Consider a complete graph, K, that has n vertices.
There is a set of edges within K that have a common property, which is that they do not share vertices anywhere on the graph. Let's call these set of edges "M".
The amount of edges in set M is limited: |M| <= n/2
Question: How many Hamiltonian cycles in graph K contain all the edges in set M? Give your answer in terms of n and |M|.
For the sake of this exercise, let's pretend graph K has 6 vertices.
My answer is that only one Hamiltonian cycle gets to cover all the edges in set M.
The reason for this is that it takes n/2 edges (in this case, 3) to surround the outer circuit of any complete graph.
Once these edges take hold you can't really make any edges inside the graph part of set M because they'd be sharing vertice with those on the outer layer.
Likewise, if you make the edges inside the graph members of set M, you won't be able to make edges on the outer circuit part of set M.
Either way, it seems to me only one Hamiltonian cycle can contain all members of set M.
Am I interpreting this correctly?