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to the sum of the degrees. Another is to ask what a given edge “contributes”
to the sum of the degrees.
102.b. To make your inductive step, think about what happens to a graph if you
delete an edge.
102.d. Suppose that instead of summing the degree of v over all vertices v, you sum
some quantity defined for each edge e over all the edges.
103. Whatever you say should be consistent with what you already know about
degrees of vertices.
108. What happens if you choose an edge and delete it, but not its endpoints?
109. One approach to the problem is to use facts that we already know about
degrees, vertices and edges. Another approach is to try deleting an edge
from a tree with more than one vertex and analyze the possible numbers of
vertices of degree one in what is left over.
111. When you get to four and especially five vertices, draw all the unlabeled trees
you can think of, and then figure out in how many different ways you can put
labels on the vertices.
112.b. Do some examples.
112.c. Is it possible for a 1 to be equal to one of the b j s?
112.d. You have seen that the sequence b determines a 1 . Does it determine any other
a j s? If you knew all the a j s and all the b j s, could you reconstruct the tree?
What are the possible values of b 1 ? b j ?
113. What vertex or vertices in the sequence b 1 , b 2 ,...,b n−1 can have degree 1?
115. If a vertex has degree 1, how many times does it appear in the Prüfer code of
the tree? What about a vertex of degree 2?
116. How many vertices appear exactly once in the Prüfer code of the tree and
how many appear exactly twice?
118. Think of selecting one edge of the tree at a time. Given that you have chosen
some edges and have a graph whose connected components are trees, what
is a good way to choose the next edge? To prove your method correct, use
contradiction by assuming there is a spanning tree tree with lower total cost.
Additional Hint: Think of selecting one edge of the tree at a time. But now
do it in such a way that one connected component is a tree and the other
connected components have just one vertex. What is a good way to make