Math 485, Graph Theory:
Midterm Exam Solutions
Stephen G. Simpson
Monday, October 27, 2008
1. (5 points) Draw a 3-regular graph with 11 vertices or prove that no such
graph exists.
Solution. By the degree-sum formula, a k-regular graph with n vertices
has kn/2 edges. It follows that kn is even. In this case kn = 3 · 11 = 33
is odd, hence no such graph exists.
2. (5 points) True or false.
(a) The number of spanning trees in the n-cycle is n. True. (This is
obvious.)
(b) Dijkstra’s algorithm applied to a connected weighted graph produces
a minimum-weight spanning tree. False. (The algorithm for a
minimum-weight spanning tree is Kruskal’s algorithm or Prim’s algorithm, not Dijkstra’s algorithm.)
(c) Every connected bipartite graph contains a cycle of even length.
False. (Any tree is a counterexample.)
(d) The Petersen graph has 25 edges. False. (It has 15 edges.)
(e) A weakly connected digraph with d+ (v) = d− (v) for all vertices v is
strongly connected. True. (These are just the Eulerian digraphs.)
3. (5 points) How many automorphisms does Q3 have? Justify your answer.
Solution. The answer is 48. In general, Qn has 2n n! automorphisms. We
can prove this by induction on n, using the fact that an automorphism of
Qn+1 is determined once we know where it carries one of the copies of Qn
which are inside it.
4. (5 points) How many spanning trees does Kn have? Justify your answer
by citing an appropriate theorem. How many spanning trees does Kn − e
have? Justify your answer.
Solution. Cayley’s Theorem says that τ (Kn ) = nn−2 . To calculate
τ (Kn − e) note that the number of edges in a spanning tree is n − 1
while the number of edges in Kn is n(n − 1)/2. Hence, the probability
1
that a randomly chosen edge e belongs to a randomly chosen spanning
tree T is (n − 1)/(n(n − 1)/2) which simplifies to 2/n. It follows that
the probability of e not belonging to T is 1 − 2/n. Thus, the number of
spanning trees not containing e is
2
nn−2 = (n − 2)nn−3 .
τ (Kn − e) = 1 −
n
5. (5 points) It is known that K4 has 16 spanning trees. Prove this using the
Matrix-Tree Theorem.
Solution.
3
τ (K4 ) = −1
−1
−1 −1 3 −1 = 16.
−1
3 6. (5 points) What is the length of a DeBruijn cycle for words of length 3 on
the alphabet {a, b}? Draw the appropriate DeBruijn digraph and use it
to construct such a cycle.
Solution. All such cycles are of length 23 = 8. The appropriate DeBruijn
digraph is
ab
|= O BBB
''
BB5
2 |||
BB |
'
B fe
g̀fe
ab ' |||
abc
d
aa aB
1
4 3
6
bb
BB
||
BB
|
B
||
8 BB ~||| 7
ba
and the DeBruijn cycles are in one-to-one correspondence with the Eulerian circuits in this digraph. For instance, the DeBruijn cycle arising from
the Eulerian circuit 1, 2, 3, 4, 5, 6, 7, 8 is aaababbb.
7. (10 points) Let G be an acyclic digraph consisting of k pairwise disjoint
uv-paths plus l pairwise disjoint vw-paths. Consider the associated 2person positional game starting at u. Under what conditions does the
first player have a winning strategy?
Solution. Which player has a winning strategy depends on the parity of
the lengths of the uv-paths and the vw-paths. Player one has a winning
strategy if at least one uv-path is even and at least one vw-path is odd.
Also, player one has a winning strategy if at least one uv-path is odd and
all vw-paths are even. Player two has a winning strategy in all other cases.
8. (10 points) The 5-ladder is this graph:
·−·−·−·−·−·−·
| | | | |
·−·−·−·−·−·−·
2
Let Ln be the n-ladder. Use the method of deleting and contracting edges
to find a recursion formula for the number of spanning trees in Ln . Use
your formula to calculate the number of spanning trees in L5 .
Solution. The method of deleting and contracting edges plus the doubleedge formula gives the 2-step recursion τ (Ln ) = 4τ (Ln−1 ) − τ (Ln−2 ).
Starting with τ (L1 ) = 1 and τ (L2 ) = 4 we obtain τ (L3 ) = 4 · 4 − 1 = 15
and τ (L4 ) = 4 · 15 − 4 = 56 and τ (L5 ) = 4 · 56 − 15 = 209.
9. (10 points) The Fibonacci numbers are 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . .
where each number is obtained by adding the previous two numbers. Let
φn be the nth Fibonacci number. Thus φ1 = 1, φ2 = 1, φ3 = 2, φ4 = 3,
φ5 = 5, φ6 = 8, φ7 = 13, etc.
(a) Write a recursion formula for φn .
(b) Solve your recursion to obtain an explicit formula for φn .
φn+1
.
(c) Calculate lim
n→∞ φn
Solution. Obviously the recursion formula for (a) is φn = φn−1 + φn−2 .
If we define φ0 = 0 then our recursion formula holds for all n ≥ 2. The
characteristic equation is α2 = α − 1 and the roots are
√
√
1− 5
1+ 5
and α2 =
.
α1 =
2
2
It remains to find the coefficients A and B such that
φn = Aαn1 + Bαn2 .
With n = 0 we have φ0 = 0 = A + B and with n = 1 we have φ1√=
Aα1 + Bα2 . Solving
these two equations for A and B we obtain A = 1/ 5
√
and B = −1/ 5, so our explicit formula for (b) is
√ !n
√ !n
1+ 5
1− 5
1
1
−√
.
φn = √
2
2
5
5
√ √ n
∼ 0 hence
Since 1−2 5 < 1 it follows that 1−2 5
φn
hence
1
∼ √
5
√ !n
1+ 5
2
√
φn+1
1+ 5
−→
φn
2
and this is our answer to (c).
3