On the chromatic numbers of signed triangular and hexagonal grids
OOn the chromatic numbers of signed triangular andhexagonal grids
Fabien JacquesLIRMM, University of Montpellier, CNRS, Montpellier, France
Abstract
A signed graph is a simple graph with two types of edges. Switching a vertex v of a signedgraph corresponds to changing the type of each edge incident to v .A homomorphism from a signed graph G to another signed graph H is a mapping ϕ : V ( G ) → V ( H ) such that, after switching any number of the vertices of G , ϕ maps every edgeof G to an edge of the same type in H . The chromatic number χ s ( G ) of a signed graph G isthe order of a smallest signed graph H such that there is a homomorphism from G to H .We show that the chromatic number of signed triangular grids is at most 10 and thechromatic number of signed hexagonal grids is at most 4. A or a signed graph G = ( V, E, s ) is a simple graph (
V, E ) with two kindsof edges: positive and negative edges. We do not allow parallel edges nor loops. The signature s : E ( G ) → {− , +1 } assigns to each edge its sign. For the concepts discussed in this article,2-edge-colored graphs and signed graphs only differ on the notion of homomorphisms.Given two 2-edge-colored graphs G and H , the mapping ϕ : V ( G ) → V ( H ) is a homomorphism if ϕ maps every edge of G to an edge of H with the same sign. This can be seen as coloring thegraph G by using the vertices of H as colors. The target graph H gives us the rules that thiscoloring must follow. If vertices 1 and 2 of H are adjacent with a positive (resp. negative) edge,then every pair of adjacent vertices in G colored with 1 and 2 must be adjacent with a positive(resp. negative) edge. Switching a vertex v of a 2-edge-colored or signed graph corresponds to reversing the signs ofall the edges that are incident to v .Given two signed graphs G and H , the mapping ϕ : V ( G ) → V ( H ) is a homomorphism ifthere is a homomorphism from G to H after switching some subset of the vertices of G and/orswitching some subset of the vertices of H . However, switching in H is unnecessary (as explainedin Section 3.3 of [1]).The chromatic number χ s ( G ) of a signed graph G is the order of a smallest signed graph H such that G admits a homomorphism to H . The chromatic number χ s ( C ) of a class of signedgraphs C is the maximum of the chromatic numbers of the graphs in the class.Homomorphisms of signed graphs were introduced by Naserasr, Rollov´a and Sopena [1]. Thistype of homomorphism allows us to generalize several classical problems such has Hadwiger’sconjecture [1, 2] and have therefore been studied by a great number of authors. Here are severalknown results on the chromatic number of some classes of signed graphs that are related to theclasses we study in this article. Theorem 1.
1. The chromatic number of signed planar graphs is at most 40 [3].2. The chromatic number of signed planar graphs with girth at least 6 is at most 6 [4]. a r X i v : . [ m a t h . C O ] D ec . The chromatic number of signed graphs with maximum degree 3 is at most 7 [5].4. The chromatic number of signed graphs with maximum average degree less than 3 is at most6 [5].5. The chromatic number of signed square grids is at most 6 [6]. In Section 2 we present our results on the chromatic number of hexagonal and triangular gridsand in Section 3 we introduce the target graphs that we use in Sections 4 and 5 to prove theseresults.
A square (resp. triangular, hexagonal) grid is a finite induced subgraph of the graph associatedwith the tiling of the plane with squares (resp. equilateral triangles, equilateral hexagons). SeeFigures 4 and 5. Since signed hexagonal grids have maximum degree 3 and therefore maximumaverage degree at most 3, we already know that their chromatic number is at most 6 by Theo-rem 1(3) or by Theorem 1(4). Moreover, signed triangular grids are planar and have thereforechromatic number at most 40 by Theorem 1(1). We improve these bounds as follows.
Theorem 2.
The chromatic number of signed hexagonal grids is 4.
Theorem 3.
The chromatic number of signed triangular grids is at most 10.
In order to prove these theorems, we will show that every signed hexagonal grid admits ahomomorphism to a target graph of order 4 we call T (see Figure 1) and that every signedtriangular grid admits a homomorphism to a target graph of order 10 called SP +9 . Constructionsof T and SP +9 are explained in Section 3. Note that it is conjectured that every signed planar graphadmits a homomorphism to SP +9 [7]. Theorem 3 brings further evidence toward this conjecture. A 2-edge-colored graph (
V, E, s ) is said to be antiautomorphic if it is isomorphic to (
V, E, − s ).A 2-edge-colored graph G = ( V, E, s ) is said to be K n -transitive if for every pair of cliques { u , u , . . . , u n } and { v , v , . . . , v n } in G such that s ( u i u j ) = s ( v i v j ) for all i (cid:54) = j , there exists anautomorphism that maps u i to v i for all i . For n = 1 or 2, we say that the graph is vertex-transitive or edge-transitive , respectively.A 2-edge-colored graph G has Property P k,n if for every sequence of k distinct vertices ( v , v , . . . , v k )that induces a clique in G and for every sign vector ( α , α , ..., α k ) ∈ {− , +1 } k there exist atleast n distinct vertices { u , u , ..., u n } such that s ( v i u j ) = α i for 1 ≤ i ≤ k and 1 ≤ j ≤ n .41 23Figure 1: The graph T .2 x x +1 x +22 x x +1 2 x +2 Figure 2: The graph SP , non-edges are negative edges.The SP has vertex set V ( SP ) = F , the field of order 9. Two vertices u and v ∈ V ( SP ), u (cid:54) = v , are connected witha positive edge if u − v is a square in F and with a negative edge otherwise. Notice that thisdefinition is consistent because 9 ≡ − F and if u − v is a squarethen v − u is also a square.Given a 2-edge-colored graph G with signature s G , we create the antitwinned graph of G denoted by ρ ( G ) as follows:Let G +1 , G − be two copies of G . The vertex corresponding to v ∈ V ( G ) in G i is denotedby v i . • V ( ρ ( G )) = V ( G +1 ) ∪ V ( G − ) • E ( ρ ( G )) = { u i v j : uv ∈ E ( G ) , i, j ∈ {− , +1 }} • s ρ ( G ) ( u i v j ) = i × j × s G ( u, v )By construction, for every vertex v of G , v − and v +1 are antitwins , the positive neighbors of v − are the negative neighbors of v +1 and vice versa. A 2-edge-colored graph is antitwinned ifevery vertex has a unique antitwin.When coloring a 2-edge-colored graph with an antitwinned graph, we say that two verticeshave the same identity if they are mapped to the same vertex or vertices that are antitwinned. Inan antitwinned 2-edge-colored graph we denote the antitwin of v with v . Lemma 4 ([8]) . Let G and H be 2-edge-colored graphs. The two following propositions areequivalent: • The 2-edge-colored graph G admits a homomorphism to ρ ( H ) . • The signed graph G (cid:48) defined by the 2-edge-colored graph G admits a homomorphism to thesigned graph H . Given a signed graph H , we define H + to be H with an added universal vertex ∞ that ispositively connected to all the other vertices (See Figure 3). We will use this construction tocreate the target graph SP +9 used in Section 5.We will use the following properties of our target graphs to prove our theorems. Lemma 5 ([9]) . The graph ρ ( SP +9 ) is vertex-transitive, edge-transitive, antiautomorphic and hasProperties P , , P , and P , . Lemma 6.
The graph ρ ( T ) has Property P , and the following Property that we call P ∗ , :Let u, v ∈ V ( ρ ( T )) such that u (cid:54) = v , u (cid:54) = v and { u, v } (cid:54) = { , } , { , } , { , } , { , } . Thereexist at least one vertex in ρ ( T ) that is a positive neighbor of both u and v . The last lemma can be checked by going through every pair of adjacent vertices in ρ ( T ).3 ∞ Figure 3: The construction of H + . In this section, we prove that the chromatic number of signed hexagonal grids equals 4. To getthis result, we first prove that 2-edge-colored hexagonal grids admit a homomorphism to the2-edge-colored graph ρ ( T ). Lemma 4 will allow us to conclude. Lemma 7.
Every 2-edge-colored hexagonal grid admits a homomorphism to the 2-edge-coloredgraph ρ ( T ) .Proof. Let G be a 2-edge-colored hexagonal grid and s be its signature. We want to show that G admits a homomorphism to ρ ( T ). We give two coordinates i and j to each of the vertices of thehexagonal grid according to Figure 4. These coordinates allow us to create an order on the vertexset of G by saying that v i,j < v k,l if i < k or i = k, j < l . v , v , v , v , v , v , v , v , v , v , v , v , v , v , v , v , v , v , v , v , Figure 4: A hexagonal grid.Let A be a 2-edge-colored graph that admits a homomorphism ϕ to B and let A (cid:48) be the graphobtained after switching A at a vertex v . By Lemma 4, A (cid:48) admits a homomorphism ϕ (cid:48) to I . Itsuffices to take ϕ (cid:48) as follows: ϕ (cid:48) ( u ) = ϕ ( u ) if u (cid:54) = vϕ ( v ) otherwiseTherefore we can, without loss of generality, start by switching G at several vertices such thatevery edge v i,j v k,l with i = k and j + 1 = l or i + 1 = k and j = l are positive (these edges arethicker in Figure 4).To do that, for every vertex v i,j such that i + j = 1 mod 2 and i ≥ • If s ( v i,j v i − ,j ) = s ( v i − ,j v i − ,j +1 ) = −
1, we switch G at v i − ,j . • If s ( v i,j v i − ,j ) = 1 and s ( v i − ,j v i − ,j +1 ) = −
1, we switch G at v i,j and v i − ,j .4 If s ( v i,j v i − ,j ) = − s ( v i − ,j v i − ,j +1 ) = 1, we switch G at v i,j .We now create a homomorphism ϕ from G to ρ ( T ) by coloring each vertex in the order definedearlier. We partition the vertices of G into two sets V and V . In V we put every vertex v i,j suchthat i + j ≡ V we put all the other vertices.When coloring a vertex v i,j in V , such a vertex is adjacent to one already colored vertex(unless i = 1 in which case it is trivial to color v i,j ). Therefore, Property P , of ρ ( T ) tells us thatthere are at least 3 possible colors for v i,j with respect to its already colored neighbor. Note thatamong these 3 available colors, there is no antitwins. Thanks to this remark, it is always possibleto choose one color as follows: • If ϕ ( v i − ,j +1 ) = 1 or 1, ϕ ( v i,j ) / ∈ { , , , } ; • If ϕ ( v i − ,j +1 ) = 4 or 4, ϕ ( v i,j ) / ∈ { , , , } ; • If ϕ ( v i − ,j +1 ) = 2 or 2, ϕ ( v i,j ) / ∈ { , , , } ; • If ϕ ( v i − ,j +1 ) = 3 or 3, ϕ ( v i,j ) / ∈ { , , , } ;When coloring a vertex v i,j in V , such a vertex is adjacent to two already colored vertex(unless i = 1 or j = 1 in which cases it is trivial to color v i,j ). Vertex v i − ,j belongs to V and wecan therefore use P ∗ , , thanks to the restrictions on ϕ ( v i − ,j ) defined earlier, to find a color foreach vertex in V .We use Lemmas 4 and 7 to prove that the chromatic number of signed hexagonal grids is atmost 4. Note that a cycle on 6 vertices with exactly one negative edges needs at least 4 color tobe colored [10]. Therefore, the chromatic number of signed hexagonal grids is 4. In this section, we prove that the chromatic number of signed triangular grids is at most 10. Toget this result, we first prove that 2-edge-colored triangular grids admit a homomorphism to the2-edge-colored graph ρ ( SP +9 ). Lemma 4 will allow us to conclude. Lemma 8.
Every 2-edge-colored triangular grid admits a homomorphism to the 2-edge-coloredgraph ρ ( SP +9 ) . v v v v n u u u u n Figure 5: A triangular grid.
Proof.
Remember that SP +9 is SP with an added vertex ∞ that is positively adjacent to everyother vertex. Let G be a 2-edge-colored triangular grid and s be the signature of G . We proceedby induction on the horizontal rows of G as depicted in Figure 5. Note that the first row of G is trivial to color. Let G (cid:48) be G without the last row. By the induction hypothesis, there is a5omomorphism ϕ (cid:48) from G (cid:48) to ρ ( SP +9 ). We now show that we can extend this homomorphism toa homomorphism ϕ from the whole graph G to ρ ( SP +9 ).Let v , v , ..., v n be the vertices of the last row of G and u , u , ..., u n be the vertices of thesecond to last row of G (the last row of G (cid:48) ). See Figure 5.By Property P , of ρ ( SP +9 ), we can find two colors (or even four but we only need two) tocolor v with respect to its already colored neighbors ( u and u ). Note that we do not take thecolor of u into account (yet).Without loss of generality, we can assume that u u is a positive edge, ϕ (cid:48) ( u ) = 0 and ϕ (cid:48) ( u ) = 1 because ρ ( SP +9 ) is edge-transitive and antiautomorphic.Suppose s ( u v ) = s ( u v ) = s ( u v ) = s ( v v ) = +1. The four colors available for v byProperty P , of ρ ( SP +9 ) with respect to the colors of u and u are 2 , ∞ , x + 2 and 2 x + 2.Since ϕ (cid:48) ( u ) = 0 and s ( u v ) = +1, the two colors available for v belong to the set: { , , x, x + 1 , x + 2 , x, x + 1 , x + 2 , ∞} If v is colored in 1, v can be colored in 2 , ∞ , x + 2 or 2 x + 2.If v is colored in 2, v can be colored in ∞ .If v is colored in x , v can be colored in ∞ or x + 2.If v is colored in x + 1, v can be colored in 2 or x + 2.If v is colored in x + 2, v can be colored in 2 x + 2.If v is colored in 2 x , v can be colored in ∞ or x + 2.If v is colored in 2 x + 1, v can be colored in 2 or 2 x + 2.If v is colored in 2 x + 2, v can be colored in x + 2.If v is colored in ∞ , v can be colored in 2.We can now see that any pair of vertices in { , , x, x + 1 , x + 2 , x, x + 1 , x + 2 , ∞} allows v to be colored in at least two colors.We can proceed in a similar manner with the following three cases and arrive to the sameconclusion: • s ( u v ) = − s ( u v ) = s ( u v ) = s ( v v ) = +1, • s ( u v ) = +1, s ( u v ) = − s ( u v ) = s ( v v ) = +1, • s ( u v ) = s ( u v ) = +1, s ( u v ) = − s ( v v ) = +1.By Lemma 4, each of these four cases also accounts for 3 other cases: the signature obtainedafter switching at v , v and both v and v . We have therefore covered all 16 (2 ) possiblesignatures of u v , u v , u v and v v .Therefore, v can be colored in at least 2 colors. Similarly, we can find at least two colors for v and so on until v n . Finally, we can arbitrarily choose one of these two colors for v n , accordinglychoose a color for v n − and so on to get a homomorphism ϕ from G to ρ ( SP +9 ).We conclude the proof of Theorem 3 by using Lemmas 8 and 4.We say that a cycle is unbalanced if it has an odd number of negative edges. Every C inthe signed triangular grid from Figure 6 is unbalanced and it can be colored with 6 colors (notethat the resulting target graph is SP +5 ). We can easily extend this construction to create a signedtriangular grid of any size such that every C is unbalanced and it can be colored with 6 colors.To do so, we can repeat a motif made of six vertices in all directions (the black vertices in Figure 6represent this motif). By Proposition 3.2 of [11], this means that every signed triangular grid suchthat every C is unbalanced can be colored with 6 colors. This is of particular interest becausewhen coloring an unbalanced C , every vertex in the cycle must have different identities (this is notthe case with a C that is not unbalanced) and therefore a graph in which every C is unbalancedis especially hard to color with few colors. Therefore, we conjecture the following:6 0 1 03 4 3 4 32 5 2 5 2 51 0 1 0 1 0 13 4 3 4 3 42 5 2 5 21 0 1 0Figure 6: A signed triangular grid in which every C is unbalanced colored with 6 colors. Conjecture 9.
The chromatic number of signed triangular grids is . Note that the chromatic number of signed triangular grids is at least 6 since a wheel on 7vertices such that every C is unbalanced cannot be colored with 5 colors. To prove it, let G be a2-edge-colored wheel with vertex set { u, u , ..., u } and center u , and let T be an antitwinned graphof vertex set { , , , , ..., , } where i is the antitwin of i such that G admits a homomorphism ϕ to T . By Proposition 3.2 of [11], it is possible to switch some of the vertices of G such thatthe outer cycle alternates between positive and negative edges and every edge incident to u ispositive. Assume w.l.o.g. that ϕ ( u ) = 0 (note that we can relabel the vertices of T if needed).We now show that we cannot color a pair of vertices ( v , v ) of G with antitwins. Suppose v and v have colors that are antitwinned. If v or v = u then we have a contradiction because v and v are adjacent. Otherwise, v and v are both positively adjacent to u which also givesus a contradiction. Therefore we can assume w.l.o.g. that we do not need to use colors 0 , , ..., ϕ ( u ) = 1, ϕ ( u ) = 2 and ϕ ( u ) = 3. We can collor u in either 1 or 4. Suppose wecolor u with 1. We have to color u in 4 and we then cannot color u . Suppose we color u with4. We can color u with 1 or 2 but both possibilities do not allow us to color u . Therefore G does not admit a homomorphism to T . We conclude by using Lemma 4. References [1] R. Naserasr, E. Rollov´a, and ´E. Sopena. Homomorphisms of signed graphs.
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