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Figure 11.7.12
THEOREM 11.7.3
Connections in Dominance-Directed Graphs
In any dominance-directed graph, there is at least one vertex from which there is a 1-step or 2-step connection to any other
vertex.
Proof Consider a vertex (there may be several) with the largest total number of 1-step and 2-step connections to other vertices in
the graph. By renumbering the vertices, we may assume that is such a vertex. Suppose there is some vertex such that there is
no 1-step or 2-step connection from to . Then, in particular, is not true, so that by definition of a
dominance-directed graph, it must be that . Next, let be any vertex such that is true. Then we cannot have
, as then would be a 2-step connection from to . Thus, it must be that . That is,
has 1-step connections to all the vertices to which has 1-step connections. The vertex must then also have 2-step connections
to all the vertices to which has 2-step connections. But because, in addition, we have that , this means that has
more 1-step and 2-step connections to other vertices than does . However, this contradicts the way in which was chosen.
Hence, there can be no vertex to which has no 1-step or 2-step connection.
This proof shows that a vertex with the largest total number of 1-step and 2-step connections to other vertices has the property
