CSE 202 Homework 2 Solution
TA: Nan Zang
Problem 1
Suppose you are given a connect graph G, with edge costs that are all distinct. Prove that
G has a unique MST.
Proof:
Suppose the graph has two different MSTs, T and T1. Then there must be some
edge (u, v) which is present in T but not in T1. The graph obtained by removing
the edge (u, v) from T consists of two connected components, which define a cut
on G. There must be some edge (u1, v1) across this cut in T1. Also, one of these
two edges must be lighter than the other, since all edge weights are distinct.
Without loss of generality, assume that w(u, v) < w(u1, v1). But then, removing
(u1, v1) from T1 and adding (u, v) gives a tree of lower cost than T1. However,
this is a contradiction with that T1
is an MST. Hence, the MST of the graph must
be unique.
Problem 2
Design a spanning network for which the most expensive edge is as cheap as possible.
A spanning tree T of G is a minimumbottleneck spanning tree if there is no spanning tree
T’ of G with a cheaper bottleneck edge.
(a) Is every minimum bottleneck tree of G a minimum spanning tree of G.
No. Every spanning tree of the following graph is a MBT, but only one of them is a
MST.
(b) Is every minimum spanning tree of G a minimum bottleneck tree of G.
Yes, Proof by Contradiction.
Suppose T is a MST but not a MBT. T’ is a MBT. The bottleneck edge in T is e.
the weight of e is larger than any edge in T’, since T’ is a MBT. Removing edge e
from T, T will have two connected components S and M. Since T’ is a spanning
tree, in T’, there must be an edge connecting these two components S and M,
denoting this edge e’. So, T{e} +{e’} is a new spanning tree with smaller weight
than T. This is a contradiction with the assumption that T is a MST.
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 Winter '08
 Graham
 Graph Theory, duv

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