MC 302 – GRAPH THEORY – 11/5/13 –HW #4 SOLUTIONS
44 points + 6 extra credit points
1.
If is a graph, the line graph of , , has one vertex for each edge in
, and two vertices in are adjacent if the corresponding edges are
incident in . For example, a graph is shown here, in which the edges
are labeled 1 to 7. Hence has vertices 1 to 7. To find the neighbors of
a vertex in , look at each endpoint and see which edges share that
endpoint. For example edge 1 is incident with edges 4 and 5 at endpoint
, and with edges 2 and 6 at vertex . So in vertex 1 has neighbors 4, 5, 2, and 6.
a. For the graph shown, draw its line graph (I find it helpful in my drawing to place the
vertices of in roughly the relative positions where they’re shown as edge-labels in above).
For parts b, c, d, and g, let G be an arbitrary graph.
b. If - is an edge of , give a formula for the degree of as a vertex in , in terms of the
degrees of and .
c. Prove that if is Eulerian, then is also Eulerian. (Note: Sometimes it’s best to think of
Eulerian graphs in terms of the definition, and other times it’s best to remember their
characterization as connected graphs in which all vertices have even degree.)
d. Prove that if is Eulerian, then is Hamiltonian. Note: To prove that is Hamiltonian, you
will have to explain how to construct a Hamiltonian cycle in .
e. Construct an example of a non-Eulerian graph whose line graph is Eulerian. Give labeled
drawings of both and .
f. Construct an example of a non-Eulerian graph whose line graph is Hamiltonian. Give labeled
drawings of both and .
g. EXTRA CREDIT #1: Find a formula for the number of edges in in terms of the degrees of the
vertices in . Your formula may use only simple arithmetic operations: sums (summations are OK
too), products, differences, quotients. Submit this problem for extra credit only if you do it
completely on your own. If you find a solution somewhere, then do not submit a solution to this
problem.
h. EXTRA CREDIT #2: Prove that if is Hamiltonian, then is Hamiltonian. As for the
preceding extra credit problem, submit this problem for extra credit only if you do it
completely on your own. If you find a solution somewhere, then do not submit a solution
to this problem.
SOLUTION TO #1: (30 pts: a-f 4 each, g-h 3XC each)
a. The drawing is at the right.
b. If - is an edge of , then deg deg counts all the edges incident to and , but it also counts the edge twice. Thus in , deg deg 2
equals deg.
c. If is Eulerian, it is connected and, by a theorem, every vertex has even degree. Since is connected,
is also connected, since a path between two edges in is induced in a natural way by a path
between one endpoint of each edge in . Furthermore, if - is any edge in , deg deg deg 2 by part (a), and this is even since it is a sum of even numbers, because every degree
in is even. Hence it follows from a theorem that is Eulerian.
d. If is Eulerian, it has a circuit that includes every edge exactly once. Since every edge in corresponds
to a vertex in , this induces in a natural way a circuit in that includes every vertex 0f exactly once, i.e., a Hamiltonian cycle in .
e. The Petersen graph, shown at the right with its line graph below, is not Eulerian
because not every vertex has even degree (in fact every degree is odd), but all
the degrees in are even, and is connected, so is Eulerian.
f. The same graph is not Hamiltonian (see proof at end of assignment), but has
a Hamiltonian cycle:
1 2 3 8 12 10 4 5 6 11 15 9 14 13 7 1
Note: The complete graph is a simpler example than the Petersen graph for
parts (e) and (f), but I wanted you to see that the Petersen graph is not Hamiltonian.
g. (Extra credit #1) If is any vertex in , then any two edges incident to are
6
adjacent to each other. In other words, the edges incident to induce a clique
5
1
(complete subgraph) in . Since has
this clique is
$%&'$%&'!"
.
#
!"
#
edges, the number of edges in
7
15
If we add this up over all the vertices in , we count
each edge of exactly once . Hence the number of edges in equals
$%&'$%&'!"
∑'∈*+
.
#
13
10
4
11
2
14
12
9
8
3
h. (Extra credit #2) Suppose , is a Hamiltonian cycle in . Then the cycle of edges in , is
a cycle of vertices in , although not necessarily a Hamiltonian cycle. To create a
Hamiltonian cycle in , begin by writing the list of vertices and edges of , in their
order on the cycle, starting and ending with an edge on the cycle. Then for each
vertex in order, replace it with any incident edges that are not already on the cycle,
otherwise just delete it. Since all edges incident with are adjacent to each other,
these added edges are pairwise adjacent, and they are also adjacent to the two
edges on , that were incident with . Once an edge is added for one of its endpoints,
do not add it again for the second endpoint. When finished we have a Hamiltonian
cycle in .
For example, the graph shown at the right has a Hamiltonian cycle. We list all its vertices and edges in order:
1-23.451
To create a Hamiltonian cycle in , we modify this list by first replacing - with 6, then replacing with 7,
deleting ., deleting , and lastly deleting (all edges incident with ., , and are already on the modified
cycle). When we are finished, we have the following Hamiltonian cycle in :
16273451
2. The table below is modified (Baton Rouge instead of New Orleans!) from the slides of 10/22/13. Below
each of you is assigned a city as your “starting vertex:”
ALB
0
Albany, NY
990
Springfield, IL
170
Boston, MA
626
Raleigh, NC
Baton Rouge, LA 1482
293
Harrisburg, PA
113
Hartford, CT
196
Trenton, NJ
ALBANY: Mohammed, Madeleine, Brennan;
BOSTON: Aimee, Siobhan, Piper;
BATON ROUGE: Alex, Conor, Pauline;
HARTFORD: Carol, Cindy;
SPR
990
0
1162
846
758
755
1048
876
BOS
170
1162
0
700
1571
383
101
279
RAL
626
846
700
0
929
373
601
431
BAT
1482
758
1571
929
0
1194
1474
1317
HRR
293
755
383
373
1194
0
285
127
HRT
113
1048
101
601
1474
285
0
183
TRE
196
876
279
431
1317
127
183
0
SPRINGFIELD: Clark, Nathan, Eleanor;
RALEIGH: Katie, Kyaw, Yaomingxin;
HARRISBURG: Michaella, Carl;
TRENTON: Nikolay, Huang;
a. Using your assigned starting vertex , apply the theorem on slide 7 of 10/22/13 to get a lower
bound on the length of a traveling salesperson tour. Your answer should show:
i. The vertex and the two incident weighted edges you used,
ii. The minimum weight spanning tree for the graph obtained by deleting your vertex (with
weights on edges), and
iii. The grand total sum of weights for your lower bound.
b. Using your assigned starting vertex, apply the Nearest Neighbor Algorithm to find a Hamiltonian
circuit. Draw the circuit you find with weights on edges, and give the circuit’s total weight.
SOLUTION TO #2: (10 pts: a-6, b-4)
• The optimal tour is 3,620 miles: ALB-BOS-HRT-TRE-RAL-BAT-SPR-HRR-ALB. I found this by writing a
little program in Mathematica that compared the lengths of all 7! 5040 Hamiltonian cycles starting
and ending at one fixed city, say ALB.
• Here are the numeric totals for parts (a) and (b) from each city. I have not shown the trees and
cycles. The greatest lower bound of 3339 used either SPR or BAT; the least upper bound of 3631 used
RAL.
ALB SPR BOS RAL BAT HRR HRT TRE
2580 3339 2580 2841 3339 2844 2593 2878
Lower bound
Nearest Neighbor 4079 3649 4085 3631 3649 4005 4045 4230
3. p. 128, #4.1.6. Prove that the “triple flyswat” has a matching with 7 edges,
but no perfect matching.
15
4
16
3
5
14
2
13
12
1
6
SOLUTION TO #3: (10 pts: proof-6, matching-4) The picture at the right shows a
7
matching with 7 edges. Here are two possible ways to prove there is no perfect
8
11
matching:
10
9
Proof 1: We prove by contradiction that there is no perfect matching, i.e., one in
which each vertex is incident to an edge in the matching. So suppose M is a perfect matching. Then vertex 1
is incident to a matching edge, and by the symmetry of the graph we may assume without loss of generality
that the edge 1-7 is in the matching. Then the edges 1-2 and 1-12 are not in the matching, which implies that
the vertices 2-6 must be matched using only the edges in the subgraph on these vertices (and similarly for
the vertices 12-16). But this subgraph has five vertices, so some vertex cannot be matched, contradicting the
assumption that M is a perfect matching.
Proof #2: By Berge’s Theorem, a matching M is maximum if and only if there is no M-augmenting path.
Considering the 7-edge matching M shown above, it is easy to see that there is no M-augmenting path, and
so it is maximum. Since a maximum matching has 7 edges, this means that G has not perfect matching, which
would have 8 edges.
Theorem. The Petersen graph is non-Hamiltonian.
Proof by contradiction (adapted from a proof by Doug West). It is easy to see that the Petersen graph
is 3-regular and has no 3-cycles or 4-cycles. Suppose it has a Hamiltonian cycle C. Number the vertices
in order on the cycle, 1 to 10. To get the original graph we must add 5 more edges so that every vertex
has degree 3 and we create no 3-cycles or 4-cycles.
First note that no vertex can be adjacent to another vertex two steps away on the cycle (because that
would create a 3-cycle) or three steps away (because that would create a 4-cycle). So every vertex is
adjacent either to another that is 4 steps away or 5 steps away (5 steps away means directly opposite
on the cycle). Next , note that no two consecutive vertices on the cycle can be adjacent to “opposite”
vertices, because that creates a 4-cycle. For example, if 1 is adjacent to 6 and 2 is adjacent to 7, that
creates the 4-cycle 1-2-7-6-1. So for any two consecutive vertices on the cycle, one of them must be
adjacent to a vertex that is four steps away. Thus we may assume without loss of generality that
vertices 1 and 5 are adjacent. Now consider vertex 6. There is no way to add an edge incident to 6
without creating a 4-cycle. This is a contradiction to the assumption that the Petersen graph has a
Hamiltonian cycle, so we conclude that it does not.
∎