March 7, 2009
Page 1
Mathematics 2156b
Second Midterm Exam
Instructions: Print your name on the SCANTRON answer sheet. Sign the SCANTRON
answer sheet, and use a PENCIL to mark your student number on the SCANTRON answer
sheet. Use a PENCIL to mark your answers to questions 1–20 on the SCANTRON answer
sheet, and as well, circle your answers on the question sheet.
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A1. In the graph G1 , how many paths are there from s to t?
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A: 3
B: 4
C: 5
D: 7
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G6
E: None of A, B, C, D
Solution:
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Thus there are 7 paths from s to t in G1 . The answer is D.
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A2. Which of the following statements concerning the graphs drawn above is true?
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(i) No two of the graphs are isomorphic.
(ii) G3 and G4 are isomorphic, and no other isomorphisms exist between
pairs of the above graphs.
(iii) G3 and G6 are isomorphic, and no other isomorphisms exist between
pairs of the above graphs.
(iv) G4 and G6 are isomorphic, and no other isomorphisms exist between
pairs of the above graphs.
(v) G3 , G4 , and G6 are all isomorphic.
A: (i)
B: (ii)
C: (iii)
D: (iv)
E: (v)
Solution: If two graphs are isomorphic, then they have identical degree sequences. The degree sequences
of G1 , G2 , G3 , G4 , G5 , and G6 , respectively, are (4, 4, 3, 3, 2), (3, 3, 3, 3, 3), (4, 3, 3, 3, 3), (4, 3, 3, 3, 3),
(3, 3, 3, 3, 3, 3), and 4, 3, 3, 3, 3). The only possiblity would be that G3 , G4 , and G6 are isomorphic. In fact,
if we redraw each of G3 and G4 , placing the vertex of degree 4 inside the 4-cycle, we obtain the diagram
of G6 . Thus G3 , G4 , and G6 are isomorphic.
The answer is E.
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A3. Exactly which of the graphs G1 , G2 , G3 , G4 , G5 and G6 are regular?
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A: G2 , G3 , G4
B: G2
C: G2 , G5
D: G5
E: G2 , G5 , G6
Solution: G1 has a vertex of degree 2 and a vertex of degree 3, so G1 is not regular. G2 = K5 is 4-regular.
Each of G3 and G4 has a vertex of degree 3 and a vertex of degree 4, so neither G3 nor G4 is regular.
G5 = K3,3 is 3-regular. Finally, G6 has a vertex of degree 3 and a vertex of degree 4, so G6 is not regular.
Thus G2 and G5 are regular, and the others are not. The answer is C.
March 7, 2009
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Mathematics 2156b
Second Midterm Exam
2
A4. Exactly which of the graphs G1 , G2 , G3 , G4 , G5 and G6 are bipartite?
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A: G1
B: G3
C: G4
D: G5
E: None of them
Solution: Recall that a non-null graph is bipartite if and only if it is without cycles of odd length. Since
each graph except G5 actually has a 3-cycle, while G5 = K3,3 , only G5 is bipartite.
The answer is D.
2
A5. What is τ (G1 ), the number of spanning trees of G1 ?
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A: 16
B: 24
C: 32
D: 40
E: None of A, B, C, D
Solution: In the computation that follows, we utilize the fact that after the first step, the graph obtained
by deleting e has the vertex of degree 1 joined by a cut-edge to the remainder of the graph, to conclude
that the number of spanning trees of G − e is the product of the number of spanning trees of the two
components that result when the cut-edge is deleted, and one of those components is a point. We also use
the fact that we can delete any loops that result from an edge contraction without changing the number
of spanning trees. Finally, we use the fact that in the “eyeglass” graph, the middle vertex is a cut-vertex,
and so the number of spanning trees of the “eyeglass” graph is the square of the number of spanning trees
of the 2-cycle.
τ
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e
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= τ (K4 ) + τ 
= 2τ (K4 ) + τ
e
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= 2τ (K4 ) + τ [•
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e
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h
= 2(44−2 ) + (2)(2) + 4 = 40.
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The answer is D.
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A6. In this question, G denotes a simple graph with n vertices and m edges. Exactly which of
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the following statements are true?
(i)
(ii)
(iii)
A: (i)
If G is connected, then n ≥ m − 1.
m ≤ n2 /2.
If G is connected and every vertex of degree greater than 1 is a
cut-vertex, then G is a tree.
B: (ii)
C: (iii)
D: (i), (ii)
E: None of A, B, C, D
Solution: (i) false. In fact, we must have m ≥
n − 1.
n
(ii) true, for in fact, we must have m ≤ 2 = n(n − 1)/2 < n2 /2.
(iii) false. For consider a 3-cycle with a vertex of degree 1 attached to each vertex of the 3-cycle. Each of
the three vertices of the 3-cycle are cut-vertices, the graph is connected, and the remaining three vertices
are all degree 1, yet the graph is not a tree.
Since only (ii) is true, the answer is B.
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A7. Exactly which of the following statements are true?
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March 7, 2009
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Mathematics 2156b
Second Midterm Exam
(i)
(ii)
(iii)
A: (i), (iii)
There exists a simple graph with 6 vertices and 14 edges.
There exists a simple 3-regular graph with 9 edges.
There exists a simple 3-regular graph with 26 edges.
B: (iii)
C: (ii)
D: (ii), (iii)
E: None of A, B, C, D
Solution: (i) true. K6 has 62 = 15 edges, so the graph obtained by removing a single edge from K6 is a
simple graph with 6 vertices and 14 edges.
(ii) true. K3,3 is a simple 3-regular graph with 6 vertices and 9 edges.
(iii) false. If G were a simple 3-regular graph with n vertices and 26 edges, then by the handshake
lemma, we would have 3n = 52. Since 3 does not divide 52, there is no such graph.
Since (i) and (ii) are true while (iii) is false, the answer is E.
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A8. Let S be a set of complete bipartite graphs on 7 vertices. Further suppose that S has the
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property that every complete bipartite graph on 7 vertices is isomorphic to one and only
one graph in S. The size of S is:
A: 3
B: 4
C: 5
D: 6
E: None of A, B, C, D
Solution: Since two complete bipartite graphs Kr,s , r ≥ s, and Km,n , m ≥ n, are isomorphic if and only
if r = m and s = n, we see that a complete collection of complete bipartite graphs on 7 vertices is given
by { Km,n | m ≥ n, m + n = 7 }. Since we must have m ≤ 7 and n ≥ 1, we have m = 6, n = 7 − 6 = 1,
m = 5, n = 7 − 5 = 2, m = 4, n = 7 − 4 = 3, and that is all.
Thus a complete collection of complete bipartite graphs on 7 vertices has 3 graphs in it. The answer is
A.
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A9. If G is a connected graph on 8 vertices in which every edge has weight 3, what is the weight
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of an optimal spanning tree for G?
A: 12
B: 21
C: 25
D: 26
E: None of A, B, C, D
Solution: Any spanning tree for G will have 7 edges, each of weight 3, so an optimal spanning tree for G
will have weight (3)(7) = 21. The answer is B.
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A10. If G is a graph with degree sequence (4, 4, 3, 3, 2, 2, 1, 1), how many edges does G have?
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A: 8
B: 10
C: 12
D: 21
E: None of A, B, C, D
Solution: By the handshake lemma, such a graph would have m edges, where 2 m = 4 + 4 + 3 + 3 + 2 +
2 + 1 + 1 = 20, so m = 10. Here is a graph with this degree sequence:
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The answer is B.
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A11. What is the maximum number of edges in a simple graph on 10 vertices with 3 components?
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A: 66
B: 33
C: 26
D: 10
E: None of A, B, C, D
2
Solution: By a result in the text, this will be the number of edges in the complete graph with 10 − (3 − 1) =
12 vertices, and this is 12
2 = 66.
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Mathematics 2156b
Second Midterm Exam
The answer is A.
2
A12. Let S be a set of connected simple graphs each having four edges. Further suppose that S
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has the property that every connected simple graph with four edges is isomorphic to one
and only one graph in S. The size of S is:
A: 2
B: 3
C: 4
D: 5
E: 6
ν Solution: By a result in the text, if G is a connected simple graph, then νG − 1 ≤ εG ≤ 2G , so we
must have νG ≤ 5. If νG ≤ 3, then 4 = εG ≤ 32 = 3, which is not possible. Thus either νG = 5 or else
νG = 4. If νG = 5, then G is connected with εG = νG − 1, so G is a tree. Thus a complete collection of
connected simple graphs with four edges will contain as a subcollection a complete collection of all trees
on 5 vertices. We can construct such a collection by organizing by maximum vertex degree. The possible
maximum vertex degrees for a tree on 5 vertices would be 2, 3, or 4. There is only one tree on 5 vertices
with all vertices of degree either 1 or 2, and that is the chain on 5 vertices. Next, consider the trees on 5
vertices with maximum vertex degree equal to 3. Then there is at least one vertex of degree 3, with the
other 4 of degree at most 3. If there were two vertices of degree 3, then there would have to be exactly
two of degree 1 and then the sum of vertex degrees would be greater than 8. Since 2εG = 8, this would
contradict the handshake lemma. Thus there is exactly one vertex of degree 3, and so there must be an
odd number of vertices of degree 1. Since there are at least two vertices of degree 1 in any tree, we must
have one vertex of degree 3 and 3 vertices of degree 1. The remaining vertex has degree 2. There is only
one (up to isomorphism) to construct such a graph from the star graph on 4 vertices: adjoin a new edge to
one of the degree 1 vertices, termininating in a new vertex. Thus up to isomorphism, there is exactly one
tree on 5 vertices with maximum vertex degree equal to 3. Finally, there is exactly one tree on 5 vertices
with maximum vertex degree 4: the star graph on 5 vertices. Thus there are 3 graphs in a complete
collection of trees on 5 vertices.
Now consider the connected simple graphs with 4 edges on 4 vertices. Those of maximum vertex degree
2 must have all four of degree 2 by the handshake lemma, and up to isomorphism, there is exactly one
such graph, C4 . For maximum degree 3, three of the four edges are incident to one of the vertices, so
the remaining edge must join two of the remaining three vertices. Up to isomorphism, only one graph is
obtained by joining any choice of two vertices of degree 1 in the four vertex star graph.
Thus a complete collection of connected simple graphs with 4 edges contains 3 trees on 5 vertices and
2 graphs on 4 vertices; namely the 4-cycle and the 3-cycle with one new vertex joined by an edge to one
of the vertices of the 3-cycle.
Since a complete collection of connected simple graphs with 4 edges contains 5 graphs, the answer is D.
2
A13. What is the maximum number of vertices in a connected simple graph with 30 edges?
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A: 15
B: 29
C: 30
D: 31
E: None of A, B, C, D
Solution: There is a result in the text
which states that a simple graph with k components, n vertices and
m edges has n − k ≤ m ≤ n−(k−1)
. Thus for k = 1, we must have n − 1 ≤ 30, and so n ≤ 31. Since a
2
tree on 31 vertices has 30 edges, the maximum number of vertices in a connected simple graph is 31.
The answer is D.
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A14. Let G be a simple graph on 10 vertices with m edges. For exactly which values of m must
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it be the case that G contains at least one 3-cycle?
(i) 20 (ii) 25 (iii) 30
A: (i), (ii)
B: (ii), (iii)
C: (i)
D: (ii)
E: (iii)
Solution: By Turan’s theorem, if G is a simple graph on 10 vertices and εG ≤ (10)2 /4 > 25, then G
March 7, 2009
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Mathematics 2156b
Second Midterm Exam
contains a 3-cycle, while there exists a simple graph on 10 vertices and 25 edges with no 3-cycle; namely
K5,5 . Thus only (iii) is guaranteed to force a 3-cycle, so the answer is E.
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A15. If G is a graph with 26 edges and degG (v) ≥ 3 for each vertex v of G, then the largest
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possible value for νG is:
A: 18
B: 17
C: 15
D: 13
Solution: By the handshake lemma, we have 3 νG ≤
P
v∈V
G
E: None of A, B, C, D
degG (v) = 2 εG = 52, so it must be that
νG ≤ ⌊ 52/3 ⌋ = 17. Is there actually a graph with 17 vertices and 26 edges? Yes, consider either of
the following graphs. Each has 17 vertices, sixteen of degree 3 and one of degree 4. Thus each has
(3(16) + 4)/2 = 26 edges.
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The answer is B.
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A16. The number of spanning trees for K2,3 is
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A: 12
B: 10
C: 40
D: 32
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E: None of A, B, C, D
Solution: The number is 23−1 32−1 = 12, so the answer is A.
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A17. How many cut-edges does the graph
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A: 0
B: 1
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C: 2
have?
D: 4
E: None of A, B, C, D
Solution: An edge is a cut-edge if and only if it is not in any cycle. The only such edge is the one whose
two endpoints have degree 3.
The answer is B.
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A18. How many cut-vertices does the graph
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A: 0
B: 1
C: 2
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have?
D: 4
E: None of A, B, C, D
Solution: Each of the two vertices of degree 3 are cut-vertices, and none of the vertices of degree 2 are
cut-vertices, so there are two cut-vertices.
The answer is C.
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A19. Exactly which of the following are true:
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(i) The cycle graph C4 is bipartite, 3-partite and 4-partite.
(ii) For any vertices u and v in a graph G, any two distinct paths from u to v
contain a cycle.
(iii) For any two vertices u and v in a connected graph G, there is a path from u to v.
March 7, 2009
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Mathematics 2156b
Second Midterm Exam
A: (i), (ii)
B: (i), (iii)
C: (ii), (iii)
D: (i), (ii), (iii)
E: None of A, B, C, D
Solution: (i) true. C4 is bipartite; that is, 2-partite. Since any 2-partite graph is k-partite for any k ≥ 2,
it follows that C4 is 2-partite, 3-partite, and 4-partite. (ii) true. This is a result in the text. (iii) true. By
definition, there is a walk between any two given vertices u and v in a connected graph G, and we know
by a result in the text that a walk of shortest length between two given vertices is a path. Thus there is
a path between any two given vertices in a connected graph.
Since all three are true, the answer is E.
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A20. Exactly which of
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(i)
(ii)
(iii)
A: (i)
the following are true?
Every non-null tree is bipartite.
If m ≥ 3, every complete m-partite graph contains a cycle.
Every bipartite graph contains a cycle.
B: (ii), (iii)
C: (i), (iii)
D: (ii)
E: None of A, B, C, D
Solution:
(i) true. Tree are acyclic, so a non-null tree has no odd cycles, hence is bipartite.
(ii) true. Since there are at least 3 cells, we may choose 3 cells, and select a vertex from each, say v1 ,
v2 , and v3 . Then since the graph is complete multi-partite, the subgraph induced by these 3 vertices is
K3 , a 3-cycle.
(iii) false. For example, K1,1 = K2 is bipartite, but has no cycle.
Since (i) and (ii) are true, while (iii) is false, the answer is E.
PART B
B1. Consider the sequence (4, 4, 3, 3, 2, 2, 1, 1).
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(a) Is there a simple graph that has this sequence as its degree sequence? If so, draw one.
If not, prove that there is no such graph.
Solution: Here are some 2 component examples to begin with:
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Next, we provide some connected examples.
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There are none with three components, as part (b) establishes.
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(b) Is there a simple graph with 3 components that has this sequence as its degree sequence? If so, draw one. If not, prove that there is no such graph.
Solution: Suppose that there is a simple graph G with three components that has degree sequence
(4, 4, 3, 3, 2, 2, 1, 1). Then each component must have at least two vertices each since the minimum
vertex degree is 1 and the graph is simple. Since G would have 8 vertices, this would mean that no
component could have more than 4 vertices each and thus G would not have any vertex of degree 4,
so this is not possible.
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B2. (a) Is Km,m eulerian for every integer m > 1? If so, prove it, while if not, explain why
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not.
Solution: A connected graph is eulerian if and only if every vertex has even degree. Since each vertex
of Km,m has degree m, it follows that Km,m is eulerian if and only if m is even.
March 7, 2009
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Mathematics 2156b
Second Midterm Exam
2
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(b) Give necessary and sufficient conditions on positive integers m and n in order that
Km,n be semi-eulerian.
Solution: A connected graph is semi-eulerian if and only if it has exactly two vertices of odd degree.
Of the vertices of Km,n , m have degree n, and n have degree m. Thus one of m or n must be two,
while the other must be an odd integer.
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B3. For any positive integers m and n with m ≥ n, what is the fewest number of edges that
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may be deleted from the complete bipartite graph Km,n to obtain a subgraph that is not
connected? Explain your answer.
Solution: If we choose a vertex of degree n and delete all n incident edges, the result is a disconnected graph
with two components, where one component is the null graph on one vertex, while the other component
is Km−1,n . Thus it is possible to obtain a disconnected subgraph by deleting n edges. Is it possible to
do so by deleting fewer than n edges? Suppose that we delete k edges with k < n. Then at least one of
the n vertices of degree m has had none of its incident edges deleted. Such a vertex is thus connected
to each of the m vertices originally of degree n. Since m ≥ n, the same observation applies to the m
vertices originally of degree n. At least one of them has had none of its incident edges deleted. Such a
vertex is connected to each of the n vertices originally of degree m. But then all m + n vertices are still
in one connected component after the k edges have been deleted. Thus deleting fewer than n edges can’t
disconnect the graph, so the fewest number of edges that may be deleted to result in a disconnected graph
is n.
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B4. (a) Draw the join of the null graph on two vertices and the chain graph on three vertices.
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Solution:
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(b) Give the degree sequence of the join of the null graph on two vertices and the chain
graph on three vertices.
Solution: The graph is shown in the solution to part (a). There is one vertex of degree 4 and four
vertices of degree 3, so the degree sequence is (4, 3, 3, 3, 3).
B5. Consider the weighted graph G given by:
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(a) How many optimal spanning trees does G have? Draw them all.
Solution:
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(b) Calculate τ (G), the number of spanning trees of G (considered as an unweighted
graph).
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Mathematics 2156b
Second Midterm Exam
Solution:
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τ (G) = τ 
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= 2 τ (K4 ) + τ 
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= 2 τ (K4 ) + τ [•
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h
•] + τ •
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= 2(42 ) + 4 + 4 = 40.
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B6. Is the graph G shown at right
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bipartite? Provide reasons for
your answer.
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Solution: G contains a cycle of length 5. Since a bipartite graph does not contain any cycles of odd length,
it follows that G is not bipartite.
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B7. (a) What is the length of a longest circuit in K8 ? Establish your answer.
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Solution: Consider C, a circuit of greatest length in K8 . The support of C is a circuit subgraph of
K8 , so must have every vertex of even degree. Since each vertex of K8 has degree 7, the greatest
possible degree of a vertex in C is 6, in which case C would have 6νC /2 = 3νC ≤ 3(8) = 24 edges. Is
there a circuit in K8 of length 24? Well, partition the 8 vertices into 4 sets of size 2. For each such
subset of size 2, remove the edge joining the two vertices of the subset. Thus we will have removed
4 edges from K8 , no two with an endpoint in commmon, so the result is a spanning subgraph H of
K8 in which each vertex has degree 6 (since every vertex has degree 7 in K8 and we removed one
incident edge from each of the 8 vertices). Since all vertices of this subgraph have even degree 6, it
has (6)(8)/2 = 24 edges and is thus connected (since H is simple, if εH > 8−1
= 21, then H is
2
connected), it follows that H is eulerian. An eulerian circuit of H is a circuit in K8 that uses 24 edges,
the maximum possible. Thus a circuit of greatest length in K8 has length 24.
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(b) How many 8-cycle subgraphs are there in K8 ? Establish your answer.
Solution: Since every pair of vertices is joined by an edge, constructing an 8-cycle in K8 is equivalent
to arranging the 8 vertices in a row, hence there are 8! 8-cycles in K8 . Now, we are not asked to
count 8-cycles, but rather 8-cycle subgraphs; that is, the subgraphs that arise as the support of an
8-cycle. Observe that for a given 8-cycle, there are 8 sequences (the given 8-cycle and 7 others) that
correspond to traversing the given 8-cycle in the same direction, varying only by their start point.
Furthermore, if we travelled any of these 8 sequences in the reverse direction, we would also obtain
an 8-cycle that has the same support as the given 8-cycle. Thus each 8-cycle subgraph is the support
of (8)(2) 8-cycles, so there are 8!/( (8)(2) ) = 7!/2 = 2520 8-cycle subgraphs of K8 .
3
B8. (a) Prove that for every
P integer n ≥ 2, for every sequence d1 ≥ d2 ≥ · · · ≥ dn of positive
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integers such that ni=1 di = 2(n − 1), dn must equal 1 and d1 ≤ n − 1.
Pn
P
Solution: First, suppose that d1 > n − 1. Then 2(n − 1) = d1 + i=2 di > n − 1 + i = 2n 1 ≥
n − 1 + n − 1 = 2(n − 1), which is impossible.
d1 ≤ n − 1. Next, suppose that dn ≥ 2, in which
Pn Thus P
n
case di ≥ 2 for every i. Then 2(n − 1) = i=1 di ≥ i=1 2 = 2n, again impossible. Thus dn = 1.
Mathematics 2156b
Second Midterm Exam
5
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March 7, 2009
Page 9
(b) Use induction on n to prove that for every
P integer n ≥ 2, and every choice of positive
integers d1 ≥ d2 ≥ · · · ≥ dn for which ni=1 di = 2(n − 1), there exists a spanning tree
of Kn with degree sequence (d1 , d2 , . . . , dn ).
Solution: Our inductiveP
hypothesis is the assertion that for every choice of positive integers d1 ≥
n
d2 ≥ · · · ≥ dn for which i=1 di = 2(n − 1), there exists a spanning tree of Kn with degree sequence
(d1 , d2 , . . . , dn ). The base case is n = 2, and by the first part, 1 ≤ di ≤ 2 − 1 = 1 for i = 1, 2, so
d1 = d2 = 1 is the only possible choice for d1 and d2 . Since K2 is a tree with degree sequence (1, 1,
the assertion is true when n = 2.
Suppose now that n ≥ 2 is an
integer for which the hypothesis is true, and let d1 ≥ d2 ≥ ·P
· · ≥ dn+1
Pn+1
n+1
be positive integers for which i=1 di = 2(n + 1 − 1) = 2n. If di = 1 for every i, then 2n = i=1 1 =
n+1, which implies that n = 1, which is not the case. Thus d1 > 1. Let m = max{ i | di > 1 }. By the
first part of the question, dn+1 = 1, so 1 < m < n + 1. We have di = 1 for m + 1 ≤ i ≤ n + 1. Subtract
1 from dm , and considerP
the integers d1 ≥ d2 ≥ · · · ≥ dm − 1 ≥ dm+1 = dm+2 = · · · = dn = 1. We
n
have 2n − 2 = dm − 1 + i=1, i6=m di , and so by hypothesis, there exists a spanning tree of Kn with
degree sequence (d1 , d2 , dm−1 , dm − 1, dm+1 , . . . , dn ). Connect vertex n + 1 to vertex m by an edge.
The result is a spanning tree of Kn+1 with degree sequence d1 , d2 , . . . , dm , . . . , dn+1 , as required. This
completes the proof of the inductive step, and the result follows now by induction.
3 Bonus: Prove that every connected non-null simple graph G has at least two vertices that are not
marks
cut-vertices.
Solution: Since G is connected, it has a spanning tree T . Since G is simple and non-null, G has at least
two vertices and thus T has at least two vertices. But then T has at least two vertices of degree 1 in T ,
and hence are not cut-vertices of T . It follows that neither is a cut-vertex of G, since adding edges will
not create cut-vertices.
Alternative, we may prove it by induction on the number of vertices of the graph. The base case would
be a connected graph with 2 vertices, and removing one vertex from the graph leaves one vertex, hence
one component. Thus neither vertex is a cut-vertex.
Suppose now that n ≥ 2 is an integer such that every connected non-null graph G with at most n
vertices has at least two vertices that are not cut-vertices, and consider a connected graph G with n + 1
vertices. If none of the vertices of G are cut-vertices, then G has at least two non-cut-vertices. Otherwise,
pick a cut-vertex v of G. Then G − v decomposes into at least two components. For each component H of
G − v, the subgraph G[VH ∪ { v }] of G that is induced by the vertex set of the component together with
v is connected and has at most n vertices and at least two vertices (v and the vertices of H), so we may
apply the induction hypothesis to this subgraph to conclude that it contains at least two non-cut-vertices.
Since it might be that v itself is a non-cut-vertex of this subgraph, we may only conclude that each such
subgraph contains at least one non-cut-vertex different from v. Since a non-cut-vertex G[VH ∪ { v }] other
than v is a non-cut-vertex of G, it follows that G has at least as many non-cut-vertices as components of
G − v. Since there are at least two components of G − v, G has at least two non-cut-vertices.
The result follows now by induction.
Instructor’s Name (Print)
Student’s Name (Print)
Student’s Signature
Student Number
THE UNIVERSITY OF WESTERN ONTARIO
LONDON
CANADA
DEPARTMENT OF MATHEMATICS
Mathematics 2156b Second Midterm Exam
March 7, 2009
9:30a.m.–12:00 noon
INSTRUCTIONS
1. BE SURE TO FILL IN THE TOP OF THIS
PAGE CORRECTLY.
2. THE EXAM CONSISTS OF TWO PARTS
(A AND B).
3. The first part of the exam (Part A: Questions
A1–A20) is MULTIPLE CHOICE. This part
is to be answered on the SCANTRON answer
sheet. As well, make sure that you circle your
selected answer on the question sheet.
The second part (Part B) has questions to be
answered in the space provided. Be sure to
SHOW ALL OF YOUR WORK, and try to
place your solution in the space provided for
that purpose.
FOR GRADING ONLY
PAGE
1
2
3
4
5
6
7
4. Print your name and your instructor’s name
on the SCANTRON answer sheet. Sign the
answer sheet, and mark your student number and section on the answer sheet. USE A
PENCIL to mark your answers to Questions
A1–A20 on the SCANTRON answer sheet.
8
9
TOTAL
5. There are two blank pages at the back of this
booklet. These may be torn off and used for
scrap paper as the need arises. We suggest
that you keep one blank page attached in case
you need extra space to answer a question. Do
not tear any other pages from the booklet.
6. Questions start on Page 1 and continue to
Page 9. BE SURE YOUR BOOKLET IS
COMPLETE.
7. CALCULATORS ARE NOT PERMITTED.
8. TOTAL MARKS = 80.
MARK