Feedback About 2014’s Exam in
MATH20902: Discrete Mathematics
Note: the distributions illustrated below have not been adjusted to take account of any Special
Circumstances.
Scaled Total (max 100)
15
5
10
Frequency
8
6
4
0
0
2
Frequency
12
20
Raw Exam Scores (max 80)
0
20
40
60
80
0
20
Marks
60
100
Marks
Figure 1: Histograms of the raw exam results (left) and for the module as a whole (right). The
latter, which is shaded to indicate degree classification boundaries, includes coursework marks and
has been scaled up slightly (see below).
General remarks
• These notes are easier to follow if you have a copy of the exam.
• I scaled the final marks up slightly as the exam seems to have been a bit too hard. Before
scaling the average for those 102 students who
– did both the coursework and the exam and
– weren’t penalised for plagiarism
was 58.5%, while after scaling it shifted to 61.8%.
• I referred 30 students to the Academic Malpractice Panel for plagiarism on the coursework
(see the Coursework Feedback for details) and 16 of them received some sort of penalty, ranging from a mark of zero for a single question up to a mark of zero for the entire coursework.
1
2
4
6
40
0
20
Frequency
20
10
8 10
0
2
4
6
8 10
0
2
4
6
8 10
Marks
Marks
B4 (N = 113)
B5 (N = 87)
B6 (N = 68)
0
5
10
Marks
20
8
0
4
Frequency
8
4
0
0
5
Frequency
12
Marks
10 15
0
Frequency
A3 (N = 122)
0
10
20
Frequency
A2 (N = 122)
0
Frequency
A1 (N = 122)
0
5
10
20
Marks
0
5
10
20
Marks
Figure 2: Histograms for the individual questions: data for the required, Section A, questions (10
marks apiece) are plotted light blue, while those for the Section B questions (best two out of three,
25 marks each) are slightly darker. There were 144 scripts in all.
Remarks about individual problems
A1 Almost all students got the 3 marks associated with the definitions, but the rest of the problem
proved rather harder than I’d hoped: although there were a few perfect answers, the average
mark for this question was only around 4.9/10. The most common mistakes arose from a
confusion about what the question asked. One needed to establish two things:
• If a saturated hydrocarbon has chemical formula Cm Hn then the numbers n and m
must satisfy n = 2m + 1.
• Given a pair of positive integers n and m satisfying n = 2m + 1, these exists at least
one saturated hydrocarbon with the chemical formula Cm Hn .
The first of these is fairly easy to establish using the Handshaking Lemma and the observations that a saturated hydrocarbon is—if viewed as a graph—a tree with hydrogen atoms
as leaves and carbon atoms as internal nodes with degree 4. The second part is most easily
proven by constructing a family of examples. One especially simple family has the m carbon
atoms arranged in a linear chain, like the vertices of a path graph, with the 2m + 1 hydrogens
attached to the carbons as in the example at left in Figure 3.
Many students tried to make proofs by induction for one or both of the statements above.
Some of those for the second statement were successful, but most of those for the first
(saturated hydrocarbon ⇒ n = 2m + 1) were not, mainly because they assumed, without
supporting argument, that all saturated hydrocarbons can be built sequentially, by repeatedly
adding a single carbon to a smaller saturated hydrocarbon. This is true, but needs an
inductive proof that starts with a saturated hydrocarbon having, say, n0 + 1 carbons and
removes a carbon to get a smaller saturated hydrocarbon.
This part of the question also elicited some odd, false claims about degree sequences and the
neighbours of a carbon atom in a saturated hydrocarbon. In fact, a carbon can have any
2
H
H
H
H
H
H C
C
C
C
C H
H
H
H
H
H
H
H HH H HH
C
H C
H C
C
C
H
H
H
H
C C H
H
C
H HH
Figure 3: Two saturated hydrocarbons: Problem A1 turns on the observation that such a molecule
is—when viewed as a graph—a tree.
number—one through four, inclusive—of other carbon atoms as neighbours: see the right
panel of Figure 3 for examples. Finally, a small minority of students got into a hopeless
muddle about the use of the symbols n and m. In the context of this problem they are,
respectively, the number of hydrogen and carbon atoms. Thus they are not the number of
vertices and edges in the tree naturally associated with a drawing of the hydrocarbon. In
light of this, it’s helpful to write the Handshaking Lemma as
X
deg(v) = |E|
v∈V
and note that trees satisfy |E| = |V | − 1.
A2 This problem turned out to be somewhat easier than the previous one, with an average of
around 5.9/10. Once again, most students got all the marks associated with the definitions
(four in this problem), though I marked them rather strictly. I saw various odd things
including:
• In the definition of connected: “. . . two vertices a and b are connected if ∃ (a, b) ∈ E”.
With this definition the only connected graphs are the complete graphs?!?
• There were a fair number of definitions of isomorphic that had missing words: these
looked like examples of ineffective memorisation.
• I didn’t deduct any marks for the following, but thought it odd. Several students defined
the degree sequence as
. . . a list of the degrees of each vertex in G(V, E) in ascending order.
It’s unclear what the word “each” is doing here, but it doesn’t really do any harm.
The main part of the problem asked whether all graphs with degree sequence (1, 1, 2, 2, 2) are
isomorphic. They aren’t, though the two examples below are, up to isomorphism, the only
graphs that share this degree sequence.
Most students who found these examples gave correct proofs of their non-isomorphism—there
were several nice proofs by contradiction—but a few people just copied out misremembered
and irrelevant bits of the lecture notes or problem sets and so got very few of the six available
marks. Weirdest of all were two students who drew the examples above, but then claimed
that all graphs with the specified degree sequence are isomorphic.
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A3 I hoped you would find this problem easy and most of you did: the average was around 8.4/10.
Aside from carelessness in copying information from the exam onto the graph in part (a),
the main ways people lost marks were by:
• drawing an undirected graph, even though the problem asked explicitly for a directed one
(one mark deducted) or drawing a graph with no edge weights and no other indication
of the times required for the various tasks;
• failing to indicate which was the critical path (it was A-D-F-G, leading to a minimum
time-to-completion of 23);
• giving a numerical table of results with some mistakes and no discussion/work to justify
partial credit.
I didn’t expect or require you to write down an explicit topological ordering on the vertices,
but if you did write one down I expected it to satisfy the rule that (vi , vj ) ∈ E =⇒ i < j
and I deducted marks if this wasn’t true. A few people tried simply to enumerate all the
paths through the graph and find the critical path by comparing their lengths. This bruteforce approach sometimes went astray (paths were missed or lengths were miscalculated),
but when it worked I awarded full credit.
B4 This was the most popular of the questions in Part B. All but 9 of the 122 students who sat
the exam tried it and the average was 12.8/25 or around 53.2%. All parts of this question
are closely related to problems that I recommended you study.
(a) In contrast to Part A, where most students gave correct definitions and got all the
associated marks, here there were lots of garbled or incorrect ones. Few students, for
example, mentioned that a spanning tree in G(V, E) should be a subgraph of G.
(b) This part of the question produced a lot of answers consisting entirely of wordless
calculation. This was just about acceptable if the final numerical answer was correct,
but I marked down heavily if any of the calculations were wrong. Common mistakes
included
• writing L = D − A without saying what D means;
• renumbering the vertices without saying anything.
The best answers explained carefully how to construct L and mentioned Tutte’s Matrix
Tree Theorem.
(c) Students who understood the question did well here, though some drew the same spreg
twice and/or missed one. A few students confused this part of the question—which is
about spregs—with the previous part, which is about spanning arborescences: the latter
are a subset of the former.
(d) A few students seemed confused about the term “relatively prime”. Two integers a
and b are said to be relatively prime if they share no prime factors or, equivalently, if
gcd(a, b) = 1.
A substantial minority of students used the Principle of Inclusion/Exclusion correctly
to compute the number of integers in the set U = {102, 103, . . . , 170} that are divisible
by at least one of 2, 3 or 7 (there are 49 of them), but then seem to have forgotten that
the question asks about those elements of U that aren’t divisible by any of those three
small primes, so that the correct answer is
|U | − 49 = 69 − 49 = 20.
As with part (b) above, there were a lot of answers consisting of wordless calculation,
many of which went amiss. I awarded such answers most of the available 6 marks if
they were right, but many fewer marks if there were any mistakes. The best answers
defined carefully three sets Xj consisting of multiples of the small primes mentioned in
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the problem and then stated the Principle of Inclusion/Exclusion, giving a formula like
this
|X1 ∪ X2 ∪ X3 | =
(|X1 | + |X2 | + |X3 |)
− (|X1 ∩ X2 | + |X1 ∩ X3 | + |X2 ∩ X3 |)
+ |X1 ∩ X2 ∩ X3 |
and finally obtained the correct numerical answer.
A couple of students actually did all the calculations twice, once for the set U1 =
{1, . . . , 101} and then again for U2 = {1, . . . , 170}. The set U in the problem is then
U2 \U1 and I guess the point of the approach is that it’s easier to count, say, the multiples
of 2 contained in U1 and U2 and then take the difference.
(e) This last part of the problem is hard, but does not have all that many marks attached
to it. I intended it to make the difference between a good mark (max possible score for
parts (a)-(d) is 20/25) and a perfect one. The idea is to find the spectrum of a matrix
derived from the graph Laplacian of the complete graph and I awarded a couple of marks
to anyone who wrote down a correct formula for this matrix. Kirchoff’s Matrix Tree
Theorem—combined with a clever observation about the matrix whose determinant one
needs to compute—then allows one to see that Kn contains nn−2 spanning trees. This
is a famous result due to Cayley and a few students seemed to know it by heart, which
earned them a couple of marks. The best answer extrapolated from some small examples
and conjectured the correct result.
B5 This was the next most popular of the Part B problems and 87 out of 122 students attempted
it. Answers varied wildly in quality, from six scripts with a score of 1 up to three with a
score of 25: the average was 10.9/25.
(a) The definitions students wrote were better than those from part (a) of B4, but there
were still an awful lot of mistakes. Examples include:
• confusion about the difference between bipartite graphs and the complete bipartite
graphs Km,n . The issue is that there are lots of bipartite graphs that aren’t complete
bipartite graphs. For example, every tree on two or more vertices is bipartite.
• confusion between Hamiltonian and Eulerian cycles. The former, which is what
the question was about, pass through every vertex exactly once.
• (incorrect) definitions of k-colouring that didn’t make any mention of the set {1, . . . , k}.
(b) As in Problem A1, a substantial minority of students didn’t seem to understand what
it takes to prove an “if and only if” result. A complete answer to the question requires
proofs of two statements:
• If a graph G is bipartite, then χ(G) = 2.
• If χ(G) = 2 then G is bipartite.
Most students only proved the first of these, though there were a few laborious (and
mostly wrong) attempts at proofs by induction.
(c) I saw many good answers to this part, including several that made a lovely argument
based on the observation that the chromatic number of an odd-length cycle is 3.
(d) Nearly everyone found a 2-colouring for H, but this only establishes that χ(H) ≤ 2.
Rather fewer students completed the proof that χ(H) = 2 by observing that H can’t
have a 1-colouring, so χ(H) ≥ 2 too.
(e) Most students said, correctly, that H is not Hamiltonian, but few offered rigorous justifications. The best answers (there were four like this) that H has 11 vertices and so any
Hamiltonian cycle would have length 11, an impossibility in a bipartite graph. Common
mistakes included:
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• confusion between criteria for Hamiltonian cycles and Eulerian ones. These students
usually mentioned the presence of a vertex of odd degree, but that’s not relevant
here.
• confusion between a Hamiltonian trail (visits every vertex) and a Hamiltonian tour
or cycle (visits every vertex and starts and finishes on the same vertex). The
Herschel graph does contain a Hamiltonian trail, but no Hamiltonian cycles.
• Misapplication of Dirac and/or Ore’s Theorems. These results are of the general
form
If all the vertices in G have high degree, then G is Hamiltonian.
Some students tried to use these result “backwards”, claiming that as all the vertices
in H have low degree, (for example, deg(v) < |V |/2 for all v ∈ V ), then H can’t
be Hamiltonian. This is a logical mistake: Dirac’s and Ore’s Theorems can only
be used to prove a graph is Hamiltonian, not to prove that it isn’t. After all, the
cycle graphs are all Hamiltonian–indeed, a cycle graph is nothing but one huge
Hamiltonian cycle—and yet all their vertices have degree 2.
B6 This was the least popular question in Part B, with only 68/122 attempts. The average was
13.3.
(a) The definitions were pretty good, though there was some muddle about subdivision.
(b) The easiest approach to this question was to make two subdivisions having different
numbers of vertices: there can be no bijection between finite sets of unequal size. But
lots of people did other things as well: there were quite a few answers that involved
sketches of two non-isomorphic graphs and a correct proof of non-isomorphism, but
in which the graphs weren’t subdivisions of the one in the question. In these cases I
awarded partial credit.
(c) This question is identical to one I suggested you study and a lot of people seem to have
taken my advice: I saw a lot of essentially perfect answers. I did, however, deduct marks
if people just wrote down the formula
|E 0 | = |E| − 1 + (k + 1)
without any explanation of why it is true.
(d) Most students who attempted this got all the marks, though there was some confusion
about the difference between a planar graph and a planar diagram. A planar graph
must, by definition, have a planar diagram, but it may have other, nonplanar ones too.
(e) Many, many students tried to compute the crossing numbers of K5 and K3,3 by looking
at the standard drawings of these graphs. This produces a wild overestimate (both
graphs have crossing number 1) and, even if it had been correct, would establish only
an upper bound. To prove that, for example, cr(K5 ) = 1 you need to:
• produce a diagram with only a single crossing (this establishes that cr(K5 ) ≤ 1)
• argue that as K5 is known to be non-planar (Kuratowski’s Theorem), it can’t have
a diagram, with zero crossings and so cr(K5 ) ≥ 1.
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