Remarks on the Thickness of K n,n,n

Remarks on the Thickness of K n,n,n Yan Yang Department of Mathematics Tianjin University, Tianjin 30007, P.R.China yanyang@tju.edu.cn Abstract The thickness θ(g) of a graph G is the minimum number of planar subgraphs into which G can be decomposed. In this paper, we provide a new upper bound for the thickness of the complete tripartite graphs K n,n,n (n 3) and obtain θ(k n,n,n ) = 3, when n 3 (mod 6). Keywords thickness; complete tripartite graph; planar subgraphs decomposition. Mathematics Subject Classification 05C10. 1 Introduction The thickness θ(g) of a graph G is the minimum number of planar subgraphs into which G can be decomposed. It was defined by Tutte [10] in 1963, derived from early work on biplanar graphs [,11]. It is a classical topological invariant of a graph and also has many applications to VLSI design, graph drawing, etc. Determining the thickness of a graph is NP-hard [6], so the results about thickness are few. The only types of graphs whose thicknesses have been determined are complete graphs [1,3], complete bipartite graphs [4] and hypercubes [5]. The reader is referred to [7,8] for more background on the thickness problems. In this paper, we study the thickness of complete tripartite graphs K n,n,n, (n 3). When n = 1,, it is easy to see that K 1,1,1 and K,, are planar graphs, so the thickness of both ones is one. Poranen proved θ(k n,n,n ) n in [9] which was the only result about the thickness of K n,n,n, as far as the author knows. We will give a new upper bound for θ(k n,n,n ) and provide the exact number for the thickness of K n,n,n, when n is congruent to 3 mod 6, the main results of this paper are the following theorems. Theorem 1. For n 3, θ(k n,n,n ) 3 + 1. Theorem. θ(k n,n,n ) = 3 when n 3 (mod 6). The proofs of the theorems Supported by the National Natural Science Foundation of China under Grant No. 11401430

In [4], Beineke, Harary and Moon determined the thickness of complete bipartite graph K m,n for almost all values of m and n. Lemm.[4] The thickness of K m,n is mn (m+n ) except possibly when m and n are odd, m n and there exists an integer k satisfying n = k(m ) m k. Lemm. For n 3, θ(k n,n,n ) 3. Proof. Since K n,n is a subgraph of K n,n,n, we have θ(k n,n,n ) θ(k n,n ). From Lemm, the thickness of K n,n (n 3) is 3, so the lemma follows. For the complete tripartite graph K n,n,n with the vertex partition (A, B, C), where A = {,..., a n 1 }, B = {,..., b n 1 } and C = {,..., c n 1 }, we define a type of graphs, they are planar spanning subgraphs of K n,n,n, denoted by G[a i b j+i c k+i ], in which 0 i, j, k n 1 and all subscripts are taken modulo n. The graph G[a i b j+i c k+i ] consists of n triangles a i b j+i c k+i for 0 i n 1 and six paths of length n 1, they are b j+1 c k+ b j+4 c k+5... i b j+3i+1 c k+3i+..., c k b j+ c k+3 b j+5... c k+3i i+1 b j+3i+..., b j c k+1 a b j+3 c k+4 a 5... b j+3i c k+3i+1 i+..., c k+1 b j+ c k+4 b j+5... i c k+3i+1 b j+3i+..., b j c k+ b j+3 c k+5... b j+3i i+1 c k+3i+..., c k b j+1 a c k+3 b j+4 a 5... c k+3i b j+3i+1 i+.... Equivalently, the graph G[a i b j+i c k+i ] is the graph with the same vertex set as K n,n,n and edge set {a i b j+i 1, a i b j+i, a i b j+i+1, a i c k+i 1, a i c k+i, a i c k+i+1 1 i n } {b j+i c k+i 1, b j+i c k+i, b j+i c k+i+1 1 i n } { b j, b j+1, a n 1 b j+n, a n 1 b j+n 1 } { c k, c k+1, a n 1 c k+n, a n 1 c k+n 1) } {b j c k, b j c k+1, b j+n 1 c k+n, b j+n 1 c k+n 1 }. Figure 1(a) illustrates the planar spanning subgraph G[a i b i c i ] of K 5,5,5.

3 c a b (a) The subgraph G 1 = G[a i b i c i ] of K 5,5,5 c c a a 3 b a (b) The subgraph G of K 5,5,5 (c) The subgraph G 3 of K 5,5,5 Figure 1 A planar subgraphs decomposition of K 5,5,5 Theorem 5. When n = 3p + (p is a positive integer), θ(k n,n,n ) p +. Proof. When n = 3p+ (p is a positive integer), we will construct two different planar subgraphs decompositions of K n,n,n according to p is odd or even, in which the number of planar subgraphs is p + in both cases. Case 1. p is odd. Let G 1,..., G p be p planar subgraphs of K n,n,n where G t = G[a i b i+3(t 1) c i+6(t 1) ], for 1 t p+1 ; and G t = G[a i b i+3(t 1) c i+6(t 1)+ ], for p+3 t p and p 3. From the structure of G[a i b j+i c k+i ], we get that no two edges in G 1,..., G p are repeated. Because subscripts in G t, 1 t p are taken modulo n, {3(t 1) (mod n) 1 t p} = {0, 3, 6,..., 3(p 1)}, {6(t 1) (mod n) 1 t p+1 p+3 } = {0, 6,..., 3(p 1)} and {6(t 1)+ (mod n) t p} = {3, 9,..., 3(p )}, the subscript sets of b and c in G t, 1 t p are the same, i.e.,

4 {i + 3(t 1) (mod n) 1 t p} = {i + 6(t 1) (mod n) 1 t p + 1 p + 3 } {i + 6(t 1) + (mod n) t p}. Furthermore, if there exists t {1,..., p} such that a i b j is an edge in G t, then a i c j is an edge in G k for some k {1,..., p}. If the edge a i b j is not in any G t, then neither is the edge a i c j in any G t, for 1 t p. From the construction of G t, the edges that belong to K n,n,n but not to any G t, 1 t p, are (t 1) 1, (t 1) 1, 1 t p (1) a n 1 (t 1), a n 1 (t 1), 1 t p () a i b i 3, a i b i, 0 i n 1 (3) a i c i 3, a i c i, 0 i n 1 (4) b i c i+3(t 1) 1, b i c i+3(t 1), 0 i n 1 and t = p + 3 (t 1) c 6(t 1) 1, (t 1) 1 c 6(t 1), 1 t p + 1 (t 1) c 6(t 1)+1, (t 1) 1 c 6(t 1)+, p + 3 (5) (6) t p and p 3 (7) Let G p+1 be the graph whose edge set consists of the edges in (3) and (5), and G p+ be the graph whose edge set consists of the edges in (1), (), (4), (6) and (7). In the following, we will describe plane drawings of G p+1 and G p+. (a) A planar embedding of G p+1. Place vertices,,..., b n 1 on a circle, place vertices a i+3 and c i+ in the middle of b i and b i+1, join each of a i+3 and c i+ to both b i and b i+1, we get a planar embedding of G p+1. For example, when p = 1, n = 5, Figure 1(b) shows the subgraph G of K 5,5,5. (b) A planar embedding of G p+. Firstly, we place vertices,,..., c n 1 on a circle, join vertex a i+3 to c i and c i+1, for 0 i n 1, so that we get a cycle of length n. Secondly, join vertex a n 1 to (t 1) for 1 t p, with lines inside of the cycle. Let l t be the line drawn inside the cycle joining a n 1 with c 6(t 1) 1 if 1 t p+1 or with c 6(t 1)+1 if p+3 t p (p 3). For 1 t p, insert the vertex (t 1) in the line l t. Thirdly, join vertex to (t 1) 1 for 1 t p, with lines outside of the cycle. Let l t be the line drawn outside the cycle joining with c 6(t 1) if 1 t p+1 or with c 6(t 1)+ if p+3 t p (p 3). For 1 t p, insert the vertex (t 1) 1 in the line l t. In this way, we can get a planar embedding of G p+. For example, when p = 1, n = 5, Figure 1(c) shows the subgraph G 3 of K 5,5,5. Summarizing, when p is an odd positive integer and n = 3p+, we get a decomposition of K n,n,n into p + planar subgraphs G 1,..., G p+.

Case. p is even. Let G 1,..., G p be p planar subgraphs of K n,n,n where G t = G[a i b i+3(t 1) c i+6(t 1)+3 ], for 1 t p ; and G t = G[a i b i+3(t 1) c i+6(t 1)+ ], for p+ t p. With a similar argument to the proof of Case 1, we can get that the subscript sets of b and c in G t, 1 t p are the same, i.e., {i + 3(t 1) (mod n) 1 t p} = {i + 6(t 1) + 3 (mod n) 1 t p p + } {i + 6(t 1) + (mod n) t p}. From the construction of G t, G p and G p+ have n edges in common, they are b i+3( p+ 1)c i+6( p+ 1)+1, 1 i n 1 and i n 4, we can delete them in one of these two graphs to avoid repetition. The edges that belong to K n,n,n but not to any G t, 1 t p, are (t 1) 1, (t 1) 1, 1 t p (8) a n 1 (t 1), a n 1 (t 1), 1 t p (9) a i b i 3, a i b i, 0 i n 1 (10) a i c i 3, a i c i, 0 i n 1 (11) b i c i 1, b i c i, b i c i+1, 0 i n 1 (1) 5 (t 1) c 6t 4, 1 t p (13) (t 1) c 6t 5, p + < t p (14) (t 1) 1 c 6t 3, 1 t < p (15) (t 1) 1 c 6t 4, p + t p (16) Let G p+1 be the graph whose edge set consists of the edges in (10), (11) and (1), and G p+ be the graph whose edge set consists of the edges in (8), (9), (13), (14), (15) and (16). We draw G p+1 in the following way. Firstly, place vertices,,,,..., b n 1, c n 1 on a circle C, join vertex c i to b i and b i+1, we get a cycle of length n. Secondly, place vertices, a,..., a n on a circle C in the unbounded region defined by the circle C such that C is contained in the closed disk defined by C, place vertices,,..., a n 1 on a circle C contained in the bounded region of C. Join a i to b i 3, b i, c i 3, and c i, join b i to c i+1. We can get a planar embedding of G p+1, so it is a planar graph. G p+ is also planar because it is a subgraph of a graph homeomorphic to a dipole (two vertices joined by some edges). For example, when p =, n = 8, Figure (c) and Figure (d) show the subgraphs G 3 and G 4 of K 8,8,8 respectively. Summarizing, when p is an even positive integer and n = 3p +, we obtain a decomposition of K n,n,n into p + planar subgraphs G 1,..., G p+.

6 Theorem follows from Cases 1 and. From the proof of Theorem 5, we draw planar subgraphs decompositions of K 5,5,5 and K 8,8,8 as illustrated in Figure 1 and Figure respectively. b 7 a 6 c 5 c 6 b a c 7 a 5 a 7 (a) The subgraph G 1 = G[a i b i c i+3] of K 8,8,8 c b a 6 c 5 b 7 c a a 5 c 6 a 7 c 7 (b) The subgraph G c c 5 b c 6 of K 8,8,8 in which G = G[a i b i+3 c i ]

7 a c7 b 7 c 6 b 7 b c c 7 b c 5 a 7 a 5 c a 6 a 7 (c) The subgraph G 3 of K 8,8,8 (d) The subgraph G 4 of K 8,8,8 Figure A planar subgraphs decomposition of K 8,8,8 Proof of Theorem 1. Because K n 1,n 1,n 1 is a subgraph of K n,n,n, θ(k n 1,n 1,n 1 ) θ(k n,n,n ), by Theorem 5, θ(k n,n,n ) p + also holds, when n = 3p or n = 3p + 1 (p is a positive integer), the theorem follows. Proof of Theorem. When n = 3p is odd, i.e., n 3 (mod 6), we decompose K n,n,n into p + 1 planar subgraphs G 1,..., G p+1, where G t = G[a i b i+3(t 1) c i+6(t 1) ], for 1 t p. With a similar argument to the proof of Theorem 5, we can get that the subscript sets of b and c in G t, 1 t p are the same, i.e., {i + 3(t 1) (mod n) 1 t p} = {i + 6(t 1) (mod n) 1 t p}. If the edge a i b j is in G t for some t {1,..., p}, then there exists k {1,..., p} such that a i c j is in G k. If the edge a i b j is not in any G t, then neither is the edge a i c j in any G t, for 1 t p. From the construction of G t = G[a i b i+3(t 1) c i+6(t 1) ], we list the edges that belong to K n,n,n but not to any G t, 1 t p, as follows. (t 1) 1, c 6(t 1) 1, 1 t p (17) a n 1 (t 1), a n 1 c 6(t 1), 1 t p (18) (t 1) c 6(t 1) 1, (t 1) 1 c 6(t 1), 1 t p (19) Let G p+1 be the graph whose edge set consists of the edges in (17), (18) and (19). It is easy to see that G p+1 is homeomorphic to a dipole and it is a planar graph. Summarizing, when p is an odd positive integer and n = 3p, we obtain a decomposition of K n,n,n into p+1 planar subgraphs G 1,..., G p+1, therefor θ(k n,n,n ) p+1. Combining this fact and Lemm, the theorem follows.

According to the proof of Theorem, we draw a planar subgraphs decomposition of K 3,3,3 as shown in Figure 3. 8 c b c a b a (a) The subgraph G 1 = G[a i b i c i ] of K 3,3,3 (b) The subgraph G of K 3,3,3 Figure 3 A planar subgraphs decomposition of K 3,3,3 For some other θ(k n,n,n ) with small n, combining Lemm and Poranen s result mentioned in Section 1, we have θ(k 4,4,4 ) =, θ(k 6,6,6 ) = 3. Since there exists a decomposition of K 7,7,7 with three planar subgraphs as shown in Figure 4, Lemm implies that θ(k 7,7,7 ) = 3. We also conjecture that the thickness of K n,n,n is 3 for all n 3. a 6 c 5 c b a c 6 (a) The subgraph G 1 = G[a i b i c i ] of K 7,7,7 a 5

9 a 6 c c 6 b c 5 a a 5 (b) The subgraph G b a a 5 c c 5 c 6 of K 7,7,7 in which G = G[a i b i+ c i+4 ] b a c 5 c 6 a 5 a 6 c (c) The subgraph G 3 of K 7,7,7 Figure 4 A planar subgraphs decomposition of K 7,7,7

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