. The graphhas m + 1 disjoint cycles of length 12 m − B , B , . . . , B m asshown in the figure. The vertices of B are called the central vertices of S m and are labelled by 1 , , . . . , m − x ∈ { , , . . . , m − } , denote by x ∗ its “opposite” vertex x + (6 m −
1) if x ∈{ , , . . . , m − } and the vertex x − (6 m −
1) if x ∈ { m, m +1 , . . . , m − } .In S m , any pair { x, x ∗ } of central vertices is connected by a central path P ( x, x ∗ ) of length 6 m − m − m − m ≥ n ≥
0, denote by Φ m ( n ) the set of all (12 m − n , n , . . . , n m − ) of nonnegative integers such that n + n + · · · + n m − = n . For every λ ∈ Φ m ( n ), denote by S m ( λ ) the graph obtainedfrom S m by replacing, for every central vertex x ∈ { , , . . . , m − } , theeight edges marked by short crossings in Fig. 20(a) by 8(1 + n x ) new edgesmarked by crossings in Fig. 20(b) (the value x + 1 in that figure is to beconsidered modulo 12 m − S m ( λ ) has m − m − B , B , . . . , B m − and three (12 m − n )-cycles B m − , B m − , B m .We want to show that for every m ≥ λ ∈ Φ m ( n ), n ≥ S m ( λ ) is an MN-graph. 31igure 19: The graph S m . Lemma 9
For every m ≥ , n ≥ and λ ∈ Φ m ( n ) , the graph S m ( λ ) − e is -planar for every edge e .Proof. If we delete an edge of a central path, then the remaining 6 m − m − B in Fig. 19.If we delete one of the edges shown in Fig. 21(a) by a thick line, then thecentral path P ( x, x ∗ ) can be drawn outside B with 6 m − m − B . If we delete one of the two edges depicted in Fig. 21(a)by a dotted line, then Fig. 21(b) shows how to place the central vertex x sothat the path P ( x, x ∗ ) can be drawn outside B with 6 m − m − B . This exhibits all possibilities for the edge e (up tosymmetries of S m ) and henceforth completes the proof. (cid:3) 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 notcontain vertices and edges of G .Denote by J m − the graph obtained from the graph S m in Fig. 19 by32igure 20: Obtaining the graph S m ( λ ).Figure 21: The central path P ( x, x ∗ ) immersed outside B .33eleting the 2-valent vertices of all central paths and by deleting all verticeslying outside the cycle B m − . Lemma 10
For every m ≥ , J m − is a PN-graph.Proof. The graph J m − contains m − L , L , . . . , L m − isomor-phic to the PN-graph G m − such that for i = 1 , , . . . , m −
1, the graph L i contains the cycles B i − , B i , and B i +1 . Consider an arbitrary 1-immersion ϕ of J m − . Suppose that in the plane embedding of the PN-graph L in ϕ , the cycle B is the boundary cycle of the outer (12 m − L determines an embedding of thesubgraph of L bounded by the cycles B and B . Since L is a PN-graph,the subgraph of L bounded by B and B lies outside the cycle B . Rea-soning similarly, we obtain that for i = 3 , , . . . , m −
3, the subgraph of thePN-graph L i bounded by B i and B i +1 lies outside B i . As a result, ϕ is aplane embedding of J m − , hence J m − is a PN-graph. (cid:3) Denote by S m ( λ ) the graph obtained from S m ( λ ), where m ≥ λ ∈ Φ m ( n ), by deleting the 2-valent vertices of all central paths. Lemma 11
For every m ≥ , n ≥ and λ ∈ Φ m ( n ) , S m ( λ ) is a PN-graph.Proof. The graph S m ( λ ) contains a subgraph G isomorphic to the PN-graph G m − n and contains a subgraph G (cid:48) homeomorphic to the PN-graph J m − .The graph G contains the cycles B m − , B m − , and B m of S m ( λ ), and thegraph G (cid:48) contains the cycles B , B , . . . , B m − of S m ( λ ) and is obtained from J m − by subdividing the edges of the cycle B m − (by using n ϕ of S m ( λ ). In ϕ ,the graph G has a plane embedding and we shall investigate in which facesof the embedding of G lie the vertices of G (cid:48) . We shall show that they all liein the face of G bounded by the (subdivided) cycle B m − .In the graph S m ( λ ) the cycles B m − and B i , i ∈ { , , . . . , m − } areconnected by 24 m − G contains all vertices of B i in its interior.Any two vertices of B i are connected by six edge-disjoint paths in G (cid:48) − B m − . Therefore:(a) No 3- or 4-gonal face of G contains any vertex of the cycles B i , i =0 , , . . . , m −
3, in its interior. 34uppose that a vertex v of G (cid:48) does not belong to the cycles B i , i =1 , , . . . , m −
2, and lies inside a 3- or 4-gonal face F of G . By constructionof G (cid:48) , the vertex v is adjacent to two vertices w and w (cid:48) of some B j , j ∈{ , , . . . , m − } . By (a), w and w (cid:48) do not lie inside F , hence they lie,respectively, in faces F and F of G adjacent to F . However, at least one of F and F is 3- or 4-gonal, contrary to (a). Therefore, no 3- or 4-gonal faceof G contains any vertex of G (cid:48) − B m − . If there is a vertex of G (cid:48) − B m − outside B m , then G (cid:48) has two adjacent vertices such that one of them is eithera vertex of B m − or lies inside B m − , and the other vertex lies outside B m ,a contradiction, since the edge joining the vertices crosses at least 5 edgesof G . This implies that all vertices of G (cid:48) − B m − lie inside the face of G bounded by B m − . Hence G (cid:48) lies inside B m − and has a proper 1-immersionin ϕ . If in this 1-immersion of G (cid:48) we ignore the 2-valent vertices on the cycle B m − of S m ( λ ), then we obtain a proper 1-immersion of the PN-graph J m − ,a contradiction. (cid:3) By the paths of S m ( λ ) associated with any central vertex x we meanthe two paths shown in Fig. 22; one of them is depicted in thick line andthe other in dashed line. Every edge of S m ( λ ) not belonging to the cycles B , B , . . . , B m is assigned a type t ∈ { , , . . . , m } as shown in Fig. 22 suchthat for i = 1 , , . . . , m , the edges of type 2 i − i are all edges lyingbetween the cycles B i − and B i , and the edges of type 2 i − i ) areincident to vertices of B i − (resp. B i ).Suppose that there is a 1-immersion ϕ of S m ( λ ). By Lemma 11, S m ( λ ) isa PN-graph. Thus, ϕ induces an embedding of this graph. We shall assumethat the outer face F of this embedding is bounded by the cycle B m . Weshall first show that F is also a face of ϕ . To prove this, it suffices to seethat no central path can enter F .Any central vertex x is separated from F by 3 m − m cycles B , B , . . . , B m and 2 m − C , C , . . . , C m , where the cycle C i consists of all edges of type i ( i = 2 , , . . . , m ). The central path P = P ( x, x ∗ ) can have at most 6 m − P cannot enter F . If P lies between B and B m in ϕ , then it must cross 2(6 m −
2) pathsassociated either with 6 m − x + 1 , x + 2 , . . . , x ∗ − m − x ∗ + 1 , x ∗ + 2 , . . . , x − m − ϕ any central path either liesinside B or crosses some edges of S m ( λ ) but does not lie entirely between B and B m . 35igure 22: The paths associated with a central vertex and the types of edges.The main goal of this section is to show that S m ( λ ) has no 1-immersions(see Theorem 4 in the sequel). Roughly speaking, the main idea of the proofis as follows. Suppose, for a contradiction, that S m ( λ ) has a 1-immersion.Every central path can have at most 6 m − m − B . Then there is a central pathwhich crosses some edges of S m ( λ ). Let P be a central path with maximumnumber of such crossings. Since P can have at most 6 m − m − P and have to “goaround” P and, in doing so, one of the paths has to cross more edges of S m ( λ ) than P does, a contradiction.Before proving Theorem 4, we need some definitions and preliminaryLemmas 12 and 13.Consider a 1-immersion of S m ( λ ) (if it exists). If a central path P = P ( x, x ∗ ) does not lie inside B , consider the sequence δ , δ , . . . , δ r ( r ≥ δ = x and δ r = x ∗ , obtained by listing the intersection points ofthe path and B when traversing the path from the vertex x to the vertex x ∗ (here δ , δ , . . . , δ r − are crossing points). By a piece of P we mean thesegment of P from δ i to δ i +1 for some i ∈ { , , . . . , r − } ; denote the pieceby P ( δ i , δ i +1 ). A piece of P with an end point x or x ∗ is called an end piece of P at the vertex x or x ∗ , respectively. An outer piece of P is every piece of P B . Clearly, either P ( δ , δ ) , P ( δ , δ ) , P ( δ , δ ) , . . . or P ( δ , δ ) , P ( δ , δ ) , P ( δ , δ ) , . . . are all outer pieces of P . The end points δ and δ (cid:48) of an outer piece Π of P partition B into two curves A and A (cid:48) such that the curve A lies inside the closed curve consisting of Π and A (cid:48) (seeFig. 23). The central vertices belonging to A and different from δ and δ (cid:48) aresaid to be bypassed by Π and P (cf. Fig. 23). Lemma 12 If P ( x, x ∗ ) bypasses neither a central vertex y nor its oppo-site vertex y ∗ , and P ( y, y ∗ ) bypasses neither x nor x ∗ , then P ( x, x ∗ ) crosses P ( y, y ∗ ) .Proof. Suppose, for a contradiction, that P ( x, x ∗ ) does not cross P ( y, y ∗ ).For every outer piece of the two paths we can replace a curve of a pathcontaining the piece by a new curve lying inside B so that the path P ( x, x ∗ )(resp. P ( y, y ∗ )) becomes a new path P (cid:48) ( x, x ∗ ) (resp. P (cid:48) ( y, y ∗ )) connectingthe vertices x and x ∗ (resp. y and y ∗ ) such that the two new paths lie inside B and do not cross each other, a contradiction. How the replacements canbe done is shown in Fig. 24, where the new curves are depicted in thick line.(Note that in Fig. 24(b), since P ( y, y ∗ ) does not bypass x , the depicted piecesbypassing x belong to P ( x, x ∗ ).) (cid:3) By a type of an outer piece of a central path we mean the maximal typeof an edge of S m ( λ ) crossed by the path.For an outer piece Π of a central path P ( x, x ∗ ), denote by b (Π) the numberof central vertices bypassed by Π, and by ∆(Π) the number of intersectionpoints of Π and S m ( λ ), including the crossings at the end points of Π (exceptif an end point is x or x ∗ ). 37igure 24: Transforming paths P ( x, x ∗ ) into paths P (cid:48) ( x, x ∗ ). Lemma 13 If Π is an outer piece of type t of a central path P , then ∆(Π) − b (Π) ≥ t − τ, where τ = 1 if Π is an end piece and τ = 0 otherwise.Proof. The piece Π crossing an edge of type t has a point separated fromthe interior of B by (cid:98) t +12 (cid:99) + t edge-disjoint cycles: B , B , . . . , B (cid:98) t +12 (cid:99)− , andthe cycles C , C , . . . , C t , where the cycle C i consists of all edges of type i ( i = 1 , , . . . , t ). Thus, Π crosses each of these cycles twice, except thatfor an end piece, we may miss one crossing with C . The piece Π bypasses b (Π) central vertices, hence Π crosses 2 b (Π) paths associated with those b (Π)vertices. Hence we obtain two inequalities∆(Π) ≥ (cid:4) t +12 (cid:5) + 2 t − τ, (1)and ∆(Π) ≥ (cid:4) t +12 (cid:5) + 2 b (Π) − τ. (2)Now add (1) and (2), apply 2 (cid:98) t +12 (cid:99) ≥ t , then divide by 2 and rearrange toget the result. (cid:3) By the type of a central path not lying (entirely) inside B we mean themaximal type of the outer pieces of the path. If t different central pathsbypass 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, weobtain that one of the central paths crosses an edge of type at least t . Hencewe have: 38D) If t different central paths bypass the same central vertex, then oneof the paths has type at least t . Theorem 4
For every m ≥ and λ ∈ Φ m ( n ) , the graph S m ( λ ) is not -planar.Proof. Consider, for a contradiction, a 1-immersion ϕ of S m ( λ ) and a path P = P ( x, x ∗ ) of maximal type t > P ) be the number of crossing points of P and S m ( λ ),and let b ( P ) be the number of distinct central vertices bypassed by P anddifferent from x and x ∗ .There are at least 6 m − − b ( P ) different pairs { y, y ∗ } ( { y, y ∗ } (cid:54) = { x, x ∗ } )of central vertices such that P does not bypass y and y ∗ ; denote by P theset of the corresponding (at least 6 m − − b ( P )) paths P ( y, y ∗ ). If P ( x, x ∗ )does not bypass y and y ∗ , then P ( y, y ∗ ) either does not bypass x and x ∗ (inthis case P ( y, y ∗ ) crosses P ( x, x ∗ ) by Lemma 12) or bypasses at least one ofthe vertices x and x ∗ . Hence, we have6 m − − b ( P ) ≤ β + γ + ε, (3)where: β is the number of paths of P that cross P and do not bypass x or x ∗ ; γ is the number of paths of P that bypass x or x ∗ and do not cross P ; ε isthe number of paths of P that cross P and bypass x or x ∗ . We are interestedin the number γ + ε of paths of P that bypass x or x ∗ .The path P has at most 6 m − m − − ∆( P ) ≥ β + ε and, by (3), we obtain γ ≥ m − − b ( P ) − ( β + ε ) ≥ ∆( P ) − b ( P ) + 1 , whence γ + ε ≥ ∆( P ) − b ( P ) + 1 + ε. (4)Let Π , Π , . . . , Π (cid:96) ( (cid:96) ≥
1) be all outer pieces of P , and let Π be of themaximal type t . We have ∆( P ) = (cid:80) (cid:96)i =1 ∆(Π i ) and b ( P ) ≤ (cid:80) (cid:96)i =1 b (Π i ) (thevertices x and x ∗ can be bypassed by P , and some central vertices can be39ypassed by P more than once). By Lemma 13, ∆(Π i ) − b (Π i ) ≥ i = 1 , , . . . , (cid:96) . Hence, by (4) and Lemma 13, we obtain γ + ε ≥ (cid:96) (cid:88) i =1 ∆(Π i ) − (cid:96) (cid:88) i =1 b (Π i ) + 1 + ε ≥ ∆(Π ) − b (Π ) + 1 + ε ≥ (2 t + 1) − τ + ε, (5)where τ = 1 if Π is an end piece and τ = 0 otherwise. If Π is not an endpiece or ε ≥
1, then, by (5), γ + ε ≥ t + 1, hence one of the vertices x and x ∗ is bypassed by at least t + 1 paths of P . Now, by (D), one of the t + 1paths has type at least t + 1, a contradiction. Now suppose that Π is anend piece at the vertex x and ε = 0. Then every path of P either crosses P or bypasses x or x ∗ , and at least 2 t paths of P bypass x or x ∗ . If no oneof the 2 t paths bypasses x , then all the 2 t ≥ t + 1 paths bypass x ∗ and, by(D), one of the paths has type at least t + 1. If one of the 2 t paths, say, P (cid:48) ,bypasses x , then P (cid:48) has an outer piece Π (cid:48) that bypasses x and does not crossΠ (since ε = 0, P (cid:48) does not cross P ). The piece Π has type t and is an endpiece at x , hence Π (cid:48) has type at least t + 1, a contradiction. (cid:3) We have shown that every graph S m ( λ ), where m ≥ λ ∈ Φ m ( n ),is an MN-graph. These graphs have order (5 m − m −
2) + 5 n . Clearly,graphs S m ( λ ) and S m ( λ ), where λ ∈ Φ m ( n ) and λ ∈ Φ m ( n ), arenonisomorphic for m (cid:54) = m and for m = m and n (cid:54) = n . Claim 2
For any integers m ≥ and n ≥ , there are at least m − (cid:0) n +12 m − m − (cid:1) nonisomorphic MN-graphs S m ( λ ) , where λ ∈ Φ m ( n ) .Proof. The automorphism group of S m is the dihedral group D m − oforder 24 m −
4. Now the claim follows by recalling a well-known fact that | Φ m ( n ) | = (cid:0) n +12 m − m − (cid:1) . (cid:3) In this section we prove that testing 1-immersibility is NP-hard. This showsthat it is extremely unlikely that there exists a nice classification of MN-graphs. 40 heorem 5
It is NP-complete to decide if a given input graph is -immersible. Since 1-immersions can be represented combinatorially, it is clear that 1-immersability is in NP. To prove its completeness, we shall make a reductionfrom a known NP-complete problem, that of 3-colorability of planar graphsof maximum degree at most four [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 isto be tested. We shall show how to construct, in polynomial time, a relatedgraph G such that G is 1-immersible if and only if G is 3-colorable. We mayassume that G has no vertices of degree less than three (since degree 1 and2 vertices may be deleted without affecting 3-colorability).To construct G , we will use as building blocks graphs which have a unique1-immersion. These building blocks are connected with each other by edgesto form a graph which also has a unique 1-immersion. Then we add someadditional paths to obtain G .We say that a 1-planar graph G has a unique -immersion if, whenevertwo edges e and f cross each other in some 1-immersion, then they crosseach other in every 1-immersion of G , and secondly, if G • is the planar graphobtained from G by replacing each pair of crossing edges e = ab and f = cd by a new vertex of degree four joined to a, b, c, d , then G • is 3-connected (andthus has combinatorially unique embedding in the plane – the one obtainedfrom 1-immersions of G ).It was proved in [13] that for every n ≥
6, the graph with 4 n vertices and13 n edges shown in Fig. 25(a) has a unique 1-immersion. (To be precise, thepaper [13] considers the graph for even values of n ≥ n ≥ U-graph . Fig. 25(b) shows a designation of the U-graph usedin what follows. In the 1-immersion of the U-graph shown in Fig. 25, thevertices 1 , , , . . . , n − , n which lie on the boundary of the outer face of thespanning embedding (the boundary is called the outer boundary cycle of the1-immersed U-graph) are called the boundary vertices of the U-graph in the1-immersion. If a graph has a U-graph as a subgraph, then the U-graph iscalled the U-subgraph of the graph.Take two 1-immersed U-graphs U and U such that each of them hasthe outer boundary cycle of length at least 7, and construct the 1-immersedgraphs shown in Figs. 26(a) and (b), respectively, where by 1 , , . . . , U and U are connected by a (1)-grid (resp. (2)-grid). The vertices labeled1 , , . . . , basic vertices of the grid and for i = 1 , , . . . ,
7, the h -pathconnecting the vertices labeled i of the ( h )-grid, h ∈ { , } , is called the basicpath of the grid connecting these vertices. Let us denote the i th basic pathby P i . The paths P i − and P i , i = 2 , , . . . ,
7, are neighboring basic paths ofthe grid. For two basic paths P = P i and P (cid:48) = P j , 1 ≤ i < j ≤
7, denote by C ( P, P (cid:48) ) the cycle of the graph in Fig. 26 consisting of the two paths and ofthe edges ( i, i + 1) , ( i + 1 , i + 2) , . . . , ( j − , j ) of the two graphs U and U .By a U-supergraph we mean every graph obtained in the following way.Consider a plane connected graph H . Now, for every vertex v ∈ V ( H ), takea 1-immersed U-graph U ( v ) of order at least 28 · deg( v ) and for any twoadjacent vertices u and w of the graph, connect U ( u ) and U ( w ) by a (1)- or(2)-grid as shown in Fig. 27 such that any two distinct grids have no basicvertices in common. We obtain a 1-immersed U-supergraph.42igure 27: Constructing a U-supergraph.Figure 28: A 1-immersion of a subgraph. Theorem 6
Every U-supergraph M has a unique 1-immersion.Proof. It suffices to show the following:(a) The graph consisting of two U-graphs connected by an ( h )-grid, h =1 ,
2, has a unique 1-immersion.(b) In every 1-immersion ϕ of M , the edges of distinct grids do not inter-sect.Note that M contains no subgraph which can be 1-immersed inside theboundary cycle of a 1-immersed U-subgraph of M in a 1-immersion of M as shown in Fig. 28 in dashed line. Hence, in every 1-immersion of M , theboundary edges of the U-subgraphs of M are not crossed.We prove (a) and (b) in the following way. We consider a 1-immersedsubgraph W of M (cf. Fig. 26) consisting of two U-graphs U and U con-nected by an ( h )-grid Γ, h ∈ { , } , 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 notcrossed by edges of other grids.Suppose, for a contradiction, that U and U are 1-immersed under ϕ asshown in Fig. 29(a) (the situation described in the figure arises when U is43igure 29: Cycles of two adjacent 1-immersed U-subgraphs.drawn clockwise and U counter-clockwise (or vice versa)). Clearly, thereare two basic paths P i and P j of Γ, 1 ≤ i < j ≤
7, which do not intersect.Then the cycle shown in Fig. 29(a) in thick line is embedded in the plane, acontradiction, since the cycle is crossed by 5 other basic paths of Γ, but thecycle has only 2 h ≤ U and U are 1-immersed as shown in Fig. 26.Suppose that in ϕ , a basic path P i of Γ crosses a basic path Q of somegrid of M exactly once. If Q = P j is a basic path of Γ, j (cid:54) = i (see Fig. 29(b)),then the closed curves C and C shown in Fig. 29(b) by dashed cycles, areembedded in the plane and each of the other five basic paths of Γ crosses anedge of C or C , a contradiction, since C and C have 2 h − ≤ Q is a basic path of a grid Γ (cid:48) different from Γ, then there is a basic path P j , j (cid:54) = i , of Γ such that P j is notcrossed by P i and Q . Hence, the cycle C ( P i , P j ) is embedded and Q crossesthe edges of the cycle exactly once. Then for every other basic path Q (cid:48) ofΓ (cid:48) , the cycle C ( Q, Q (cid:48) ) crosses C ( P i , P j ) at least twice and Q (cid:48) crosses P i or P j (the edges of different U-subgraphs do not intersect). We have that the 7basic paths of Γ (cid:48) cross P i and P j , a contradiction. Hence, in ϕ , if two basicpaths intersect, then they intersect twice. In particular, only basic paths of(2)-grids can intersect.Now we claim the following:(E) If in ϕ two neighboring basic paths P i − and P i of the (2)-grid Γ donot intersect, then the edge e joining the middle vertices of P i − and P i liesinside the embedded cycle C ( P i − , P i ).Indeed, if e lies outside C ( P i − , P i ), then the 4-cycle shown in Fig. 29(c)in thick line is crossed by 5 basic paths P r , r (cid:54) = i − , i , a contradiction.44uppose that in ϕ , a basic path of Γ crosses a basic path of a grid Γ (cid:48) twice (that is, Γ and Γ (cid:48) are (2)-grids). Since the number of basic paths ofΓ is odd (namely, 7), it can not be that every basic path of Γ crosses someother basic path of Γ twice. Then there is a basic path of Γ which is notcrossed by other basic paths of Γ. Hence, if Γ = Γ (cid:48) (resp. Γ (cid:54) = Γ (cid:48) ), then thereare two neighboring basic paths P i − and P i of Γ such that one of them,say, P i − , is crossed twice by some basic path Q of Γ (resp. Γ (cid:48) ), and theother basic path P i (cid:54) = Q is not crossed by Q . Then the cycle C ( P i − , P i ) isembedded. By (E), the edge e joining the middle vertices of P i − and P i liesinside C ( P i − , P i ). Denote by C and C the two embedded adjacent 4-cycleseach of which consists of e and edges of the 6-cycle C ( P i − , P i ). The middlevertex of Q lies outside C ( P i − , P i ) and the two end vertices of Q lie inside C and C , respectively. The end vertices of Q belong to two U-subgraphsconnected by Γ (cid:48) . Since the edges of the two U-subgraphs do not cross theedges of C and C , we obtain that one of the U-subgraphs lies inside C and the other lies inside C , a contradiction, since the two U-subgraphs areconnected by at least four basic paths different from Q , P i − , and P i . Hence,no basic path of Γ crosses some other basic path twice.We conclude that the basic paths of the grids connecting U-subgraphs donot intersect.Now it remains to show that if Γ is a (2)-grid, then the edges joining themiddle vertices of the basic paths of Γ are not crossed. Consider any twoneighboring basic paths P i − and P i of Γ. The cycle C ( P i − , P i ) is embeddedand, by (E), the edge e joining the middle vertices of P i − and P i lies inside C ( P i − , P i ). It is easy to see that for every edge e (cid:48) of G not belonging toU-subgraphs and different from e and the edges of C ( P i − , P i ), in the graph G − e (cid:48) the end vertices of e (cid:48) are connected by a path which consists of edgesof U-subgraphs and basic paths of grids and which does not pass throughthe vertices of C ( P i − , P i ). Now, if the edge e is crossed by some other edge e (cid:48) , then e (cid:48) is not an edge of a U-subgraph, the end vertices of e (cid:48) lie insidethe cycles C and C , respectively (where C and C are defined as in thepreceding paragraph) whose edges are not crossed by edges of U-subgraphsand basic paths, a contradiction.Therefore, the edges of M do not intersect and that the graph W has aunique 1-immersion. This completes the proof of the theorem. (cid:3) Now, given a plane graph G every vertex of which has degree 3 or 4, weconstruct a graph G such that G is 3-colorable if and only if G is 1-immersible.45igure 30: Constructing the graph G (1) .To obtain G , we proceed as follows. First we construct a subgraph G (1) of G such that G (1) has a unique 1-immersion. The graph G (1) is obtained from aU-supergraph W by adding some additional vertices and edges. By inspectionof the subsequent figures which 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 thatgiven the 1-immersion of W , the additional vertices and edges can be placedin 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 G we place some new additional paths “between” 1-immersedU-subgraphs of G (1) . Notice that due to circle inversion one may assume thatin a 1-planar drawing of G each U-subgraph is drawn in such a way that theouter boundary cycle is containing all other vertices of the U-subgraph insidethe 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 U ( F ) and replace every vertex v by a vertex-block B ( v ) as shown at the top of Fig. 30. At the bottom of Fig. 30 weshow the designation of a (1)-grid used at the top of the figure and at what46ollows. The vertex-block B ( v ) has a unique 1-immersion and is obtainedfrom a U-supergraph by adding some additional vertices and edges. Fig. 30shows schematically the boundary of B ( v ) and Fig. 33 shows B ( v ) in moredetail. For a k -valent vertex v of G , 3 ≤ k ≤
4, the vertex-block B ( v ) has 3 k boundary vertices labeled clockwise as a, b, c, a, b, c, . . . , a, b, c ; these verticesdo not belong to U-subgraphs of B ( v ). In Figs. 30 and 33 we only show thecase of a 3-valent vertex v ; for a 4-valent vertex the construction is analogous– there are three more boundary vertices labeled a, b, c , respectively.We say that vertex-blocks B ( v ) and B ( w ) are adjacent if v and w areadjacent vertices of G .Figure 31: The pending paths connecting adjacent vertex-blocks.The graph G is obtained from G (1) if we take a collection of additionaldisjoint paths of length ≥ pending paths ) and identifythe end vertices of every path with two vertices, respectively, of G (1) . Thegraph G (1) has a unique 1-immersion and the edges of the U-subgraphs of G (1) can not 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 toobtain a 1-immersion of G . Every pending path connects either boundaryvertices of adjacent vertex-blocks or vertices of the same vertex-block.The graph G is such that for any two adjacent vertex-blocks, there areexactly three pending paths connecting the vertices of the vertex-blocks. Thepaths have length 3 and are shown in Fig. 31; we say that these pending pathsare incident with the two vertex-blocks. Each of the three pending pathsconnects the boundary vertices labeled by the same letter: a , b , or c . For h ∈ { a, b, c } , the pending path connecting vertices labeled h is called the ( h ) -path connecting the two vertex-blocks. In Fig. 31 the ( h )-path, h ∈ { a, b, c } ,47s labeled by the letter h .Denote by G (2) the graph obtained from G (1) if we add all triples ofpending paths connecting vertex-blocks B ( v ) , B ( w ), for all edges vw ∈ E ( G ).Figure 32: Pending paths of an h -family.The graph G also satisfies the properties stated below. The pendingpaths connecting vertices of the same vertex-block B ( v ) are divided intothree families called, respectively, the a -, b -, and c -families of B ( v ). Giventhe 1-immersion of G (1) , for every h ∈ { a, b, c } , the h -family of B ( v ) has thefollowing properties:(i) Every path P of the h -family admits exactly two embeddings in theplane such that we obtain a 1-immersion of G (2) ∪ P .(ii) The h -family consists of paths P , P , . . . , P n such that the graph G (2) ∪ P ∪ P ∪· · ·∪ P n has exactly two 1-immersions. In the two 1-immersions,every path P i uses its two embeddings. In one of the 1-immersions,paths of the h -family cross all ( h )-paths incident with B ( v ). In theother 1-immersion, the paths of the h -family do not cross any ( h )-pathincident with B ( v ).Fig. 32 shows fragments of the two 1-immersions of the union of G (2) andthe pending paths of an h -family. In the figure, each of the depicted (in thick48ine) six edges of the family, they are labeled by 1 , , . . . ,
6, respectively, usesits two embeddings in one of the two 1-immersions.If in a 1-immersion of G , paths of an h -family of B ( v ), h ∈ { a, b, c } ,cross ( h )-paths incident with B ( v ), then we say that the h -family of B ( v ) is activated in the 1-immersion of G .Figure 33: A vertex-block and the activated h -families.Figs. 33 and 34 show a vertex-block B ( v ) and the h -families of the vertex-block ( h = a, b, c ) in the case where v is 3-valent (the generalization for a4-valent vertex v is straightforward). The pending paths of the three h -families are shown by thick lines and the three families are activated. Toavoid cluttering a figure, Fig. 33 contains a fragment denoted by R which isgiven in more detail in Fig. 34. Recall that the grey areas in Figs. 33 and 34represent U-graphs.In Figs. 33 and 34 we use designations of some fragments of B ( v ); thedesignations are given at the left of Fig. 35 and the corresponding fragmentsare given at the right of Fig. 35 (that is, the connections between the greyareas in Figs. 33 and 34 consist of seven basic paths). The reader can easilycheck that for every pending path P of the three families, there are exactly49igure 34: The fragment R of the vertex-block in Fig. 33.two ways to embed the path so that we obtain a 1-immersion of G (2) ∪ P .The vertex-block B ( v ) contains a 2-path connecting the vertices labeled 0 inFig. 34 and a 1-path connecting the vertices labeled 1 in Fig. 34; we call thepaths the (0)- and (1)-blocking paths, respectively). For every h ∈ { a, b, c } ,exactly one pending path of the h -family of B ( v ) crosses a blocking path: thepending path has length 33, crosses the (1)-blocking (resp. (0)-blocking) pathwhen the h -family is activated (resp. not activated), and the pending path ineach of its two embeddings crosses exactly one pending path of each of theother two families. Fig. 34 shows the two embeddings of the pending 33-pathof the b -family (one of them is in thick line, the other, when the family isnot activated, is in dashed line). Note that Fig. 34 shows something thatis not a 1-immersion, since all three families of paths are activated, and the(1)-blocking path is crossed three times.50igure 35: The designations of fragments of the vertex-block B ( v ).Denote by G (2) v the union of G (2) and the paths of all three h -families of B ( v ). Now the reader can check that B ( v ) and the h -families of B ( v ) areconstructed in such a way that the following holds:(F) In every 1-immersion of G (2) v (and, hence, of G ) exactly one h -familyof B ( v ) is activated, and for each h ∈ { a, b, c } , there is a 1-immersion of G (2) v in which the h -family of B ( v ) is activated.By construction of G , if G has a 1-immersion, then in the 1-immersion forevery vertex v of G , exactly one h -family ( h ∈ { a, b, c } ) of B ( v ) is activatedand taking Fig. 31 into account we obtain that in the 1-immersion the h -families of the vertex-blocks adjacent to B ( v ) are not activated.Now take a 1-immersion of G (if it exists) and assign every vertex v of G a color h ∈ { a, b, c } such that the h -family of B ( v ) is activated in the1-immersion of G . We obtain a proper 3-coloring of G with colors { a, b, c } .51ake a proper 3-coloring of G (if it exists) with colors { a, b, c } and forevery vertex v of G , if h ( v ) is the color of v , take the h ( v )-family of B ( v ) to beactivated and the other two families not to be activated. By the constructionof G , and by the mentioned properties of 1-immersions of its subgraphs B (2) v ,it follows that we obtain a 1-immersion of G .When constructing G , we choose the order of every U-subgraph suchthat every boundary vertex of the U-subgraph is incident with an edge notbelonging 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 U ( F )is bounded by a constant multiple of k . Similarly, for each v ∈ V ( G ), theunion of B ( v ) and its three h -families has constant size. Therefore, the wholeconstruction of G can be carried over in linear time. This completes the proofof Theorem 5. k -planarity testing for multigraphs A graph drawn in the plane is k - immersed in the plane ( k ≥
1) if any edge iscrossed by at most k other edges (and any pair of crossing edges cross onlyonce). 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 asto obtain a proof that k -planarity testing ( k ≥
2) for multigraphs is NP-complete. Below we give only a sketch of the proof, the reader can easily fillin the missing details.Denote by G ( k ), G ( k ), and G (1) ( k ), respectively, the multigraphs ob-tained from the graphs G , G , and G (1) if we replace every edge by k paralleledges. For an edge e of the multigraphs denote by H ( e ) the set consistingof e and all other k − e . Denote by ϕ the unique plane1-immersion of G (1) , and by ϕ k the plane k -immersion of G (1) ( k ) obtainedfrom ϕ if we replace every edge of G (1) by k parallel edges. Lemma 14
The multigraph G (1) ( k ) , k ≥ , has a unique plane k -immersion.Proof. We consider an arbitrary plane k -immersion ψ of G (1) ( k ) and showthat ψ is ϕ k .First we show that if edges e and e of G (1) ( k ) cross in ψ , then each edgeof H ( e ) intersects every edge of H ( e ). Suppose, for a contradiction, thatan edge e (cid:48) of H ( e ) does not intersect e (see Fig. 36(a)). Consider the 2-cell D whose boundary consists of the edges e and e (cid:48) . Since e and e (cid:48) can have52igure 36: Different plane 1-immersions of G (1) .at most 2 k 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 hasdifferent positions in the two different plane 1-immersions), a contradiction.Hence, each edge of H ( e ) intersects every edge of H ( e ). Delete k − k parallel edges. We obtain a plane 1-immersion of G (1) , that is, ϕ . Hence ψ is ϕ k . (cid:3) The graph G ( k ) is obtained from G (1) ( k ) if we add the pending pathsof G where every edge is replaced by k parallel edges. Now, considering apending path of an h -family, we have (see Fig. 37, where each thick edgerepresents k parallel edges) that if e (cid:48) ∈ H ( e ), then each of the edges e and e (cid:48) is already crossed by k edges of H ( e ) and H ( e ), respectively, thus theedges of the pending path incident with the vertex v can not cross edges e and e (cid:48) , a contradiction. Hence all edges of H ( e ) cross either H ( e ) or H ( e ).As a result, the h -families and the other pending paths of G ( k ) “behave”in the same way as in G . We conclude that G has a plane 1-immersion ifand only if G ( k ) has a plane k -immersion. Since G has a plane 1-immersionif and only if G has a proper 3-coloring, we get that k -planarity testing formultigraphs is NP-complete.If we restrict ourselves to simple graphs only, then to have a proof anal-ogous to the proof of Theorem 5 we need simple graphs that have a uniqueplane k -immersion ( k ≥ h -family in G ( k ). Acknowledgement.
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