Is there a minimum spanning tree that does not contain the min/max weighted edge?

Question

If we have an (arbitrary) connected undirected graph G, whose edges have distinct weights,

1. does every MST of G contains the minimum weighted edge?
2. is there an MST of G that does not contain the maximum weighted edge?

Also, I'm more thankful if someone can give a hint of the key things one must keep in mind when dealing with such MST questions.

This is a homework problem. Thanks.

Answer 1

is there an MST of G that does not contain the maximum weighted edge?

There may be, but there doesn't have to be. Consider a 4-vertex graph as follows:

[A]--{2}--[B]
 |         |
 |         |
{1}       {3}
 |         |
 |         |
[C]-{50}--[D]

The minimum spanning tree consists of the edge set {CA, AB, BD}. The maximum edge weight is 50, along {CD}, but it's not part of the MST. But if G were already equal to its own MST, then obviously it would contain its own maximum edge.

does every MST of G contains the minimum weighted edge?

Yes. MSTs have a cut property. A cut is simply a partition of the vertices of the graph into two disjoint sets. For any cut you can make, if the weight of an edge in that cut is smaller than the weights of the other edges in the cut, then this edge belongs to all MSTs in the graph. Because you guaranteed that the edge weights are distinct, you have also guaranteed that there is an edge which is smaller than all other edges.

Also, I'm more thankful if someone can give a hint of the key things one must keep in mind when dealing with such MST questions.

Your best bet is to reason about things using the properties of MSTs in general, and to try to construct specific counterexamples which you think will prove your case. I gave an instance of each line of reasoning above. Because of the cut and cycle properties, you can always determine exactly which edges are in an MST, so you can systematically test each edge to determine whether or not it's in the MST.