At least how many edges of Kn complete graph do we have to remove not to have a Hamilton circuit in the remaining graph?

For Kn complete graph,

The
total number of vertices = n

The total number of edges =
n(n-1)/2

By definition, a Hamilton circuit will have each
vertex (except for the start vertex) visited only once.

For
a Hamilton circuit to exist, the remaining graph would have to be an n-order Cycle
graph, Cn, which has exactly n edges.

To avoid having a Cn,
there should be n-1 edges left in the remaining graph.

As a
result, the least number of edges to be removed from a Kn graph would
be:

n(n-1)/2 - (n-1)

= (n-1)
[n/2 - 1]

=
(n-1)(n-2)/2