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Proper 1-immersions of graphs triangulating the plane
Discrete Mathematics 313 (2013) 2673–2686
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Proper 1-immersions of graphs triangulating the plane Vladimir P. Korzhik ∗ National University of Chernivtsi, Chernivtsi, Ukraine Institute of Applied Problems of Mechanics and Mathematics of National Academy of Science of Ukraine, Lviv, Ukraine
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Article history: Received 10 April 2012 Received in revised form 19 July 2013 Accepted 26 July 2013 Available online 28 August 2013 Keywords: Topological graph Crossing edges 1-planar graph 1-immersion
abstract In this paper we study what planar graphs are ‘‘rigid’’ enough such that they can not be drawn on the plane with few (1, 2, or 3) crossings per edge. A graph drawn on the plane is k-immersed in the plane if each edge is crossed by at most k other edges. By a proper k-immersion of a graph we mean a k-immersion of the graph in the plane such that there is at least one crossing point. We give a characterization (in terms of forbidden subgraphs) of 4-connected graphs which triangulate the plane and have a proper 1-immersion. We show that every planar graph has a proper 3-immersion. © 2013 Elsevier B.V. All rights reserved.
1. Introduction In this paper we study what planar graphs are ‘‘rigid’’ enough such that they cannot be drawn on the plane with few (1, 2, or 3) crossings per edge. A graph drawn on the plane is k-immersed in the plane if each edge is crossed by at most k other edges (this plane drawing of the graph is called a k-immersion of the graph in the plane). In what follows, by a k-immersion of a graph we will mean a k-immersion of the graph in the plane. We may assume that in a k-immersion of a graph any pair of crossing edges crosses only once, that adjacent edges do not cross and that no edge crosses itself. We take this assumption as a part of the definition of k-immersions since this limits the number of possible cases when discussing k-immersions. A graph is k-planar if it can be k-immersed. The notion of a 1-immersion of a graph was introduced by Ringel [18] when trying to color the vertices and faces of a plane graph so that the adjacent or incident elements receive distinct colors. Little is known about 1-planar graphs (see [2,4,5,1,6–15,17–21]). One of the basic questions has been how to recognize 1-planar graphs. Recently it was shown [16] that testing 1-planarity is NP-complete. By a proper k-immersion of a graph we mean a k-immersion with at least one crossing point. The planar graphs that have no proper 1-immersions are called the PN-graphs, for short. The notion of a PN-graph was introduced in [16] where PN-graphs were used to construct infinite families of minimal non-1-planar graphs (a graph G is minimal non-1-planar if G is non-1-planar but the graph G − e is 1-planar for every edge e of G). An important point in introducing PN-graphs is that the graphs can be of use when we want to check whether a graph is 1-planar or not. Every 3-connected PN-graph has a unique 1-immersion, namely, its unique plane embedding. Hence, if a graph G contains a 3-connected PN-graph H as a subgraph, then to check whether G has a 1-immersion we consider the plane embedding of H and check whether it is possible to add the vertices V (G) \ V (H ) and the edges E (G) \ E (H ) in such a way as to obtain a 1-immersion of G. Such an approach can work well when H is a spanning subgraph of G and |E (G)| − |E (H )| is small enough.
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Correspondence to: Bogomoltsa St. 3/5, Chernivtsi, 58001, Ukraine. E-mail address: [email protected].
0012-365X/$ – see front matter © 2013 Elsevier B.V. All rights reserved. http://dx.doi.org/10.1016/j.disc.2013.07.019
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Fig. 1. Two PN-graphs.
Fig. 2. A wheel Wn (w1 w2 . . . wn ) with the center v .
A class of PN-graphs was determined in [16] and some graphs of the class were constructed. Two PN-graphs of the class are shown in Fig. 1. This class was introduced with a particular purpose of constructing minimal non-1-planar graphs and seems to be rather special (the definition of the class includes a list of 8 conditions which any graph of the class has to satisfy and, in addition, three forbidden subgraphs). In the present paper we consider a rather more general class of PN-graphs. In Section 3 we prove Theorem 1 which characterizes (in terms of forbidden subgraphs) 4-connected PN-graphs which triangulate the plane. In Section 4 we show (Theorem 2) that every planar graph has a proper 3-immersion. We expect that there is much more to be discovered by following the themes of the paper. Problem 1. Determine all planar graphs that do not have a proper 1-immersion. Is it possible to characterize such planar graphs in terms of forbidden subgraphs? The following problem is possibly more difficult than it first appears. Problem 2. Is it true that every planar graph has a proper 2-immersion? 2. Preliminaries 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-selfintersecting) cycle C of G, if x lies in the interior (resp. exterior) of C , and does not lie on C . Two cycles of a graph are adjacent if they share a common edge. For convenience, we will denote a cycle as (v1 v2 . . . vn ) where vi and vi+1 are adjacent vertices for i = 1, 2, . . . , n (here vn+1 = v1 ). Since we assume that in 1-immersions, adjacent edges do not cross each other and no edge crosses itself, every 3-cycle of a 1-immersed graph is embedded in the plane. Hence, given a 3-cycle of a 1-immersed graph, we can speak about its interior and exterior. Throughout this section we will deal with 1-immersed graphs only. When a 1-immersion of a graph G is clear from the context, we shall identify vertices, edges, cycles and subgraphs of G with their image in R2 under the 1-immersion. Then by a face of a 1-immersion of G we mean any connected component of R2 \ G. By using Möbius transformations combined with homeomorphisms of the plane it is always possible to exchange the interior and exterior of any embedded cycle and it is possible to change any face of a given 1-immersion into the outer face of a 1-immersion. Formally, we have the following observation (which we will use without referring to it every time): (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 ′ with the same number of crossings as ϕ , in which C is embedded and all vertices of G, which lie inside C in ϕ , lie outside C in ϕ ′ and vice versa. By the wheel Wn (w1 w2 . . . wn ) of order n with the center v we mean the graph shown in Fig. 2. In what follows we will often use the following observation: Observation 1. If a graph G triangulates the plane, then a vertex v of G has degree n (v is an n-vertex, for short) if and only if G has as a subgraph a wheel of order n with the vertex v as the center of the wheel. Hence, every vertex of a 4-connected graph triangulating the plane is the center of a wheel of order at least 4.
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Fig. 3. Cycles of G.
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Fig. 4. Adjacent cycles of G.
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Fig. 5. An edge crossing the common edge of two adjacent 3-cycles.
In a 1-immersion of a graph G, if we consider the crossing points as new vertices, then we have an embedded graph G. Every cycle of G is embedded in the plane, so we can speak about the interior of the cycle. An edge of G is normal if it connects two vertices of G. It is easy to see that the following statement holds. (B) For a cycle D of G (where G is a 4-connected graph), if the sum of the number of vertices of G belonging to D and the number of normal edges belonging to D is less than 4, then if there are vertices of G outside (resp. inside) D, then there are no vertices of G inside (resp. outside) D; otherwise, G has a 3-cut, a contradiction. Observation 2. By (B), for any of the cycles of G shown in Fig. 3(a) (resp. Fig. 3(b)), there are no vertices of G inside (resp. outside) the cycle. In the figures, as in what follows, a cross inside (resp. outside) a cycle of G means that, in view of (B), there are no vertices of G inside (resp. outside) the cycle. Observation 3. Suppose that in a 1-immersion of a 4-connected graph G there is a fragment shown in Fig. 4 where the vertex
v is not adjacent to the vertices that lie inside the cycle (v ab1c ) in Fig. 4(a)–(b) and inside the cycle (v ab1c2) in Fig. 4(c) (in the figures, as in what follows, we label crossing points by integers). Suppose that the edges v a, ab, and v c in Fig. 4(a), the edges v a, ab, and bc in Fig. 4(b), the edges v a and ab in Fig. 4(c) are not crossed by other edges in the 1-immersion. Now we claim that there are no vertices inside (v ab1c ) in Fig. 4(a)–(b) and inside (v ab1c2) in Fig. 4(c). Indeed, consider, for example, the case in Fig. 4(a). If there are vertices inside the cycle (v ac ), then there is a 3-cut (including a and c), a contradiction. Then no edge crosses ac and if there are vertices inside (ab1c ), then {a, b, c } is a 3-cut, a contradiction. The cases in Fig. 4(b) and (c) are treated analogously. 3. Proper 1-immersions of 4-connected planar triangulations Denote by T4 the class of all 4-connected planar graphs which have a triangular embedding in the plane. First we need Lemmas 1–3 which give us some properties of proper 1-immersions of graphs of T4 . The proofs of the lemmas are based on straightforward observations and case analysis, and do not use any known special techniques or results. Lemma 1. Suppose that a graph G ∈ T4 has a proper 1-immersion ϕ . In ϕ , if edges e1 and e2 of G cross, then there is a 3-cycle C containing e1 such that one endvertex of e2 lies inside C and the other endvertex of e2 lies outside C . Proof. If e1 and e2 cross, then Fig. 5(a)–(e) show all possible cases of how the two 3-cycles containing e1 are embedded in the plane and which of the two 3-cycles has one endvertex of e2 inside and outside (the 3-cycle is denoted C ). In the case in Fig. 5(f), any endvertex of e1 is the center of a wheel of order 3, a contradiction, hence the case cannot take place. Lemma 2. Suppose that a graph G ∈ T4 without adjacent 4-vertices has a proper 1-immersion ϕ . In ϕ , suppose an edge vw crosses an edge yz of a 3-cycle C = (xyz ) such that the vertex v is inside and w is outside C (see Fig. 6(a)). Suppose that there is more than one vertex of G outside C . Then v is the only vertex lying inside C .
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Fig. 6. The vertices a and b are either both inside or both outside the cycle C .
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Fig. 7. Exactly two vertices are inside C .
Proof. Let (vw a) and (vw b) be the two 3-cycles of G containing the edge vw . If a, b ∈ {x, y, z }, then the edge ab is contained in three 3-cycles (namely, (xyz ), (v ab), and (w ab)), a contradiction. Hence {a, b} ̸⊂ {x, y, z }. Case 1: a, b ̸∈ {x, y, z }. If a and b lie outside C (Fig. 6(b)), then, taking Observation 2 into account, we obtain that v is the only vertex inside C . Suppose, for a contradiction, that a and b lie inside C (Fig. 6(c)). Then, by (A) and taking Observation 2 into account, we obtain that w is the only vertex outside C , a contradiction. Suppose, for a contradiction, that a lies inside and b lies outside C (see Fig. 7(a)). First, suppose, for a contradiction, that a and v are the only vertices inside C . Then the vertex a can have degree 4 only (see Fig. 7(b)). If v has degree 4, then a and v are adjacent 4-vertices, a contradiction. If v has degree 6, then v is the center of W3 (xyz ), a contradiction. Hence, v has degree 5. If there are xv and yv , then xy belongs to three 3-cycles, a contradiction. If there are xv and z v (see Fig. 7(c)), then x is the center of W4 (yav z ), hence x has degree 4, a and x are adjacent 4-vertices, a contradiction. If there are yv and z v (see Fig. 7(d)), then y is the center of W4 (xz v a), hence y has degree 4, a and y are adjacent 4-vertices, a contradiction. Now suppose, for a contradiction, that there are at least three vertices inside C . An edge e crosses av (see Fig. 7(a)), otherwise {a, v, x} or {a, v, y} is a 3-cut, a contradiction. If both endvertices of e lie inside C , then the two 3-cycles containing e pass the vertices a and v , respectively, thus v is the center of a wheel of order 3, a contradiction. Hence, one endvertex of e is either x or y, and the other endvertex of e lies inside C . Subcase 1.1: e = yu (see Fig. 8(a)). There is ux (otherwise, the two 3-cycles containing e pass the vertices a and v , respectively, thus v is the center of W3 (ayu), a contradiction), and there is either the 2-path yau or the 2-path yv u The edge au belongs to two 3-cycles: (awv) and (aw t ). It is easy to see that t ∈ {x, y}. Suppose, for a contradiction, that there is aw y (see Fig. 8(b); in this figure, as in what follows, a dashed line connecting two vertices means that G does not have an edge joining the two vertices). Since xy belongs to exactly two 3-cycles, there is no ax. Since vw belongs to exactly two 3-cycles, there is no yv . Hence yu belongs to the 3-cycles (yux) and (yau) (see Fig. 8(c)). If there is uv (see Fig. 8(d)), then a is the center of W4 (uvw y), hence a has degree 4, and v and u are the only vertices inside C which are adjacent to a. Then, taking Observation 3 into account, we obtain that there are no vertices inside (xua1) and (x3v u), and that a, u, and v are the only vertices inside C , thus a and u are adjacent 4-vertices, a contradiction. Hence, there is no uv , and since av belongs to two 3-cycles (one of them is (avw)), a 2-path ac v crosses xu. If c lies inside (xua1) (see Fig. 8(e)), then the only way for each of xu and v c to belong to two 3-cycles is that there are edges xc , cu, and xv . But then a is the center of W5 (uc vw y), hence a has degree 5, and u, c, v are the only vertices inside C which are adjacent to a. Then, taking Observation 3 into account, we obtain that there are no vertices inside (xcua), hence a, v , u, c are the only vertices inside C , thus u and c are adjacent 4-vertices, a contradiction. If c lies inside (x3v 2u) (see Fig. 8(f)), then the only way for each of xu and ac to belong to two 3-cycles is that there are edges xc and cu. But then u is the center of W4 (xcay), hence u has degree 4, and a, v, c are the only vertices inside C which are adjacent to u. Then, taking Observation 3 into account, we obtain that there are no vertices inside (x3v 2uc ), hence a, v , u, c are the only vertices inside C , thus u and c are adjacent 4-vertices, a contradiction. Now suppose, for a contradiction, that there is (aw x) (see Fig. 9(a)).
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Fig. 8. More than two vertices are inside C and there is (aw y).
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Fig. 9. More than two vertices are inside C and there is (aw x).
There are no ay (otherwise, xy belongs to three 3-cycles, a contradiction), hence the only way for yu to belong to two 3-cycles is that there is a 2-path yv u (see Fig. 9(b)). Then there is no xv (otherwise, xy belongs to three 3-cycles, a contradiction).
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Fig. 10. More than two vertices are inside C and there is xu.
Fig. 11. Two edges crossing a 3-cycle.
If there is au, then u is the center of W4 (xav y), hence u has degree 4, and a and v are the only vertices inside C which are adjacent to u. Then, taking Observation 3 into account, we obtain that a, v , and u are the only vertices inside C , hence a and u are adjacent 4-vertices, a contradiction. Hence there is no au, and since av belongs to two 3-cycles (one of them is (avw)), a 2-path ac v crosses xu. If c lies inside (xu2a) (see Fig. 9(c)), then the only way for each of xu and v c to belong to two 3-cycles is that there are edges xc and cu. But then a is the center of W4 (xc vw), hence a has degree 4, and v and c are the only vertices inside C which are adjacent to a. Then, taking Observation 3 into account, we obtain that there are no vertices inside (xc v 2a), hence a, v, u, c are the only vertices inside C , thus u and c are adjacent 4-vertices, a contradiction. If c lies inside (x3v u) (see Fig. 9(d)), then the only way for each of xu and ac to belong to two 3-cycles is that there are edges xc and cu. But then u is the center of W4 (xc v y), hence u has degree 4, and a, v, c are the only vertices inside C which are adjacent to u. Then, taking Observation 3 into account, we obtain that there are no vertices inside (x3v uc ), hence a, v , u, c are the only vertices inside C , thus u and c are adjacent 4-vertices, a contradiction. Subcase 1.2: e = xu (see Fig. 10(a)). The reader can easily check that the foregoing proof that the case in Fig. 10(b) cannot take place is also a proof that the case in Fig. 10(c) cannot take place as well the two proofs have exactly the same text but different figures, namely, Fig. 10(b) and (c), respectively. But Fig. 10(a) is obtained from Fig. 10(c) by relabeling vertices, hence the case in Fig. 10(a) cannot take place also. Case 2: the vertex a belongs to C , and b does not belong to C . Here we need the following claim. Claim 1. In ϕ , if edges e1 and e2 cross the edges of a 3-cycle C = (xyz ) (see Fig. 11), then e1 and e2 have a common endvertex. Proof. Suppose, for a contradiction, that e1 and e2 do not have a common endvertex. Then any 3-cycle containing e1 does not share a common edge with any 3-cycle containing e2 . If each of the two 3-cycles containing ei (i ∈ {1, 2}) passes a vertex of C , then an edge of C belongs to three 3-cycles, a contradiction. Hence, for each of the edges e1 and e2 , at least one of the 3-cycles containing the edge crosses an edge of C , that is, the edge yz, a contradiction. Now we consider the case when a = y and b lies outside C (see Fig. 12(a)). If no edge crosses xy and there are at least two vertices inside C , then {x, y, v} is a 3-cut, a contradiction. Hence, an edge e′ crosses xy. By Claim 1, e′ has a common endvertex with both vw and v b, thus e′ is adjacent to v (see Fig. 12(b)), and v is the only vertex inside C . Now suppose, for a contradiction, that a = y and b lies inside C (see Fig. 12(c)). Using (A) and taking into account the foregoing case where b lies outside C , we obtain that w is the only vertex outside C , a contradiction. In the case where a = x and b lies outside C (see Fig. 12(d)) we show analogously to the case in Fig. 12(a) that v is the only vertex inside C . Using (A) and taking into account the case in Fig. 12(d), we obtain in the case in Fig. 12(e) that w is the only vertex outside C , a contradiction. This completes the proof of the lemma. Lemma 3. Suppose that a graph G ∈ T4 without adjacent 4-vertices has a proper 1-immersion ϕ . In ϕ , if there is exactly one vertex v inside a 3-cycle C , then there can be only two cases shown in Fig. 13. Proof. If v is the only vertex inside C , then, since G is 4-connected, at least one vertex incident with v crosses an edge of C . If v is adjacent to all vertices of C , the v is the center of a wheel of order 3, a contradiction. Hence v has degree 4 or 5. If v has degree 5, then v is adjacent to two vertices of C , and three edges incident with v cross edges of C (see Fig. 14(a)). If v has degree 4, then there are exactly three cases shown in Fig. 14(b)–(d), respectively.
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Fig. 12. The vertex a lies on the cycle C .
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Fig. 13. All possible cases when there is exactly one vertex inside a 3-cycle.
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Fig. 14. A vertex v is inside a 3-cycle C .
The case in Fig. 14(a). Consider Fig. 15(a). If there are yw and v z, then yz belongs to three 3-cycles, a contradiction. Suppose, for a contradiction, that there is yw and there is no z w (see Fig. 15(b)). There is no ya (otherwise, v y belongs to three 3-cycles, a contradiction). Since v is the center of a wheel of order 5, the vertices a and b are adjacent, and w is adjacent either to b (see Fig. 15(c); in this case z is adjacent to a) or to a (see Fig. 15(d); in this case z is adjacent to b). In Fig. 15(c) there are two vertices inside and two vertices outside the 3-cycle (av z ), contrary to Lemma 2, hence this case cannot take place. Consider the case in Fig. 15(d). The edge xy belongs to two 3-cycles (xyz ) and (xyt ). It is easy to see that either t ∈ {w, b} or t lies outside (abz1w). If t = w (see Fig. 15(e)), then y is the center of W4 (xz vw), hence y has degree 4. The edge xw belongs to two 3-cycles: (xw y) and (xwt ′ ). It is easy to see that t ′ ∈ {a, b}. If t ′ = a (there is ax), then w is the center of W4 (xyv a), hence w and y are adjacent 4-vertices, a contradiction. If t ′ = b (there is bx), then z is the center of W4 (xyv b), hence z and y are adjacent 4-vertices, a contradiction. If t = b (see Fig. 15(f)), then x is the center of W3 (bzy), a contradiction. It remains to consider the case in Fig. 15(g) where t lies outside (abz1w). There is no zt (otherwise, x is the center of W3 (tyz ), a contradiction. Since xz belongs to two 3-cycles, there must be xb. Now z is the center of W4 (xbv y), hence z has degree 4. The edge xt belongs to two 3-cycles: (xty) and (xtt ′ ). It is easy to see that t ′ ∈ {a, b}. If t ′ = a (there is ax), then b is the center of W4 (xz v a), hence b and z are adjacent 4-vertices, a contradiction. If t ′ = b (there is tb), then x is the center of W4 (tyzb), hence x and z are adjacent 4-vertices, a contradiction.
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Fig. 15. A 5-vertex v inside a 3-cycle.
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Fig. 16. A 4-vertex v inside a 3-cycle (the case in Fig. 14(b)).
Hence, we obtain that when v has degree 5 there can be the case in Fig. 13(a) only. The case in Fig. 14(b). Consider Fig. 16(a). If there are yw and z w , then yz belongs to three 3-cycles, a contradiction. Since vw belongs to two 3-cycles, there is aw and there is one of the two edges yw and z w . Suppose, for a contradiction, that there is yw and there is no w z (see Fig. 16(b)). Since v is the center of a wheel of order 4, there is az (see Fig. 16(c)). The edge xy belongs to two 3-cycles: (xyz ) and (xyt ). It is easy to see that either t ∈ {a, w} or t lies outside (az1w). If t = a (there is ay), then v is the center of a wheel of order 3, a contradiction. If t = w (there is xw ), then y is the center of W4 (xz vw), thus y and v are adjacent 4-vertices, a contradiction. Hence, t lies outside (az1w) (see Fig. 16(d)). Taking Observation 2 into account, we obtain that t is the only vertex outside (az1w), thus w has degree 4 (there must be w t)), and v and w are adjacent 4-vertices, a contradiction. Hence, we obtain that when v has degree 4, then in the case in Fig. 14(b) there can be the case in Fig. 13(b) only. The case in Fig. 14(c). We prove that the case in Fig. 17(a) cannot take place. First we consider the case when v is the center of W4 (axbw) (see Fig. 17(b)). Suppose, for a contradiction, that the case in Fig. 17(b) takes place. The edge yz belongs to two 3-cycles: (xyz ) and (yzt ). It is easy to see that either t ∈ {b, w} or t lies outside (xbw a). If t = b (and yb crosses aw ), then x is the center of W3 (byz ), a contradiction. Suppose that t = w (see Fig. 17(c)). There is no ay (otherwise, a is the center of W4 (xvw y), hence a and v are adjacent 4-vertices, a contradiction). Analogously, there is no bz. If there is xw , then x is the center of W3 (bw a), a contradiction. We see that the only way for each of the edges xy and xz to belong to two 3-cycles is shown in Fig. 17(d), but then y and z are adjacent 4-vertices, a contradiction. Now suppose that t lies outside (xbw a) (see Fig. 17(e)). If there is xt, then x is the center of W3 (yzt ), a contradiction. Now we see that the only way for each of the edges xy and xz to belong to two 3-cycles is that there are ay and bz (see Fig. 17(f)). There is no yw (otherwise, a is the center of W4 (xvw y), hence a and v are adjacent 4-vertices, a contradiction). Analogously, there is no w z. Then w has degree 4 (there is w t), and w and v are adjacent 4-vertices, a contradiction. In the case where v is the center of W4 (abw x) (see Fig. 17(g)), taking Observation 2 into account, we obtain that the adjacent vertices y and z can have degree 4 only (there are ay, yw, w z, and zb), a contradiction. Hence, we obtain that the case in Fig. 14(c) cannot take place. The case in Fig. 14(d). We prove that the case in Fig. 18(a) cannot take place. Since v is the center of a wheel of order 4, there are exactly three cases shown in Fig. 18(b)–(d).
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Fig. 17. A 4-vertex v inside a 3-cycle (the case in Fig. 14(c)).
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Fig. 18. A 4-vertex v inside a 3-cycle (the case in Fig. 14(d)).
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Fig. 19. The graph L(n), n ≥ 4.
Suppose, for a contradiction, that the case in Fig. 18(b) takes place. If there is ax, then y is the center of W4 (axz v), hence y and v are adjacent vertices, a contradiction. Analogously we show that there is no xb. Then the only way for each of the edges xy and xz to belong to two 3-cycles is that for each of the edges, a 3-cycle containing the edge crosses ab, a contradiction. In the cases in Fig. 18(c) and (d), taking Observation 2 into account, we obtain that x has degree 4 (there are ax and xb), and then z is the center of W4 (xyv a), hence v and z are adjacent vertices, a contradiction. Hence, we obtain that the case in Fig. 14(d) cannot take place. This completes the proof of the lemma. Consider the planar graphs L(n) for n ≥ 4, M (n) for n ≥ 1, P (n) for n ≥ 1, and Q (n) for n ≥ 2 shown in Figs. 19(a), 20(a), 21(a), and 22(a), respectively. The graph Q (n) in Fig. 22(a) is defined up to the mirror reflection. If in Figs. 20(a), 21(a), and 22(a) we identify the vertices x and y (and identify the edges v x and v y, the edges w x and w y), we obtain the planar graphs M (n) for n ≥ 1, P (n) for n ≥ 1, and Q (n) for n ≥ 2, respectively.
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Fig. 20. The graph M (n), n ≥ 1.
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Fig. 21. The graph P (n), n ≥ 1.
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Fig. 22. The graph Q (n), n ≥ 2.
Theorem 1. A graph G ∈ T4 has a proper 1-immersion if and only if at least one of the following conditions holds: (a) G has adjacent 4-vertices; (b) G is one of the graphs L(n) for n ≥ 4, M (n) for n ≥ 2, P (n) for n ≥ 2, and Q (n) for n ≥ 3 (c) G has as a subgraph one of the graphs M (n) for n ≥ 2, P (n) for n ≥ 2, and Q (n) for n ≥ 3 Proof. If G has adjacent 4-vertices, then G has a proper 1-immersion (see Fig. 23).
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Fig. 23. Adjacent 4-vertices.
Fig. 24. An end 3-cycle.
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Fig. 25. 3-cycles of type A and B.
The graphs L(n), M (n), P (n), Q (n), M (n), P (n), Q (n) are 3-connected, hence they have a unique plane embedding. If G has a subgraph F which is one of the graphs M (n), P (n), and Q (n), then, since G is 4-connected, in the plane embedding of G there are no vertices inside the faces of the plane embedding of F . Hence, if in the plane embedding of G we replace the plane embedding of F by the proper 1-immersion of F (as shown in Figs. 20–22), we obtain a proper 1-immersion of G. It is easy to see that each of the graphs M (1), P (1), and Q (2) has adjacent 4-vertices not belonging to the boundary of the outer 4-face of the plane embedding of the graph. Hence, if G is one of the graphs M (1), P (1), and Q (2), or if G has as a subgraph one of the graphs M (1), P (2), and Q (1), then G has adjacent 4-vertices. Now to prove the theorem it suffices to prove the following statement: (C) suppose that a graph G ∈ T4 without adjacent 4-vertices has a proper 1-immersion ϕ . Then either (b) or (c) of Theorem 1 holds. From here on we prove (C). Keeping (A) in mind and taking Lemma 2 into account, from now on by the interior of a 3-cycle C we will mean a component of R2 \ C which contains at most one vertex of G. In ϕ , a 3-cycle is crossed if at least one of its edges is crossed. A crossed 3-cycle with no vertex inside is called an end 3-cycle (see Fig. 24). A crossed 3-cycle C = (xyz ) with the only one vertex v inside shown in Fig. 25(a) (resp. Fig. 25(b)) is called a 3-cycle of type A (resp. type B). It is important to keep in mind that since vw belongs to two 3-cycles, there are the edges aw and bw in the case in Fig. 25(a), and the edge aw in the case in Fig. 25(b). Notice that the vertex w in Fig. 25 does not lie inside any 3-cycle. Indeed, if w lies inside a 3-cycle D, then either D contains yz (a contradiction, since yz belongs to exactly two 3-cycles) or D passes v (a contradiction, since then D has at least two vertices inside, contrary to Lemma 2). Taking Lemma 3 into account, we obtain that in ϕ , every crossed 3-cycle is of type A or B. In ϕ , two adjacent crossed 3-cycles are cross-adjacent (through an edge e) if the following holds: (i) the 3-cycles share the common edge e and do not cross each other; (ii) an edge crosses e; (iii) for each of the 3-cycles, if the 3-cycle contains exactly one vertex inside, then the vertex does not belong to the other 3-cycle. Notice that, the adjacent 3-cycles (xyz ) and (v yz ) in Fig. 25 are not cross-adjacent (in spite of the fact that the edge vw crosses the common edge yz of the two 3-cycles).
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Fig. 26. Cross-adjacent 3-cycle of type B and end 3-cycle.
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Fig. 27. A 3-cycle of type A is cross-adjacent to an end 3-cycle or a 3-cycle of type B.
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Fig. 28. Cross-adjacent 3-cycles of type A.
We have the following: (D) The type A 3-cycle in Fig. 25(a) is cross-adjacent (through the edges xy and xz, respectively) to exactly two other 3-cycles. The type B 3-cycle in Fig. 25(b) is cross-adjacent (through the edge xy) to exactly one other 3-cycle. Now we consider how 3-cycles can be cross-adjacent. If two end 3-cycles are cross-adjacent (see Fig. 5(f)), then any vertex of the 3-cycles is the center of a wheel of order 3, a contradiction. If a type B 3-cycle and an end 3-cycle are cross-adjacent (see Fig. 26), then the vertex y is the center of W4 (xz v a), hence y and v are adjacent 4-vertices, a contradiction. If two type B 3-cycles are cross-adjacent, then the only vertices inside the 3-cycles are adjacent, and we obtain adjacent 4-vertices, a contradiction. The case when a type A 3-cycle and an end 3-cycle are cross-adjacent is shown in Fig. 27(a). A case when a type A 3-cycle and a type B 3-cycle are cross-adjacent is shown in Fig. 27(b). The case in Fig. 27(c) cannot take place since there are two vertices inside and two vertices outside the 3-cycle (v bw), contrary to Lemma 2. A case when two type A 3-cycles are cross-adjacent is shown in Fig. 28(a). The case in Fig. 28(b) cannot take place since there are two vertices inside and two vertices outside the 3-cycle (v bw), contrary to Lemma 2. In the case in Fig. 28(c), since bc belongs to two 3-cycles, there is v c, and we obtain that x and z are adjacent 4-vertices, a contradiction. Now we see that all possible cases of how two 3-cycles can be cross-adjacent are shown in Fig. 27(a) and (b), and Fig. 28(a). Taking into account (D), we obtain that in ϕ , if there is only one vertex inside a 3-cycle C , then either (X1) or (X2) holds: (X1) C is an element of a cyclic sequence (C1 , C2 , . . . , Cn ), n ≥ 4, of type A 3-cycles such that Ci and Ci+1 are cross-adjacent for i = 1, 2, . . . , n (here Cn+1 = C1 ). (X2) C is an element of a sequence C1 , C2 , . . . , Cn (n ≥ 3), of 3-cycles such that: Ci and Ci+1 are cross-adjacent for i = 1, 2, . . . , n − 1; the cycles C2 , C3 , . . . , Cn−1 have type A; each of C1 and Cn either has type B or is an end 3-cycle (Note that if both C1 and Cn are end cycles, then n ≥ 4, since there are adjacent 4-vertices for n = 3).
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Fig. 29. Obtaining a proper 3-immersion of a planar graph.
In the case (X1), taking Fig. 28(a) into account, we obtain a 1-immersed graph L(n), n ≥ 4 shown in Fig. 19(b), that is, G is L(n), n ≥ 4. In the case (X2), taking Figs. 27(a), 27(b), and 28(a) into account, we obtain a 1-immersed graph which is one of the 1-immersed graphs shown in Figs. 20(b), 21(b), and 22(b), where the vertices x and y may be identified (it means that the edges v x and v y are identified, and the edges w x and w y are identified also). Hence either G is one of the graphs M (n) for n ≥ 2, P (n) for n ≥ 2, and Q (n) for n ≥ 3, or G has as a subgraph one of the graphs M (n) for n ≥ 2, P (n) for n ≥ 2, and Q (n) for n ≥ 3. This completes the proof of the theorem. One can check that: (i) each of the graphs M (n) for n ≥ 2, P (n) for n ≥ 2, and Q (n) for n ≥ 3 has a 4-valent vertex; (ii) each of the plane graphs M (n) for n ≥ 2, P (n) for n ≥ 2, and Q (n) for n ≥ 3 shown in Figs. 20(b), 21(b), and 22(b), respectively, has a 4-valent vertex not belonging to the boundary (v xw y) of the outer 4-face. Hence, we have the following: Corollary 1. Suppose that a 4-connected graph G triangulates the plane. If G is not L(n), n ≥ 5, and if G has no 4-vertices, then G has no proper 1-immersions. 4. Proper 3-immersions By a k-face we mean a k-gonal face. Theorem 2. Every connected planar graph has a proper 3-immersion. Proof. It suffices to consider connected planar graphs whose vertices have degree at least 3. Borodin [3] proved that any plane embedding of such a graph has one of the following vertices and faces: (a) (b) (c) (d)
a 5-vertex v incident with four 3-faces; a 4-vertex v incident with a 3-face; a 3-vertex v incident with a 3-face or a 4-face; a 5-face incident with four 3-vertices. Fig. 29(a)–(d) shows how for each of the cases (a)–(d), respectively, we can obtain a proper 3-immersion of the graph.
References [1] R. Bodendiek, H. Schumacher, K. Wagner, Bemerkungen zu einen Sechsfarbenproblem von G. Ringel, Abh. Math. Sem. Univ. Hamburg 53 (1983) 41–52. [2] O.V. Borodin, Solution of Ringel’s problem about vertex bound colouring of planar graphs and colouring of 1-planar graphs, Metody Discrete Analiz. 41 (1984) 12–26 (in Russian). [3] O.V. Borodin, A structural theorem about piane graphs and its application to labeling, Diskret. Mat. 51 (1992) 60–65 (in Russian).
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[4] O.V. Borodin, A new proof of the 6-color theorem, J. Graph Theory 19 (1995) 507–521. [5] O.V. Borodin, A.V. Kostochka, A. Raspaud, E. Sopena, Acyclic colouring of 1-planar graphs, Discrete Anal. Oper. Res. 6 (1999) 20–35. [6] F. Brandenburg, D. Eppstein, A. Gleiner, M. Goodrich, K. Hanauer, J. Reislhuber, On the density of maximal 1-planar graphs, in: Lecture Notes in Computer Science, vol. 7704, 2013, pp. 327–338. [7] Zhi-Zhong Chen, Approximation algorithms for independent sets in map graphs, J. Algorithms 41 (2001) 20–40. [8] Zhi-Zhong Chen, New bounds on the number of edges in a k-map graph, in: Lecture Notes in Computer Science, vol. 3106, 2004, pp. 319–328. [9] Zhi-Zhong Chen, Enory Grigni, Christos H. Papadimitriou, Planar map graphs, in: Proceedings of the Thirtieth Annual ACM Symposium on Theory of Computing, 1998, Dallas, Texas, pp. 514–523. May 24–26. [10] Zhi-Zhong Chen, M. Grigni, C.H. Papadimitriou, Map graphs, J. ACM 49 (2002) 127–138. [11] Zhi-Zhong Chen, M. Kouno, A linear-time algorithm for 7-coloring 1-plane graphs, Algorithmica 43 (2005) 147–177. [12] P. Eades, Seok-Hee Hong, Naoki Katoh, G. Liotta, P. Schweitzer, Yusuke Suzuki, Testing maximal 1-planarity of graphs with a rotation system in linear time, in: Lecture Notes in Computer Science, vol. 7704, 2013, pp. 339–345. [13] P. Eades, G. Liotta, Right angle crossing graphs and 1-planarity, Discrete Appl. Math. 161 (2013) 961–969. [14] I. Fabrici, T. Madaras, The structure of 1-planar graphs, Discrete Math. 307 (2007) 854–865. [15] V.P. Korzhik, Minimal non-1-planar graphs, Discrete Math. 308 (2008) 1319–1327. [16] V.P. Korzhik, Bojan Mohar, Minimal obstructions for 1-immersions and hardness of 1-planarity testing, J. Graph Theory 72 (2013) 30–70. [17] J. Pach, G. Toth, Graphs drawn with few crossings per edge, Combinatorica 17 (1997) 427–439. [18] G. Ringel, Ein Sechsfarbenproblem auf der Kugel, Abh. Sem. Univ. Hamburg 29 (1965) 107–117. [19] H. Schumacher, Zur Struktur 1-planarer Graphen, Math. Nachr. 125 (1986) 291–300. [20] Y. Suzuki, Re-embeddings of maximum 1-planar graphs, SIAM J. Discrete Math. 24 (2010) 1527–1540. [21] C. Thomassen, Rectilinear drawings of graphs, J. Graph Theory 12 (1988) 335–341.
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Discrete Mathematics journal homepage: www.elsevier.com/locate/disc
Proper 1-immersions of graphs triangulating the plane Vladimir P. Korzhik ∗ National University of Chernivtsi, Chernivtsi, Ukraine Institute of Applied Problems of Mechanics and Mathematics of National Academy of Science of Ukraine, Lviv, Ukraine
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Article history: Received 10 April 2012 Received in revised form 19 July 2013 Accepted 26 July 2013 Available online 28 August 2013 Keywords: Topological graph Crossing edges 1-planar graph 1-immersion
abstract In this paper we study what planar graphs are ‘‘rigid’’ enough such that they can not be drawn on the plane with few (1, 2, or 3) crossings per edge. A graph drawn on the plane is k-immersed in the plane if each edge is crossed by at most k other edges. By a proper k-immersion of a graph we mean a k-immersion of the graph in the plane such that there is at least one crossing point. We give a characterization (in terms of forbidden subgraphs) of 4-connected graphs which triangulate the plane and have a proper 1-immersion. We show that every planar graph has a proper 3-immersion. © 2013 Elsevier B.V. All rights reserved.
1. Introduction In this paper we study what planar graphs are ‘‘rigid’’ enough such that they cannot be drawn on the plane with few (1, 2, or 3) crossings per edge. A graph drawn on the plane is k-immersed in the plane if each edge is crossed by at most k other edges (this plane drawing of the graph is called a k-immersion of the graph in the plane). In what follows, by a k-immersion of a graph we will mean a k-immersion of the graph in the plane. We may assume that in a k-immersion of a graph any pair of crossing edges crosses only once, that adjacent edges do not cross and that no edge crosses itself. We take this assumption as a part of the definition of k-immersions since this limits the number of possible cases when discussing k-immersions. A graph is k-planar if it can be k-immersed. The notion of a 1-immersion of a graph was introduced by Ringel [18] when trying to color the vertices and faces of a plane graph so that the adjacent or incident elements receive distinct colors. Little is known about 1-planar graphs (see [2,4,5,1,6–15,17–21]). One of the basic questions has been how to recognize 1-planar graphs. Recently it was shown [16] that testing 1-planarity is NP-complete. By a proper k-immersion of a graph we mean a k-immersion with at least one crossing point. The planar graphs that have no proper 1-immersions are called the PN-graphs, for short. The notion of a PN-graph was introduced in [16] where PN-graphs were used to construct infinite families of minimal non-1-planar graphs (a graph G is minimal non-1-planar if G is non-1-planar but the graph G − e is 1-planar for every edge e of G). An important point in introducing PN-graphs is that the graphs can be of use when we want to check whether a graph is 1-planar or not. Every 3-connected PN-graph has a unique 1-immersion, namely, its unique plane embedding. Hence, if a graph G contains a 3-connected PN-graph H as a subgraph, then to check whether G has a 1-immersion we consider the plane embedding of H and check whether it is possible to add the vertices V (G) \ V (H ) and the edges E (G) \ E (H ) in such a way as to obtain a 1-immersion of G. Such an approach can work well when H is a spanning subgraph of G and |E (G)| − |E (H )| is small enough.
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Correspondence to: Bogomoltsa St. 3/5, Chernivtsi, 58001, Ukraine. E-mail address: [email protected].
0012-365X/$ – see front matter © 2013 Elsevier B.V. All rights reserved. http://dx.doi.org/10.1016/j.disc.2013.07.019
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Fig. 1. Two PN-graphs.
Fig. 2. A wheel Wn (w1 w2 . . . wn ) with the center v .
A class of PN-graphs was determined in [16] and some graphs of the class were constructed. Two PN-graphs of the class are shown in Fig. 1. This class was introduced with a particular purpose of constructing minimal non-1-planar graphs and seems to be rather special (the definition of the class includes a list of 8 conditions which any graph of the class has to satisfy and, in addition, three forbidden subgraphs). In the present paper we consider a rather more general class of PN-graphs. In Section 3 we prove Theorem 1 which characterizes (in terms of forbidden subgraphs) 4-connected PN-graphs which triangulate the plane. In Section 4 we show (Theorem 2) that every planar graph has a proper 3-immersion. We expect that there is much more to be discovered by following the themes of the paper. Problem 1. Determine all planar graphs that do not have a proper 1-immersion. Is it possible to characterize such planar graphs in terms of forbidden subgraphs? The following problem is possibly more difficult than it first appears. Problem 2. Is it true that every planar graph has a proper 2-immersion? 2. Preliminaries 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-selfintersecting) cycle C of G, if x lies in the interior (resp. exterior) of C , and does not lie on C . Two cycles of a graph are adjacent if they share a common edge. For convenience, we will denote a cycle as (v1 v2 . . . vn ) where vi and vi+1 are adjacent vertices for i = 1, 2, . . . , n (here vn+1 = v1 ). Since we assume that in 1-immersions, adjacent edges do not cross each other and no edge crosses itself, every 3-cycle of a 1-immersed graph is embedded in the plane. Hence, given a 3-cycle of a 1-immersed graph, we can speak about its interior and exterior. Throughout this section we will deal with 1-immersed graphs only. When a 1-immersion of a graph G is clear from the context, we shall identify vertices, edges, cycles and subgraphs of G with their image in R2 under the 1-immersion. Then by a face of a 1-immersion of G we mean any connected component of R2 \ G. By using Möbius transformations combined with homeomorphisms of the plane it is always possible to exchange the interior and exterior of any embedded cycle and it is possible to change any face of a given 1-immersion into the outer face of a 1-immersion. Formally, we have the following observation (which we will use without referring to it every time): (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 ′ with the same number of crossings as ϕ , in which C is embedded and all vertices of G, which lie inside C in ϕ , lie outside C in ϕ ′ and vice versa. By the wheel Wn (w1 w2 . . . wn ) of order n with the center v we mean the graph shown in Fig. 2. In what follows we will often use the following observation: Observation 1. If a graph G triangulates the plane, then a vertex v of G has degree n (v is an n-vertex, for short) if and only if G has as a subgraph a wheel of order n with the vertex v as the center of the wheel. Hence, every vertex of a 4-connected graph triangulating the plane is the center of a wheel of order at least 4.
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Fig. 3. Cycles of G.
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Fig. 4. Adjacent cycles of G.
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Fig. 5. An edge crossing the common edge of two adjacent 3-cycles.
In a 1-immersion of a graph G, if we consider the crossing points as new vertices, then we have an embedded graph G. Every cycle of G is embedded in the plane, so we can speak about the interior of the cycle. An edge of G is normal if it connects two vertices of G. It is easy to see that the following statement holds. (B) For a cycle D of G (where G is a 4-connected graph), if the sum of the number of vertices of G belonging to D and the number of normal edges belonging to D is less than 4, then if there are vertices of G outside (resp. inside) D, then there are no vertices of G inside (resp. outside) D; otherwise, G has a 3-cut, a contradiction. Observation 2. By (B), for any of the cycles of G shown in Fig. 3(a) (resp. Fig. 3(b)), there are no vertices of G inside (resp. outside) the cycle. In the figures, as in what follows, a cross inside (resp. outside) a cycle of G means that, in view of (B), there are no vertices of G inside (resp. outside) the cycle. Observation 3. Suppose that in a 1-immersion of a 4-connected graph G there is a fragment shown in Fig. 4 where the vertex
v is not adjacent to the vertices that lie inside the cycle (v ab1c ) in Fig. 4(a)–(b) and inside the cycle (v ab1c2) in Fig. 4(c) (in the figures, as in what follows, we label crossing points by integers). Suppose that the edges v a, ab, and v c in Fig. 4(a), the edges v a, ab, and bc in Fig. 4(b), the edges v a and ab in Fig. 4(c) are not crossed by other edges in the 1-immersion. Now we claim that there are no vertices inside (v ab1c ) in Fig. 4(a)–(b) and inside (v ab1c2) in Fig. 4(c). Indeed, consider, for example, the case in Fig. 4(a). If there are vertices inside the cycle (v ac ), then there is a 3-cut (including a and c), a contradiction. Then no edge crosses ac and if there are vertices inside (ab1c ), then {a, b, c } is a 3-cut, a contradiction. The cases in Fig. 4(b) and (c) are treated analogously. 3. Proper 1-immersions of 4-connected planar triangulations Denote by T4 the class of all 4-connected planar graphs which have a triangular embedding in the plane. First we need Lemmas 1–3 which give us some properties of proper 1-immersions of graphs of T4 . The proofs of the lemmas are based on straightforward observations and case analysis, and do not use any known special techniques or results. Lemma 1. Suppose that a graph G ∈ T4 has a proper 1-immersion ϕ . In ϕ , if edges e1 and e2 of G cross, then there is a 3-cycle C containing e1 such that one endvertex of e2 lies inside C and the other endvertex of e2 lies outside C . Proof. If e1 and e2 cross, then Fig. 5(a)–(e) show all possible cases of how the two 3-cycles containing e1 are embedded in the plane and which of the two 3-cycles has one endvertex of e2 inside and outside (the 3-cycle is denoted C ). In the case in Fig. 5(f), any endvertex of e1 is the center of a wheel of order 3, a contradiction, hence the case cannot take place. Lemma 2. Suppose that a graph G ∈ T4 without adjacent 4-vertices has a proper 1-immersion ϕ . In ϕ , suppose an edge vw crosses an edge yz of a 3-cycle C = (xyz ) such that the vertex v is inside and w is outside C (see Fig. 6(a)). Suppose that there is more than one vertex of G outside C . Then v is the only vertex lying inside C .
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Fig. 6. The vertices a and b are either both inside or both outside the cycle C .
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Fig. 7. Exactly two vertices are inside C .
Proof. Let (vw a) and (vw b) be the two 3-cycles of G containing the edge vw . If a, b ∈ {x, y, z }, then the edge ab is contained in three 3-cycles (namely, (xyz ), (v ab), and (w ab)), a contradiction. Hence {a, b} ̸⊂ {x, y, z }. Case 1: a, b ̸∈ {x, y, z }. If a and b lie outside C (Fig. 6(b)), then, taking Observation 2 into account, we obtain that v is the only vertex inside C . Suppose, for a contradiction, that a and b lie inside C (Fig. 6(c)). Then, by (A) and taking Observation 2 into account, we obtain that w is the only vertex outside C , a contradiction. Suppose, for a contradiction, that a lies inside and b lies outside C (see Fig. 7(a)). First, suppose, for a contradiction, that a and v are the only vertices inside C . Then the vertex a can have degree 4 only (see Fig. 7(b)). If v has degree 4, then a and v are adjacent 4-vertices, a contradiction. If v has degree 6, then v is the center of W3 (xyz ), a contradiction. Hence, v has degree 5. If there are xv and yv , then xy belongs to three 3-cycles, a contradiction. If there are xv and z v (see Fig. 7(c)), then x is the center of W4 (yav z ), hence x has degree 4, a and x are adjacent 4-vertices, a contradiction. If there are yv and z v (see Fig. 7(d)), then y is the center of W4 (xz v a), hence y has degree 4, a and y are adjacent 4-vertices, a contradiction. Now suppose, for a contradiction, that there are at least three vertices inside C . An edge e crosses av (see Fig. 7(a)), otherwise {a, v, x} or {a, v, y} is a 3-cut, a contradiction. If both endvertices of e lie inside C , then the two 3-cycles containing e pass the vertices a and v , respectively, thus v is the center of a wheel of order 3, a contradiction. Hence, one endvertex of e is either x or y, and the other endvertex of e lies inside C . Subcase 1.1: e = yu (see Fig. 8(a)). There is ux (otherwise, the two 3-cycles containing e pass the vertices a and v , respectively, thus v is the center of W3 (ayu), a contradiction), and there is either the 2-path yau or the 2-path yv u The edge au belongs to two 3-cycles: (awv) and (aw t ). It is easy to see that t ∈ {x, y}. Suppose, for a contradiction, that there is aw y (see Fig. 8(b); in this figure, as in what follows, a dashed line connecting two vertices means that G does not have an edge joining the two vertices). Since xy belongs to exactly two 3-cycles, there is no ax. Since vw belongs to exactly two 3-cycles, there is no yv . Hence yu belongs to the 3-cycles (yux) and (yau) (see Fig. 8(c)). If there is uv (see Fig. 8(d)), then a is the center of W4 (uvw y), hence a has degree 4, and v and u are the only vertices inside C which are adjacent to a. Then, taking Observation 3 into account, we obtain that there are no vertices inside (xua1) and (x3v u), and that a, u, and v are the only vertices inside C , thus a and u are adjacent 4-vertices, a contradiction. Hence, there is no uv , and since av belongs to two 3-cycles (one of them is (avw)), a 2-path ac v crosses xu. If c lies inside (xua1) (see Fig. 8(e)), then the only way for each of xu and v c to belong to two 3-cycles is that there are edges xc , cu, and xv . But then a is the center of W5 (uc vw y), hence a has degree 5, and u, c, v are the only vertices inside C which are adjacent to a. Then, taking Observation 3 into account, we obtain that there are no vertices inside (xcua), hence a, v , u, c are the only vertices inside C , thus u and c are adjacent 4-vertices, a contradiction. If c lies inside (x3v 2u) (see Fig. 8(f)), then the only way for each of xu and ac to belong to two 3-cycles is that there are edges xc and cu. But then u is the center of W4 (xcay), hence u has degree 4, and a, v, c are the only vertices inside C which are adjacent to u. Then, taking Observation 3 into account, we obtain that there are no vertices inside (x3v 2uc ), hence a, v , u, c are the only vertices inside C , thus u and c are adjacent 4-vertices, a contradiction. Now suppose, for a contradiction, that there is (aw x) (see Fig. 9(a)).
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Fig. 8. More than two vertices are inside C and there is (aw y).
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Fig. 9. More than two vertices are inside C and there is (aw x).
There are no ay (otherwise, xy belongs to three 3-cycles, a contradiction), hence the only way for yu to belong to two 3-cycles is that there is a 2-path yv u (see Fig. 9(b)). Then there is no xv (otherwise, xy belongs to three 3-cycles, a contradiction).
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Fig. 10. More than two vertices are inside C and there is xu.
Fig. 11. Two edges crossing a 3-cycle.
If there is au, then u is the center of W4 (xav y), hence u has degree 4, and a and v are the only vertices inside C which are adjacent to u. Then, taking Observation 3 into account, we obtain that a, v , and u are the only vertices inside C , hence a and u are adjacent 4-vertices, a contradiction. Hence there is no au, and since av belongs to two 3-cycles (one of them is (avw)), a 2-path ac v crosses xu. If c lies inside (xu2a) (see Fig. 9(c)), then the only way for each of xu and v c to belong to two 3-cycles is that there are edges xc and cu. But then a is the center of W4 (xc vw), hence a has degree 4, and v and c are the only vertices inside C which are adjacent to a. Then, taking Observation 3 into account, we obtain that there are no vertices inside (xc v 2a), hence a, v, u, c are the only vertices inside C , thus u and c are adjacent 4-vertices, a contradiction. If c lies inside (x3v u) (see Fig. 9(d)), then the only way for each of xu and ac to belong to two 3-cycles is that there are edges xc and cu. But then u is the center of W4 (xc v y), hence u has degree 4, and a, v, c are the only vertices inside C which are adjacent to u. Then, taking Observation 3 into account, we obtain that there are no vertices inside (x3v uc ), hence a, v , u, c are the only vertices inside C , thus u and c are adjacent 4-vertices, a contradiction. Subcase 1.2: e = xu (see Fig. 10(a)). The reader can easily check that the foregoing proof that the case in Fig. 10(b) cannot take place is also a proof that the case in Fig. 10(c) cannot take place as well the two proofs have exactly the same text but different figures, namely, Fig. 10(b) and (c), respectively. But Fig. 10(a) is obtained from Fig. 10(c) by relabeling vertices, hence the case in Fig. 10(a) cannot take place also. Case 2: the vertex a belongs to C , and b does not belong to C . Here we need the following claim. Claim 1. In ϕ , if edges e1 and e2 cross the edges of a 3-cycle C = (xyz ) (see Fig. 11), then e1 and e2 have a common endvertex. Proof. Suppose, for a contradiction, that e1 and e2 do not have a common endvertex. Then any 3-cycle containing e1 does not share a common edge with any 3-cycle containing e2 . If each of the two 3-cycles containing ei (i ∈ {1, 2}) passes a vertex of C , then an edge of C belongs to three 3-cycles, a contradiction. Hence, for each of the edges e1 and e2 , at least one of the 3-cycles containing the edge crosses an edge of C , that is, the edge yz, a contradiction. Now we consider the case when a = y and b lies outside C (see Fig. 12(a)). If no edge crosses xy and there are at least two vertices inside C , then {x, y, v} is a 3-cut, a contradiction. Hence, an edge e′ crosses xy. By Claim 1, e′ has a common endvertex with both vw and v b, thus e′ is adjacent to v (see Fig. 12(b)), and v is the only vertex inside C . Now suppose, for a contradiction, that a = y and b lies inside C (see Fig. 12(c)). Using (A) and taking into account the foregoing case where b lies outside C , we obtain that w is the only vertex outside C , a contradiction. In the case where a = x and b lies outside C (see Fig. 12(d)) we show analogously to the case in Fig. 12(a) that v is the only vertex inside C . Using (A) and taking into account the case in Fig. 12(d), we obtain in the case in Fig. 12(e) that w is the only vertex outside C , a contradiction. This completes the proof of the lemma. Lemma 3. Suppose that a graph G ∈ T4 without adjacent 4-vertices has a proper 1-immersion ϕ . In ϕ , if there is exactly one vertex v inside a 3-cycle C , then there can be only two cases shown in Fig. 13. Proof. If v is the only vertex inside C , then, since G is 4-connected, at least one vertex incident with v crosses an edge of C . If v is adjacent to all vertices of C , the v is the center of a wheel of order 3, a contradiction. Hence v has degree 4 or 5. If v has degree 5, then v is adjacent to two vertices of C , and three edges incident with v cross edges of C (see Fig. 14(a)). If v has degree 4, then there are exactly three cases shown in Fig. 14(b)–(d), respectively.
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Fig. 12. The vertex a lies on the cycle C .
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Fig. 13. All possible cases when there is exactly one vertex inside a 3-cycle.
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Fig. 14. A vertex v is inside a 3-cycle C .
The case in Fig. 14(a). Consider Fig. 15(a). If there are yw and v z, then yz belongs to three 3-cycles, a contradiction. Suppose, for a contradiction, that there is yw and there is no z w (see Fig. 15(b)). There is no ya (otherwise, v y belongs to three 3-cycles, a contradiction). Since v is the center of a wheel of order 5, the vertices a and b are adjacent, and w is adjacent either to b (see Fig. 15(c); in this case z is adjacent to a) or to a (see Fig. 15(d); in this case z is adjacent to b). In Fig. 15(c) there are two vertices inside and two vertices outside the 3-cycle (av z ), contrary to Lemma 2, hence this case cannot take place. Consider the case in Fig. 15(d). The edge xy belongs to two 3-cycles (xyz ) and (xyt ). It is easy to see that either t ∈ {w, b} or t lies outside (abz1w). If t = w (see Fig. 15(e)), then y is the center of W4 (xz vw), hence y has degree 4. The edge xw belongs to two 3-cycles: (xw y) and (xwt ′ ). It is easy to see that t ′ ∈ {a, b}. If t ′ = a (there is ax), then w is the center of W4 (xyv a), hence w and y are adjacent 4-vertices, a contradiction. If t ′ = b (there is bx), then z is the center of W4 (xyv b), hence z and y are adjacent 4-vertices, a contradiction. If t = b (see Fig. 15(f)), then x is the center of W3 (bzy), a contradiction. It remains to consider the case in Fig. 15(g) where t lies outside (abz1w). There is no zt (otherwise, x is the center of W3 (tyz ), a contradiction. Since xz belongs to two 3-cycles, there must be xb. Now z is the center of W4 (xbv y), hence z has degree 4. The edge xt belongs to two 3-cycles: (xty) and (xtt ′ ). It is easy to see that t ′ ∈ {a, b}. If t ′ = a (there is ax), then b is the center of W4 (xz v a), hence b and z are adjacent 4-vertices, a contradiction. If t ′ = b (there is tb), then x is the center of W4 (tyzb), hence x and z are adjacent 4-vertices, a contradiction.
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Fig. 15. A 5-vertex v inside a 3-cycle.
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Fig. 16. A 4-vertex v inside a 3-cycle (the case in Fig. 14(b)).
Hence, we obtain that when v has degree 5 there can be the case in Fig. 13(a) only. The case in Fig. 14(b). Consider Fig. 16(a). If there are yw and z w , then yz belongs to three 3-cycles, a contradiction. Since vw belongs to two 3-cycles, there is aw and there is one of the two edges yw and z w . Suppose, for a contradiction, that there is yw and there is no w z (see Fig. 16(b)). Since v is the center of a wheel of order 4, there is az (see Fig. 16(c)). The edge xy belongs to two 3-cycles: (xyz ) and (xyt ). It is easy to see that either t ∈ {a, w} or t lies outside (az1w). If t = a (there is ay), then v is the center of a wheel of order 3, a contradiction. If t = w (there is xw ), then y is the center of W4 (xz vw), thus y and v are adjacent 4-vertices, a contradiction. Hence, t lies outside (az1w) (see Fig. 16(d)). Taking Observation 2 into account, we obtain that t is the only vertex outside (az1w), thus w has degree 4 (there must be w t)), and v and w are adjacent 4-vertices, a contradiction. Hence, we obtain that when v has degree 4, then in the case in Fig. 14(b) there can be the case in Fig. 13(b) only. The case in Fig. 14(c). We prove that the case in Fig. 17(a) cannot take place. First we consider the case when v is the center of W4 (axbw) (see Fig. 17(b)). Suppose, for a contradiction, that the case in Fig. 17(b) takes place. The edge yz belongs to two 3-cycles: (xyz ) and (yzt ). It is easy to see that either t ∈ {b, w} or t lies outside (xbw a). If t = b (and yb crosses aw ), then x is the center of W3 (byz ), a contradiction. Suppose that t = w (see Fig. 17(c)). There is no ay (otherwise, a is the center of W4 (xvw y), hence a and v are adjacent 4-vertices, a contradiction). Analogously, there is no bz. If there is xw , then x is the center of W3 (bw a), a contradiction. We see that the only way for each of the edges xy and xz to belong to two 3-cycles is shown in Fig. 17(d), but then y and z are adjacent 4-vertices, a contradiction. Now suppose that t lies outside (xbw a) (see Fig. 17(e)). If there is xt, then x is the center of W3 (yzt ), a contradiction. Now we see that the only way for each of the edges xy and xz to belong to two 3-cycles is that there are ay and bz (see Fig. 17(f)). There is no yw (otherwise, a is the center of W4 (xvw y), hence a and v are adjacent 4-vertices, a contradiction). Analogously, there is no w z. Then w has degree 4 (there is w t), and w and v are adjacent 4-vertices, a contradiction. In the case where v is the center of W4 (abw x) (see Fig. 17(g)), taking Observation 2 into account, we obtain that the adjacent vertices y and z can have degree 4 only (there are ay, yw, w z, and zb), a contradiction. Hence, we obtain that the case in Fig. 14(c) cannot take place. The case in Fig. 14(d). We prove that the case in Fig. 18(a) cannot take place. Since v is the center of a wheel of order 4, there are exactly three cases shown in Fig. 18(b)–(d).
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Fig. 17. A 4-vertex v inside a 3-cycle (the case in Fig. 14(c)).
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Fig. 18. A 4-vertex v inside a 3-cycle (the case in Fig. 14(d)).
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Fig. 19. The graph L(n), n ≥ 4.
Suppose, for a contradiction, that the case in Fig. 18(b) takes place. If there is ax, then y is the center of W4 (axz v), hence y and v are adjacent vertices, a contradiction. Analogously we show that there is no xb. Then the only way for each of the edges xy and xz to belong to two 3-cycles is that for each of the edges, a 3-cycle containing the edge crosses ab, a contradiction. In the cases in Fig. 18(c) and (d), taking Observation 2 into account, we obtain that x has degree 4 (there are ax and xb), and then z is the center of W4 (xyv a), hence v and z are adjacent vertices, a contradiction. Hence, we obtain that the case in Fig. 14(d) cannot take place. This completes the proof of the lemma. Consider the planar graphs L(n) for n ≥ 4, M (n) for n ≥ 1, P (n) for n ≥ 1, and Q (n) for n ≥ 2 shown in Figs. 19(a), 20(a), 21(a), and 22(a), respectively. The graph Q (n) in Fig. 22(a) is defined up to the mirror reflection. If in Figs. 20(a), 21(a), and 22(a) we identify the vertices x and y (and identify the edges v x and v y, the edges w x and w y), we obtain the planar graphs M (n) for n ≥ 1, P (n) for n ≥ 1, and Q (n) for n ≥ 2, respectively.
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Fig. 20. The graph M (n), n ≥ 1.
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Fig. 21. The graph P (n), n ≥ 1.
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Fig. 22. The graph Q (n), n ≥ 2.
Theorem 1. A graph G ∈ T4 has a proper 1-immersion if and only if at least one of the following conditions holds: (a) G has adjacent 4-vertices; (b) G is one of the graphs L(n) for n ≥ 4, M (n) for n ≥ 2, P (n) for n ≥ 2, and Q (n) for n ≥ 3 (c) G has as a subgraph one of the graphs M (n) for n ≥ 2, P (n) for n ≥ 2, and Q (n) for n ≥ 3 Proof. If G has adjacent 4-vertices, then G has a proper 1-immersion (see Fig. 23).
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Fig. 23. Adjacent 4-vertices.
Fig. 24. An end 3-cycle.
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Fig. 25. 3-cycles of type A and B.
The graphs L(n), M (n), P (n), Q (n), M (n), P (n), Q (n) are 3-connected, hence they have a unique plane embedding. If G has a subgraph F which is one of the graphs M (n), P (n), and Q (n), then, since G is 4-connected, in the plane embedding of G there are no vertices inside the faces of the plane embedding of F . Hence, if in the plane embedding of G we replace the plane embedding of F by the proper 1-immersion of F (as shown in Figs. 20–22), we obtain a proper 1-immersion of G. It is easy to see that each of the graphs M (1), P (1), and Q (2) has adjacent 4-vertices not belonging to the boundary of the outer 4-face of the plane embedding of the graph. Hence, if G is one of the graphs M (1), P (1), and Q (2), or if G has as a subgraph one of the graphs M (1), P (2), and Q (1), then G has adjacent 4-vertices. Now to prove the theorem it suffices to prove the following statement: (C) suppose that a graph G ∈ T4 without adjacent 4-vertices has a proper 1-immersion ϕ . Then either (b) or (c) of Theorem 1 holds. From here on we prove (C). Keeping (A) in mind and taking Lemma 2 into account, from now on by the interior of a 3-cycle C we will mean a component of R2 \ C which contains at most one vertex of G. In ϕ , a 3-cycle is crossed if at least one of its edges is crossed. A crossed 3-cycle with no vertex inside is called an end 3-cycle (see Fig. 24). A crossed 3-cycle C = (xyz ) with the only one vertex v inside shown in Fig. 25(a) (resp. Fig. 25(b)) is called a 3-cycle of type A (resp. type B). It is important to keep in mind that since vw belongs to two 3-cycles, there are the edges aw and bw in the case in Fig. 25(a), and the edge aw in the case in Fig. 25(b). Notice that the vertex w in Fig. 25 does not lie inside any 3-cycle. Indeed, if w lies inside a 3-cycle D, then either D contains yz (a contradiction, since yz belongs to exactly two 3-cycles) or D passes v (a contradiction, since then D has at least two vertices inside, contrary to Lemma 2). Taking Lemma 3 into account, we obtain that in ϕ , every crossed 3-cycle is of type A or B. In ϕ , two adjacent crossed 3-cycles are cross-adjacent (through an edge e) if the following holds: (i) the 3-cycles share the common edge e and do not cross each other; (ii) an edge crosses e; (iii) for each of the 3-cycles, if the 3-cycle contains exactly one vertex inside, then the vertex does not belong to the other 3-cycle. Notice that, the adjacent 3-cycles (xyz ) and (v yz ) in Fig. 25 are not cross-adjacent (in spite of the fact that the edge vw crosses the common edge yz of the two 3-cycles).
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Fig. 26. Cross-adjacent 3-cycle of type B and end 3-cycle.
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Fig. 27. A 3-cycle of type A is cross-adjacent to an end 3-cycle or a 3-cycle of type B.
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Fig. 28. Cross-adjacent 3-cycles of type A.
We have the following: (D) The type A 3-cycle in Fig. 25(a) is cross-adjacent (through the edges xy and xz, respectively) to exactly two other 3-cycles. The type B 3-cycle in Fig. 25(b) is cross-adjacent (through the edge xy) to exactly one other 3-cycle. Now we consider how 3-cycles can be cross-adjacent. If two end 3-cycles are cross-adjacent (see Fig. 5(f)), then any vertex of the 3-cycles is the center of a wheel of order 3, a contradiction. If a type B 3-cycle and an end 3-cycle are cross-adjacent (see Fig. 26), then the vertex y is the center of W4 (xz v a), hence y and v are adjacent 4-vertices, a contradiction. If two type B 3-cycles are cross-adjacent, then the only vertices inside the 3-cycles are adjacent, and we obtain adjacent 4-vertices, a contradiction. The case when a type A 3-cycle and an end 3-cycle are cross-adjacent is shown in Fig. 27(a). A case when a type A 3-cycle and a type B 3-cycle are cross-adjacent is shown in Fig. 27(b). The case in Fig. 27(c) cannot take place since there are two vertices inside and two vertices outside the 3-cycle (v bw), contrary to Lemma 2. A case when two type A 3-cycles are cross-adjacent is shown in Fig. 28(a). The case in Fig. 28(b) cannot take place since there are two vertices inside and two vertices outside the 3-cycle (v bw), contrary to Lemma 2. In the case in Fig. 28(c), since bc belongs to two 3-cycles, there is v c, and we obtain that x and z are adjacent 4-vertices, a contradiction. Now we see that all possible cases of how two 3-cycles can be cross-adjacent are shown in Fig. 27(a) and (b), and Fig. 28(a). Taking into account (D), we obtain that in ϕ , if there is only one vertex inside a 3-cycle C , then either (X1) or (X2) holds: (X1) C is an element of a cyclic sequence (C1 , C2 , . . . , Cn ), n ≥ 4, of type A 3-cycles such that Ci and Ci+1 are cross-adjacent for i = 1, 2, . . . , n (here Cn+1 = C1 ). (X2) C is an element of a sequence C1 , C2 , . . . , Cn (n ≥ 3), of 3-cycles such that: Ci and Ci+1 are cross-adjacent for i = 1, 2, . . . , n − 1; the cycles C2 , C3 , . . . , Cn−1 have type A; each of C1 and Cn either has type B or is an end 3-cycle (Note that if both C1 and Cn are end cycles, then n ≥ 4, since there are adjacent 4-vertices for n = 3).
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Fig. 29. Obtaining a proper 3-immersion of a planar graph.
In the case (X1), taking Fig. 28(a) into account, we obtain a 1-immersed graph L(n), n ≥ 4 shown in Fig. 19(b), that is, G is L(n), n ≥ 4. In the case (X2), taking Figs. 27(a), 27(b), and 28(a) into account, we obtain a 1-immersed graph which is one of the 1-immersed graphs shown in Figs. 20(b), 21(b), and 22(b), where the vertices x and y may be identified (it means that the edges v x and v y are identified, and the edges w x and w y are identified also). Hence either G is one of the graphs M (n) for n ≥ 2, P (n) for n ≥ 2, and Q (n) for n ≥ 3, or G has as a subgraph one of the graphs M (n) for n ≥ 2, P (n) for n ≥ 2, and Q (n) for n ≥ 3. This completes the proof of the theorem. One can check that: (i) each of the graphs M (n) for n ≥ 2, P (n) for n ≥ 2, and Q (n) for n ≥ 3 has a 4-valent vertex; (ii) each of the plane graphs M (n) for n ≥ 2, P (n) for n ≥ 2, and Q (n) for n ≥ 3 shown in Figs. 20(b), 21(b), and 22(b), respectively, has a 4-valent vertex not belonging to the boundary (v xw y) of the outer 4-face. Hence, we have the following: Corollary 1. Suppose that a 4-connected graph G triangulates the plane. If G is not L(n), n ≥ 5, and if G has no 4-vertices, then G has no proper 1-immersions. 4. Proper 3-immersions By a k-face we mean a k-gonal face. Theorem 2. Every connected planar graph has a proper 3-immersion. Proof. It suffices to consider connected planar graphs whose vertices have degree at least 3. Borodin [3] proved that any plane embedding of such a graph has one of the following vertices and faces: (a) (b) (c) (d)
a 5-vertex v incident with four 3-faces; a 4-vertex v incident with a 3-face; a 3-vertex v incident with a 3-face or a 4-face; a 5-face incident with four 3-vertices. Fig. 29(a)–(d) shows how for each of the cases (a)–(d), respectively, we can obtain a proper 3-immersion of the graph.
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