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[Math.CO] 28 Sep 2006 Distance-Residual Graphs Distance-residual graphs PrimoˇzLukˇsiˇc IMFM, University of Ljubljana, Ljubljana Slovenia [email protected] TomaˇzPisanski IMFM, University of Ljubljana, Ljubljana and University of Primorska, Koper Slovenia [email protected] September 15, 2006 Abstract If we are given a connected finite graph G and a subset of its vertices V0, we define a distance-residual graph as a graph induced on the set of vertices that have the maximal distance from V0. Some properties and examples of distance-residual graphs of vertex-transitive, edge-transitive, bipartite and semisymmetric graphs are shown. The relations between the distance-residual graphs of product graphs and their factors are shown. Keywords: distance-residual graphs, product graphs, vertex-transitive graphs, edge-transitive graphs, semisymmetric graphs, Gray graph Mathematics Subject Classification: 05C12 1 Introduction arXiv:math/0609810v1 [math.CO] 28 Sep 2006 Let G be a connected finite graph and let V0 ⊂ V (G) be a nonempty subset of vertices of G. We may form a distance partition P (G,V0) of G with respect to V0, where P (G,V0)= {V0,V1,...Vr} and V0 ∪ V1 ∪···∪ Vr = V (G), Vi ∩ Vj = ∅, for i =6 j, Vi =6 ∅, for all i. The classes Vi are defined recursively as Vi+1 := {v ∈ V (G) \ (V0 ∪ V1 ∪···∪ Vi) | ∃u ∈ Vi :[u, v] ∈ E(G)}. 1 The set Vi contains all vertices of G that have a minimal distance of i to the vertices of V0 where the distance d(u, v) between two vertices u and v is defined as the shortest path between them. Therefore d(v, vi) ≥ i for v ∈ V0 and vi ∈ Vi, and there exists a vertex v0 ∈ V0 for which d(v0, vi)= i. We are interested in induced subgraphs hVii defined by the distance classes, particu- larly in the subgraph with the lowest index RG := hV0i, which we call the root, and the subgraph with highest index Res(G, RG) := hVri, which we call the distance-residual ∼ graph or the distance residual. When the root consists of a single vertex, i.e. RG = K1, ∼ the residual is called a vertex residual. When RG = K2, the residual is called an edge residual. With respect to the definition of the distance residuals, all of the graphs in the paper will be nontrivial, simple, finite, and in most cases connected. Also some standard labels for some known graphs will be used: Kn for complete graphs, Km,n for complete bipartite graphs, Cn for cycles, Pn for paths, and mKn for a disjoint union of m complete graphs on n vertices. The motivation for the definition of distance-residual graphs was in extending the definition of distance sequence which is an ordered list where the i-th element equals the number of vertices at distance i from the selected root. The distance sequence therefore presents only the number of vertices at a distance i from the root but we are also interested in the induced subgraphs on those vertices, especially on the set farthest away from the root. In the next section we present some properties of the distance-residual graphs with the focus on the vertex- and edge-transitive graphs [1, 2], bipartite graphs, and semisym- metric graphs [8, 21]. In section 3 we show how the distance residuals of product graphs for some well-known products depend on the distance residuals of their factors. We con- clude with some open questions regarding distance residuals and other distance related problems. 2 Properties We can only define a distance-residual graph of a connected graph but any graph, connected or not, can be a distance-residual graph. Theorem 2.1 Let H be an arbitrary graph and n ∈ N. Then there exists a connected graph G with the root RG of order n such that H is isomorphic to Res(G, RG). Proof: Let us choose an arbitrary graph RG of order n with the property V (RG) ∩ V (H) = ∅. Graph G is constructed as follows: The vertex set V (G) consists of V (H) ∪ V (RG). There are two types of edges in G. All the original edges of H and RG remain edges in G and for each vertex v ∈ V (H) and each vertex r ∈ V (RG) there is an edge between them. Clearly, the distance partition is given by V0 = V (RG) and V1 = V (H), therefore Res(G, RG)= H. Most of the interesting cases occur for vertex-transitive and edge-transitive graphs. 2 2.1 Vertex- and edge-transitive graphs Lemma 2.1 Let G be a connected vertex-transitive graph. Then all of its vertex residu- als are isomorphic, i.e. they do not depend on the choice of the vertex for the root. The converse is not true even if the graph is regular. Proof: The first part is obvious because if the vertex residuals would not be isomorphic, the graph would not have a transitive automorphism group. The graph in Figure 1 proves that the converse is not true. It is built from two copies of K4 by joining some of their their edges (see [16] for a definition of joining) and is therefore 3-regular. All of its vertex residuals are isomorphic (to K1) but it is not vertex-transitive because the automorphism, which would map a vertex from K4 to a vertex created by joining edges, does not exist. Figure 1: Non-vertex-transitive cubic graph with isomorphic vertex residuals. Vertex residual is also connected to the growth of the graph (see [30]) because its order is the leading coefficient of the growth polynomial of the graph at its root. This is equivalent to taking the last element of the distance sequence of the root. We know that there exist graphs that are growth-regular, i.e. their growth function is independent of the root vertex, but not vertex-transitive. From the above proof we can also see that such graphs can have isomorphic vertex residuals and still not be vertex- transitive. The growth function of the graph in Figure 1 is namely 1+3x+6x2 +5x3 +x4 independent of the vertex we take for the root. Lemma 2.2 Let G be a connected edge-transitive graph. Then all of its edge residuals are isomorphic, i.e. they do not depend on the choice of the two adjacent vertices for the root. The converse is not true. Proof: The first part is similar as in the prior lemma. As a counterexample of the converse we present the graph in Figure 2 that has all of its edge residuals isomorphic (to K1) but it is clearly not edge-transitive. 3 Figure 2: Non-edge-transitive graph with isomorphic edge residuals. Lemma 2.3 Let G be a connected edge-transitive graph and L(G) its line graph. Then all of the distance residuals Res(L(G),K1) are isomorphic. Proof: The line graph of a connected edge-transitive graph is a connected vertex- transitive graph. The rest follows from Lemma 2.1. We present the distance residuals of some well-known graphs. Example 2.1 ∼ ∼ Res(C2k,K1) = K1 Res(C2k,K2) = K2 ∼ ∼ Res(C2k+1,K1) = K2 Res(C2k+1,K2) = K1 ∼ ∼ Res(Kn,K1) = Kn−1 Res(Kn,Kk) = Kn−k Example 2.2 The Petersen graph P (5, 2) is vertex-transitive, edge-transitive and also distance-transitive, therefore it has isomorphic distance residuals for some roots: ∼ Res(P (5, 2),K1) = C6, ∼ Res(P (5, 2),K2) = 2K2, ∼ Res(P (5, 2),P3) = 2K1, ∼ Res(P (5, 2),P4) = K1, ∼ Res(P (5, 2),C5) = C5. 2.2 Bipartite graphs For bipartite graphs a special relation between vertex and edge residuals holds. But before we present it, we shall define the distance between vertices and subgraphs. Let d(R, v) be the distance from R ⊂ G to vertex v ∈ V (G), i.e. d(R, v) := minr∈V (R) d(r, v), and let d(R,H) := minv∈V (H) d(R, v) be the distance between two subgraphs of graph G. If graph G is connected, the distances are well defined. We will often need the distance between the root and the distance residual of some connected graph G, so we will denote d(RG,Res(G, RG)) as dRG . The distance namely depends on the root of the graph. 4 In the next theorem we use the above notation to shorten the writing for the distance d(h{u}i,Res(G, h{u}i)) to du and similarly to shorten d(h{u, v}i,Res(G, h{u, v}i)) to du,v. ∼ Theorem 2.2 Let G 6= K2 be a connected bipartite graph with partitions P1 and P2. Let u ∈ P1 and v ∈ P2 be two neighbors from G. Then the edge residual equals h V (Res(G, h{u}i)) ∪ V (Res(G, h{v}i)) i , if du = dv; Res(G, h{u, v}i)= Res(G, h{u}i), if d = d + 1; u v Res(G, h{v}i), if dv = du +1. There are no other possibilities. Proof: First we prove that the three cases above are the only ones possible. Let us assume that du > dv + 1 and take a vertex r ∈ V (Res(G, h{u}i)). From dv ≥ d(v,r) we get du > d(v,r) + 1. But that means there is a shorter path between u and r via v (u and v are neighbors) which is a contradiction. The proof is similar if we assume dv > du + 1. From this we can also deduce that d(u,s) = d(v,s) ± 1 for any vertex s ∈ V (G); because the graph is bipartite, the distances cannot be equal.
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