Shortly after solving the monopolar partition problem for line graphs, Jing Huang and I realized that our solution could be used to solve the “precoloured” version of the problem, and then further extended to claw-free graphs. Jing presented our result at the French Combinatorial Conference and the proceedings have now been published in Discrete Mathematics.

I’ve published a new paper in the SIAM Journal on Discrete Mathematics! The work is the result of the research term I took as an undergraduate in the summer of 2009. It studies the edge versions of the monopolar and polar partition problems, and presents a linear-time solution to both.

I am grateful to NSERC for funding my work with a Undergraduate Student Research Award, and to my supervisor and coauthor Jing Huang.

Earlier this year, I presented the first results of what would become my master’s thesis at the International Workshop on Combinatorial Algorithms. The paper, coauthored with Jing Huang and Xuding Zhu, has now been published in the LNCS proceedings. It studies the computational complexity of the following problem: in a given graph, is there a matching which breaks all cycles of a given length?

I am grateful to NSERC for funding this research with a Alexander Graham Bell Canada Graduate Scholarship.

I was recently told that European hotels are subject to a reduced VAT rate. They must have a big lobby.

My most recent talk in UVic’s discrete math seminar presented three poetic proofs by Adrian Bondy… and three actual poems summarizing the ideas in each one.

Ore’s Theorem

A red-blue $K_n$:
bluest Hamilton circuit
lies fully in $G$.

Ore’s theorem states:

Let $G$ be a simple graph on $n \geq 3$ vertices such that $d(u) + d(v) \geq n$ for any nonadjacent $u, v$. Then $G$ contains a Hamilton cycle.

Bondy’s proof is roughly as follows. Colour the edges of $G$ blue and add red edges to make a complete graph. The complete graph on $n \geq 3$ vertices has no shortage of Hamilton cycles, so choose one and label it $v_1 v_2 \ldots v_n v_1$.

Consider the blue neighbours of $v_1$ on the cycle, and then move one vertex to the right along the cycle so you’re looking at the set

There are $d_G(v_1)$ of these vertices, and if $v_1 v_2$ is red (i.e. $v_1$ and $v_2$ are nonadjacent in $G$), then the theorem’s hypothesis tells us that

Since $v_2$ has only $n-1-d_G(v_2)$ red neighbours and is not itself in $S$, that means at least one vertex $v_{i+1} \in S$ is a blue neighbour of $v_2$. We can now replace $v_1 v_2$ and $v_i v_{i+1}$ in the cycle with the two blue edges $v_1 v_2$ and $v_2 v_{i+1}$ to get a Hamilton cycle of $K_n$ with at least one more blue edge than we had before.

We can make the same argument again and again until we have a Hamilton cycle of $K_n$ with no red edges. The cycle being blue means it lies entirely in $G$, which proves the theorem.

Brooks’ Theorem

Greedily colour,
ensuring neighbours follow
all except the last.

Choose the last vertex wisely:
friend of few or of leaders.

Brooks’ Theorem says:

If $G$ is connected and is not an odd cycle or a clique, then $\chi(G) \leq \Delta(G)$.

If $G$ is not regular, Bondy colours it as follows. Let $r$ be a vertex of smaller-than-maximum degree. If we consider the vertices in the reverse order of a depth-first search rooted at $r$, each vertex other than $r$ has at least one neighbour later in the order and at most $d(v)-1$ neighbours earlier. Greedily colouring in this order will assign one of the first $d(v) \leq \Delta(v)$ colours to each $v \neq r$, and at least one colour remains available for $r$ since its degree is strictly smaller than $\Delta(G)$.

If $G$ is regular, there are a few cases. If it has a cut vertex, we can break the graph into two (non-regular) parts, colour each of them separately, and put them back together. If it has a depth-first tree that branches at some point, then Bondy constructs an ordering similar to the one for the non-regular case: if $x$ has distinct children $y, z$ in the tree, $G$ can be ordered so $y$ and $z$ come first, $x$ comes last, and every other vertex has at least one neighbour later on in the order. A greedy colouring according to this order uses at most $\Delta(G)$ colours.

Finally, if $G$ is regular, 2-connected, and all of its depth-first trees are paths, it turns out that $G$ must be a chordless cycle, a clique, or a complete bipartite graph. Since bipartite graphs have $\chi(G) = 2 \leq \Delta(G)$, the only exceptions to the theorem are odd cycles and cliques.

Vizing’s Theorem

Induction on $n$.
Swap available colours
and find SDR.

Vizing’s theorem says:

For any simple graph $G$, the edge chromatic number $\chi'(G)$ is at most $\Delta(G)+1$.

The proof is by induction on the number of vertices; pick a vertex $v$ and start with a $(\Delta(G)+1)$-edge-colouring of $G-v$.

Ideally, we could just colour the edges around $v$ and extend the edge-colouring to one of $G$. Those colours would need to be a system of distinct representatives (SDR) for the sets of colours still available at each of the neighbours of $v$.

If we can’t find a full SDR, then we can still try to colour as many of $v$‘s edges as possible. Then, if an edge $uv$ is still uncoloured, we can follow a proof of Hall’s Theorem to get a set of coloured edges $u_1 v, u_2 v, \ldots, u_k v$ such that:

• $uv$ can’t be coloured because there are fewer than $k+1$ total colours among those available at $u$ and $u_1, u_2, \ldots, u_k$ in the colouring of $G-v$, but
• if you un-colour any one of the edges and swap around the colours of the other ones, it would become possible to colour $uv$.

Note also that

• regardless of how we assign $\Delta(G)+1$ colours to edges of $G$, each vertex will have at least one colour left unused by the edges around it.

The first and third facts put together tell us that some colour (say, blue) is available at two vertices among $u$ and $u_1, u_2, \ldots, u_k$.

The second fact implies that any colour (e.g. red) still available at $v$ can’t be available at any of the $u_i$, as otherwise we could swap around the colours to colour $uv$ and then use red for $u_i v$.

Look at the subgraph $H$ of $G$ formed by the red and blue edges — it’s made of paths and even cycles. We know that $v$ is incident with a blue edge but not a red one, so it’s the end of a path in $H$.

We also know that there are two vertices among $u, u_1, u_2, \ldots, u_k$ that are incident with a red edge but not a blue one; those are also ends of paths in $H$. The paths might be the same as each other, and one might be the same as $v$‘s path, but the important thing is that we have a red-blue path where at least one of the ends is a neighbour $u'$ of $v$ and the other end is any vertex other than $v$.

We can now modify the colouring of $G-v$ to swap red and blue along this path, which frees up red at $u'$. If $u'$ is $u$ itself, then we can colour $uv$ red; otherwise, we can uncolour $u'v$, swap colours around to colour $uv$, and use red for $u'v$.

Either way, we’ve shown that the modified colouring of $G-v$ has a larger set of distinct representatives than the original colouring. By repeating this process, we eventually find a colouring of $G-v$ that has a full SDR — and therefore can be extended to a full $(\Delta(G)+1)$-edge-colouring of $G$.

If you’ve heard of Erdős numbers, Erdős-Bacon numbers, and the fact that Queen lead guitarist Brian May has a PhD, you may have wondered whether Brian May has a well-defined Erdős number.

As a matter of fact, he does! I traced down a collaboration path of length seven through a 1972 paper he published in Nature.

This beats the best previous attempt I found, a path of length eight through a popular science book cowritten by May. It gives him an Erdős-Bacon number of at most 10 (and an Erdős-Bacon-Sabbath number of at most 11).

Did you hear the one about the geology costume contest?

The gneiss guise finished last.

One of my favourite video games of all time is the inexplicable Katamari Damacy. Its quirky premise involves, as Wikipedia puts it, “a diminutive prince rolling a magical, highly adhesive ball around various locations, collecting increasingly larger objects until the ball has grown great enough to become a star.” In other words, it’s the most successful game ever made about exponential growth.

Katamari makes you explore a world at many different scales, all in the same level. You might start by dodging mice under a couch; just a few minutes later, you’re rolling up the family cat, the furniture, and everything else in the room. It’s an even better playable version of the Powers of 10 video, made possible by the differential equation:

You make your magically sticky katamari bigger by rolling stuff up; the bigger you are, the bigger the things you can pick up. So we would expect the radius $R$ of the katamari to grow at a rate which roughly proportional to $R$ itself. The exact rate of change is governed by some function $s(t) \approx k$, which depends on how good you are at finding a route filled with objects of just the right size for you to pick up. The solution to this differential equation

gives a formula for the katamari’s size at a given time $t$.

How justified are we in saying that $s(t)$ is roughly constant? I charted the minute-to-minute progress of four let’s players on YouTube. If the exponential model is a good one, then katamari size should trace out a straight line on a log scale. And so it is:

The runs keep up a remarkably consistent exponential pace, with a couple visible exceptions — one at the end of the level, when the world starts running out of stuff, and one at roughly the ten-minute mark, when a couple of the players struggled to find items at the right scale to roll up.

I’m not sure if this proves anything other than the fact that I like to do strange things in my spare time. But if you’re a calculus teacher with a bit of time and a PlayStation 2, I suspect this would make a very interesting three-act problem for your class.

Based on corpus data, over half of the words in a typical page of English text has four or fewer letters, with the average word length being slightly less than five.

import pandas

wd = pandas.read_csv(
"eng-ca_web_2002_1M-words.txt",
"\t",
header=None,
names=["Rank", "Word", "Frequency"]
)

wd.Word = wd.Word.apply(str).str.lower()
wd = (
wd.query("Frequency > 5") # Remove rare words,
.query("Word.str.match('^[A-Za-z]\*$')") # words with non-letters,
.query("Word.str.contains('[aeiouy]')") # abbrvns w/o vowels,
.groupby("Word") # and duplicates
.sum()
)

wd["Length"] = pandas.Series(wd.index, index=wd.index).str.len()

sm = wd.groupby("Length")

result = pandas.DataFrame({
"NormalizedCount": sm.Frequency.count() / sm.Frequency.count().sum(),
"NormalizedFrequency": sm.Frequency.sum() / sm.Frequency.sum().sum()
})

According to an old piece of email forwarding-spam, it’s easy to read text even if you scramble all but the first and last letters in each word. But the truth is a bit more complicated.

The ancient meme reads:

Aoccdrnig to rseearch at Cmabrigde Uinervtisy, it deosn’t mttaer in waht oredr the ltteers in a wrod are, the olny iprmoetnt tihng is taht the frist and lsat ltteer be at the rghit plcae. The rset can be a toatl mses and you can sitll raed it wouthit porbelm. Tihs is bcuseae the huamn mnid deos not raed ervey lteter by istlef, but the wrod as a wlohe.

The form of this paragraph appears at first glance to provide direct evidence of its own “azanmig” claim. But something’s a little fishy: a lot of the words aren’t actually scrambled. Short words aren’t affected much if at all by the message’s middle-muddling, and most English words are short!

Matt Davis, an actual researcher at the University of Cambridge, wrote an informative response to the meme:

There are elements of truth in this, but also some things which scientists studying the psychology of language (psycholinguists) know to be incorrect.

I’m going to list some of the ways in which I think that the author(s) of this meme might have manipulated the jumbled text to make it relatively easy to read. This will also serve to list the factors that we think might be important in determining the ease or difficulty of reading jumbled text in general.

1. Short words are easy.
2. Function words (the, be, and, you etc.) stay the same - mostly because they are short words.
3. Of the 15 words in this sentence, there are 8 that are still in the correct order. However, as a reader you might not notice this since many of the words that remain intact are function words, which readers don’t tend to notice when reading.
4. Transpositions of adjacent letters are easier to read than more distant transpositions.
5. None of the words that have reordered letters create another word.
6. Transpositions were used that preseve the sound of the original word (e.g. toatl vs ttaol for total).
7. The text is reasonably predictable.

If you want to test your own permutation powers against realistic examples, I whipped up a bookmarklet that you can use to scramble the words on any website you want to challenge!