HW 3 comments

Max: 6.7
Min: 3.1
Average: 5.39
Median: 5.55
 
Comments:
 
A. Some realize that they need a tree but did not write down the number of edges needed to be
removed.

B.  Some mistakenly think that "any two vertices are joined by
exactly one simple path" means "every path in G is simple".  Some
assume G has two simple paths joining two vertices and prove there exists
a cycle (instead of the converse).  Some assume there is a cycle and then
claim that since a cycle is not simple, the given assumption is
contradicted.
 
C. Majority overlooks the case of a trivial tree for part (c).
 
D. In part (a), it would be nicer to have those vertices of odd degrees specified, instead of
just saying "there are more than 3 vertices of odd degree in the graph".
 
E. Many overlook the case that the original graph G can be disconnected. Some wrongly asserted
that since there is an Eulerian path, G' is connected and so does G.
 
F. A direct corollary of this question would be that the existence of open and closed Eulerian
path are mutually exclusive. Some used this fact in the proof and claimed that they could
deduce this from Theorem 2.6.1. But Theorem 2.6.1 only gave a necessarily and sufficient
condition for the existence of a closed Eulerian path.