New Upper Bounds for the Heights of Some Light Subgraphs in 1-Planar Graphs with High Minimum Degree

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1 New Upper Bounds for the Heights of Some Light Subgraphs in 1-Planar Graphs with High Minimum Degree Xin Zhang, Jian-Liang Wu, Guizhen Liu To cite this ersion: Xin Zhang, Jian-Liang Wu, Guizhen Liu. New Upper Bounds for the Heights of Some Light Subgraphs in 1-Planar Graphs with High Minimum Degree. Discrete Mathematics and Theoretical Computer Science, DMTCS, 011, Vol. 13 no. 3 (3), pp <hal > HAL Id: hal Submitted on 13 May 01 HAL is a multi-disciplinary open access archie for the deposit and dissemination of scientific research documents, whether they are published or not. The documents may come from teaching and research institutions in France or abroad, or from public or priate research centers. L archie ouerte pluridisciplinaire HAL, est destinée au dépôt et à la diffusion de documents scientifiques de nieau recherche, publiés ou non, émanant des établissements d enseignement et de recherche français ou étrangers, des laboratoires publics ou priés.

2 Discrete Mathematics and Theoretical Computer Science DMTCS ol. 13:3, 011, 9 16 New Upper Bounds for the Heights of Some Light Subgraphs in 1-Planar Graphs with High Minimum Degree Xin Zhang Jian-Liang Wu Guizhen Liu School of Mathematics, Shandong Uniersity, Jinan 50100, P. R. China receied 5 th May 010, reised 5 th July 011, accepted 1 st September 011. A graph is 1-planar if it can be drawn on the plane so that each edge is crossed by at most one other edge. In this paper, it is shown that each 1-planar graph of minimum degree 6 contains a copy of -cycle with all ertices of degree at most 19. In addition, we also show that the complete graph K is light in the family of 1-planar graphs of minimum degree 7, with its height at most 11. Keywords: 1-planar graphs, light graphs, height. 1 Introduction Throughout this paper, all graphs are finite, simple and undirected. We use V (G), E(G), δ(g) and (G) to denote the ertex set, the edge set, the minimum degree and the maximum degree of a graph G. Let e(g) = E(G) and (G) = V (G). For planar graphs, we use F (G) to denote the face set of G and let f(g) = F (G). A ertex (face) of degree k is called a k-ertex (face) and a ertex (face) of degree at least k is called a k -ertex (face). For undefined concepts we refer the reader to []. Let G be a family of graphs and let H be a connected graph such that each member of G contains a subgraph isomorphic to H. Denote h(h, G ) and w(h, G ) respectiely to be the smallest integers with the property that each graph G G contains a subgraph K H such that max x V (K) {d G (x)} h(h, G ) and x V (K) {d G(x)} w(h, G ). These two parameters h(h, G ) and w(h, G ) are called the height and the weight of H in the family G. If they are finite, then we say H is light in G, otherwise we say H is heay in G. A graph is 1-planar if it can be drawn on the plane so that each edge is crossed by at most one other edge. The notion of 1-planar was introduced by Ringel [1] while studying the simultaneous ertex-face coloring of planar graphs; in the mentioned paper, he proed that each 1-planar graph is 7-colorable. Now this bound was improed to 6 (being sharp) by Borodin [3, ]. Borodin et al. [5] also proed that each 1-planar graph is acyclically 0-colorable. In addition, the list analogue of ertex coloring of 1-planar Supported by NSFC grants , , , RFDP grant and GIIFSDU grant yzc sdu.zhang@yahoo.com.cn, xinzhang@mail.sdu.edu.cn jlwu@sdu.edu.cn gzliu@sdu.edu.cn c 011 Discrete Mathematics and Theoretical Computer Science (DMTCS), Nancy, France

3 10 X. Zhang, J.-L. Wu and G. Liu graphs was inestigated in [1] by Albertson and Mohar. In [1], Wang and Lih proed that each 1-planar graph is list 7-colorable. Recently, Zhang et al. showed that each 1-planar graph G with maximum degree is -edge colorable if 10 [17], or 9 and G contains no chordal 5-cycles [16], or 8 and G contains no chordal -cycles [15], or 7 and G contains no 3-cycles [19]; it is -edge choosable and ( 1)-total choosable if 1 [0]; it is ( 1)-edge choosable and ( )-total choosable if 16 [0]. The generalization of the total coloring of 1-planar graphs, the so called (p, 1)-total labelling problem, is studied in []. Although 1-planar graphs is an interesting family, it is still little explored comparing to the family of planar graphs. In general, the family of 1-planar graphs has many fundamental aspects which are different from the planar ones. It is well-known that the planarity test for a graph is polynomial. But for 1-planarity, it is extremely unlucky that its recognizing is NP-complete [10]. This hardness may come from the fact that the set of 1-planar graphs is not closed under taking minors and thus the 1-planarity cannot be characterized by forbidding some minors. On the other hand, 1-planar graphs also hae some similar local structures as planar graphs. For example, the maximum possible number of edges in a 1-planar graph G is (G) 8 and there are many 1-planar graphs that attain this bound [7]. This implies that the maximum possible minimum degree of a 1-planar graph is 7. Note that eery planar graph is 5-degenerate. So each 1-planar graph with minimum degree 6 or 7 cannot be planar. Howeer, such 1-planar graphs may hae many nice and similar properties as planar graphs. In particular, we wonder whether a 1-planar graph with minimum degree 6 or 7 has a similar behaior as a planar graph with high minimum degree concerning the local structures. In the Fifth Workshop of Graph Embeddings and Maps on Surfaces held at Tále in the Summer of 009, Madaras [11] reported some new results on the light subgraphs in 1-planar graphs with minimum degree 6 and 7, as well as in 1-planar graphs with minimum degree 5 and girth. They (Hudák and Madaras) proed that each 1-planar graph with minimum degree 6 contains a -cycle with all ertices of degree at most 71, and this upper bound was later decreased to 7 [9]. They also proed the existence of a complete graph K with all ertices of degree at most 13 in 1-planar graphs with minimum degree 7 [9, 11]. Some other local structures were extensiely studied by many authors including [6 8, 10, 13, 18, 1]. In this paper, we improe Hudák and Madaras s upper bounds for the heights of C and K in the family of 1-planar graphs with minimum degree 6 and 7, respectiely. We proe Theorem 1 Each 1-planar graph with minimum degree 6 contains a copy of C with all ertices of degree at most 19. Theorem Each 1-planar graph with minimum degree 7 contains a copy of K with all ertices of degree at most 11. Howeer, we still do not know whether or not these bounds in Theorems 1 and are best possible. We leae this as an open problem for further research. Preliminaries In the rest of this paper, we always assume that G is a 1-planar graph with minimum degree 6 or 7, and it is drawn on a plane so that (1) eery edge is crossed by at most one another edge and () the number of crossings is as small as possible. The associated plane graph G of G is the plane graph that is obtained from G by turning all crossings of G into new -alent ertices. By (1), no two -ertices are

4 Light subgraphs in 1-planar graphs 11 adjacent in G. A 3-face in G is called true if it is not incident with -ertices and false otherwise. A big face denotes a face of degree at least. For conenience, we use some specialized notations during the proofs in the next section. Let be a -ertex in G and let 1,, 3, be its neighbors in clockwise order. By f i denote the face incident with i and i1 in G, where the subtraction and addition on subscripts are taken modulo. We call a -ertex closed if d(f 1 ) = d(f ) = d(f 3 ) = d(f ) = 3, and open otherwise. It is easy to see that if d(f i ) = 3, then i i1 E(G ). In this case, let f i be the other face incident with the edge i i1. If d(f i ) = 3, then the third ertex on the boundary of f i, which is different from i and i1, will be denoted by i. So i is a -ertex if and only if f i is false, in which case we denote the neighbor of i (or i1 ) in G, such that the edge connecting them in G contains the crossing point i, to be i (or i1, respectiely). Namely, i i and i1 i1 are two edges in G that cross each other at the point i. Denote the face incident with the path x i x i x i1 (or x i1x i x i ) in G by fi L (or fi R, respectiely). The proofs of the two theorems are carried out by contradiction and their beginnings are just the same. Suppose that G is a counterexample to the theorem. Consider the associated plane graph G of G. Using the well-known Euler formula (G ) e(g ) f(g ) = and the relation V (G ) d() = f F (G ) d() = e(g ), we hae V (G ) (d() 10) f F (G ) (3d(f) 10) = 0. Define an initial charge w on V (G ) F (G ) by w() = d() 10 if V (G ) and w(f) = 3d(f) 10 if f F (G ). Thus we hae x V (G ) F (G ) w(x) = 0. In order to proe Theorem 1 and Theorem, we shall design some discharging rules so that after discharging the new charge w (x) of eery element x V (G ) F (G ) is nonnegatie. Since our rules only moe charge around and do not affect the total charges, this leads to a contradiction at the end and completes the proofs. 3 Proofs 3.1 Proof of Theorem 1 The proof follows the strategy described in Section. Suppose that G is a counterexample to Theorem 1. Then each copy of C contained in G has a ertex of degree at least 0; we call such a ertex big. The ertices of degree from 6 to 19 are called intermediate ertices. In the following. we proceed with the discharging method and the initial charges are redistributed according to the following rules. Note that all the faces as mentioned in the rules are faces in G. R1. Each 6 -ertex sends 1 3 to each of its incident faces. R. Each -ertex sends 1 3 to each of its incident false 3-faces. R3. Each big face sends 1 to each of its incident -ertices. R. Let α = [xyz] be a true 3-face haing a common edge yz with a false 3-face β = [uyz]. If x is a big ertex, then x sends 3 to u through yz. R5. Let α = [xyz] and β = [uyz] be two adjacent false 3-faces and z be a -ertex. If y is a big ertex, then y sends to z. R6. Let α = [xyz] be a false 3-face haing a common edge yz with a big face β. If y is a big ertex and z is a -ertex, then y sends 1 to z.

5 1 X. Zhang, J.-L. Wu and G. Liu R7. Let α = [xyz] and β = [uyz] be two adjacent false 3-faces, x be a big ertex, z be a -ertex, and y, u be two intermediate ertices. Suppose that yu is incident with another false 3-face γ = [yuw] and yy crosses uu in G at w. If yu E(G) and u is an intermediate ertex, or y u E(G) and y is an intermediate ertex, then x sends 1 3 to w through yz and yu. R8. Let α be a big face haing a common edge xy with a false 3-face β = [xyz]. If z is a -ertex, then α sends 3 to z through xy. R9. Let α be a big face haing a common edge xy with a false 3-face β = [xyz]. If x is a -ertex, y, z are both intermediate ertices, and yz is incident with another false 3-face γ = [yzu], then α sends 1 3 to u through xy and yz. By the aboe rules, one can proe the following two claims that are quite useful. Claim 1. For a big face f, if R3 is applied to f once, then R9 can be executed at most once. Proof. Suppose that f is a big face with three incident ertices x, y, z such that xy, yz E(G ) and y is a -ertex. Then xz E(G) and we can suppose that xx crosses zz in G at y. By R3, there must be a transfer from f to y now. If R3 is applied to f once and R9 is executed at least twice, then xz, x z E(G) and all of x, x, z, z are intermediate ertices. This implies the existence of a -cycle xx zz in G with all its ertices intermediate, a contradiction. Hence R9 can be executed at most once. Claim. For a big ertex x, if R5 is applied to x once, then R7 can be executed at most once. Proof. Let xyz be a false 3-face with a big ertex x and a -ertex z. Suppose that xx crosses yy in G at z. If R5 is applied to x once, then xy E(G). Now we proe that R7 cannot be executed twice. For otherwise, x yz, x y z are both false 3-faces and y, y, x are all intermediate ertices. Furthermore, x y is incident with another 3-face x yu such that u y and u is an intermediate ertex by R7. Consequently, the four intermediate ertices x, y, y, u form a -cycle in G, a contradiction. We now check that the final charge of eery ertex and face in G is nonnegatie. Case 1. Suppose that f is a 3-face. Then by R1 and R, each ertex incident with f sends 1 3 to f. This implies that w (f) = w(f) = 1 1 = 0. Case. Suppose that f is a big face. Let a be the number of transfers from f to its incident -ertices by R3 and let b be the number of transfers from f to its non-incident -ertices by R8. Since no two -ertices are adjacent in G, a d(f) b and f is incident with at least d(f) 6 -ertices. By Claim from f by R9. Therefore, 1, if there is a transfer of 1 from f by R3, then there is a total transfer at most 1 3 w (f) w(f) 1 3 d(f) (1 1 3 )a 3 b 19 6 d(f) d(f) 0 0 by R1 for d(f). Case 3. Suppose that is a -ertex. Then w() = d() 10 =. Subcase 3.1. Suppose that is incident with at least three big faces. Then by R and R3, w () w() = 3 > 0. Subcase 3.. Suppose that is incident with two big faces and two false 3-faces. We follow the definitions in Section and split our proofs into two subcases. Subcase Without loss of generality, suppose that f 1, f 3 are big faces and f, f are false 3-faces. Then α = [ 1 3 ] is a -cycle and thus at least one of the ertices on the boundary of α, say 1, is a big ertex. This implies that will receie 1 from 1 by R6. Finally, by R and R3, we also hae w () w() = 1 3 > 0. Subcase 3... In this case, we suppose that f 1, f are big faces and f 3, f are false 3-faces. If is (a b) 19 6 d(f) 10 3 d(f) = adjacent to at least two big ertices, then at least one of them will send charge to by R6. Thus we still hae w () w() = 1 3 > 0 by R, R3 and R6. So we suppose that is adjacent to

6 Light subgraphs in 1-planar graphs 13 at most one big ertex. If one of 1, 3 and is big, then the redistributed charge of can be calculated as before, hence, it is nonnegatie. If none of 1, 3 and is big (namely, all of them are intermediate), then will not send charge to by any of the rules no matter is big or intermediate. Now, consider the face f 3. If it is a big face, then it will send 3 to through 3 by R8. If f 3 is 3-face, denoted by 3 3 (note that 3 1, ), then 3 must be a big ertex while f 3 is true, that is because forms a -cycle in G with 1, 3, being intermediate. In such a case, 3 will send 3 to through 3 by R. If f 3 is false, then 3 is a -ertex. Now consider the faces f3 L and f3 R. If f3 L is big face, then it will send 1 3 to through 3 by R9. Otherwise f3 L must be a false 3-face, denoted by 3 3 (note that 1, ). In this case 1 3 is a -cycle in G with 1, 3, being intermediate, which implies that must be a big ertex. Then by R7, will send 1 3 to through 3. By a similar discussion on f3 R, will receie another 1 3 from f 3 R or 3 through 3 (again, note that 3 1, and if 3 =, then f3 R must be a big face). Consequently, using the transfers through 3, receies totally 1 3 = 3. By symmetry, the same discussion can be applied to the calculation of the transfers through 1. Thus, by R and R3, we still hae w () w() = 3 > 0. Subcase 3.3. Suppose that is incident with one big faces and three false 3-faces. Without loss of generality, we assume that f 1 is big and the other three faces are false. Since 1 3 forms a -cycle in G, must be adjacent to at least one big ertex. If is adjacent to at least two big ertices, then by R5 and R6, receies at least 1 = from its big neighbors. Together with R and R3, we hae w () w() = 0. Now we consider the case when is adjacent to exactly one big ertex. If 3 or is big, then by R5, a transfer of from 3 or to will be put into practice. This implies that w () w() = 0. If 1 or is big, say, then will send 1 to by R6. By the same argument as in Subcase 3.., will also receie 3 through the edge 3 and another 3 through the edge 1. In total, by R and R3, we shall also hae w () w() = 1 3 > 0. Subcase 3.. Suppose that is incident with four false 3-faces. Since 1 3 is a -cycle in G, must be adjacent to at least one big ertex. If is adjacent to at least two big ertices, then by R and R5, we hae w () w() 1 3 = 3 > 0. If is adjacent to exactly one big ertex, say, then will send to by R5. Again, by the same argument as in Subcase 3.., will totally receie 3 = 3 through the edges 1 and 3. This implies that w () w() = 0 by R and R3. Case. Suppose that is an intermediate ertex. Then sends out charges only by R1, which implies that w () w() 1 3 d() = d() d() 30 3d() = 3 0 for d() 6. Case 5. Suppose that is a big ertex. Let F denote the subgraph induced by the faces which are incident with. Then F can be decomposed into many parts, each of which is one of the following fie clusters in Figure 1, and any two parts of which are adjacent only if they hae a common edge w such that w is a 6 -ertex. The hollow ertices in Figure 1 are -ertices and the solid ones are 6 -ertices; all the faces marked by are big faces and there is at least one big face contained in the clusters of type, and 5. Let n i be the number of clusters of type i contained in F. By their definitions, one can easily obsere that n 1 n n 3 3n n 5 d(). If there is a cluster of type 1 in F, then will send to y by R5. Meanwhile, will also send out at most 1 3 through either xy or yz by R7 and Claim. If there is a cluster of type in F, then will send 1 to y by R6 and send out at most 1 3 through xy by R7. If there is a cluster of type 3 in F, then will send out at most 3 though xy by R. If there is a cluster of type in F, then will send 1 to each of y and u by R6 and totally send out at most 1 3 = 3 through xy and uw by R7. If there is a cluster of type 5 in F, then will not send out charges ia this cluster by any of the rules. Therefore, together with R1, we hae w () w() 1 3 d() ( 1 3 )n 1 (1 1 3 )n 3 n 3 ( 1 3 )n =

7 1 X. Zhang, J.-L. Wu and G. Liu x... y z x y x y type1 type type3 x y... type u w... type5 Fig. 1: Fie types of cluster 5 3 d() (n 1n n 3 3n n 5 )(n 1 n n 7 6 n 5) 5 3 d() 10 7 d() 0 6d() = 0 for d() Proof of Theorem The proof also follows the strategy described in Section. Let G be a counterexample to Theorem. So each copy of K in G contains a 1 -ertex, which is called big ertex. The ertices of degree from 7 to 11 are called intermediate ertices. Let t() be the number of 3-faces that are incident with. Then is closed if and only if t() = (recall the definition of closed ertices in Section ). Let c() (or c(f)) be the number of -ertices that are adjacent to (or incident with) (or f). By c 1 () and c (), respectiely, we denote the number of closed -ertices and open -ertices adjacent to. Obiously we hae c 1 () c () = c(). Claim 1. If d() = 7, then t() 6 if c() =, t() if c() = 5, t() if c() 6. Proof. This follows directly from the fact that no two -ertices are adjacent in G. Claim. If d() 8, then t() c() 3 d() and t() c() d(). Proof. If c() d() d() d(), then this holds triially. So we assume that c(). Beginning with, each unit of increase on c() will imply two units of decrease on the largest possible number of the 3-faces that are incident with, that is because no two -ertices are adjacent in G. Therefore, t() d() (c() d() d() ) if d() is een, and t() d() 1 (c() ) if d() is odd. In either case, t() c() d(). Since c() d(), we hae t() c() d() c() 3 d(). Claim 3. If d() 1, then c 1 () c () d(). Proof. Since no two -ertices are adjacent in G, we hae c 1 () d() d(). If c(), then, then each unit of increase on c() from c 1 () c () = c 1 () c() d(). If c() > d() implies the existence of at least two open -ertices that are adjacent to. This implies that d() c () (c() d() ) (c 1() c ()) d(). Therefore, we hae c 1 () c () d(). We complete the proof of Theorem by applying the following rules of discharging. R1. Each big face sends 1 to each of its incident -ertices. R. Each intermediate ertex sends 1 3 to each of its adjacent -ertices.

8 Light subgraphs in 1-planar graphs 15 R3. Each big ertex sends to each of its adjacent open -ertices, and 9 to each of its adjacent closed -ertices. R. Let α = [xyz] be a true 3-face. Then each of x, y, z sends 1 3 to α. R5. Let α = [xyz] be a false 3-face. If x is a -ertex, then x sends 1 9 while both y and z sends 9 to α. We now check that the final charge of eery ertex and face in G is nonnegatie. Case 1. Suppose that f is a 3-face. By R and R5, one can obsere that w (f) = w(f)1 = 11 = 0. Case. Suppose that f is a big face. Since no two -ertices are adjacent in G, c(f) d(f). Therefore, by R1, we hae w (f) w(f) d(f) 5d(f) 0 0 for d(f). Case 3. Suppose that is a -ertex. If is incident with at least two big faces, then by R1, R, R3 and R5, w () w() = 10 9 > 0. If is incident with exactly one big face, then we still hae w () w() = 0. If is incident with no big faces, then is closed. Meanwhile, 1,, 3 and forms a K in G, which implies that at least one of them is a big ertex, say 1. Then 1 will send 13 9 to by R3. Since, 3 and are either intermediate or big, each of them shall send at least 1 3 to by R and R3. Together with R5, we also hae w () w() = 0. Case. Suppose that is an intermediate ertex. Subcase.1. Suppose that d() = 7. If c() 3, then can be incident with at most six false 3-faces. So by R, R and R5, we hae w () w() = 0. If c() =, then by Claim 1, t() 6. It follows that w () w() = 0. If c() = 5, then t() by Claim 1, which implies that w () w() = 5 9 > 0. If c() 6, then t() by Claim 1. Consequently, we still hae w () w() 1 3 c() 9 t() = 7 9 > 0. Subcase.. Suppose that 8 d() 11. By R, R, R5 and Claim, we hae w () w() 1 3 c() 9 t() d() 10 9 (c() t()) d() d() 30 d() = 3 > 0 for d() 8. Case 5. Suppose that is a big ertex. If c() d(), then by R3, R and R5, w () w() 13 9 c() 13 9d() d() 10 9 d() 5d() 60 9d() = 6 0 for d() 1. On the other hand, if c() d(), then by R3, R, R5, Claim and Claim 3, we hae w () w() 13 9 c 1() 1 3 c () 9 t() = d() (c 1()c ()) 9 t() 7 9 c 1() d() d() 9 (d() c 1() c ()) 7 9 c 1() = 7 9 d() c 1() 8 9 c () 7 9 d() c() 7 9 d() d() 60 18d() = 6 0 for d() 1. References [1] M. O. Albertson and B. Mohar. Coloring ertices and faces of locally planar graphs. Graphs Combin., :89 95, 006. [] J. A. Bondy and U. S. R. Murty. Graph Theory with Applications. American Elseier Pub. Co., New York, [3] O. V. Borodin. Solution of Ringel s problems on the ertex-face coloring of plane graphs and on the coloring of 1-planar graphs. Metody Diskret. Analiz, 1:1 6, 198. (in Russian). [] O. V. Borodin. A new proof of the 6 color theorem. J. Graph Theory, 19():507 51, [5] O. V. Borodin, A. V. Kostochka, A. Raspaud, and E. Sopena. Acyclic colouring of 1-planar graphs. Discrete Appl. Math., 10:9 1, 001.

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