Moore Graph - Moore Graphs As Cages

Moore Graphs As Cages

Instead of upper bounding the number of vertices in a graph in terms of its maximum degree and its diameter, we can calculate via similar methods a lower bound on the number of vertices in terms of its minimum degree and its girth (Erdõs, Rényi & Sós 1966). Suppose G has minimum degree d and girth 2k+1. Choose arbitrarily a starting vertex v, and as before consider the breadth-first search tree rooted at v. This tree must have one vertex at level 0 (v itself), and at least d vertices at level 1. At level 2 (for k > 1), there must be at least d(d-1) vertices, because each vertex at level 1 has at least d-1 remaining adjacencies to fill, and no two vertices at level 1 can be adjacent to each other or to a shared vertex at level 2 because that would create a cycle shorter than the assumed girth. In general, a similar argument shows that at any level 1 ≤ ik, there must be at least d(d-1)i vertices. Thus, the total number of vertices must be at least

In a Moore graph, this bound on the number of vertices is met exactly. Each Moore graph has girth exactly 2k+1: it does not have enough vertices to have higher girth, and a shorter cycle would cause there to be too few vertices in the first k levels of some breadth first search tree. Therefore, any Moore graph has the minimum number of vertices possible among all graphs with minimum degree d and diameter k: it is a cage.

For even girth 2k, one can similarly form a breadth-first search tree starting from the midpoint of a single edge. The resulting bound on the minimum number of vertices in a graph of this girth with minimum degree d is

(The right hand side of the formula instead counts the number of vertices in a breadth first search tree starting from a single vertex, accounting for the possibility that a vertex in the last level of the tree may be adjacent to d vertices in the previous level.) Thus, the Moore graphs are sometimes defined as including the graphs that exactly meet this bound. Again, any such graph must be a cage.

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