The 6-girth-thickness of the complete graph

The $g$-girth-thickness $\theta(g,G)$ of a graph $G$ is the minimum number of planar subgraphs of girth at least $g$ whose union is $G$. In this paper, we determine the $6$-girth-thickness $\theta(6,K_n)$ of the complete graph $K_n$ in almost all cases. And also, we calculate by computer the missing value of $\theta(4,K_n)$.


Introduction
In this paper, all graphs are finite and simple. A graph in which any two vertices are adjacent is called a complete graph and it is denoted by K n if it has n vertices. If a graph can be drawn in the Euclidean plane such that no inner point of its edges is a vertex or lies on another edge, then the graph G is called planar. The girth of a graph is the size of its shortest cycle or ∞ if it is acyclic. It is known that an acyclic graph of order n has size at most n − 1 and a planar graph of order n and finite girth g has size at most g g−2 (n − 2), see [8].
The thickness θ(G) of a graph G is the minimum number of planar subgraphs whose union is G. Equivalently, it is the minimum number of colors used in any edge coloring of G such that each set of edges in the same chromatic class induces a planar subgraph.
The concept of the thickness was introduced by Tutte [19]. The problem to determine the thickness of a graph G is NP-hard [15], and only a few of exact results are known, for instance, when G is a complete graph [2,5,6], a complete multipartite graph [7,11,18,21,22] or a hypercube [14].
Generalizations of the thickness for the complete graphs also have been studied such that the outerthickness θ o , defined similarly but with outerplanar instead of planar [12], and the Sthickness θ S , considering the thickness on a surface S instead of the plane [4]. The thickness has many applications, for example, in the design of circuits [1], in the Ringel's earth-moon problem [13], or to bound the achromatic numbers of planar graphs [3]. See also [16].
In [17], the g-girth-thickness θ(g, G) of a graph G was defined as the minimum number of planar subgraphs of girth at least g whose union is G. Indeed, the g-girth thickness generalizes the thickness when g = 3 and the arboricity number when g = ∞.
This paper is organized as follows. In Section 2, we obtain the 6-girth-thickness θ(6, K n ) of the complete graph K n getting that θ(6, K n ) equals n+2

3
, except for n = 3t + 1, t ≥ 4 and n = 2, for which θ(6, K 2 ) = 1. In Section 3, we show that there exists a set of 3 planar triangle-free subgraphs of K 10 whose union is K 10 . The decomposition was found by computer and, as a consequence, we disproved the conjecture that appears in [17] about the missing case of the 4-girth-thickness of the complete graph.
2 Determining θ(6, K n ) A planar graph of n vertices with girth at least 6 has size at most 3(n − 2)/2 for n ≥ 6 and size at most n − 1 for 1 ≤ n ≤ 5, therefore, the 6-girth-thickness θ(6, K n ) of the complete graph K n is at least for n ≥ 6, as well as, n+2 3 for n ∈ {1, 3, 4, 5}. We have the following theorem.
Proof. To begin with, Figure 1 displays equality for n = 2, 4, 7, 10 with θ(6, K n ) = 1, 2, 3, 4, respectively. The rest of the cases for 1 ≤ n ≤ 10 are obtained by the hereditary property of the induced subgraphs. We remark that the decomposition of K 10 was found by computer using the database of the connected planar graphs of order 10 that appears in [9]. Now, we need to distinguish two main cases, namely, when t is even or t is odd for n = 3t, that is, when n = 6k and n = 6k + 3 for k ≥ 2. The cases n = 6k − 1 and n = 6k + 2, i.e., for n = 3t + 1, are obtained by the hereditary property of the induced subgraphs, that is, Therefore, the case of n = 6k shows a decomposition of K 6k into 2k + 1 planar subgraphs of girth at least 6, while the case of n = 6k+3 shows a decomposition of K 6k+3 into 2k+2 planar subgraphs of girth at least 6. Both constructions are based on the planar decomposition of K 6k of Beineke and Harary [5] (see also [2,6,20]) but we use the combinatorial approach given in [3]. Then, for the sake of completeness, we give a decomposition of K 6k in order to obtain its usual thickness. In the remainder of this proof, all sums are taken modulo 2k.
We recall that complete graphs of even order 2k are decomposable into a cyclic factorization of Hamiltonian paths, see [10]. Let G x be a complete graph of order 2k, label its vertex set V (G x ) as {x 1 , x 2 , . . . , x 2k } and let F x i be the Hamiltonian path with edges , see Figure 2.
x i+k x i+k+1 Let G u , G v and G w be the complete subgraphs of K 6k having 2k vertices each of them and such that G w is K 6k \(V (G u )∪V (G v )). The vertices of V (G u ), V (G v ) and V (G w ) are labeled as {u 1 , u 2 , . . . , u 2k }, {v 1 , v 2 , . . . , v 2k } and {w 1 , w 2 , . . . , w 2k }, respectively.
Let x be an element of {u, v, w}. Take the cyclic factorization Hamiltonian paths and denote as P x i and P x i+k the subpaths of F x i containing k vertices and the leaves x i and x i+k , respectively. We define the other leaves of P x i and P x i+k as f (x i ) and f (x i+k ), respectively and according to the parity of k, that is (see Figure 2), We remark that the set of edges {x i x i+k : 1 ≤ i ≤ k} is the same set of edges that Now, we construct the maximal planar subgraphs G 1 , G 2 ,...,G k and a matching G k+1 with 6k vertices each in the following way. Let G k+1 be the perfect matching with the edges u j u j+k , v j v j+k and w j w j+k for j ∈ {1, 2, . . . , k}.
For each i ∈ {1, 2, . . . , k}, let G i be the spanning planar graph of K 6k whose adjacencies are given as follows: we take the 6 paths, P u i , P u i+k , P v i , P v i+k , P w i and P w i+k and insert them in the octahedron with the vertices u i , u i+k , v i , v i+k , w i and w i+k as is shown in Figure 2 (Left).
The vertex x j of each path P x j is identified with the vertex x j in the corresponding triangle face and join all the other vertices of the path with both of the other vertices of the triangle face, see Figure 3 (Right).
Step 1. For each i ∈ {1, . . . , k}, remove the six edges of the triangles u i v i w i and u i+k v i+k w i+k .
Step 2. For each i ∈ {1, . . . , k}, divide the obtained subgraph into two subgraphs H 1 i and H 2 i as follows: The maximum matching of P x i incident to the vertex f (x i ) belongs to H 1 i (see dotted subgraph in Figure 4) while the maximum matching of P x i+k incident to the vertex f (x i+k ) belongs to H 2 i .
Next, the rest of the edges joined to the vertices of the paths P x i and P x i+k , in an alternative way from the exterior region to the region with the vertices {u i , v i , w i }, belong to H 1 i and H 2 i respectively, such that the edges f (w i )u i+k , f (v i )w i+k and f (u i )v i+k belong to H 1 i and the edges f (w i )v i+k , f (v i )u i+k and f (u i )w i+k belong to H 2 i , see Figure  4. Step 3. Consider the removed edges in Step 1, add the edges f (v i+k )f (u i+k ) and Figure 5. The rest of the edges removed in Step 1 are added to G k+1 getting the subgraph H k+1 which is the union of the paths Figure 5: Subgraphs H 1 i and H 2 i for the Case 1.
Consider the set of planar subgraphs {G 1 , G 2 , . . . , G k+1 } of K 6k which is described above as well as Step 1 and 2 of the previous case.
Step 3. Add three vertices u, v and w in the subgraphs H 1 i and H 2 i , for each i ∈ {1, . . . , k}, and the edges uw i , uf (v i+k ), vu i , vf (w i+k ), wv i , wf (u i+k ) into H 1 i as well as the edges uw i+k , uf (v i ), vu i+k , vf (w i ), wv i+k , wf (u i ) into H 2 i , see Figure 6. Figure 6: Subgraphs H 1 i and H 2 i for Case 2.
Step 4. On one hand, remains to define the adjacencies between u, v, w and all the adjacencies between u and u i , v and v i , w and w i , for each j ∈ {1, . . . , k}. On the other hand, the edges of the graph G k+1 together with the removed edges of the Step 1 form a set of triangle prisms which we split into two subgraphs called H 1 k+1 and H 2 k+1 in the following way: a) The adjacency vw is in H 1 k+1 while the adjacencies uv and uw are in H 2 k+1 , see Figure 7.
b) The set of adjacencies vv j+k , ww j , ww j+k and uu j+k are in H 1 k+1 while the set of adjacencies vv j , and uu j are in H 2 k+1 , for each j ∈ {1, . . . , k}, see Figure 7. c) The subgraph H 1 k+1 contains the adjacencies v j+k v j , v j u j , u j w j and w j+k u j+k (a set of subgraphs P 4 ∪ K 2 ) and the subgraph H 2 k+1 contains the adjacencies u j u j+k , u j+k v j+k , v j+k w j+k ,w j+k w j and w j v j (a set of subgraphs P 6 ) for all j ∈ {1, . . . , k}, see Figure 7.
By the small cases and the two main cases, the theorem follows.
3 The 4-girth thickness of K 10 In [17], Rubio-Montiel gave a decomposition of K n into θ(4, K n ) = n+2 4 triangle-free planar subgraphs, except for n = 10. In that case, it was bounded by 3 ≤ θ(4, K 10 ) ≤ 4 and conjectured that the correct value was the upper bound. Using the database of the connected Figure 7: Partial subgraphs H 1 k+1 and H 2 k+1 .
planar graphs of order 10 that appears in [9] and the SageMath program, we found two decompositions of K 10 into 3 planar subgraphs of girth at least 4 illustrated in Figure 8. In summary, the correct value of θ(4, K n ) was the lower bound and then, we have the following theorem.