Difference of facial achromatic numbers between two triangular embeddings of a graph

A facial $3$-complete $k$-coloring of a triangulation $G$ on a surface is a vertex $k$-coloring such that every triple of $k$-colors appears on the boundary of some face of $G$. The facial $3$-achromatic number $\psi_3(G)$ of $G$ is the maximum integer $k$ such that $G$ has a facial $3$-complete $k$-coloring. This notion is an expansion of the complete coloring, that is, a proper vertex coloring of a graph such that every pair of colors appears on the ends of some edge. For two triangulations $G$ and $G'$ on a surface, $\psi_3(G)$ may not be equal to $\psi_3(G')$ even if $G$ is isomorphic to $G'$ as graphs. Hence, it would be interesting to see how large the difference between $\psi_3(G)$ and $\psi_3(G')$ can be. We shall show that the upper bound for such difference in terms of the genus of the surface.


Introduction
In this paper, we consider finite and undirected graph.A graph is called simple if it has no loops and multiple edges.We mainly focus on simple graphs unless we particularly mention it.An embedding of a graph G on a surface F is a drawing of G on F with no pair of crossing edges.Technically, we regard an embedding as injective continuous map f : G → F, where G is regarded as a one-dimensional topological space.We sometime consider that G is already mapped on a surface and denote its image by G itself to simplify the notation, while if we deal with two or more embeddings of G on a surface, we denote them by f 1 (G), f 2 (G), . . . to distinguish them.
The faces of a graph G embedded on a surface F are the connected components of the open set F − G.We denote by V (F ) the set of vertices in the boundary of a face F of G, and by F(G) the set of faces of G.A triangulation on a surface F is an embedding of a graph on F so that each face is bounded by a 3-cycle.A graph G is said to have a triangulation on a surface, if G is embeddable on the surface as a triangulation.
A (vertex) k-coloring of a graph G is a map c : V (G) → {1, 2, . . ., k}.A kcoloring c of G is proper if c(u) = c(v) whenever two vertices u and v are adjacent.For a subset S ⊆ V (G), we denote by c(S) the set of colors of the vertices in S. Colorings of graphs embedded on surfaces with facial constraints have attracted a lot of attention.In particular, facially-constrained colorings of plane graphs were overviewed by Czap and Jendrol' [2].Many facially-constrained colorings can be translated into colorings of some kind of hypergraphs, called "face-hypergraphs".The face-hypergraph H(G) of a graph G embedded on a surface is the hypergraph with vertex-set V (G) and edge-set {V (F ) : F ∈ F(G)}, whose concept was introduced in [6].
A complete k-coloring of a graph G is a proper k-coloring such that each pair of k-colors appears on at least one edge of G.The achromatic number of G is the maximum integer k such that G has a complete k-coloring.This notion was introduced by Harary and Hedetniemi [4], and has been extensively studied (see [5] for its survey).Recently, Matsumoto and the second author [7] introduced a new facially-constrained coloring, called the "facial complete coloring", which is an expansion of the complete coloring.A k-coloring, which is not necessarily proper, of a graph G embedded on a surface is facially t-complete if for any t-element subset X of the k colors, there is a face F of G such that X ⊆ c(V (F )).The maximum integer k such that G has a facial t-complete k-coloring is the facial t-achromatic number of G, denoted by ψ t (G).It seems to be natural to consider facial t-complete colorings for graphs embedded on a surface so that each face is bounded by a cycle of length t.
We should notice that the facial t-achromatic number of an embedded graph depends on the embedding of the graph in general.That is, if a graph G has two distinct embeddings f 1 (G) and f 2 (G) on F, then ψ t (f 1 (G)) may not be equal to ψ t (f 2 (G)).Hence, it would be interesting to see how large the difference between ψ t (f 1 (G)) and ψ t (f 2 (G)) can be.In this paper, we focus on facial 3-complete colorings of triangulations on a surface from this point of view, and show the upper bound for such difference as follows.
Theorem 1.Let G be a graph which has two triangulations f 1 (G) and f 2 (G) on a surface F, and let g be the Euler genus of F. If F is orientable, then If F is non-orientable, then Note that we can easily construct a triangulation on each surface so that its facial 3-achromatic number is an arbitrarily large, while Theorem 1 implies that the difference of the facial 3-achromatic numbers between two triangulations f 1 (G) and f 2 (G) on a given surface, which is obtained from the same graph G, can be bounded by a constant.
On the other hand, the upper bounds in Theorem 1 do not seem to be sharp.Unfortunately, we have no construction of a graph which has two triangulations on a surface whose facial 3-achromatic numbers differ.So one may suspect that ψ t (f 1 (G)) = ψ t (f 2 (G)) whenever a graph G has two triangulations f 1 (G) and f 2 (G) on a surface.However, we do not believe that.Actually, we shall show in Section 5, the non-simple graphs having two triangulations on a surface whose facial 3achromatic numbers differ (the definition of the facial complete coloring can be extended to non-simple graphs naturally).Hence, we hope that there exist such graphs for simple graphs.
We introduce some useful lemmas in Section 3 to prove Theorem 1 in Section 4. Before these sections, we would like to introduce some related results dealing with other facially-constrained colorings.For not only the facial complete coloring but also other facially-constrained colorings, the possibility of such a coloring depends on the embedding in general.We survey some results from this point of view in the next section.

Related results
A rainbow coloring (or a cyclic coloring) is a coloring of a graph G embedded on a surface so that each face is rainbow, that is, any two distinct vertices on its boundary have disjoint colors.The minimum integer n such that G has a rainbow n-coloring is the rainbowness of G, denoted by rb(G).An antirainbow coloring (or a valid coloring) is a coloring of a graph G embedded on a surface so that no face is rainbow.The maximum integer n such that G has a surjective antirainbow n-coloring is the antirainbowness of G, denoted by arb(G).
Let G be the graph consisting of m ≥ 3 cycles of length 3 with one common vertex, which has two embeddings f 1 (G) and f 2 (G) on the sphere as shown in Fig. 1.Then G has 2m + 1 vertices.
As there is a face incident with all vertices in f 1 (G), we have rb(f It is also easy to see that rb(f 2 (G)) = 5.Hence, the difference between rb(f 1 (G)) and rb(f 2 (G)) is 2m − 4. It implies that the rainbowness of a graph embedded on a surface depends on the embedding.Moreover, such a difference can be arbitrarily large.On the other hand, it is easy to see that rb(G) = χ(G) for every triangulation on a surface, where χ(G) is the chromatic number of G.This implies that the rainbowness of a triangulation does not depend on the embedding.
Ramamurthi and West [11] observed that for the above two embeddings f 1 (G) and f 2 (G) of G on the sphere, arb(f 1 (G)) = m + 1 and arb(f 2 (G)) = 3m/2 .Then the difference of these antirainbownesses is m/2−1 , and hence the antirainbowness of a graph embedded on a surface also depends on the embedding.Ramamurthi and West [11] conjectured this difference is the maximum difference for two embeddings of a graph on the sphere, that is, for every planar graph G of order n, there is no pair of embeddings of G on the sphere whose antirainbownesses differ from at least (n − 2)/4 .
Arocha, Bracho and Neumann-Lara [1] studied the antirainbow 3-colorability of triangulations obtained from complete graphs, which they called the tightness.They proved that the complete graph of order 30 has both of a tight triangulation and an untight one on the same surface.This implies that the antirainbowness of triangulations depends on the embedding.As the generalization of their work, Negami [10] introduced the looseness of a triangulation G on a surface, which corresponds to arb(G) + 2.He proved that for any graph having two triangulations f 1 (G) and A weak coloring of a graph G embedded on a surface is a coloring of G such that no face is monochromatic, that is, all vertices on its boundary have the same color.Note that a weak coloring of an embedded graph corresponds to a proper coloring of its face-hypergraph.The weak chromatic number of G, denoted by χ w (G), is the minimum integer k such that G has a weak k-coloring.Kündgen and Ramamurthi [6] studied weak colorings of graphs embedded on surfaces from various viewpoints and conjectured that for each positive integer k, there is a graph that has two different embeddings on the same surface whose weak chromatic numbers differ by at least k.Recently, the first author and Noguchi [3] answered this conjecture affirmatively in two ways.They first constructed two distinct embeddings of a simple graph on a surface such that one of them has a weak 2-coloring but the other has arbitrarily large weak chromatic number.They second showed that there are non-simple graphs G having two triangulations f 1 (G) and f 2 (G) on a surface with χ w (f

Cycles in a triangulation
To prove Theorem 1, we give some notations and introduce some lemmas.
Let G be a graph and H be a subgraph of G.An edge not in H but with both ends in H is called chord of H.A subgraph H of a graph G is induced if H has no chord.An H-bridge is a subgraph of G induced by a chord of H, or a component of G − V (H) together with all edges joining it to H.In an H-bridge, a vertex belongs to V (H) is called a vertex of attachment.Note that any two H-bridges are edge-disjoint and meet only the common vertices of attachment.(See [9] for more details of H-bridges.)Lemma 2. Let G be a triangulation on a surface, and Proof.Let C = uvw be a facial cycle of G bounded by three vertices u, v and w.Suppose that C is not contained in H. Since H consists of vertex-disjoint cycles and has no chord, C meets at most one cycle of H. Suppose that C meets C 1 at a vertex, say u, and v, w ∈ V (C i ) for any 1 ≤ i ≤ k.If v and w belongs to different H-bridges in G, then the edge vw joins these H-bridges, a contradiction.Hence, v and w belongs to the same H-bridge in G.It implies that all vertices and edges around C i belongs to one H-bridge in G. Suppose that C meets none of C 1 , C 2 , . . ., C k .Then it is clear that u, v and w belong to the same H-bridge in G. Therefore, there is only one H-bridge in G.
Let G be a graph embedded on a surface F. A cycle C of G is contractible if it bounds a disk in F, and separating if it separates F into two parts.We say that C is 2-sided if it divides its annular neighbourhood into two parts, and is 1-sided otherwise.Note that a non-separating cycle of G must be non-contractible, and if a separating cycle C of G is not facial then there are at least two C-bridges in G. Lemma 3. Let G be a graph which has two triangulations f 1 (G) and f 2 (G) on a surface, and For two disjoint cycles C 1 and C 2 of a graph embedded on a surface F, cut the surface F along them.When one of the component of the resulting surface is an annulus with boundary components C 1 and C 2 , we say that C 1 and C 2 are homotopic.
We introduce two lemmas about sets of pairwise non-homotopic cycles.The second lemma closely follows from the proof of [8,Proposition 3.7], which corresponds to the first one.However, to keep the paper self-contained, we give its proof.
Lemma 4 (Malnič and Mohar [8]).Let G be a graph embedded on a surface F, and let g be the Euler genus of F. Let Γ be a set of pairwise disjoint, non-contractible and pairwise non-homotopic cycles of G.If F is orientable, then Lemma 5. Let G be a graph embedded on a non-orientable surface F of Euler genus g. let Γ 1 (resp.Γ 2 ) be a set of pairwise disjoint, non-contractible and pairwise nonhomotopic 1-sided (resp.2-sided) cycles of G. Then |Γ 1 | ≤ g and Proof.It is easy to see that this lemma holds for g ≤ 2. Hence, we may assume that g ≥ 3.Moreover, we may assume that Γ 1 is maximal, that is there is no 1-sided cycle in G disjoint from Γ 1 .Cutting F along the cycles in Γ 1 , we obtain a connected surface, denoted by F , which has |Γ 1 | boundary components.Thus, χ(F We may also assume that Γ 2 is maximal, that is, all 2-sided cycles in G disjoint from Γ 2 is contractible or homotopic to some element of Γ 2 .Cut F along the cycles in Γ 2 .Then F is separated into some connected surfaces, denoted by F 1 , F 2 , . . ., F k .Note that they are all compact and with non-empty boundary.We denote by b(∂F i ) the number of boundary components of F i for 1 ≤ i ≤ k.Since each cycle in Γ 2 gives rise to two boundary components, we have . ., F * k be the surfaces obtained from F 1 , F 2 , . . ., F k by pasting a disk to each boundary component.By the maximality of Γ 2 , F * i is the sphere or the projective plane for 1 ≤ i ≤ k.We denote by n s and n p the numbers of the spheres and the projective planes among F * i 's, respectively.Then we have n p ≤ g and Then F i is an annulus.If two cycles of Γ 2 corresponding to the boundary components F i are the same, then F must be the Klein bottle, a contradiction.Thus, these two cycles are different from each other.However, in this situation, they are homotopic in F, a contradiction.Therefore, we may assume that b(∂F i ) ≥ 3. It implies that 3n s + n p ≤ 2|Γ 2 |.
Since χ(F) is equal to the sum of all F i 's, we have

Proof of Theorem 1
Proof of Theorem 1. Suppose that Then, every triple of k-colors appears in some face of f 1 (G).On the other hand, some triples do not appear in the faces of f 2 (G).Let T be a set of triples in k colors such that any triple in T does not appear in the faces of f 2 (G), and for any pair of triples T and T in T , T ∩ T = ∅.Moreover, we choose T so that |T | is as large as possible.Let T 1 , T 2 , . . ., T m be the triples in T , and so |T | = m.By the maximality of T , we can choose k − 3m colors so that every triple in these colors appear in some face of f 2 (G).It implies that f 2 (G) has a facial 3-complete (max{3, k − 3m})-coloring.Then, . ., C m } be a set of facial cycles in f 1 (G) such that c(V (C i )) = T i for 1 ≤ i ≤ m.Since every C i is not facial in f 2 (G), it follows from Lemma 3 that every C i is non-contractible in f 2 (G).
We first consider the case when the surface F is homeomorphic to one of the sphere, the projective plane, and the torus.Suppose that F is the sphere.All cycles in G is contractible, and hence C = ∅.Actually, it follows Lemma 3 that f 1 (G) and f 2 (G) are essentially equivalent embeddings.(In general, Whitney [13] showed that every 3-connected planar graph has essentially unique embedding in the sphere.)Suppose that F is the projective plane.There is no pair of disjoint non-contractible cycles in f 2 (G), and hence m ≤ 1. Suppose that F is the torus.All non-contractible and pairwise disjoint cycles in G are pairwise homotopic.Then, all cycles in C are pairwise homotopic by Lemma 4, and hence it follows from Claim 6 that m ≤ 3.
Second, suppose that F is an orientable surface of genus at least two.If m > 9g − 9, then there are at least four pairwise homotopic cycles in C by Lemma 4, which contradicts Claim 6.Hence, we have m ≤ 9g − 9. Finally, suppose that F is a non-orientable surface of genus at least two.If m > 7g − 9, then there are at least 6g − 8 2-sided cycles in C, and hence some four of them are pairwise homotopic by Lemma 5, which contradicts Claim 6.Therefore, in any case, the desired inequality holds.

Facial complete colorings of non-simple graphs
In this section, we consider graphs which may have multiple edges.We denote by K n the complete graph of order n, and denote by K m n the non-simple graph obtained from K n by replacing each edge with m multiple edges.
The first author [3] constructed two triangulations f 1 (G) and f 2 (G) obtained from the graph G = K 6m−1 12m on a surface for any positive integer m.The weak chromatic numbers of these triangulations differ by at least 2m, and hence his construction gives an affirmatively answer of Kündgen and Ramamurthi's conjecture [6, Conjecture 8.1] (see also Section 2 in this paper).We now show that the facial 3-achromatic numbers of these triangulations also differ.
For details of constructions of f 1 (G) and f 2 (G), see [3,Section 3].The facehypergraph H(f 1 (G)) of f 1 (G) is isomorphic to a complete 3-uniform hypergraph.That is, the triangulation f 1 (G) has exactly |V (G)| 3 faces and there is a face bounded by each triple of vertices.(Such a triangulation is called complete, whose notion was defined in [6].)Then it is easy to see that ψ 3 (f 1 (G)) = |V (G)| = 12m.
Let T be a triangulation on a surface obtained from K 12m (by Ringel's Map Color Theorem [12], K 12m has a triangulation on a surface).The edge-set of H(f 2 (G)) coincides with that of H(T ) by ignoring the multiplicity of the edge-sets.It implies that ψ 3 (f 2 (G)) = ψ 3 (T ).Suppose that T is facially 3-complete k-colorable.Then, T must have at least k 3 faces, and hence we obtain the following inequality: Then, ψ 3 (f 2 (G)) ≤ 7m + 2 (this bound might be loose), and hence we have Since G is isomorphic to K 6m−1 12m , both of two triangulations f 1 (G) and f 2 (G) are embedded on a surface of Euler genus (m − 1)(m − 2)(2m + 3)/3.It implies that for

Fig. 1 :
Fig. 1: Two embeddings of G on the sphere.

Claim 6 .
There are at most three pairwise homotopic cycles of C in f 2 (G).Proof.Suppose that C 1 , C 2 , C 3 and C 4 are pairwise homotopic in f 2 (G), and appear on the annulus bounded by C 1 and C 4 in this order.Thus, the union C 2 ∪ C 4 separates C 1 from C 3 , and hence there are no chords of C 1 ∪ C 3 .Similarly, C 1 ∪ C 3 also separates C 2 from C 4 .It implies that there are at least two C 1 ∪ C 3 -bridges in G. On the other hand, since both of C 1 and C 3 are facial in f 1 (G) and C 1 ∪ C 3 has no chord, it follows from Lemma 2 that there is only one C 1 ∪ C 3 -bridge in G, a contradiction.Therefore, there are at most three pairwise homotopic cycles of C in f 2 (G).Now we shall give the upper bound for |T | = m, which induces the upper bound for |ψ 3