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The Marking Process 1 Initially, assign marker F to each v in V 2 Each v exchanges its open neighbor set N(v) with all its neighbors 3 Each v assigns its marker m(v) to T if there exist two unconnected neighbors In the example of Figure 201, N(u) = {v, y}, N(v) = {u, w, y}, N(w) = {v, x}, N(y) = {u, v}, and N(x) = {w} After Step 2 of the marking process, vertex u has N(v) and N(y), B has N(u), N(w), and N(y), w has N(v) and N(x), y has N(u) and N(v), and x has N(w) Based on Step 3, only vertices v and w are marked T Clearly, each vertex knows distance-2 neighborhood information after Step 2 of the marking process, ie, neighbor information of its neighbors.

Summary

The cost of checking the connectivity of two neighbors is upper bounded by 2 (G) or simply 2, where is the degree of graph G, ie, (G) = max{|N(v)||v V} There are |N(v)|(|N(v)| 1)/2 possible pairs of neighbors of vertex v, which is upper bounded by 2 Therefore, the cost of the marking process at each vertex is O( 2) The amount of message exchange at each vertex is also O( ), which corresponds to the number of neighbors 2032 Properties Assume that V is the set of vertices that are marked T in V, ie, V = {v|v V, m(v) = T} The induced graph G is the subgraph of G induced by V , ie, G = G[V ] The following two theorems show that G is a connected dominating set of G.

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Theorem 1 Given a graph G = (V, E) that is connected but not completely connected, the vertex subset V , derived from the marking process, forms a dominating set of G Proof: Randomly select a vertex v in G We show that v is either in V (a set of vertices in V that are marked T) or adjacent to a vertex in V Assume that v is marked F If there is at least one neighbor marked T, the theorem is proved If all its neighbors are marked F, we consider the following two cases Case 1: All the other vertices in G are neighbors of v Based on the marking process and the fact that m(v) = F, all these neighbors must be pairwise connected, ie, G is completely connected This contradicts the assumption that G is not completely connected.

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Case 2: There is at least one vertex u in G that is not adjacent to vertex v Construct a shortest path path, (v, v1, v2, , u), connecting vertices v and u Such a path always exists since G is a connected graph Note that v2 is u when v and u are distance-2 apart in G, ie, dG(v, u) = 2 Also, v and v2 are disconnected; otherwise, (v, v2, , u) is a shorter path connecting v and u Based on the marking process, vertex v1, with both v and v2 as its neighbors, must be marked T Again this contradicts the assumption that neighbors of v are all marked F When the given graph G is completely connected, all vertices are marked F This is desirable, because if all vertices are directly connected, there is no need for gateway hosts..

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