Guest wrote:The answer is 7.
Probably the easiest way to go about it is to inductively (on the number of vertices) construct all edge minimal graphs (where an edge indicates the vertices are of distance 1).
On 1 vertex there is one graph (just the vertex itself).
On 2 vertices there is one graph (just the pair of vertices with no edge between them (remember we are looking for edge minimal graphs)).
On 3 vertices there is one graph (an edge together with an isolated vertex).
On 4 vertices there are two graphs, which we can deduce by building on the 3 vertex case. If we add an extra vertex there are only a few ways we could join it to the 3 vertex case:
a) No edges from the new vertex. This forces all the other edges to be present, resulting in the minimal graph which is a triangle together with an isolated vertex.
b) One edge, from the new vertex to the isolated vertex. We don't need to add any other edges to make this work, the graph of two separate edges is edge minimal.
c) One edge, from the new vertex to an end point of an edge. This forces the edge from the other end of the "3-vertex case edge" to the isolated vertex to be present, which means this graph contains case (b) so is not edge minimal.
d) two edges, one to the isolated vertex, the other to the end point of the existing edge. Not minimal because it contains case (b).
e) two edges, both to the end points of the existing edge. Not minimal contains case (a).
f) three edges. Not minimal contains case (a).
On 5 vertices there are two graphs. A triangle with an edge, and a pentagon. I'll leave the reasoning as an exercise for you to do.
On 6 vertices there are two graphs. One of them is two triangles, I'll leave the other one for you to find.
On 7 vertices there is one graph, and on 8 vertices there are no graphs, I'll leave these as an exercise for you to do.
Hope this helped,
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