91Ó°ÊÓ

First, consider three vertices and try to create every possible unique tournament. With three vertices (2^3=8 edges), the possible graph arrangements are three. These are: 1) All vertices form a cycle clockwise or anti-clockwise. 2) One vertex is connected to all other vertices in one direction. 3) A vertex is connected to another vertex, and the second vertex is connected to the third one. These three tournaments are nonisomorphic as there is no bijective relation preserving the order between them.

Using the same method for four vertices, attempt to create unique tournaments. With four vertices, there can be potentially 2^(4(4-1)/2) = 64 directed graphs. Yet, only four of these are non-isomorphic tournaments: 1) All vertices form a cycle in one direction. 2) One vertex is connected to all other vertices in one direction, and the others form a cycle.3) A vertex is connected to all others in one direction, the second vertex is connected to the remaining two, and the third vertex is connected to the last one. 4) Two pairs of vertices where each pair is connected in both directions.

The in-degree and out-degree of a vertex in a graph are the number of heads and tails, respectively, incident to this vertex. In tournament 2: Vertex 1's indegree=0, outdegree=3. Vertex 2's indegree=1, outdegree=2. Vertex 3's indegree=2, outdegree=1. Vertex 4's indegree=3, outdegree=0. In tournament 3: Vertex 1's indegree=0, outdegree=3. Vertex 2's indegree=1, outdegree=2. Vertex 3's indegree=2, outdegree=1. Vertex 4's indegree=3, outdegree=0. In tournament 4: Vertex 1's indegree=1, outdegree=2. Vertex 2's indegree=2, outdegree=1. Vertex 3's indegree=1, outdegree=2. Vertex 4's indegree=2, outdegree=1. Each vertex of every tournaments matches the in-degree and out-degree summing to (n-1), where n is the total number of vertices.