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MC 302 – GRAPH THEORY –SOLUTIONS TO HW #3 – 10/27/11
42 points plus 4 extra credit points
1.
pp. 56-57, #2.2.3 (be sure to justify your answer!) G has 4 connected components and 24 edges –
what’s the maximum number of vertices that G can have?
SOLUTION TO #1 (4 pts):
By a theorem (Thm. 2.9 in the text), q
n -
(G), where q, n and
(G) are
the
numbers
of
edges,
vertices
and
components,
resp.,
of
G.
Thus
n
q
+
(G)
=
24 + 4 = 28, i.e., G can have at most 28 vertices.
2.
pp. 56-57, #2.2.4. Prove that a connected graph G with
n
vertices and
e
edges is unicyclic if and
only if
n
=
e
. Hint: For sufficiency, you might want to argue the contrapositive.
SOLUTION TO #2 (10 pts):
Necessity (
):
We assume that the connected graph G is unicyclic and prove that
n = e.
So suppose
G is connected and has a unique cycle C. Choose any edge
f
=
u-v
on C. Since the remaining section
of C is a path from
u
to
v
, it can replace
f
in any walk from one vertex of G to another. Thus G\
f
is still
connected and is now acyclic. Thus G\
f
is a tree, so by a theorem it has
e = n
-1 edges. Therefore G
has
e =n
edges, proving necessity.
Sufficiency (
): (Proof by contrapositive)
We prove the contrapositive, namely, we show that if G
is not unicyclic, then n
e. If G is not unicyclic, then G has either no cycles or more than one cycle. If
G has no cycles, then since it’s connected, G is a tree. Thus n = e – 1
e. If G has more than one cycle,
say it has distinct cycles C
1
and C
2
, choose an edge e
1
in C
1
but not in C
2
. Then G\e
1
is still connected,
since any path using e
1
could instead use the rest of the cycle C
1
. But G\e
1
still has the cycle C
2
, thus
G\e
1
is not a tree and so has more than n-1 edges, which implies that G has more than n edges.
Sufficiency (
): (Direct proof)
Assume G is connected with
e = n
edges. Then G cannot be a tree,
by the theorem that says a tree has
n-
1 edges. So, since G is connected, it must have a cycle C. We
must show that C is unique, i.e., if C and C’ are both cycles in G, then C = C’. So suppose
C
and
C’
are
both cycles in
G
, and let
f
be an edge of
C
. Then
G\f
is still connected, and it has
n-1
edges, so
G\f
is a
tree. Thus, removing
f
must have disconnected C’ as well, so the edge
f
is in both cycles. If the
endpoints of
f
are
u
and
v
, then
C\f
and
C’\f
are both paths between
u
and
v
in the tree
G\f
. By the
unique path property of trees, it follows that
C\f = C’\f
, and thus
C = C’
, proving that
G
has a unique
cycle.
3.
For the graph shown at the right,
a.
Use Prim’s Algorithm starting at vertex
A
to build
a minimum-weight spanning tree. Be sure to
specify the order in which you add the edges to
your tree, and to give the total weight of the
tree.