The crossing number of K 1,4,n

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1 Discrete Mathematics 308 (2008) The crossing number of K 1,4,n Yuanqiu Huang, Tinglei Zhao Department of Mathematics, Normal University of Hunan, Changsha , PR China Received 15 June 2004; received in revised form 22 April 2006; accepted 18 December 2006 Available online 19 April 2007 Abstract In this paper, we prove that the crossing number of the complete tripartite graph K 1,4,n is n(n 1). Our result also shows that the crossing number of the Cartesian product S 4 S n is n(n 1), where S n denotes the star K 1,n for arbitrary integer n Published by Elsevier B.V. Keywords: Drawing; Complete tripartite graphs; Crossing number 1. Introduction For graph theory terminology not defined here we direct the reader to [2]. Adrawing of a (undirected) graph G = (V, E) is a mapping f that assigns to each vertex in V a distinct point in the plane and to each edge uw in E a continuous arc (i.e., a homeomorphic image of a closed interval) connecting f (u) and f(v), not passing through the image of any other vertex. In addition, we impose the following conditions on a drawing: (1) no three edges have an interior point in common, (2) if two edges share an interior point p, then they cross at p, and (3) any two edges of a drawing have only a finite number of crossings (common interior points). The crossing number, cr(g), of a graph G is the minimum number of edge crossings in any drawing of G. Let be a drawing of the graph G. We denote the number of crossings in by cr (G). For more on the theory of crossing numbers, we refer the reader to [4]. Computing the crossing number of graphs is a classical problem (see [7 or 4]), and yet it is also an elusive one. In fact, Garey and Johnson [5] have proved that in general the problem of determining the crossing number of a graph is NP-complete (the reader can also refer to two recent results on complexity of the crossing number of graphs in [6,8], respectively). At present the classes of graphs whose crossing numbers have been determined are very scarce, and there are only some classes of special graphs whose crossing numbers are known. For example, these include the complete graph, K n, for small n [20], the complete bipartite graph, K m,n, for small m and any n [10,20], certain generalized Petersen graphs [17], line graphs [9,15], and some Cartesian product of special graphs [1,3,11,13,14,18,19]. The crossing number of the complete bipartite graphs K m,n was computed by Kleitman [10], for all m 5 and all n. More precisely, he proved that if m 6, cr(k m,n ) = Z(m, n), This work is supported by the key project of the Education Department of Hunan Province of China (05A037). addresses: hyqq@public.cs.hn.cn (Y. Huang), tlzhao@hunnu.edu.cn (T. Zhao) X/$ - see front matter 2007 Published by Elsevier B.V. doi: /j.disc

2 Y. Huang, T. Zhao / Discrete Mathematics 308 (2008) where Z(m, n) denotes the Zarankiewicz conjecture, and [ m Z(m, n) = 2 ] [ ] m 1 [n ] [ ] n (For any real number x, [x] denotes the maximum integer not greater than x). On the crossing number of complete tripartite graphs, there are fewer results. Particularly, Asano in [16] proved that [ n ] cr(k 1,3,n ) = Z(4,n)+ 2 and cr(k 2,3,n ) = Z(5,n)+ n. In this paper, using Kleitman s theorem, we determine the crossing number of tripartite graph K 1,4,n. Our method is simple and differs from that used in [16]. The main result of this paper is the following theorem. Theorem. cr(k 1,4,n ) = n(n 1). Note that in [12] Klešč proves that the crossing number of K 1,4,n plus two additional edges, and hence our result should be presented as a strengthening of his. In [13], Klešč presents a table which lists the crossing numbers of Cartesian products of paths, cycles, and stars. Some of these are unknown, including the Cartesian product S 4 S n, where S n denotes the star K 1,n for arbitrary integer n 1. From our Theorem above, the crossing number of the Cartesian product S 4 S n is obtained. Corollary. cr(s 4 S n ) = n(n 1). 2. The proofs of the theorem and the corollary Proof of Theorem. Let K 1,4,n be a complete tripartite graph with the vertex partition (X,Y,Z), where X ={x}, Y ={y i 1 i 4}, and Z={z i 1 i n}. First, we will display a drawing of K 1,4,n in the plane such that cr (K 1,4,n )= n(n 1). The desired drawing is constructed as follows: (i) (x) = (ε, ε) where ε is a sufficiently small positive number, (y 1 ) = (0, 1), (y 2 ) = (0, 2), (y 3 ) = (0, 1), and (y 4 ) = (0, 2); (ii) for any 1 i n, ifiis even then (z i ) = (i/2, 0),ifiis odd then (z i ) = ( (i + 1)/2, 0); (iii) the image of each edge is a straight line segment. For example, we draw (K 1,4,5 ) in Fig. 1 below. For convenience, in Figs. 1 3 we denote a vertex or edge in the drawing simply by r instead of the image of r, (r). Fig. 1. A drawing for K 1,4,5.

3 1636 Y. Huang, T. Zhao / Discrete Mathematics 308 (2008) Fig. 2. Edges and vertices adjacent to (x). Fig. 3. A drawing of K 5,n+1 obtained from (K 1,4,n ). Then it is not difficult to see that cr (K 1,4,n ) = Z(5,n)+ 2[n/2]=n(n 1), and so cr(k 1,4,n ) n(n 1). Thus, in order to prove the theorem, we need only to prove that cr (K 1,4,n ) n(n 1) for any drawing of K 1,4,n. Assume to contrary that there exists a drawing of K 1,4,n satisfying cr (K 1,4,n ) < n(n 1). In the proof that follows, our method is to construct a drawing of the complete bipartite graph K 5,n+1 so that cr (K 5,n+1 )<Z(5,n+ 1), a contradiction to the fact that cr(k 5,n+1 ) = Z(5,n+ 1). Let e i denote the edge xy i of K 1,4,n, i = 1, 2, 3, 4. Without loss of generality, assume that under the drawing, the clockwise order of these four images (e i ) around (x) is (e 1 ) (e 2 ) (e 3 ) (e 4 ). Since the graph K 1,4,n has an additional n edges f j = z j x incident with x (1 j n). Let A i denote the set of all those images (f j ), each of which lies in the angle α i formed between (e i ) and (e i+1 ), where the indices are read modulo 4 (see Fig. 2). We note that A 1 + A 2 + A 3 + A 4 =n.again, we see that in the plane R 2, there exists a circular neighborhood around (x), N( (x), ε) ={s R 2 : s (x) <ε}, where ε is a sufficiently small positive number, such that for any other edge e of K 1,4,n not incident with x, (e) N( (x), ε) =. We now consider two cases. Case 1: Assume n is even. We consider arbitrarily a pair A 1 and A 3 or A 2 and A 4, say A 1 and A 3. Without loss of generality, assume A 1 A 3. In the following we produce a complete bipartite graph K 5,n+1 together with its drawing. Step 1: Add a new vertex z n+1 in some location of (e 2 ) N( (x), ε). Step 2: For all 1 i 4, remove the partition of (e i ) lying in N( (x), ε) (do not remove the vertex z n+1 ). Step 3: Connect z n+1 to each vertex in { (x), (y 1 ), (y 2 ), (y 3 ), (y 4 )} in such a way as described in Fig. 3.

4 Y. Huang, T. Zhao / Discrete Mathematics 308 (2008) For example, connect z n+1 to (x) along the section of (e 2 ) N( (x), ε); connect z n+1 to (y 2 ) first along the section of (e 2 ) N( (x), ε) and then along the original section (e 2 ) outside of N( (x), ε). Again, the way of connecting z n+1 to (y 4 ) is first by successively traversing through the angles α 1 and α 4 (near to (x)), and then along the original section of (e 4 ) outside N( (x), ε). Thus, we obtain a drawing of the complete bipartite graph K 5,n+1. It is easy to see that cr (K 5,n+1 ) = cr (K 1,4,n ) + 2 A 1 + A 2 + A 4. Since A 1 A 3,wehave cr (K 5,n+1 ) cr (K 1,4,n ) + 4 A i =cr (K 1,4,n ) + n. i=1 Again, by our assumption that cr (K 1,4,n ) < n(n 1), it follows that cr (K 5,n+1 )<n 2. But, on the other hand we have that cr(k 5,n ) = Z(5,n+ 1) = n 2 when n is even. This is a contradiction. Case 2: Assume n is odd. First, if A 1 = A 3 and A 2 = A 4, then it follows that n = 4 i=1 A i =2( A 1 + A 2 ). This contradicts that n is odd. Therefore, either A 1 = A 3,or A 2 = A 4. Without loss of generality, let A 1 = A 3, and moreover let A 3 A Completely analogous to Case (1) above, we can obtain a drawing of the complete bipartite graph K 5,n+1 such that cr (K 5,n+1 ) = cr (K 1,4,n ) + 2 A 1 + A 2 + A 4. Thus, we have cr (K 5,n+1 ) = cr (K 1,4,n ) + 2 A 1 + A 2 + A 4 4 cr (K 1,4,n ) + A i 1 i=1 = cr (K 1,4,n ) + n 1. Similarly, since cr (K 1,4,n ) < n(n 1) by our assumption, we get that cr (K 5,n+1 )<n 2 1. This also contradicts that cr(k 5,n+1 ) = Z(5,n+ 1) = n 2 1 when n is odd. By the contradictions derived in Cases (1) and (2), we have shown that cr (K 1,4,n ) n(n 1) for any good drawing. Thereby, from the arguments above we finish the proof of the theorem. Proof of Corollary. Ignoring a 2-degree in a graph means that we remove the vertex and also its two incident edges, and add an edge connected to its original two adjacent vertices. In the following we first shall see that S 4 S n is homeomorphic to K 1,4,n.Ifn = 1, by the definition of the Cartesian product S 4 S n is the following graph depicted in the left of Fig. 4. Fig. 4. The Cartesian product S 4 S 1 and S 4 S 2.

5 1638 Y. Huang, T. Zhao / Discrete Mathematics 308 (2008) Fig. 5. A copy of S n and the Cartesian product S 4 S n (n 3). If we ignore each of these 2-degree vertices denoted by 1, 2, 3 and 4, then the resulting graph is isomorphic to K 1,4,1. That is to say, S 4 S 1 is homeomorphic to K 1,4,1.Ifn = 2, S 4 S 2 is shown as in the right of Fig. 4. Similarly, ignoring all the 2-degree vertices in S 4 S 2 results in the graph isomorphic to K 1,4,2, and thus S 4 S 2 is homeomorphic to K 1,4,2. For other general value n 3, by the definition S 4 S n is displayed in the right of Fig. 5. Also, if we ignore all the 2-degree vertices in S 4 S n, the resulting graph is isomorphic to K 1,4,n, and thus S 4 S n is homeomorphic to K 1,4,n. Since two homeomorphic graphs have the same crossing number, by our Theorem we get that cr(s 4 S n )=cr(k 1,4,n )= n(n 1) for any n 1, proving the corollary. Acknowledgment The authors are very grateful to the anonymous referees for many comments and suggestions, which are very helpful to improve the presentation of this paper. References [1] L.W. Beineke, R.D. Ringeisen, On the crossing numbers of products of cycles and graphs of order four, J. Graph Theory 4 (1980) [2] J.A. Bondy, M.S.R. Murty, Graph Theory with Application, Elsevier, New York, [3] A.M. Dean, R.B. Richter, The crossing number of C 4 C 4, J. Graph Theory 19 (1995) [4] P. Erdös, R.K. Guy, Crossing number problems, Amer. Math. Mon. 80 (1973) [5] M.R. Garey, D.S. Johnson, Crossing number is NP-complete, SIAM J. Algebraic Discrete Methods 4 (1983) [6] M. Grohe, Computing crossing number in quadratic time, in: Proceedings of the 33rd ACM Symposium on Theory of Computing STOC 01, 2001, pp [7] F. Harary, Graph Theory, Addison-Wesley, Reading, MA, [8] P. Hliněný, Crossing number is hard for cubic graphs, in: Math Foundations of Computer Science MFCS 2004, Lecture Notes in Computer Science, vol. 3153, Springer, Berlin, 2004, pp [9] S. Jendrol, M. Klešč, On graphs whose line graphs have crossing number one, J. Graph Theory 37 (2001) [10] D.J. Kleitman, The crossing number of K 5,n, J. Combinatorial Theory 9 (1970) [11] M. Klešč, The crossing numbers of products of path and stars with 4-vertex graphs, J. Graph Theory 18 (6) (1994) [12] M. Kle s c, On the crossing number of products of stars and graphs of oder five, Graphs Combin. 17 (2001) [13] M. Klešč, The crossing numbers of Cartesian products of paths with 5-vertex graphs, Discrete Math. 233 (2001) [14] M. Kle s c, R.B. Richter, I. Stobert, The crossing number of C 5 C n, J. Graph Theory 22 (3) (1996) [15] V.R. Kulli, D.G. Akka, L.W. Beineke, On the line graphs with crossing number 1, J. Graph Theory 3 (1979) [16] K. Asano, The crossing number of K 1,3,n and K 2,3,n, J. Graph Theory 10 (1980) 1 8. [17] D. McQuillan, R.B. Richter, On the crossing numbers of certain generalized Petersen graphs, Discrete Math. 104 (1992) [18] R.B. Richter, G. Salazar, The crossing number of C 6 C n, Australas. J. Combin. 23 (2001) [19] R.B. Richter, C. Thomassen, Intersection of curve systems and the crossing number of C 5 C 5, Discrete Comput. Geom. 13 (1995) [20] D.R. Woodall, Cyclic-order graphs and Zarankiewicz s crossing number conjecture, J. Graph Theory 17 (1993)