In this storify, we collect the tweets for the hashtag #math3033, emails to the lecturer (and responses) and other bits and pieces for math3033 graph theory in semester 1 2016/17. (Please read from the bottom to the top, for my responses to correspond correctly to the tweets and emails collected.)
Storified by JW Anderson ·
Mon, Jan 23 2017 20:28:48
[email 2017.01.23] I have the following questions regarding Graph Theory exam:
1) Will we have Random Graphs on the exam/ Are they examinable?
2) Will we see any invariants we haven't covered in the lectures?
3) Where could I find more exercises of the same type as the ones on the exam excluding the ones from Lecture Notes, Past Papers and formative feedback questions or are those sufficient?
[answer 2017.01.23] 1) The basic material on random graphs is examinable, through Exercise 24.11 on page 105 but not beyond this.
2) Yes, it is possible that you might be asked about the basic properties of and calculating relating to an invariant we haven't covered in lecture.
3) My best suggestion here is to make use of the internet or the University library. There are many on line and print versions of graph theory textbooks. But some care is needed. Not all questions in a graph theory textbook are going to be relevant.
My view is that the exercises provided through the notes, the past papers, the in class tests and the formative feedback questions are sufficient, but I also think that math is a practice based activity, and the more practice the better.
[email 2017.01.23] Is question 1c from the 15/16 paper an error?
The question is to prove that \overline{G \ x} and \overline{G} \ x are equal.
xy is an edge of \overline{G \ x}
but xy is not an edge of \overline{G} \ x
so \overline{G \ x} and \overline{G} \ x have different edge sets
so \overline{G \ x} and \overline{G} \ x are not equal.
Apologies if you have already addressed this.
[answer 2017.01.23] This is one of the questions we covered in the revision session last week, the recording of which is available in the recorded sessions of the MATH3033 Blackboard site. The question is correctly worded.
To address your specific point, unfortunately xy is not an edge of \overline{G \ x}. By the definition of removing a vertex, x is not a vertex of G \ x and so x cannot be a vertex of \overline{G \ x}, since the vertex sets of \overline{G \ x} and of G \ x are the same.
In other words, we remove all of the edges adjacent to x from G when we construct G \ x, but none of these edges make a reappearance in \overline{G \ x}, since we've also removed x from the vertex set.
[response email 2017.01.23] That makes sense, I got confused about it because the solutions say that xy is an edge of \overline{G \ x}.
[response answer 2017.01.23] My apologies for the typo and for the confusion caused. I inadvertently missed out the word 'not' before 'an' in the sentence 'so xy is an edge of \overline{G \ x}.' in the second sentence in the third paragraph of the solution.
For greater clarity, I should also have added 'since no edges of G adjacent to x are edges of either G \ x or its complement \overline{G \ x} since neither contains x in its vertex set' as part of my solution.
[email 2017.01.22] I can only come to the revision session from three until four on Wednesday, is it ok if I come one hour late?
Also would it be possible to go over an exercise on graph probability ? ( Exercise 24.1 or 24.9 for example or any other one really)
[answer 2017.01.22] Of course, and I'll be happy to.
[email 2017.01.19] Thank you for scheduling the revision sessions, the first was very helpful! Unfortunately I have an exam during the second one, will you be recording it so I can access it before the exam the following morning?
[answer 2017.01.20] Of course
[email 2017.01.06] Could you allow an extension for me to answer the formative feedback 8 over the weekend before you post the solutions ? (I have been busy this week with CRM, which is due Tuesday)
[answer 2017.01.07] Of course. In fact, it is my intention to talk through this question during the revision sessions, rather than post a solution.
[email 2016.12.25] Could you give me some guidance on how to prove Exercise 8.4 ?
[answer 2016.12.26] I'll be working through Exercise 8.4 during the revision sessions, but as a hint in terms of moving forward, take any vertex x in V(G) and start walking through the graph from x. Since the degree of every vertex is 2, there are 2 choices of direction of movement from x. Choose one.
As we walk through the graph, at each vertex, there is only one choice of further movement. We keep walking through the graph. At some point, since our graph is finite, we need to come back to our starting vertex, and this describes a cycle.
[email 2016.12.11] I’ve been reading back through the notes, and noticed something odd in the proof of Euler’s formula on page 96. In the induction hypothesis, we assume the statement holds for all SCaF graphs with fewer than n vertices. However, we never seem to use that information in the subsequent argument, since H and G both have the same number of vertices. This seems to make the entire proof invalid. Have I missed something?
[answer 2016.12.11] You are right that the proof as it stands needs some work. The induction should be over the number of edges, rather than the number of vertices. I am in the process of revising it, but thanks, good catch.
[email 2016.12.06] Regarding my confusion in graph theory today, I am happy that your statement is correct, and I think it's a strict inequality in interesting cases. Essentially, we want to say
P(complete subgraph) = P(some subset of vertices forms a complete graph) = P(this set does OR that set does OR this other set does OR....)
where we take in all possible sets of vertices which might induce a complete subgraph.
Time to drop a standard probability thing: P(A or B) = P(A) + P(B) - P(A and B). In particular this extends naturally to P(A or B or ... or Z) is less than or equal to P(A) + P(B) + ... + P(Z) (because every time we break a piece off we subtract something positive). In particular we get a strict inequality if it's possible for more than one set to give a complete graph, which happens whenever k<n.
[answer 2016.12.08] Thanks for this. And this is actually sort of why I only worried about getting the inequality, rather than working out exactly what the probability might be. For the result we're after, we only need the inequality, and this is a not uncommon phenomenon in mathematics, of getting the strength of result we need but not going further than that.
[email 2016.12.05] I’m wondering about some results in chapter 24:
1) Lemma 24.6: I think the proof is correct but in the lemma itself it has to be (\binom{n-1}{k} p^k p^{n-1-k})^n because currently it is more like the expected number of vertices with degree k and the (\binom{n-1}{k}) is still missing.
2) Lemma 24.12 and 24.13: Again I think that the proofs are correct but not the results because like this (in Lemma 24.12) \binom{n}{k}(1-p)^{\binom{k}{2}} is just the probability that the independence number is equal to k but the probabilities that the independence number is equal to k+1,…,n are missing. So with the estimate in the proof the current result should be multiplied by (n-k). The same holds for lemma 24.13.
Further you said today that the in class test 2 will cover all the material in the notes to theorem 24.9. (including). I assume that the exercises 24.10 and 24.11 are also relevant and from lemma 24.12 on the notes aren’t relevant for the test?
[answer 2016.12.06] Answers to be given in the lecture on 6 December. And you are right that exercises 24.10 and 24.11 are relevant, and that from Lemma 24.12 will not be covered on the in class test or the final examination, at least not directly.
[email 2016.12.05] Hey professor,
thought I'd send in a typo spotting for the notes.
on page 104, exercise 25.2, it says:
'independence number(G)=clique number(G complement))'
I believe that is an extra bracket on the RHS of the equation.
See you later
[answer 2016.12.05] correct you are.