THE CROSSING NUMBERS OF PRODUCTS WITH CYCLES

The crossing numbers of Cartesian products of all graphs of order at most four with cycles are known. The crossing numbers of Cartesian products G2Cn for several graphs G on five and six vertices and the cycle Cn are also given. In this paper, we extend these results by determining crossing numbers of Cartesian products G2Cn for some specific six vertex graphs G and for some fixed number n = 3, 4, 5.


INTRODUCTION
Let G be a simple graph with vertex set V and edge set E. A drawing of the graph in the plane is called a good drawing if and only if no edge crosses itself, no two edges cross more than once, and no two edges incident with the same vertex cross.The crossing number cr(G) of a graph G is the minimum number of crossings of edges in a drawing of G in the plane such that no three edges cross in a point.A drawing with minimum number of crossings is always a good drawing.
It is very difficult to establish the crossing number of a given graph.So, the crossing numbers are known only for a few families of graphs.Most of these graphs are Cartesian products of special graphs.The Cartesian product G 1 2 G 2 of graphs G 1 and G 2 has vertex set V (G 1 2 G 2 ) = V (G 1 ) 2V (G 2 ) and edge set E(G 1 2 G 2 ) = {{(u i , v j ), (u k , v h )} : (u i = u k and {v j , v h } ∈ E(G 2 )) or ({u i , u k } ∈ E(G 1 ) and v j = v h )}.
Let C n be the cycle of length n, P n be the path of length n, and S n be the star isomorphic to K 1,n .Harary et al. [9] conjectured that the crossing number of C m 2C n is (m − 2)n, for all m, n satisfying 3 ≤ m ≤ n.This has been proved only for m, n satisfying m ≤ 7 [1], [4], [17], [18], [19].It was recently proved by Glebsky and Salazar [8] that the crossing number of C m 2C n equals its long-conjectured value at least for n ≥ m(m + 1).Beineke and Ringeisen in [2] as well as Jendrol' and Ščerbová in [10] determined the crossing numbers of the Cartesian products of all graphs on four vertices with cycles.Klešč in [11], [12], [13], [14], Klešč, Richter and Stobert in [15], and Klešč and Kocúrová in [16] gave the crossing numbers of G 2C n for several graphs G of order five.
We are interested in the crossing numbers of Cartesian products of graphs on six vertices with cycles.Except for the star S 5 , in [6] there are given the crossing numbers of G 2C n for all five-edge graphs G on six vertices.In [7],the values of crossing numbers for sevetal Cartesian products of cycles and six-edge graphs G on six vertices are presented.In [7] and [5] are given the crossing numbers for Cartesian products of cycles and two seven-edge graphs G on six vertices.In this paper, we give the crossing number of the Cartesian products G 2C n for two graphs G on six vertices and fixed number n.In [6] there is presented only upper bound 4 n for the crossing numbers of Cartesian products of star on six vertices with cycles S 5 2C n obtained from the drawing of the graph S 5 2C n for n ≥ 3. We suppose that the upper bound in [6] is stated for n ≥ 6.This bound is lower for n = 3, 4, 5.
In the next text we determine that cr(S 5 2C 3 ) = 4 and cr(S 5 2C 4 ) = 8.The hypothesis about lower bound for n = 5, using the drawing of the graph S 5 2C 5 , is 16.
At least -formulate the hypothesis.
In this section, we give the crossing numbers of the Cartesian products G 2C 3 , G 2C 4 and G 2C 5 for the graph G shown in Fig. 2. We prove, that cr(G 2C 3 ) = 5, cr(G 2C 4 ) = 10 and cr(G 2C 5 ) = 14.Fig. 3 shows the drawing of the graph G 2C n in which the edges of every subgraph isomorphic to G are crossed exactly three times.Hence, the crossing number of G 2C n for n ≥ 6 is at most 3n, we conjecture that it is exactly 3n.

a
Fig. 2 The graph G Let D be a good drawing of the graph G.We denote the number of crossings in D by cr D (G).Let G i and G j be edge-disjoint subgraphs of G.We denote by cr D (G i , G j ) the number of crossings among edges of G i and edges of G j , and by cr D (G i ) the number of crossings between edges of G i in D.
Assume n ≥ 3, and consider the graph G 2C n in the following way: it has 6n vertices and edges that are the edges in the n copies G i , i = 0, 1, . . ., n − 1, and in the six cycles of length n.For i = 0, 1, . . ., n − 1, let a i and b i be the vertices of G i of degree one, c i the vertex of degree four and let d i , e i and f i be the vertices of G i of degree two (see Fig. 3).Thus, for x ∈ {a, b, c, d, e, f }, the n-cycle C x n is induced by the vertices x 0 , x 1 . . ., x n−1 .
Proof.Fig. 4 shows the good drawing of the graph G 2C 3 with five crossings, thus cr(G 2C 3 ) ≤ 5.
Fig. 4 The drawing of the graph G 2C 3 Assume that there is a good drawing of G 2C 3 with at most 4 crossings and let D be such a drawing.The subgraph C c 3 ∪ T d ∪ T e ∪ X f of the graph G 2C 3 is isomorphic to the graph C 4 2C 3 and cr(C 4 2C 3 ) = 4 (see [19]).Thus, in D there is no crossing on the edges of T a ∪ T b .The planar subdrawing of T a ∪ T b induced by D is unique within isomorphism and divides the plane into two triangular and three hexagonal regions in such a way that there is no region with all three vertices c 0 , c 1 , and c 2 on its boundary.So, an edge of T d crosses in D an edge of T a ∪ T b , which contradicts the assumption that no edge of T a ∪ T b is crossed. 2 Theorem 3.2.cr(G 2C 4 ) = 10.
Proof.In Fig. 5 there is a good drawing of G 2C 4 with ten crossings, thus cr(G 2C 3 ) ≤ 10.Assume that there is a good drawing of G 2C 4 with at most 9 crossings and let D be such a drawing.The graph G 2C 4 contains the subgraph C c 4 ∪ T d ∪ T e ∪ X f which is isomorphic to the graph C 4 2C 4 and cr(C 4 2C 4 ) = 8 (see [4]).Thus, in D there is at most one crossing on the edges of T a ∪ T b .Consider the subgraph T a ∪ T b of the graph G 2C 4 and let D be its subdrawing induced by D.
First, suppose that cr D (T a ∪ T b ) = 0.As T a ∪ T b is subdivision of the planar graph P 1 2C 4 , the planar subdrawing of T a ∪ T b induced by D is unique within isomorphism and divides the plane into two quadrangular and four hexagonal regions in such a way that there are at most two of the vertices c 0 , c 1 , c 2 , and c 3 on the boundary of every region.So, in D, the edges of T d cross the edges of T a ∪ T b at least twice and it contradicts our assumption.
Next, let cr D (T a ∪ T b ) = 1.The subgraph T a ∪ T b is obtained from C 4 2 P 1 by an elementary subdivision of every edge joining two 4-cycles C a 4 and C b 4 and for the graph C 4 2 P 1 there is no good drawing with exactly one crossing, because for any two edges which cross each other one can find two vertex-disjoint cycles such that crossed edges are in different cycles.Therefore two vertex-disjoint cycles cannot cross only once, the only one crossing in D is between an edge incident with a vertex of degree two and an edge of the cycle C a Proof.In the drawing of the graph G 2C 5 in Fig. 6 one can easily see that cr(G 2C 5 ) ≤ 14.
Fig. 6 The drawing of the graph G 2C 5 Assume that there is a good drawing of the graph G 2C 5 with at most 13 crossings and let D be such a drawing.The graph G 2C 5 contains the graph C 4 2C 5 as a subgraph and cr(C 4 2C 5 ) = 10 (see [2]).Thus, in D there are at most three crossings on the edges of T a ∪ T b .Consider the subgraph T a ∪ T b of the graph G 2C 5 and let D be its subdrawing induced by D. In case 2, assume that cr D (T a ∪ T b ) = 1.As the subgraph T a ∪ T b is obtained from P 1 2C 5 by elementary subdivision of every edge joining two 5-cycles C a 5 and C b 5 , and therefore for the graph C 5 2 P 1 there is no good drawing with exactly one crossing (because for any two edges which cross each other one can find two vertex-disjoint cycles such that crossed edges are in different cycles and two vertex-disjoint cycles cannot cross only once), the only one crossing in D is between an edge incident with a vertex of degree two and an edge of the cycle C a

Fig. 3
Fig.3The drawing of the graph G 2C n

4 or the cycle C b 4 . 2 Theorem 3 . 3 .
In this case, the cycle C a 4 or the cycle C b 4 separates in D some vertex c i of the cycle C c 4 from the other vertices of C c 4 .Hence, C c 4 crosses in D the edges of T a ∪ T b at least twice and this contradiction completes the proof.cr(G 2C 5 ) = 14.
First, assume that cr D (T a ∪ T b ) = 0.As T a ∪ T b is a subdivision of the planar graph P 1 2C 5 , the subdrawing D of T a ∪ T b induced by D divides the plane into two regions without vertices of C c 5 on their boundaries and into five regions having two vertices of C c 5 on the boundary of every region.If, in D, the cycle C d 5 is placed in a region of D with fewer than two vertices of C c 5 on its boundary, then cr D (T a ∪ T b , T d ) ≥ 5.If C d 5 is placed in a region with two vertices of C c 5 on the boundary, then one vertex of C c 5 is separated from C d 5 by at least two vertex-disjoint cycles.Hence, cr D (T a ∪ T b , T d ) ≥ 4. If the cycle C d 5 crosses the edges of T a ∪ T b two or three times, then it is placed in two regions of D with at most three vertices of C c 5 on their boundaries and, in D, the edges of T d cross the edges of T a ∪ T b at least four times.If there are four vertices of C c 5 on the boundaries of the regions in D in which C d 5 is placed in D, the edges of C d 5 cross the edges of T a ∪ T b at least four times.

5 or the cycle C b 5 .
In this case, the cycle C a 5 or the cycle C b 5 separates in D some vertex c i of the cycle C c 5 from the other vertices of C c 5 .Hence, C c 5 crosses in D the edges of T a ∪ T b at least twice.The removing of the separated vertex c i of the cycle C c 5 from D we have the drawing without crossings.This drawing divides the plane in such a way that there are at most two vertices of C c 5 on the boundary of every region.As the vertex c i is in D separated from the other vertices of C c 5 , in the subdrawing D of T a ∪ T b with one crossings there are at most two vertices of C c 5 on the boundary of a region.If the cycle C d 5 of T d crosses the 2-connected subgraph T a ∪ T b , it crosses T a ∪ T b at least two times.Otherwise C d 5 is in D placed in one region in the view of the subdrawing of T a ∪ T b and at least two edges of T d joining C d 5 with the vertices of C c 5 cross the edges of T a ∪ T b .So, in this case, again, there are more than three crossings on the edges of T a ∪ T b .It is a contradiction.In case 3, assume that cr D (T a ∪ T b ) ≥ 2. Then at least one subgraph T d or T e does not cross in D the edges of T a ∪ T b .Without loss of generality, let T d does not cross the edges of T a ∪T b .So, cr D (T a ∪T b , T d ) = 0.In this case, cr D (T a , T d ) = 0 and cr D (T b , T d ) = 0.As T a ∪ T d is a subdivision of the planar graph P 1 2C 5 , the subdrawing D of T a ∪ T d divides the plane into several regions without vertices of C c 5 on their boundaries and into regions, which have exactly two vertices of C c 5 on the boundary of one region.Fig. 7 shows the drawing D in which possible crossings among the edges of T a are inside the left disc bounded by the dotted cycle and possible crossings among the edges of T d are inside the right disc bounded by the dotted cycle.

7
The subdrawing of the subgraph T a ∪ T d We can suppose that if, in D, an edge not incident with a vertex of C a 5 or C d 5 passes through one of these two discs, then it crosses the edges of T a ∪ T d at least twice.Consider now a subgraph T b .Both C b 5 and T a ∪ T d are 2-connected graphs and so, cr D (C b 5 , T a ∪ T d ) = 1.If, in D, the cycle C b 5 is placed in a region of D with fewer than two vertices of C c 5 on its boundary, then cr D (T a ∪ T d , T b ) ≥ 4. If C b 5 is