## Superflip composed with fourspot

Somehow, it slipped past my radar that Tomas Rokicki and Morley Davidson have established the maximum number of quarter-turns necessary to solve a Rubik’s cube (the more widely-popularised figure of 20 was for the half-turn metric). That is to say, any of the six faces can be rotated by 90° clockwise or anticlockwise in a single move.

We can consider the Cayley graph, which is 12-regular (since at each position, there are 12 possible moves one can make) and bipartite; this is most easily seen by noting that a quarter-turn induces an odd permutation (a 4-cycle) on the eight vertices of the cube. Then the maximum number of moves necessary is, by definition, the diameter of the Cayley graph.

Anyway, the diameter of the graph is 26, and surprisingly there are very few positions which take 26 moves to solve (compared with billions of distance-20 positions in the half-turn metric). Indeed, it is conjectured that there are only three such positions, or one up to isomorphism: the so-called superflip composed with fourspot. It is worth explaining this terminology.

The superflip is the unique non-identity element in the centre of the Rubik’s cube group where every cell remains in its original position, but with all 12 edge cells reversed. Since it commutes with every element, the order of composing it with fourspot is not important.

The fourspot is a common pattern where four of the face cells undergo a derangement, and all other cells remain unchanged. Somehow, the fourspot has acquired something of a cult following, and even boasts its own music video:

In other news, the Erdos discrepancy problem was recently solved by Terry Tao. There is a natural way to turn this into a two-player game. Specifically, Alice and Bob take turns colouring positive integers (cobalt blue for Alice and moss green for Bob), and the game terminates when there exists a progression {n, 2n, 3n, …, kn} (for some positive integers k and n) such that the difference between the number of cobalt blue elements and moss green elements in that progression exceeds d (a predetermined constant that determines the difficulty of winning).

Also, there’s a petition for LEGO to produce a Lovelace and Babbage themed set. You’re strongly encouraged to support the petition; more information is available over at the Aperiodical. This is currently in the limelight due to Ada Lovelace’s impending 200th birthday (this December); watch this cp4space for more details of Lovelace-themed events in the coming months…

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## Coverings, convolutions and Corbynatorics

Amidst Saturday’s turmoil, the following problem came into consideration:

There are plans to build a nuclear power station on an initially empty 12-by-12 chessboard. Doing so would require an empty 4-by-3 or 3-by-4 rectangular region of squares. To foil this plan, Jeremy decides to erect coal mines on squares of the chessboard (with each mine occupying a single square) to ensure there is no sufficiently large empty rectangle. What is the minimum number of coal mines necessary for Jeremy to prevent the construction of the nuclear power station?

With little effort, one can obtain a lower bound of 12 by partitioning the chessboard into twelve disjoint rectangles, noting that each one must contain at least one mine:

Similarly, we can establish an upper bound of 16 by placing a mine at (x, y) if and only if x and y are both divisible by 3. Can this bound be improved? In particular, note that it is never wise to place a coal mine near the edge of the board, since we can move it inwards without reducing its effectiveness for blocking potential nuclear power plants. Without loss of generality, we will therefore only consider arrangements where the mines are in the 8-by-8 ‘reasonable region’ in the middle of the board:

One reasonable approach is to place mines on the boundary of this region, and then concentrate on the centre of the board. This reduces the upper bound from 16 to 12, and is therefore optimal:

What about a larger board? In particular, what is the lowest density of mines that can be placed on the infinite plane to leave no 4-by-3 or 3-by-4 rectangle unoccupied? Again we have a lower bound of 1/12 and an upper bound of 1/9. The upper bound can actually be reduced to 1/10 by using the following lattice:

It appears that it would be difficult, if not impossible, to improve upon this. Observe that if an empty 3-by-3 box occurs in any valid configuration, then there must be a coal mine in each of the four surrounding 1-by-3 boxes. Moreover, the lattice shown above achieves this bound, and has no 3-by-3 boxes containing more than one mine.

But how would one prove optimality? Given an arrangement of mines, we define a scoring function over $\mathbb{Z}^2$ by means of a convolution. Specifically, each mine contributes a score to its own square and those surrounding it as follows:

That is to say, each mine contributes a score of 20 to its own cell, 15 to the four edge-adjacent cells, 16 to the four other vertex-adjacent cells, 4 to the four cells at a distance of 2 in a horizontal or vertical direction, and 5 to each of the eight cells a knight’s move away from the mine.

Then the score associated with a given square is simply the sum of the contributions from nearby mines.

This may appear to be a seemingly arbitrary scoring function, but it was chosen to exhibit the following properties:

• The score associated with the density-1/10 lattice is uniform across all cells (and equal to 20);
• For any valid configuration (one with no 3-by-4 or 4-by-3 empty rectangle), every cell gets a reasonably high score (in this case, the minimum possible is 15).

Since the total score contributed by each mine is 200, we can determine the density by dividing the average score by 200. Hence, proving that the density-1/10 lattice is optimal is equivalent to the following statement:

The average score in any valid configuration must be at least 20.

It transpires that this is true, and can be obtained by locally perturbing the scoring function (without violating conservation of total score) so that no cell has a score below 20. This is a finite check, and sufficiently small that it can be performed without the assistance of a computer.

Suppose we have a cell with a score strictly less than 20. It is easy enough to show that one of the following two possibilities must arise:

• Case I: The score is 19, and the cell is positioned between two mines like so:

• Case II: The cell belongs to precisely one empty 3-by-3 region, which is surrounded by four mines.

We deal with all instances of Case I before Case II, by incrementing each ’19’ and decrementing the two cells which are edge-adjacent to the ’19’ and vertex-adjacent to the neighbouring mine. This actually decreases the overall score, so (if we can show that the average score is still at least 20) any configuration with an example of Case I is suboptimal.

Case II is more complicated to address than Case I, because each of the four mines surrounding the 3-by-3 region can be in any of three possible positions. Moreover, empty 3-by-3 regions can potentially overlap slightly. Nevertheless, if we consider only the portion of the 3-by-3 region which belongs to no other 3-by-3 regions, the average score is still at least 20, with equality if and only if the mines are arranged to resemble a portion of the density-1/10 lattice. This is true even after we perform any reductions associated with Case I.

Convincing yourself is a matter of testing each of the possible sub-cases for Case II (there are at most 81, and this quickly reduces when you take symmetry into account).

### What have convolutions ever done for us?

In addition to solving this rather niche problem in Corbynatorial geometry, convolutions have a wealth of other applications.

• Even if we restrict ourselves to considering convolutions of an arbitrary function on a square grid with a small finitely-supported kernel, this is useful for image-processing. Common edge-detection algorithms use this concept, right up to the sophisticated convolutional neural networks underlying Google’s image recognition software. These are explained very clearly in various places.
• A continuous analogue is taking a convolution of measurable functions over $\mathbb{R}^n$ or the torus obtained by quotienting by a lattice. Gaussian kernels are often used.
• More generally, one can convolve continuous functions over an arbitrary compact group. In the cases where the group is Abelian, this can be computed quickly by means of a Fourier transform (although this is typically overkill when the kernel has small finite support).
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## ∂ eclipse

There was a solar eclipse today.

Certain regions (including the Faroe Islands, familiar to anyone who has listened to the Shipping Forecast) landed in the umbra, experiencing a total eclipse. I was slightly less fortunate, landing in the penumbra (thereby only seeing a partial eclipse, which itself was largely obscured by cloud cover).

### Pi day and the media

Since citizens of the United States feel the need to use Middle-Endian date formats (mm/dd/yy, instead of the standard yyyy-mm-dd format), and because sequences of digits can be interpreted as decimal digits irrespective of whether or not they actually are, the 14th March 2015 was proclaimed ‘pi day’. Consequently, after a 10.5-mile run, the founders of Oligomath baked a pie containing blackberries, blueberries and raspberries. Unfortunately we didn’t photograph the pie, so here’s a plan view of the run instead:

As one would expect, this has been covered in an extensive barrage of posts in the Aperiodical, including an ode* to constrained writing by Alex Bellos.

* the type of poem, rather than differential equation. Feel free to write an ode to ordinary differential equations.

A sesquimonth ago, the same Alex Bellos invited me down to London to watch a media screening of X+Y. I then wrote a review for his Guardian column, attracting a similar amount of controversy as Noa Lessof-Gendler’s review of a particular coffee house in Cambridge. If you enjoyed the latter, you may also want to read her brother’s more positive synopsis of the Rado Graph.

### Distributed searching

Around the same time, I launched a distributed search I had been working on since August. It collects data from people running a particular Golly script to simulate millions of random initial seeds in a variety of cellular automata rules. It’s currently gathering around 8 * 10^9 objects per day, depending on how many machines are running the script, and has already found previously-undiscovered patterns. We’ve also had a few other surprises, such as these pairs of interacting spaceships at the bottom of this list:

The third column gives the total number of occurrences so far in the census. So whilst over 19 billion gliders have made an appearance, and millions of each of the other ‘standard spaceships’, there are only a handful of occurrences of other moving objects (in this case, pairs of interacting standard spaceships).

If you want to get involved, you can download the requisite software (Golly, Python, and the search script) from here. In order to maximise your machine’s potential, run one instance of the search program per CPU. For instance, if you have a quad-core computer, run four instances of Golly, each running the apgsearch script.

The script will prompt you for the number of soups to search between successive uploads (default: 5000000), the rule to use (default: B3/S23 = Conway’s Game of Life), and the seed symmetry (default: C1 = no symmetry). You can leave all of these parameters unchanged.

Happy searching…

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## Proto-Penrose tilings

A while ago, Roger Penrose famously filed a lawsuit against Kimberly Clark when his wife discovered that a roll of quilted lavatory paper was adorned with his aperiodic tiling. Although the tiling occurs naturally in quasicrystals, Penrose was probably the first person to discover and abstractly formalise the pattern. Nevertheless, there have been several historical ‘near-misses’ which came close, and indeed would have yielded isomorphic tilings if applied recursively. Since we’ve already mentioned Penrose, it seems only natural to approach this in inverse chronological order.

### Kepler’s monsters

It is unknown as to exactly why Johannes Kepler investigated this particular tiling, although one sensible suggestion is that he was trying to construct a tiling of the plane involving only shapes with D10 symmetry. For example, ten regular pentagons neatly fit around a regular decagon of the same edge length, and it is possible to continue the pattern further if pentacles are also allowed:

Unfortunately, however, it is not possible to continue this process forever. Pretty soon one is required to have decagons overlapping, as in this particular drawing by Kepler:

Kepler’s original drawing of the Kepler’s Monsters tiling.

With the assistance of these fused decagons, termed monsters by Kepler, it is possible to create various periodic and nonperiodic tilings of the plane. Craig Kaplan explored many variations on this theme in his article, with the Keplerian problem of finding a tiling using only (finitely many distinct) shapes of D10 symmetry remaining unsolved.

Even earlier were several attempts by Albrecht Dürer, whom you may know from his Melencolia I engraving inter alia. These featured pentagons and rhombi, similar to Penrose’s original tiling but with less sophisticated matching rules incapable of enforcing aperiodicity.

### Girih tiles

I was in the Persian section of the Victoria and Albert Museum during the post-Christmas weekend at the end of last year, and happened to notice various geometric designs. Some of these were periodic carvings of knotwork, not unlike similar examples in the Alhambra in Moorish Spain.

Others were more exciting. One particularly ornate example features angles commensurate with the internal angles of a pentagon, and bears a striking similarity to the aforementioned tilings of Penrose, Kepler and Dürer:

Apparently these are formed from a set of five so-called Girih tiles, and patterns of a similar nature occurred throughout the Islamic world since the Middle Ages. In particular, I am intrigued to see the more sophisticated Girih tilings featuring patterns on two scales, where the small-scale pattern is created by applying a set of subdivision rules to the large-scale pattern. This is analogous to the method by which Penrose tilings can be constructed, as you will know from my online demonstration:

This surprising and fascinating connection between Islamic architecture and quasicrystalline tilings was first discovered and investigated by Paul Steinhardt and Peter Lu, the latter of whom presented an exposition on the subject at the Harvard Physics Colloquium:

If you found this interesting, there is a talk on early Islamic mathematics by Dr. Bursill-Hall at 16:00 today in Meeting Room 3 of the Centre of Mathematical Sciences.

### Miscellany

Finally, I have a few late items of news. Stuart Gascoigne has just sent in the first 8192 2-adic valuations of polylogarithms. The ‘spike’ in the last row (associated with 8192) penetrates significantly deeper into the grid, with Q(3, 8181) being the first integer of the form 64n + 53:

Here are some other recent highlights:

• Eugenia Cheng has committed a third instance of gastro-mathematical marketing, namely a formula for the perfect doughnut.
• Alex Bellos and I attended a pre-release viewing of X+Y, an action thriller following the romantic life of a British olympiad contestant. Expect a review from us next month.
• Jeffrey Ventrella has been investigating self-similar fractal curves for a while, and recently sent me this analysis of fractal curves based on Gaussian and Eisenstein integers.
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## Oligomath 2: When are polylogarithms integers?

Consider the following series, where k and r are positive integers (and r is strictly greater than 1). Here Li_-is a polylogarithm.

We shall pose the following question:

For what values of k and r is Q(r, k) an integer?

It is possible to show that Q(r, k) is necessarily rational and that the denominator divides a power of r − 1. We do this by induction on k, expressing Li_-k as a rational function using the following recurrence relation and applying the product rule:

In particular, for small values of k, we obtain the following:

It is immediately evident that if r = 2, Q(r, k) must necessarily be an integer. On the other hand, if r ≥ 4, Q(r, k) can never be an integer. This leaves only the boundary case of r = 3, which is surprisingly non-trivial.

### Which values of Q(3, k) are integers?

The answer is ‘some of them’.

Recalling that the denominator is necessarily a power of two, we can determine whether Q(3, k) is an integer by examining its 2-adic valuation and seeing whether it is positive. (The 2-adic valuation v2(n) of an integer n is defined to be the largest k such that 2^k divides n. This is extended to rationals by setting v2(p/q) = v2(p) − v2(q).)

I decided to compute v2(Q(3, k)) for all k between 0 and 2047. The results are summarised in the grid below, where non-integers are red, integers are green, and the saturation of the colour corresponds to the absolute value of the 2-adic valuation:

There are lots of obvious patterns, but many of them seem to break eventually. In particular, one may initially have conjectured that all numbers of the form 64n + 56 are non-integers, but this pattern breaks down at the pale green square corresponding to 1528. Indeed, there appears to be a ruler sequence of spikes originating from the right-hand side of the grid, which (extrapolated in the natural way) would eventually hit every residue class and therefore break every congruence pattern.

So far, Matei Mandache and Sahl Khan established the following facts:

• If k is a power of two, then Q(3, k) is not an integer;
• If k is one less than a power of two (and k ≥ 3), then Q(3, k) is an integer.

### Stirling numbers of the second kind

Since there seems to be no obvious pattern, we decided to investigate the expression for the polylogarithm in terms of Stirling numbers of the second kind:

To compute the 2-adic valuation of a sum, it would certainly help to compute the 2-adic valuation of each term in the sum. It would then be possible to obtain either a lower bound or the exact value* of the 2-adic valuation of the sum by applying the following rules (known to anyone who’s played Gabriella Cirulli’s 2048 game):

• If v2(a) = v2(b), then v2(a + b) > v2(a);
• Otherwise, v2(a + b) = min(v2(a), v2(b)).

* We obtain the exact value if the number of terms of minimal v2 is odd; otherwise, we merely obtain a lower bound.

By multiplicativity, it suffices to know the 2-adic valuations of the following:

• v2(i!) = i − (sum of binary digits of i)
• v2(2^(i+1)) = i + 1
• v2(S(k + 1, i + 1)) = ???

Fortunately for us, there exists a paper on arXiv entitled:

The 2-adic valuations of Stirling numbers of the second kind

Considering the fact that we want to know the 2-adic valuations of Stirling numbers of the second kind, this paper seemed rather promising. However, upon closer inspection, it is by no means exhaustive (many values are not covered by the paper), so is unlikely to answer our question definitively.

So, as usual, let the collaborations begin!

Posted in Oligomath | 5 Comments

## Oligomath 1: Deletion and duplication

Collaborative mathematics tends to be extremely fruitful. This is epitomised by the recent ‘polymath’ projects, culminating in a nice proof of density Hales-Jewett along with the frighteningly impressive results on bounded gaps between primes. On a much smaller scale, several of us were contemplating a bunch of interesting and completely unrelated problems we devised in the downstairs kitchen of a converted house in Burrell’s Field, Trinity College, Cambridge.

We’ve made non-trivial progress on many of these, and even succeeded in solving one! It was a natural question, which grew out of a much easier problem on an IMO shortlist:

Suppose you begin with a finite graph, and are allowed to apply operations of the following form:

• Deletion: Choose a vertex of odd degree and delete it (along with all edges incident with it).
• Duplication: Produce an identical copy of the graph, and connect each vertex in the original graph to its corresponding vertex in the clone.

Show that, after finitely many operations, one can reach a graph containing no edges.

This is not terribly difficult. It did, however, lead us to consider several related questions (which, at the time of proposal, were all completely new and unsolved). Observe that neither of the allowed operations can decrease the number of connected components of the graph, so if the original graph G has n connected components, then it is impossible to end up with fewer than n vertices. This prompts the following question:

Is this bound tight? Specifically, given a graph G with n connected components, can we necessarily reduce it to the empty graph on n vertices?

It is not too difficult to show that this is equivalent to the question where n = 1. In particular, every time we perform a duplication operation, we then immediately reduce any two-vertex components to single vertices by deletion. Then we can just concentrate on each component sequentially in isolation, without undoing any of our ‘previous work’. (More formally, we can induct on the value of n.)

Hence the problem reduces to:

Can every connected graph be reduced to a single vertex?

In the next few sections, we shall present a proof in the order we devised it.

### Tackling the square

It was quite informative to consider some special cases. Any tree can quickly be destroyed by repeatedly removing leaves until we are left with a single vertex. We then decided to consider the square, since it’s the simplest graph which is not a tree (chromatic number is considered to be more important than size, since it cannot increase under either of the operations). After much experimentation, Matei Mandache found the following sequence of operations (which is provably optimal):

Using this as a lemma, we were then able to reduce any cycle, and more generally any connected graph with $E \leq V + 1$. This was something of a casebash, however, and clear that it would not generalise to yield a full proof. Surprisingly, this ad hoc case of reducing a square to a single vertex does turn out to be essential to our proof!

After having reduced the cube, it seemed natural to try the other Platonic solids. The tetrahedron is quite trivial (we can delete a vertex to yield a triangle, duplicate it to obtain a triangular prism, and remove two non-adjacent vertices to give a tree). The dodecahedron is slightly more convoluted, and this particular desynthesis embodies the method applicable to reducing any cycle:

According to Donald Knuth, if a conjecture about all graphs seems plausible, it’s often the case that the Petersen graph is a counter-example. But in this case, the Petersen graph is really easy to reduce (it doesn’t require any duplication operations):

### The hypercube and unduplication

We realised that it would be very useful if we developed a method to reduce an arbitrary hypercube, since induced hypercubes appear as the result of performing repeated duplication operations. The first step in reducing a degree-(2k) hypercube must be to duplicate it to form a degree-(2k + 1) hypercube, since the former has no vertices of odd degree. Consequently, the first thing to consider was the degree-5 hypercube (with 32 vertices). Unfortunately, these things get very messy very quickly:

A more helpful way to represent this is as a ‘meta-cube’, each vertex of which is a ‘meta-vertex’ corresponding to a square of four ordinary vertices. Then we can delete an entire meta-vertex by deleting two non-adjacent vertices, followed by the other two vertices (the so-called Illingworth manoeuvre). Indeed, in any meta-graph, we can delete any meta-vertex of odd degree. And trivially we can duplicate the entire meta-graph. This has the following consequence:

The product G × C4 of a graph G with a square C4 can be reduced to a single square C4 if the graph G can be reduced to a single vertex.

And since we already know how to reduce a square to a single vertex, we obtain the following lemma:

If G can be reduced to a single vertex, then so can the product of G with a square C4.

Naturally, this can be generalised by iterated application of itself:

If G can be reduced to a single vertex, then so can the product of G with an arbitrary hypercube.

We shall think of this from an equivalent, but much more liberating, perspective:

If we grant ourselves the operation of ‘unduplication’, then the set of graphs we can reduce to a single vertex does not increase.

Henceforth, we shall thus allow unduplication.

### Consequences of unduplication

Suppose we have two adjacent vertices of even degree, for example any two adjacent vertices in the octahedron. We can duplicate the graph so that these two adjacent vertices become a square of four vertices of odd degree. Then, we can remove that square by the Illingworth manoeuvre of deleting two non-adjacent vertices followed by the other two vertices. Finally, unduplicate the graph to recover the original graph minus those two vertices:

So we now have a fourth operation, namely deleting two adjacent vertices of even degree. I now claim that two of our four operations can together reduce any connected graph to a single vertex:

• Removal of a vertex of odd degree.
• Removal of two adjacent vertices of even degree.

Suppose instead that there is a counter-example, and consider a minimal counter-example (i.e. one with the fewest vertices). This must have the following properties:

• Every odd vertex is a cut-point (since otherwise we could remove that odd vertex without disconnecting the graph);
• Removing any pair of adjacent even vertices must also disconnect the graph.

Consider a graph where these properties hold. We’ll distinguish two cases, namely those where every vertex is even, and those where there exists at least one odd vertex. In each case, we shall obtain a contradiction.

### Case I: At least one odd vertex exists

For each odd vertex, we can define its ‘depth’ to be the minimum order of the connected components produced when that vertex is deleted. Now take a vertex v of minimum depth; the minimal connected component produced by deleting v must necessarily consist only of even vertices (since otherwise any of those odd vertices must have lower depth). So our counter-example must resemble this:

We’ll concentrate entirely in the region on the right, where everything has even degree. Matei Mandache had the idea of separating this into strata according to their distance from v.

If the final stratum contains two vertices connected to each other, then we can delete those two adjacent vertices without disconnecting the graph. Hence, assume without loss of generality that the final stratum contains no edges.

Then every vertex in the final stratum must be connected to at least two vertices in the penultimate stratum. Consequently, we can remove one vertex in the final stratum and its neighbour in the penultimate stratum without disconnecting the graph (since everything in the final stratum is now connected to at least one vertex in the penultimate stratum).

This contradicts our assumption of minimality.

### Case II: All vertices have even degree

This is actually amenable to the same proof technique as before. We let v be an arbitrary vertex, and again stratify the graph so that it resembles this:

Now we can use exactly the same proof as in the previous case.

The result follows.

### What other questions can we ask?

We can ask the more general question as to whether, given graphs G and H, we can get from G to H using those two operations. Note that in general, we no longer have the freedom of unduplication or the removal of two adjacent vertices of even degree.

Clearly, we can get from G to H only if H is an induced subgraph of the product of G with a hypercube. Is the converse also true? Again, this is a strictly harder question than the one that we solved in the downstairs kitchen of ‘V’ House, but it should be quite fun to solve.

Let the collaborations begin! #oligomath

Posted in Oligomath | 4 Comments

## Holyhedron update

A while ago, I mentioned the concept of holyhedra: polyhedra, all of whose faces are multiply-connected. The first known example had 78 585 627 faces, and was later superseded by a highly symmetric example with as few as 492 faces. This is far from optimal, though, as yesterday I received an e-mail from Nathan Ho describing a 12-faced holyhedron!

This leaves open the question of the minimum number of faces a holyhedron can possess. We know that for the deceptively similarly named golyhedra, the answer is 11 as demonstrated by Alexey Nigin’s wonderfully minimal example:

The corresponding lower bound of 11 can be derived from the following simple derivation:

• By numerical considerations, the number of faces can only be 0 or 3 (modulo 4).
• If an orthogonal polyhedron has 8 or fewer faces, then one of the three axes has at most two faces orthogonal to it. Clearly, they must point in opposite directions and have equal area, so the polyhedron cannot be a golyhedron.

### Other projects

In other news, Hans Havermann has attacked the ménage prime problem with a computer search, exhausting all ménage numbers with fewer than 110 000 digits. Consequently, the expected number of ménage primes with fewer than 1000000 digits is about 0.195.

Also, I mentioned a while ago about the project to create a 31c/240 spaceship in Conway’s Game of Life. Dave Greene finally implemented my shield bug design, although changing one of the details resulted in the completed spaceship looking nothing like a shield bug whatsoever!

It’s an immense 934856 by 290486 pixels in size, making it (by area) much larger than the 17c/45 caterpillar that began the tradition of naming these large cumbersome track-laying spaceships after insects. The (23,5)c/79 and (13,1)c/31 projects appear to have been abandoned.

On an unrelated topic, a discussion with Tim Hutton and Simon Rawles in a pub in Cambridge has culminated in them investigating various embeddings of surfaces by crocheting them. Most recently Tim has produced a Chen-Gackstatter surface in this manner:

Another interesting blog by a computer scientist is that of Susan Stepney. One of her recent posts featured a video of a rendition of a particular music video in the Alexandroff Corridor of Newnham College, Cambridge, which Trinity has recently reproduced in an attempt to encourage the artist to perform at the forthcoming (149th) May Ball:

This has recently gained quite a lot of media attention across Europe, appearing in both a Spanish and a Croatian newspaper. Further bulletins as events warrant…

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