TEST2/MAD3305 Page 2 of 4
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2. (15 pts.)
For the graph G below, determine the cut-vertices,
bridges, and blocks of G. List the cut-vertices and bridges in
the appropriate places, and provide carefully labelled sketches
of the blocks.
Cut-vertices:
c, d, f, and i
Bridge(s):
ci and fg
Block(s):
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3. (10 pts.)
Below, provide a proof by induction on the order
of the graph G that every nontrivial connected graph G has a
spanning tree.
[Hint: If the order of the graph G is at least
3, Theorem 1.10 implies that G has a vertex v with G - v
connected.] //
Let S(k) be the following assertion: "For any graph G, if G is connected
with k vertices, then G has a spanning tree T."
If G is a connected graph of order 2, then G is K
2
which is a tree and
there is nothing to show. Thus S(2) is true, and we get a basis for the
induction with no real work.
To deal with the induction step, we must provide a proof of the proposition
(
∀
k
≥
2)( S(k)
⇒
S(k+1) ). To show this, we let k
≥
2 be fixed and arbitrary.
We need to show for this k that S(k)
⇒
S(k+1). Suppose that S(k) is true. To
show that S(k+1) follows, we may assume as true the hypothesis of S(k+1), namely
that we are dealing with G, an arbitrary connected graph with k+1 vertices. We
need to show that G has a spanning tree. Since G has at least 3 vertices,
Theorem 1.10 implies that there are two vertices in G, say u and v, with G - u
and G - v connected. We shall focus on G - v. G - v is a connected graph with
k
≥
2 vertices. From the induction hypothesis, S(k), with G - v replacing G, it
follows that G - v has a spanning tree, T
0
say. Since G is connected, v is
adjacent to at least one of the vertices of G - v. Pick such a vertex, say w,
and let T = (V(G), E(T
0
)
∪
{vw}). Then T is a spanning tree for G. Since G was an
arbitrary connected graph of order k+1, S(k+1) follows. Thus, S(k)
⇒
S(k+1).
Since k was arbitrary, we have (
∀
k
≥
2)( S(k)
⇒
S(k+1) ). We may now apply the
principle of induction. [Hello, modus ponens.]