Computer Engineering(CE)
Question 1: Algorithms
August 2013
Solution 1
A MBST tree is not always a MST.
Proof: Suppose we have a spanning tree $ T $ which is a MBST of graph \mathbf{G}, and $ T $ is also a MST. The bottle neck is edge $ e $ with weight $ w $. Then I have another vertex $ X $ that will be connected to the graph \mathbf{G} to form a new graphs \mathbf{G'}. $ \mathbf{G'} = \mathbf{G} +X $. And the possible edges that connect graph \mathbf{G} and new vertex $ X $ has weights $ w_1, w_2, ... W_x $, and each is no greater than the bottle neck weight $ w $. Using any of the edges from $ $w_1, w_2, ... W_x $, I get a new tree $ T' $. S $ T' $ is still a spanning tree, and its largest weight remains $w$, so $T'$ is also an MBST. However, among these edges that I can choose to span the tree, only the one that has the smallest weight will be a MST, by definition. So an MBST for a graph \mathbf{G'} is not a minimum spanning tree for $ \mathbf{G'} $. End of proof.
