The problem of deciding whether an arbitrary graph G has a homomorphism into a given graph H has been widely studied and has turned out to be very difficult. Hell and Nešetril proved that the decision problem is NP-complete unless H is bipartite. We consider a restricted problem where G is an arbitrary triangle-free hexagonal graph and H is a Kneser graph or its induced subgraph. We give an explicit construction which proves that any triangle-free hexagonal graph has a homomorphism into one-vertex deleted Petersen graph

Janez Žerovnik

Petra Šparl

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MATHEMATICAL COMMUNICATIONS 391Math. Commun., Vol. 14, No. 2, pp. 391-398 (2009)Explicit homomorphisms of hexagonal graphs to one vertexdeleted Petersen graph∗Petra Sˇparl1 and Janez Zˇerovnik2,†1 Faculty of Organizational Sciences, University of Maribor, Kidricˇeva cesta 55a, SI-4 000Kranj, Slovenia2 Faculty of Mechanical Engineering, University of Ljubljana, Asˇkercˇeva 6, SI-1 000Ljubljana, SloveniaReceived March 28, 2009; accepted October 17, 2009Abstract. The problem of deciding whether an arbitrary graph G has a homomorphisminto a given graph H has been widely studied and has turned out to be very difficult.Hell and Nesˇetril proved that the decision problem is NP-complete unless H is bipartite.We consider a restricted problem where G is an arbitrary triangle-free hexagonal graphand H is a Kneser graph or its induced subgraph. We give an explicit construction whichproves that any triangle-free hexagonal graph has a homomorphism into one-vertex deletedPetersen graph.AMS subject classifications: 05C60, 05C15, 05C85, 94C15, 68R10Key words: homomorphism, H-coloring, triangle-free hexagonal graph1. IntroductionGraph homomorphisms in the current sense were first studied by Sabidussi in thelate fifties and early sixties. This was followed by much activity in different areas, see[3] and the references therein. One of the interesting questions is to study whetheran arbitrary graph G has a homomorphism into a fixed graph H. Since an n-coloringof a graph G is a homomorphism of G to Kn, the term H-coloring of G has beenemployed to describe the existence of a homomorphism of a graph G into a graph H.In such a case the graph G is said to be H-colorable. Many authors have studied thecomplexity of the H-coloring problem. A key result was given by Hell and Nesˇetrilin 1990 [7]. They proved that the H-coloring problem is NP-complete, if H is anon-bipartite graph and polynomial otherwise, assuming P6=NP. Several restrictionsof the H-coloring problem have been studied in [3]. The case when the input graphG is restricted to have a degree bounded by a small constant is investigated in [1].An n-[k] coloring of a graph G is a multicoloring of vertices of G, where everyvertex v ∈ G is assigned a subset of k different colors from the set of n colors, suchthat subsets assigned to any two adjacent vertices are disjoint. If there exists ann-[k]coloring of G, we say that G is n-[k] colorable. In terms of homomorphisms,∗Work is in part supported by ARRS, research agency of Slovenia and authors are part timeresearchers at the Institute of Mathematics, Physics and Mechanics, Ljubljana, Slovenia.†Corresponding author. Email addresses: petra.sparl@fov.uni-mb.si (P. Sˇparl),janez.zerovnik@imfm.uni-lj.si (J. Zˇerovnik)http://www.mathos.hr/mc c©2009 Department of Mathematics, University of Osijek392 P. Sˇparl and J. Zˇerovnikan n-[k]coloring is equivalent to a homomorphism to the Kneser graph K(n, k) [3].A motivation for interest in n-[k]colorings of hexagonal graphs comes from its rela-tion to frequency assignment problems. A fundamental problem concerning cellularnetworks is to assign sets of frequencies (colors) to transmitters (vertices) in orderto avoid unacceptable interferences [4]. The number of frequencies demanded at atransmitter may vary between transmitters. In a usual cellular model, transmittersare centers of hexagonal cells and the corresponding adjacency graph is a subgraphof the infinite triangular lattice. An integer p(v) is assigned to each vertex of thetriangular lattice and it is called the demand of the vertex v. If the demand isuniform, p(v) = k and the set of available frequencies is n, then the problem isequivalent to the problem of the existence of an n-[k]coloring. For more detailssee [11]. Namely, the existence of an algorithm for n-[k]coloring of a hexagonalgraph implies the existence of an algorithm for multicoloring of a weighted hexag-onal graph with competitive ratio n2k . For more details see [6]. Several results onn-[k]colorings of hexagonal graphs have been given recently. In [5] authors provedthat every triangle-free hexagonal graph is 5-[2]colorable, indicating the existenceof a distributed algorithm to find such a coloring. In [15] a 2-local algorithm for5-[2]coloring was given. The 5-[2]colorability also follows from the fact that anytriangle-free hexagonal graph is C5-colorable [14]. The existence of 7-[3]coloring ofany triangle-free hexagonal graph is proved in [2, 5]. Reference [13] gives an ele-gant idea that implies the existence of 14-[6]colorings. However, it is unclear howto design an algorithm for 7-[3]coloring from these proofs. McDiarmid and Reedconjectured that every triangle-free hexagonal graph is 9-[4]colorable [12], and theconjecture is still open. It is clear that the question for the existence of (2k + 1)-[k]colorings has in general a negative answer for k > 4. On the other hand, planargraphs with large enough odd girth admit homomorphisms to Kneser graphs. Moreprecisely, a planar graph with girth at least 10k − 7 has a homomorphism to theKneser graph K(2k+1, k) [9]. Thus, for example, if the odd girth of a planar graphis at least 13, then this implies that the graph has a homomorphism to the Petersengraph K(5, 2). On the other hand, it is known that planar graphs with odd girth2k+1 that cannot be (2k + 1)-[k]coloured exist for arbitrarily large k [16].In this note we investigate a restricted H-coloring problem, where H is a Knesergraph K(n, k) and G an arbitrary so-called hexagonal graph, which is an inducedsubgraph of a triangular lattice. We will consider only triangle-free hexagonalgraphs, because a hexagonal graph which contains a triangle is obviously notK(5, 2)-colorable, since graph K(5, 2) does not contain a triangle as a subgraph. Theabove mentioned results on n-[k]colorings of hexagonal graphs in terms of K(n, k)-colorability are the following. An arbitrary triangle-free hexagonal graph is K(5, 2)-colorable by [6, 15], K(7, 3)-colorable by [2, 5] and K(14, 6)-colorable by [13]. Fi-nally, McDiarmid and Reed conjectured that every triangle-free hexagonal graph isK(9, 4)-colorable.Recall that the existence of an explicit homomorphism gives rise to an efficientalgorithm, see [14]. While explicit homomorphisms and algorithms are known forK(5, 2) [6, 15, 14], there are only existence proofs without construction for K(7, 3)[5, 13].We will focus on the case where G is a triangle-free hexagonal graph and HExplicit homomorphisms of hexagonal graphs 393is the one-vertex deleted induced subgraph of K(5, 2), also known as the Petersengraph. As it is well known that the Petersen graph is vertex transitive, all one-vertex deleted subgraphs are isomorphic. We will use the notation Kx(5, 2) for theone vertex deleted subgraphs of the Petersen graph.In this paper we proveTheorem 1. Any triangle-free hexagonal graph is Kx(5, 2)-colorable. There is ahomomorphism that only depends on the embedding of the graph into the infinitelattice and can be computed in linear time.More precisely, assuming that we are given the coordinates of the vertices in thetriangular lattice, the homomorphism is given as a closed expression. We sketch anidea how to evaluate the expressions for all vertices in linear time. This result is oftheoretical interest because it is an improvement of [14]; however, a true challengeis to construct an explicit homomorphism for the K(7, 3) case.A straightforward corollary of Theorem 1 is K(5, 2)-colorability, a result whichalso follows directly from [5, 6, 15].The rest of the paper is organized as follows. In Section 2 some basic definitionsand results are recalled. In Subsection 3.1 a construction for finding a homomor-phism of any triangle-free hexagonal graph to Kneser graph Kx(5, 2) is given, andin Subsection 3.2 the correctness of one construction is proved. In Section 4 someremarks and open problems are given.2. PreliminariesLet G and H be graphs. A function ϕ : G→ H is a homomorphism from G to H ifit is an adjacency preserving mapping from V (G) to V (H), namely a mapping forwhich [ϕu, ϕv] ∈ E(H) whenever [u, v] ∈ E(G). We write simply ϕ : G → H andu 7→ ϕ(u).The path of length n will be denoted by Pn = (v0, v1, v2, ..., vn). Let u and v betwo vertices of a graph G. The distance between u and v in G is the length of theshortest path from u to v in G which will be denoted as distG(u, v).Given two positive integers n and k, Kneser graph K(n, k) is the graph whosevertices represent the k-subsets of the set {1, ..., n}, and where two vertices are con-nected if and only if they correspond to disjoint subsets. GraphK(n, k) therefore has(nk)vertices and is regular of degree(n−kk). It is obvious that Kneser graph K(n, k)is connected for n > 2k. Note that the well known Petersen graph is isomorphicto Kneser graph K(5, 2), see Figure 1. Later we will give a homomorphism to theone-vertex deleted subgraph in which the vertex corresponding to the set {1, 2} isdeleted.Recall that hexagonal graphs are induced subgraphs of a triangular lattice. Ver-tices of a triangular lattice may be described as follows. The position of each vertexis an integer linear combination x~p + y~q of two vectors ~p = (1, 0) and ~q = ( 12 ,√32 ).Thus, we may identify the vertices of a triangular grid with pairs (x, y) of integers.Two vertices are adjacent when the Euclidean distance between them is one. There-fore each vertex (x, y) has six neighbors (x ± 1, y), (x, y ± 1), (x + 1, y − 1) and(x−1, y+1). For simplicity, we will refer to the neighbors as R (right), L (left), UR394 P. Sˇparl and J. Zˇerovnik(up-right), DL (down-left), DR (down-right) and UL (up-left), respectively. There isan obvious 3-coloring of the infinite triangular lattice which gives rise to a partition{1,2}{3,4} {4,5}{1,3}{2,5}{1,4} {2,4}{1,5} {2,3}{3,5}Figure 1. Petersen graph P ∼= K(5, 2)of a vertex set of any triangular lattice graph into three independent sets I0, I1and I2. The partition can be decided from the coordinates by the rule: a vertexwith coordinates (x, y) is in the independent set Ii where i(x, y) = x + 2y (mod3). According to the partition I0, I1 and I2, vertices are assigned their base colors,which are denoted by R (red), B (blue) and G (green), respectively. Figure 2 showsan example of a triangle-free hexagonal graph with 3-coloring.R BBGGBRG R BRBGGRGBGRGBRFigure 2. RBG-coloring of a triangle-free hexagonal graphLet us define two integer functions, B(x, y) = 5− i(x, y) and f1(v) = B(x(v), y(v)),defined on the triangular grid and on the vertices of a hexagonal graph, respectively.The image of both functions is the set {3, 4, 5}. More precisely,- if the base color of v is red, then f1(v) = 5,- if the base color of v is blue, then f1(v) = 4,- if the base color of v is green, then f1(v) = 3.Explicit homomorphisms of hexagonal graphs 3953. The main resultIn this section an explicit construction of a homomorphism from an arbitrary triangle-free hexagonal graph to one vertex deleted Petersen graph Kx(5, 2) is given.Let G be a hexagonal graph without triangles. Recall that for each vertex v, itsposition in the triangular grid is given by coordinates denoted by (x(v), y(v)).A vertex v ∈ G is said to be suitable if it has neither L, nor RD nor RU neighborin the triangular lattice. All eight possible suitable vertices are presented in Figure3, where suitable vertices are colored black. It is clear that the set of suitable verticesis independent. In the sequel the set of all suitable vertices of a graph G will bedenoted as SG. The intuition behind the partition of the vertex set to suitable andnot suitable vertices is to break odd cycles. It is essential that unsuitable verticesinduce bipartite subgraph whose connected components have a special structure, seethe definition of tristars below.Figure 3. All possible suitable verticesA path (x0, x1, ..., xn) is left (resp. rightup, rightdown) if xi+1 is the L (resp. RU,RD) neighbor of xi, for i = 0, 1, ..., n − 1, in a triangular lattice. A tristar is theunion of one left, one rightup and one rightdown path emerging from a commonorigin x, named a center of a tristar or shortly a center. We allow that one, two oreven all three paths of a tristar are of length zero. In the second and third case thetristar is isomorphic to a path and to an isolated vertex, respectively. An exampleof a tristar with center x is depicted in Figure 4.xFigure 4. A tristar3.1. The constructionThe input of the algorithm is a triangle-free hexagonal graph where each vertexknows its base color. The output of the algorithm is a homomorphism F : V (G)→V (Kx(5, 2)). Homomorphism F is the assignment F (v) = {f1(v), f2(v)}, wheref1(v) is defined at the end of Section 2 and f2(v) : V (G) → {1, 2} is given by thefollowing rules:396 P. Sˇparl and J. Zˇerovnik1. if v is a suitable vertex, thenf2(v) = B(x(v), y(v) + 1).2. if v is not a suitable vertex, i.e. v ∈ G\SG, thenf2(v) = distG(u, x) mod 2 + 1,where distG(u, x) is the distance of v from the center x of the tristar.Note that the only connected components of a graph G\SG are tristars. Therefore,every vertex v ∈ V (G)\SG belongs to exactly one tristar Tv of a graph G\SG, withexactly one center x ∈ Tv, which means that the distance distG(u, x) is well defined.3.2. Correctness of the constructionRecall that the only connected components of V (G)\SG are tristars (where pathsand isolated vertices are special cases of a tristar).Remark 1. Note that by the definition of base colors (see page 394) the base colorsof neighbors of suitable vertices are fixed. Therefore, we have only three differentpossibilities, which are shown in Figure 4.RRRRB BBBGGGGFigure 4. Base colors of a suitable vertex and its neighborsWe have to show that for an arbitrary triangle-free hexagonal graph G the mappingF : G → V (Kx(5, 2)) is a homomorphism. Therefore, we have to show that for anarbitrary edge uv ∈ E(G) it holds F (u)F (v) ∈ E(Kx(5, 2)).Let uv be an edge in a graph G. Since suitable vertices are independent, we haveonly two different cases. One of the vertices is suitable or none of the vertices issuitable.Case 1: Without loss of generality, let u ∈ SG and v /∈ SG. By Remark 1 and Figure4 we have three different possibilities.- If the base color of u is red, then the base color of v is blue and u and v are mappedas follows: F (u) = {5, 3}, F (v) = {4, f2(v)}.- If the base color of u is blue, then the base color of v is green and u and v aremapped as follows: F (u) = {4, 5}, F (v) = {3, f2(v)}.- If the base color of u is green, then the base color of v is red and u and v aremapped as follows: F (u) = {3, 4}, F (v) = {5, f2(v)}.Since f2(v) ∈ {1, 2}, it holds F (u) ∩ F (v) = ∅ for all three possibilities, which isExplicit homomorphisms of hexagonal graphs 397equivalent to F (u)F (v) ∈ E(K(5, 2)).Case 2: Let u, v /∈ SG. Vertices u and v have different base colors, since they areneighbors. This yields f1(u) 6= f1(v). Let Tv = Tu be a tristar containing u and v.Let x be a center of a tristar Tv. Two different cases are possible.- One of the vertices u and v is a center, let u = x. In this case we have distG(u, x) =0, which implies f2(u) = 1 and distG(v, x) = 1, which implies f2(v) = 2.- None of the vertices u and v is a center (i.e. u 6= x and v 6= x). In this case itholds |distG(v, x)− distG(u, x)| = 1 and f2(u) 6= f2(v).So, we have again F (u) ∩ F (v) = ∅, which implies F (u)F (v) ∈ E(K(5, 2)).Finally, note that F (v) = {1, 2} never occurs in the construction, which meansF (u)F (v) ∈ E(Kx(5, 2)) in both cases. Hence, we haveProposition 1. Let G be a triangle-free hexagonal graph. Then there exists a ho-momorphism F : V (G)→ V (Kx(5, 2)).4. Towards an algorithmA straightforward implementation of the construction runs in linear time. Roughlyspeaking, in the first round one can compute f1(v) for all the vertices and decidefor each vertex whether it is suitable or not and whether it is a center of tristaror not. In the second round, one can assign the colors f2(v) easily if the verticesare visited in the following order: suitable vertices and centers first, then the othervertices starting from the tristar centers.Hence the construction given can be implemented in linear time as stated inTheorem 1.Unfortunately, the construction sketched above does not give a distributed algo-rithm, since there may be arbitrary long segments of tristars in the graph G\SG,where vertices have to know the distance to the center of a tristar. In practice, itis very important that an algorithm is distributed. In particular, it is desirable thatthe communication paths are short. An algorithm is called k-local if the computa-tion at a vertex depends only on information at vertices of distance at most k [8].Reference [15] gives a 2-local distributed algorithm for 5-[2]coloring that implies theexistence of a 2-local distributed algorithm for K(5, 2)-coloring. Hence it would beinteresting to see if for reasonably small k a k-local algorithm for Kx(5, 2)-coloringof triangle-free hexagonal graphs is possible to design.Moreover, the existence of an algorithm for K(7, 3)-coloring and K(9, 4)-coloringof an arbitrary triangle-free hexagonal graph and its best locality is still left as anopen problem.AcknowledgementThe authors wish to thank the anonymous referees for their constructive remarks.398 P. Sˇparl and J. ZˇerovnikReferences[1] A.Galaccio, P.Hell, J. 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Explicit homomorphisms of hexagonal graphs to one vertex deleted Petersen graph

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