The thickness of K 1,n,n and K 2,n,n

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1 Abstract ISSN (printed edn.), ISSN (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 15 (2018) (Also available at The thickness of K 1,n,n and K 2,n,n Xia Guo School of Mathematics, Tianjin University, Tianjin, P.R.China Yan Yang School of Mathematics, Tianjin University, Tianjin, P.R.China Received 26 October 2016, accepted 12 June 2018, published online 9 July 2018 The thickness of a graph G is the minimum number of planar subgraphs whose union is G. In this paper, we obtain the thickness of complete 3-partite graph K 1,n,n, K 2,n,n and complete -partite graph K 1,1,n,n. Keywords: Thickness, complete 3-partite graph, complete -partite graph. Math. Subj. Class.: 05C10 1 Introduction The thickness θ(g) of a graph G is the minimum number of planar subgraphs whose union is G. It was first defined by W. T. Tutte [7] in 1963, then a few authors obtained the thickness of hypercubes [5], complete graphs [1, 2, 8] and complete bipartite graphs [3]. Naturally, people wonder about the thickness of the complete multipartite graphs. A complete k-partite graph is a graph whose verte set can be partitioned into k parts, such that every edge has its ends in different parts and every two vertices in different parts are adjacent. Let K p1,p 2,...,p k denote a complete k-partite graph in which the ith part contains p i (1 i k) vertices. For the complete 3-partite graph, Poranen proved θ(k n,n,n ) n 2 in [6], then Yang [10] gave a new upper bound for θ(kn,n,n ), i.e., θ(k n,n,n ) n and obtained θ(kn,n,n ) = n+1 3, when n 3 (mod 6). And also Yang [9] gave the thickness number of K l,m,n (l m n) when l + m 5 and showed that θ(k l,m,n ) = l+m 2 when l + m is even and n > 1 2 (l + m 2)2 ; or l + m is odd and n > (l + m 2)(l + m 1). In this paper, we obtain the thickness of complete 3-partite graph K 1,n,n and K 2,n,n, and we also deduce the thickness of complete -partite graph K 1,1,n,n from that of K 2,n,n. Supported by the National Natural Science Foundation of China under Grant No Corresponding author. addresses: guoia@tju.edu.cn (Xia Guo), yanyang@tju.edu.cn (Yan Yang) cb This work is licensed under

2 356 Ars Math. Contemp. 15 (2018) The thickness of K 1,n,n In [3], Beineke, Harary and Moon gave the thickness of complete bipartite graphs K m,n for most value of m and n, and their theorem implies the following result immediately. Lemma 2.1 ([3]). The thickness of the complete bipartite graph K n,n is n + 2 θ(k n,n ) =. In [], Chen and Yin gave a planar decomposition of the complete bipartite graph K p,p with p + 1 planar subgraphs. Figure 1 shows their planar decomposition of K p,p, in which {,..., p } = U and {,..., p } = V are the 2-partite verte sets of it. Based on their decomposition, we give a planar decomposition of K 2,n,n with p + 1 subgraphs when n 0 or 3 (mod ) and prove the following lemma. p {i 3, i 2 } U 1 r i=1,i r r 1 p {i 3, i 1 } V 1 r i=1,i r 1 r 1 r 2 r 3 5 r 3 r 2 p {i 2, i } V 2 r i=1,i r ur 2 3 p {i 1, i } U 2 r i=1,i r r (a) The graph G r (1 r p). p 1 p p 1 p (b) The graph G p+1. Figure 1: A planar decomposition of K p,p. Lemma 2.2. The thickness of the complete 3-partite graph K 1,n,n and K 2,n,n is when n 0 or 3 (mod ). n + 2 θ(k 1,n,n ) = θ(k 2,n,n ) =, Proof. Let the verte partition of K 2,n,n be (X, U, V ), where X = { 1, 2 }, U = {,..., u n } and V = {,..., v n }.

3 X. Guo and Y. Yang: The thickness of K 1,n,n and K 2,n,n 357 When n 0 (mod ), let n = p (p 1). Let {G 1,..., G p+1 } be the planar decomposition of K n,n constructed by Chen and Yin in []. As shown in Figure 1, the graph G p+1 consists of n paths of length one. We put all the n paths in a row, place verte 1 on one side of the row and the verte 2 on the other side of the row, join both 1 and 2 to all vertices in G p+1. Then we get a planar graph, denote it by Ĝp+1. It is easy to see that {G 1,..., G p, Ĝp+1} is a planar decomposition of K 2,n,n. Therefore, we have θ(k 2,n,n ) p + 1. Since K n,n K 1,n,n K 2,n,n, combining it with Lemma 2.1, we have p + 1 = θ(k n,n ) θ(k 1,n,n ) θ(k 2,n,n ) p + 1, that is, θ(k 1,n,n ) = θ(k 2,n,n ) = p + 1 when n 0 (mod ). When n 3 (mod ), then n = p + 3 (p 0). When p = 0, from [9], we have θ(k 1,3,3 ) = θ(k 2,3,3 ) = 2. When p 1, since K n,n K 1,n,n K 2,n,n K 2,n+1,n+1, according to Lemma 2.1 and θ(k 2,p,p ) = p + 1, we have p + 2 = θ(k n,n ) θ(k 1,n,n ) θ(k 2,n,n ) θ(k 2,n+1,n+1 ) = p + 2. Then, we get θ(k 1,n,n ) = θ(k 2,n,n ) = p + 2 when n 3 (mod ). Summarizing the above, the lemma is obtained. Lemma 2.3. There eists a planar decomposition of the complete 3-partite graph K 1,p+2,p+2 (p 0) with p + 1 subgraphs. Proof. Suppose the verte partition of the complete 3-partite graph K 1,n,n is (X, U, V ), where X = {}, U = {,..., u n } and V = {,..., v n }. When n = p + 2, we will construct a planar decomposition of K 1,p+2,p+2 with p + 1 planar subgraphs to complete the proof. Our construction is based on the planar decomposition {G 1, G 2,..., G p+1 } of K p,p given in [], as shown in Figure 1 and the reader is referred to [] for more details about this decomposition. For convenience, we denote the verte set p i=1,i r {i 3, i 2 }, p i=1,i r {i 1, i }, p i=1,i r {i 3, i 1 } and p i=1,i r {i 2, i } by U1 r, U2 r, V1 r and V2 r respectively. We also label some faces of G r (1 r p), as indicated in Figure 1, for eample, the face 1 is bounded by r 3 r v j r 1 in which v j is some verte from V1 r. In the following, for 1 r p + 1, by adding vertices, p+1, p+2, p+1, p+2 and some edges to G r, and deleting some edges from G r such edges will be added to the graph G p+1, we will get a new planar graph Ĝr such that {Ĝ1,..., Ĝp+1} is a planar decomposition of K 1,p+2,p+2. Because r 3 and r 1 in G r (1 r p) is joined by 2p 2 edge-disjoint paths of length two that we call parallel paths, we can change the order of these parallel paths without changing the planarity of G r. For the same reason, we can do changes like this for parallel paths between r 1 and r, r 2 and r, r 3 and r 2. We call this change by parallel paths modification for simplicity. All the subscripts of vertices are taken modulo p, ecept that of p+1, p+2, p+1 and p+2 (the vertices we added to G r ). Case 1. When p is even and p > 2. (a) The construction for Ĝr, 1 r p, and r is odd. Step 1: Place the verte in the face 1 of G r, delete edges r 3 r and r r 1 from G r. Do parallel paths modification, such that r+6 U1 r, r+1 V1 r and r 3, r 1,

4 358 Ars Math. Contemp. 15 (2018) r, r 3, r 2, r 1 are incident with a common face which the verte is in. Join to r 3, r 1, r, r 3, r 2, r 1 and r+6, r+1. Step 2: Do parallel paths modification, such that r+11, r+12 U2 r are incident with a common face. Place the verte p+1 in the face, and join it to both r+11 and r+12. Step 3: Do parallel paths modification, such that r+7, r+8 U2 r are incident with a common face. Place the verte p+2 in the face, and join it to both r+7 and r+8. Step : Do parallel paths modification, such that r+10, r+12 V2 r are incident with a common face. Place the verte p+1 in the face, and join it to both r+10 and r+12. Step 5: Do parallel paths modification, such that r+6, r+8 V2 r are incident with a common face. Place the verte p+2 in the face, and join it to both r+6 and r+8. (b) The construction for Ĝr, 1 r p, and r is even. Step 1: Place the verte in the face 3 of G r, delete edges r r 3 and r 3 r 2 from G r. Do parallel paths modification, such that r+7 U2 r, r+ V2 r and r 3, r 2, r, r 2, r 1, r are incident with a common face which the verte is in. Join to r 3, r 2, r, r 2, r 1, r and r+7, r+. Step 2: Do parallel paths modifications, such that r+5, r+6 U1 r, r+1, r+2 U1 r, r+5, r+7 V1 r, r+1, r+3 V1 r are incident with a common face, respectively. Join p+1 to both r+5 and r+6, join p+2 to both r+1 and r+2, join p+1 to both r+5 and r+7, join p+2 to both r+1 and r+3. Table 1 shows how we add edges to G r (1 r p) in Case 1. The first column lists the edges we added, the second and third column lists the subscript of vertices, and we also indicate the verte set which they belong to in brackets. subscript edge case Table 1: The edges we add to G r (1 r p) in Case 1. r is odd r is even u j r 3, r 1, r r + 6 (U r 1 ) r 3, r 2, r r + 7 (U r v j r 3, r 2, r 1 r + 1 (V1 r ) r 2, r 1, r r + (V r p+1 u j r + 11, r + 12 (U r r + 5, r + 6 (U r 1 ) p+2 u j r + 7, r + 8 (U r r + 1, r + 2 (U r 1 ) p+1 v j p+2 v j r + 10, r + 12 (V2 r ) r + 5, r + 7 (V r 1 ) r + 6, r + 8 (V2 r ) r + 1, r + 3 (V r 1 ) (c) The construction for Ĝp+1. From the construction in (a) and (b), the subscript set of u j that u j is an edge in Ĝr for some r {1,..., p} is {r 3, r 1, r, r + 6 (mod p) 1 r p, and r is odd} {r 3, r 2, r, r + 7 (mod p) 1 r p, and r is even} = {1,..., p}.

5 p 5 X. Guo and Y. Yang: The thickness of K 1,n,n and K 2,n,n 359 The subscript set of u j that p+1 u j is an edge in Ĝr for some r {1,..., p} is {r + 11, r + 12 (mod p) 1 r p, and r is odd} {r + 5, r + 6 (mod p) 1 r p, and r is even} = {r 3, r 2, r 1, r 1 r p, and r is even}. Using the same procedure, we can list all the edges incident with, p+1, p+2, p+1 and p+2 in Ĝr (1 r p), so we can also list the edges that are incident with, p+1, p+2, p+1 in K 1,p+2,p+2 but not in any Ĝr (1 r p). Table 2 shows the edges that belong to K 1,p+2,p+2 but not to any Ĝr, 1 r p, in which the the fourth and fifth rows list the edges deleted form G r (1 r p) in step one of (a) and (b), and the sith row lists the edges of G p+1. The Ĝp+1 is the graph consists of the edges in Table 2, Figure 2 shows Ĝp+1 is a planar graph. Table 2: The edges of Ĝp+1 in Case 1. edges subscript p+1, p+1, p+1 u j, p+1 v j j = r 3, r 2, r 1, r, p + 2 (r = 1, 3,..., p 1) p+2, p+2, p+2 u j, p+2 v j j = r 3, r 2, r 1, r, p + 1 (r = 2,,..., p) r 3 r, r r 1 r = 1, 3,..., p 1 r r 3, r 3 r 2 r = 2,,..., p u j v j j = 1,..., p + 2 p+1 p+2 p 7 p p 5 p 6 p 2 p 3 p p 1 p 2 p 3 p 1 p 7 p p 6 p p+1 p+2 Figure 2: The graph Ĝp+1 in Case 1. A planar decomposition {Ĝ1,..., Ĝp+1} of K 1,p+2,p+2 is obtained as above in this case. In Figure 3, we draw the planar decomposition of K 1,18,18, it is the smallest eample for the Case 1. We denote verte u i and v i by i and i respectively in this figure. Case 2. When p is odd and p > 3. The process is similar to that in Case 1. (a) The construction for Ĝr, 1 r p, and r is odd. Step 1: Place the verte in the face 1 of G r, delete edges r 3 r and r r 1 from G r, for 1 r p, and delete from G 1 additionally.

6 360 Ars Math. Contemp. 15 (2018) (a) The graph Ĝ1. (b) The graph Ĝ (c) The graph Ĝ3. (d) The graph Ĝ (e) The graph Ĝ5. Figure 3: A planar decomposition of K 1,18,18.

7 X. Guo and Y. Yang: The thickness of K 1,n,n and K 2,n,n 361 For 1 < r < p, do parallel paths modification to G r, such that r+6 U1 r, r+1 V1 r and r 3, r 1, r, r 3, r 2, r 1 are incident with a common face which the verte is in. Join to r 3, r 1, r, r 3, r 2, r 1 and r+6, r+1. Similarly, in G 1, join to,,,,,, and 0 U1 1, V1 1. In G p, join to p 3,p 1,p,p 3,p 2,p 1 and U p 1. Step 2: For 1 r < p, do parallel paths modification to G r, such that r+11, r+12 U2 r, r+7, r+8 U2 r, r+10, r+12 V2 r and r+6, r+8 V2 r are incident with a common face, respectively. Join p+1 to both r+11 and r+12, join p+2 to both r+7 and r+8, join p+1 to both r+10 and r+12, join p+2 to both r+6 and r+8. Similarly, in G p, join p+1 to, U p 1, join p+2 to, U p 2, join p+1 to, V p 2, join p+2 to, V p 1. (b) The construction for Ĝr, 1 r p, and r is even. Step 1: Place the verte in the face 3 of G r, delete edges r r 3 and r 3 r 2 from G r, 1 r p 1. Do parallel paths modification to G r, 1 r < p 1, such that r+7 U2 r, r+ V2 r and r 3, r 2, r, r 2, r 1, r are incident with a common face which the verte is in. Join to r 3, r 2, r, r 2, r 1, r and r+7, r+. Similarly, in G p 1, join to p 7, p 6, p, p 6, p 5, p and U p 1 2, p V p 1 2. Step 2: Do parallel paths modifications, such that r+5, r+6 U1 r, r+1, r+2 U1 r, r+5, r+7 V1 r, r+1, r+3 V1 r are incident with a common face, respectively. Join p+1 to both r+5 and r+6, join p+2 to both r+1 and r+2, join p+1 to both r+5 and r+7, join p+2 to both r+1 and r+3. Table 3 shows how we add edges to G r (1 r p) in Case 2. (c) The construction for Ĝp+1. With a similar argument to that in Case 1, we can list the edges that belong to K 1,p+2,p+2 but not to any Ĝr, 1 r p, in this case, as shown in Table. Then Ĝ p+1 is the graph that consists of the edges in Table. Figure shows Ĝp+1 is a planar graph. Therefore, {Ĝ1,..., Ĝp+1} is a planar decomposition of K 1,p+2,p+2 in this case. p+1 p p 3 p p 1 p 2 p 5 p 6 p 7 p p 3 p 1 p 2 p 5 p 6 p 7 p p p+1 p+2 Figure : The graph Ĝp+1 in Case 2.

8 362 Ars Math. Contemp. 15 (2018) Table 3: The edges we add to G r (1 r p) in Case 2. subscript case r is odd r is even edge u j r 3, r 1, r r + 6, r p (U r 1 ) 2, r = p (U r 1 ) r 3, r 2, r r + 7, r p 1 (U r 7, r = p 1 (U r v j r 3, r 2, r 1, 5, r = 1 r + 1, r 1, p (V r 1 ) r 2, r 1, r r + (V r p+1u j r + 11, r + 12, r p (U r 5, 6, r = p (U r 1 ) r + 5, r + 6 (U r 1 ) p+2u j r + 7, r + 8 (U r r + 1, r + 2 (U r 1 ) p+1v j p+2v j r + 10, r + 12, r p (V r 6, 8, r = p (V r r + 6, r + 8, r p (V r 5, 7, r = p (V r 1 ) r + 5, r + 7 (V r 1 ) r + 1, r + 3 (V r 1 ) Table : The edges of Ĝp+1 in Case 2. edges subscript p+1, p+1u j j = r 3, r 2, r 1, r, 7, 8, p + 2 (r = 3, 5, 7,..., p) p+1, p+1v j j = r 3, r 2, r 1, r, 5, 7, p + 2 (r = 3, 5, 7,..., p) p+2, p+2u j j = r 3, r 2, r 1, r, 5, 6, p + 1 (r = 1,, 6, 8,..., p 1) p+2, p+2v j j = r 3, r 2, r 1, r, 6, 8, p + 1 (r = 1,, 6, 8,..., p 1), r 3r, rr 1 r = 1, 3,..., p rr 3, r 3r 2 r = 2,,..., p 1 u jv j j = 1,..., p + 2

9 X. Guo and Y. Yang: The thickness of K 1,n,n and K 2,n,n 363 Case 3. When p 3. When p = 0, K 1,2,2 is a planar graph. When p = 1, 2, 3, we give a planar decomposition for K 1,6,6, K 1,10,10 and K 1,1,1 with 2, 3 and subgraphs respectively, as shown in Figure 5, Figure 6 and Figure 7. Figure 5: A planar decomposition of K 1,6, u1 0 v Figure 6: A planar decomposition of K 1,10,10. Lemma follows from Cases 1, 2 and 3.

10 36 Ars Math. Contemp. 15 (2018) u 9 0 u u u u10 v u3 v v Figure 7: A planar decomposition of K 1,1,1. p+1 2 p 1 p 2 p 3 p p 1 p 2 p 3 p 1 p+1 Figure 8: The graph Ĝp+1 in Case 1.

11 X. Guo and Y. Yang: The thickness of K 1,n,n and K 2,n,n 365 Theorem 2.. The thickness of the complete 3-partite graph K 1,n,n is n + 2 θ(k 1,n,n ) =. Proof. When n = p, p + 3, the theorem follows from Lemma 2.2. When n = p + 1, n = p + 2, from Lemma 2.3, we have θ(k 1,p+2,p+ p + 1. Since θ(k p,p ) = p + 1 and K p,p K 1,p+1,p+1 K 1,p+2,p+2, we obtain p + 1 θ(k 1,p+1,p+1 ) θ(k 1,p+2,p+ p + 1. Therefore, θ(k 1,p+1,p+1 ) = θ(k 1,p+2,p+ = p + 1. Summarizing the above, the theorem is obtained. 3 The thickness of K 2,n,n Lemma 3.1. There eists a planar decomposition of the complete 3-partite graph K 2,p+1,p+1 (p 0) with p + 1 subgraphs. Proof. Let (X, U, V ) be the verte partition of the complete 3-partite graph K 2,n,n, in which X = { 1, 2 }, U = {,..., u n } and V = {,..., v n }. When n = p + 1, we will construct a planar decomposition of K 2,p+1,p+1 with p + 1 planar subgraphs. The construction is analogous to that in Lemma 2.3. Let {G 1, G 2,..., G p+1 } be a planar decomposition of K p,p given in []. In the following, for 1 r p + 1, by adding vertices 1, 2, p+1, p+1 to G r, deleting some edges from G r and adding some edges to G r, we will get a new planar graph Ĝr such that {Ĝ1,..., Ĝp+1} is a planar decomposition of K 2,p+1,p+1. All the subscripts of vertices are taken modulo p, ecept that of p+1 and p+1 (the vertices we added to G r ). Case 1. When p is even and p > 2. (a) The construction for Ĝr, 1 r p. Step 1: When r is odd, place the verte 1, 2 and p+1 in the face 1, 2 and 5 of G r respectively. Delete edges r 3 r and r 1 r 2 from G r. When r is even, place the verte 1, 2 and p+1 in the face 3, and 5 of G r, respectively. Delete edge r r 3 and r 2 r 1 from G r. Step 2: Do parallel paths modifications, then join 1, 2, p+1 and p+1 to some u j and v j, as shown in Table 5. (b) The construction for Ĝp+1. We list the edges that belong to K 2,p+1,p+1 but not to any Ĝr, 1 r p, as shown in Table 6. Then Ĝp+1 is the graph that consists of the edges in Table 6. Figure 8 shows Ĝ p+1 is a planar graph. Therefore, {Ĝ1,..., Ĝp+1} is a planar decomposition of K 2,p+1,p+1 in this case. In Figure 9, we draw the planar decomposition of K 2,, it is the smallest eample for the Case 1. We denote verte u i and v i by i and i respectively in this figure.

12 366 Ars Math. Contemp. 15 (2018) Table 5: The edges we add to G r (1 r p) in Case 1. subscript edge case r is odd r is even 1u j r 1, r r + 5 (U r 1 ) r 3, r 2 r + 8 (U r 1v j r 3, r 1 r + 1 (V r 1 ) r 2, r r + (V r 2u j r 1, r r + 3 (U r r 3, r 2 r + 2 (U r 1 ) 2v j r 2, r r + 7 (V r 1 ) r 3, r 1 r + 6 (V r p+1v j r 2, r 1 p+1u j r +, r + 8 (U r r 11, r 7 (U r 1 ) Table 6: The edges of Ĝp+1 in Case 1. edges subscript 1u j j = r 2, r + 3, p + 1 (r = 1, 3,..., p 1) 1v j 2u j j = r 7, r, p + 1 (r = 2,,..., p) 2v j p+1v j j = r 3, r (r = 1, 2,..., p) p+1u j j = r 2, r 1 (r = 1, 2,..., p) r 3r, r 2r 1 r = 1, 3,..., p 1 r 3r, r 2r 1 r = 2,,..., p u jv j j = 1,..., p + 1

13 X. Guo and Y. Yang: The thickness of K 1,n,n and K 2,n,n (a) The graph Ĝ1. (b) The graph Ĝ (c) The graph Ĝ3. (d) The graph Ĝ (e) The graph Ĝ5. Figure 9: A planar decomposition of K 2,,.

14 368 Ars Math. Contemp. 15 (2018) Case 2. When p is odd and p > 3. (a) The construction for Ĝr, 1 r p. Step 1: When r is odd, place the verte 1, 2 and p+1 in the face 1, 2 and 5 of G r respectively. Delete edges r 3 r and r 1 r 2 from G r. When r is even, place the verte 1, 2 and p+1 in the face 3, and 5 of G r, respectively. Delete edge r r 3 and r 2 r 1 from G r. Step 2: Do parallel paths modifications, then join 1, 2, p+1 and p+1 to some u j and v j, as shown in Table 7. Table 7: The edges we add to G r (1 r p) in Case 2. subscript case r is odd r is even edge 1u j r 1, r r + 5, r p (U r 1 ) 1, r = p (U r 1 ) r 3, r 2 r + 8, r p 1 (U r 8, r = p 1 (U r 1v j r 3, r 1 r + 1, r p (V r 1 ) r 2, r r + (V r 2u j r 1, r r + 3, r p (U r 8, r = p (U r r 3, r 2 r + 2 (U r 1 ) 2v j r 2, r r + 7, r p (V r 1 ) 3, r = p (V r 1 ) r 3, r 1 p+1v j r 2, r 1 r + 6, r p 1 (V r 6, r = p 1 (V r p+1u j r +, r + 8, r p (U r, r = p (U r r 11, r 7 (U r 1 ) (b) The construction for Ĝp+1. We list the edges that belong to K 2,p+1,p+1 but not to any Ĝr, 1 r p, as shown in Table 8. Then Ĝp+1 is the graph that consists of the edges in Table 8. Figure 10 shows Ĝ p+1 is a planar graph. Therefore, {Ĝ1,..., Ĝp+1} is a planar decomposition of K 2,p+1,p+1 in this case. Case 3. When p 3. When p = 0, K 2,1,1 is a planar graph. When p = 1, 2, 3, we give a planar decomposition for K 2,5,5, K 2,9,9 and K 2,13,13 with 2, 3 and subgraphs respectively, as shown in Figure 11, Figure 12 and Figure 13. Summarizing Cases 1, 2 and 3, the lemma follows.

15 X. Guo and Y. Yang: The thickness of K 1,n,n and K 2,n,n 369 Table 8: The edges of Ĝp+1 in Case 2. edges subscript 1u j j = 2, r + 3, r + 6, p + 1 (r = 1, 3,..., p 2) 1v j j = 2,, r + 3, r + 6, p + 1 (r = 1, 3,..., p 2) 2u j j = 1, 2, 9, r, r + 1, p + 1 (r =,..., p 1) 2v j j = 1, 8, 9, r, r + 1, p + 1 (r =,..., p 1) p+1v j j = r 3, r (r = 1, 2,..., p) p+1u j j = r 2, r 1, p 7 (r = 1, 2,..., p) r 3r, r 2r 1 r = 1, 3,..., p r 3r, r 2r 1 r = 2,,..., p 1 u jv j j = 1,..., p + 1 vp+1 2 p 5 p 6 p 2 p 1 p 7 p p p 3 p 5 p 6 p 2 p 1 p 7 p p p 3 1 p+1 Figure 10: The graph Ĝp+1 in Case u5 v u v3 u3 1 Figure 11: A planar decomposition K 2,5,5.

16 370 Ars Math. Contemp. 15 (2018) v v5 v Figure 12: A planar decomposition K 2,9,9.

17 X. Guo and Y. Yang: The thickness of K 1,n,n and K 2,n,n 371 u10 3 u u u7 1 u u u u Figure 13: A planar decomposition of K 2,13,13.

18 372 Ars Math. Contemp. 15 (2018) Theorem 3.2. The thickness of the complete 3-partite graph K 2,n,n is n + 3 θ(k 2,n,n ) =. Proof. When n = p, p + 3, from Lemma 2.2, the theorem holds. When n = p + 1, from Lemma 3.1, we have θ(k 2,p+1,p+1 ) p + 1. Since θ(k p,p ) = p + 1 and K p,p K 2,p+1,p+1, we have p + 1 = θ(k p,p ) θ(k 2,p+1,p+1 ) p + 1. Therefore, θ(k 2,p+1,p+1 ) = p + 1. When n = p + 2, since K p+3,p+3 K 2,p+2,p+2, from Lemma 2.1, we have p + 2 = θ(k p+3,p+3 ) θ(k 2,p+2,p+. On the other hand, it is easy to see θ(k 2,p+2,p+ θ(k 2,p+1,p+1 ) + 1 = p + 2, so we have θ(k 2,p+2,p+ = p + 2. Summarizing the above, the theorem is obtained. The thickness of K 1,1,n,n Theorem.1. The thickness of the complete -partite graph K 1,1,n,n is n + 3 θ(k 1,1,n,n ) =. Proof. When n = p + 1, we can get a planar decomposition for K 1,1,p+1,p+1 from that of K 2,p+1,p+1 as follows. (1) When p = 0, K 1,1,1,1 is a planar graph, θ(k 1,1,1,1 ) = 1. When p = 1, 2 and 3, we join the verte 1 to 2 in the last planar subgraph in the planar decomposition for K 2,5,5, K 2,9,9 and K 2,13,13 which was shown in Figure 11, 12 and 13. Then we get the planar decomposition for K 1,1,5,5, K 1,1,9,9 and K 1,1,13,13 with 2, 3 and planar subgraphs respectively. (2) When p, we join the verte 1 to 2 in Ĝp+1 in the planar decomposition for K 2,p+1,p+1 which was constructed in Lemma 3.1. The Ĝp+1 is shown in Figure 8 or 10 according to p is even or odd. Because 1 and 2 lie on the boundary of the same face, we will get a planar graph by adding edge 1 2 to Ĝp+1. Then a planar decomposition for K 1,1,p+1,p+1 with p + 1 planar subgraphs can be obtained. Summarizing (1) and (2), we have K 1,1,p+1,p+1 p + 1. On the other hand, from Lemma 2.1, we have θ(k p+1,p+1 ) = p + 1. Due to K p+1,p+1 K 1,1,p,p K 1,1,p+1,p+1, we get So we have p + 1 θ(k 1,1,p,p ) θ(k 1,1,p+1,p+1 ). θ(k 1,1,p,p ) = θ(k 1,1,p+1,p+1 ) = p + 1. When n = p + 3, from Theorem 3.2, we have θ(k 2,p+2,p+ = p + 2. Since K 2,p+2,p+2 K 1,1,p+2,p+2 K 1,1,p+3,p+3 K 1,1,(p+1),(p+1), and the ideas from the previous case establish, we have p + 2 θ(k 1,1,p+2,p+ θ(k 1,1,p+3,p+3 ) θ(k 1,1,(p+1),(p+1) ) = p + 2, which shows θ(k 1,1,p+2,p+ = θ(k 1,1,p+3,p+3 ) = p + 2. Summarizing the above, the theorem follows.

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