Scientific Linux - navigation blocked

I was browsing the Scientific Computing World article A Universe of Data on the computing resources at CERN, when my attention was caught by mention of a version of Linux that I had not heard of before called Scientific Linux. What's so special about that? So I duly Googled the string "Scientific Linux" and the top hit was www.scientificlinux.org described as:

Scientific Linux - Welcome to Scientific Linux (SL) - 13:58Is a Linux release put together by Fermilab, CERN, and various other labs and universities around the world ready tuned for experimenters.
www.scientificlinux.org/ - 26k - Cached - Similar pages - Note this

So I then followed the www.scientificlinux.org link to get this:

I have never seen this sort of warning before, and I do a lot of internet browsing. I didn't explore any further because unexpected things on the internet always spook me, and by playing very safe I have managed to avoid all the nasty problems that I regularly hear about from other people.

I am reasonably sure that the problem is a trivial misconfiguration of the www.scientificlinux.org site, rather than a malicious attempt by Internet Explorer to try to prevent people from visiting this particular site. Surely, it couldn't be the case that a www.scientificlinux.org is so self-righteous that they shoo away Internet Explorer users? Or am I just being paranoid?

Dropping the Baton

There seem to have been rather a lot of dropped batons in the relay races at the Olympics (e.g. see here).

It set me thinking about where I might have seen this sort of thing happening elsewhere, and I realised that dropping the baton is like annihilating the vacuum state.

How so?

The simplest possible algebra that one can use to model the process of baton-passing goes like this:

a^† increments (by 1) the number of hands holding the baton
a decrements (by 1) the number of hands holding the baton

|0> is the "vacuum" state where the baton has one hand holding it
a^†|0> is the state where the baton has two hands holding it

a|0> = 0 is the annihilation of the vacuum where the baton has zero hands holding it, i.e. a state from which there is no way to recover

Note that it is important to define the vacuum state as corresponding to one (rather than zero) hand holding the baton, otherwise the algebra (i.e. annihilation of the vacuum) doesn't correctly model the dropping of the baton. Thus the counting of hands holding the baton is really a measure of how many excess hands are holding the baton, because the case of one hand is actually the ground (or vacuum) state in a relay race.

Most of the time the state is |0>, and during a successful handover of the baton it passes through the transition state a^†|0>, after which it returns to the state |0>. However, during an unsuccessful handover of the baton it goes to the state a|0> which is 0, where the vacuum has been annihilated.

Successful handover: a a^†|0> = |0>
Unsuccessful handover: a^†a|0> = 0

The order in which the a and a^† operations are applied is important, and is neatly summarised by how their commutator a a^† - a^†a acts on |0> (take the difference of the above equations).

(a a^† - a^†a)|0> = |0>

A stronger form of this result is the operator relation

a a^† - a^†a = 1

This relation takes note of the fact that there are n ways of applying a to the state (a^†)ⁿ |0> (i.e. choose from 1 of n excess hands to decrement by 1 the number of excess hands holding the baton), but there is only 1 way of applying a^† to the state (a^†)ⁿ |0>. The case n=0 is when the vacuum gets annihilated by application of a.

The Olympic athletes who dropped the baton were the victims of a a^† - a^†a = 1 (rather than 0). I wonder whether they saw it that way.

Ethel the Aardvark Goes Quantity Surveying

Here is some Sunday afternoon entertainment.

There was a "cheese shop sketch" before the famous Cheese Shop Sketch that we all remember. I was reminded about it whilst browsing the Wikipedia entry for Marty Feldman.

Here it is (text copied from here). The customer is played by Marty Feldman and the shop assistant by John Cleese.

Assistant: Good morning, sir.
Customer: Good morning. Can you help me? Do you have a copy of 'Thirty Days In the Samarkand Desert with a Spoon' by A.E.J. Elliott?
Assistant: Um ... well, we haven't got it in stock, sir.
Customer: Never mind. How about 'A Hundred and One Ways to Start a Monsoon'?
Assistant: ... By ... ?
Customer: An Indian gentleman whose name eludes me for the moment.
Assistant: I'm sorry, I don't know the book, sir.
Customer: Not to worry, not to worry. Can you help me with 'David Copperfield'?
Assistant: Ah, yes. Dickens ...
Customer: No.
Assistant: ... I beg your pardon?
Customer: No, Edmund Wells.
Assistant: ... I think you'll find Charles Dickens wrote 'David Copperfield', sir.
Customer: No, Charles Dickens wrote 'David Copperfield' with two 'p's. This is 'David Coperfield' with one 'p' by Edmund Wells.
Assistant: (a little sharply) Well in that case we don't have it.
Customer: Funny, you've got a lot of books here.
Assistant: We do have quite a lot of books here, yes, but we don't have David Coperfield' with one 'p' by Edmund Wells. We only have 'David Copperfield' with two 'p's by Charles Dickens.
Customer: Pity - it's more thorough than the Dickens.
Assistant: More thorough?
Customer: Yes ... I wonder if it's worth having a look through all your 'David Copperfields'...
Assistant: I'm quite sure all our 'David Copperfields' have two 'p's.
Customer: Probably, but the first edition by Edmund Wells also had two 'p's. It was after that they ran into copyright difficulties.
Assistant: No, I can assure you that all our 'David Copperfields' with two 'p's are by Charles Dickens.
Customer: How about 'Grate Expectations?
Assistant: Ah yes, we have that ...
He goes to fetch it and returns to the counter.
Customer: ... That's 'G-r-a-t-e Expectations', also by Edmund Wells.
Assistant: I see. In that case, we don't have it. We don't have anything by Edmund Wells, actually - he's not very popular.
Customer: Not 'Knickerless Nickleby'? That's K-n-i-c-k-e-r
Assistant: No!
Customer: Or 'Quristmas Quarol 'with a Q?
Assistant: No, definitely ... not.
Customer: Sorry to trouble you.
Assistant: Not at all.
Customer: I wonder if you have a copy of 'Rarnaby Budge'?
Assistant: (rather loudly) No, as I say, we're right out of Edmund Wells.
Customer: No, not Edmund Wells - Charles Dikkens.
Assistant: Charles Dickens?
Customer: Yes.
Assistant: You mean 'Barnaby Rudge'.
Customer: No, 'Rarnaby Budge' by Charles Dikkens ... that's Dikkens with two 'k's, the well-known Dutch author.
Assistant: No, no - we don't have 'Rarnaby Budge' by Charles Dikkens with two 'k's the well-known Dutch author, and perhaps to save time I should add right away that we don't have 'Carnaby Fudge' by Daries Tikkens, nor 'Stickwick Stapers' by Miles Pikkens with four Ms and a silent Q, why don't you try the chemist?
Customer: I did. They sent me here.
Assistant: (making a mental note) ... Did they?
Customer: I wonder if you have ... 'The Amazing Adventures of Captain Gladys Stoat-Pamphlet and her Intrepid Spaniel Stig among the Giant Pygmies of Corsica', Volume Two.
Assistant: No, we don't have that one. Well, I mustn't keep you standing around all day ..
Customer: I wonder if ...
Assistant: No, no, we haven't got it. I'm closing for lunch now anyway.
The assistant moves rapidly away from the counter.
Customer: ... But I thought I saw it over there.
The assistant checks and turns slowly.
Assistant: ... What?
Customer: Over there.
He indicates a bookshelf.
Customer: 'Olsen's Standard Book of British Birds'.
Assistant: (very suspiciously) 'Olsen's Standard Book of British Birds'?
Customer: Yes.
Assistant: ... 0-l-s-e-n?
Customer: Yes!
Assistant: B-i-r-d-s?
Customer: Yes!
Assistant: Well, we do have that one, yes.
He goes and takes the book off a shelf.
Customer: ... The expurgated version, of course.
Assistant: ... I'm sorry, I didn't quite catch that.
Customer: The expurgated version.
Assistant: The expurgated version of 'Olsen's Standard Book of British Birds'?
Customer: Yes. The one without the gannet.
Assistant: The one without the gannet?! They've all got the gannet it's a standard bird, the gannet, it's in all the books.
Customer: Well I don't like them. They've got long nasty beaks! And they wet their nests.
Assistant: But ... but you can't expect them to produce a special edition for gannet-haters!
Customer: I'm sorry, I specially want the one without the gannet.
The assistant is speechless.
Assistant: All right!
He suddenly tears out the relevant page.
Assistant: Anything else?
Customer: Well, I'm not too keen on robins.
Assistant: Right! Robins, robins ...
He tears that one out too and slams the book on the counter.
Assistant: No gannets, no robins - there's your book!
Customer: I can't buy that. It's torn.
Assistant: ... So it is! He tosses it into the bin.
Customer: I wonder if you've got ...
Assistant: Go on! Ask me another.
Customer: How about 'Biggles Combs his Hair'?
Assistant: No, no, we haven't got that one, funny. Try me again.
Customer: 'The Gospel According to Charlie Drake'?
Assistant: No ...
Customer: Have you got 'Ethel the Aardvark Goes Quantity-Surveying'?
Assistant: No, no, we haven't ... which one?
Customer: 'Ethel the Aardvark Goes Quantity-Surveying'.
Assistant: 'Ethel the Aardvark'?! I've seen it! We've got it!!
He dashes to a bookshelf, finds it, and holds it up triumphantly.
Assistant: Here! Here!!! 'Ethel the Aardvark Goes Quantity Surveying'. Now - buy it!
He slams it on the desk. The customer stares in horror!
Customer: ... I haven't got enough money on me.
Assistant: (quickly) I'll take a deposit!
Customer: I haven't got any money on me.
Assistant: I'll take a cheque!
Customer: I haven't got a cheque book!
Assistant: It's all right, I've got a blank one!
Customer: I don't have a bank account!!
Assistant: ... All right!! I'll buy it for You!
He rings the purchase up and pays for it himself. He gives the change to the customer.
Assistant: There we are, there's your change - that's for the taxi home ...
Customer: Wait! Wait! Wait!
Assistant: What? What? What?!!!
Customer: ... I can't read ...
Assistant: Right! Sit!! ...
He sits the customer down on his knees and starts to read aloud.
Assistant: 'Ethel the Aardvark was trotting down the lane one lovely summer day, trottety-trottety-trot, when she saw a nice Quantity-Surveyor ...

Twerp Bollickagh - the website

Twerp Bollickagh

It could be the title of a journal!

I suspect that many of the experimentally accurate theoretical "predictions" given in a "Grand Unified Theory" that is available online (search for the string "The calculated relations between the lepton masses") were arrived at by exhaustive numerology, i.e. by searching through a large number of simple expressions to find the ones that gave the required results. Then a "proof" of each of these results was reverse-engineered using pseudo-physical explanations rather than using rigorous maths.

Let me show you an example of what I mean.

This "GUT" gives some simple expressions for various mass ratios. There is even an expression for the ratio of the neutron mass to the electron mass, which depends only on the electomagnetic coupling strength (i.e. fine structure constant) and not on the strong interaction strength. How do the quarks and gluons in the neutron know how to interact in order to give this amazing result?

The expression given by this "GUT" for the muon to electron mass ratio (i.e. mμ/me) is

(α^(-2) / 2π)^(2/3) (1 + 2π α^2 / 2) / (1 + α/2)

which produces a value 206.76828 that closely corresponds to the experimentally observed value 206.76827.

Let's see whether it is possible to "derive" this result by an exhaustive search of all simple expressions of this general type. The parameterisation that I will use is the most general form that is suggested by the mass ratio quoted above

f[{a1, a2, a3, a4}, {b1, b2, b3, b4}, {c1, c2, c3, c4}, {d1, d2, d3, d4}, α]] =
(a1/a2)^(a3/a4) (α)^(b1/b2) (2π α^2)^(b3/b4) (1 + (c1/c2)(α) + (c3/c4)(2π α^2)) / (1+ (d1/d2)(α) + (d3/d4)(2π α^2))

where all of the parameters are integers which are grouped in pairs to form rational fractions. To compute numerical results I inserted specific values for these parameters (avoiding singular cases), I use α=0.00729735, and a target mass ratio mμ/me=206.76827.

I then computed f[{a1, a2, a3, a4}, {b1, b2, b3, b4}, {c1, c2, c3, c4}, {d1, d2, d3,
d4}, α]] for all parameter values in the following small ranges (I have been rather cavalier and restricted the ranges to save time):
{a1, 1, 2}, {a2, 1, 2}, {a3, 1, 3}, {a4, 1, 3},
{b1, 0, -3, -1}, {b2, 1, 3}, {b3, 0, -3, -1}, {b4, 1, 3},
{c1, 0, 2}, {c2, 1, 2}, {c3, 0, 2}, {c4, 1, 2},
{d1, 0, 2}, {d2, 1, 2}, {d3, 0, 2}, {d4, 1, 2}
There is some repetition of trial solutions here, but this doesn't matter.

I then selected from this large set of trial solutions all of the cases that predicted a value for mμ/me that lay within 0.01 of the target value, and here they are (in decreasing order of goodness of fit) with the prediction errors shown in square brackets:

(1/(2π α^2))^(2/3) (1 + π α^2) / (1 + α/2) [0.0000110213]
(1/(2π α^2))^(2/3) (1 + 2π α^2) / (1 + α/2 + π α^2) [0.000130968]
(1/(2π α^2))^(2/3) (1 + α/2 + π α^2) / (1 + α) [0.00261802]
(1/(2π α^2))^(2/3) (1 + α/2 + 2π α^2) / (1 + α + π α^2) [0.0027371]

The best fit solution at the top of this list is the same as the one given by the "GUT".

What do we conclude from this little exercise?

It is really easy to do exhaustive searches to find best-fit solutions. The above fit works as well as it does because it starts with two different quantities (α) and (2π α^2) (where π is not a rational fraction), and combines them in various ways using lots of rational fractions to tailor the combination, which then leads to a dense set of candidate solutions from which the best-fit solution can then be picked.

Unless you happened to pick the physically correct parametric form to search over (Balmer got lucky with atomic spectra, but that is not to be used as a justification for this approach), then there is no physical significance to solutions that are obtained in this way. If you hedge your bets by searching over a large set of parametric forms, then you will almost certainly find many solutions that have a good fit to the target value, but this doesn't guarantee that any of them is physically significant. Interestingly, a related problem occurs in the context of the Landscape.

The approach used in this "GUT" is numerology, pure and simple. Of course, I only suspect that this is the way that the above expression for mμ/me was "derived"; I can't prove that this is the case.

Holiday in South West Cornwall

I have been neglecting this blog; my previous posting was more than half a year ago. Not only that, but the broken basin that I described in my previous posting has still not been fixed!

I have just returned from a few weeks camping in South West Cornwall (the area known as West Penwith), which is an area that I enjoy visiting because it is like leaving your "baggage" at home and going away to "the edge of the world". Here is a small sample of some of the photographs that I took whilst I was there.

The West Penwith area is rather exposed to the Atlantic weather systems, so you have to pitch your tent away from bushes that look like this:

Much of the local granite is beautifully weathered. I wonder how long it takes for granite exposed to the elements to begun to look like this:

As usual, I tried to do a bit of maths to exercise my mind but I kept losing factors of π, so I ended up just banging the rocks together:

To satisfy my curiosity, I visited the famous tin mine engine houses at the Botallack Crowns Mine, which are much more precariously located than they seem in this photograph (someone needs to invent a simple photographic system that gives the viewer a full 3D spatial awareness, like a feel for the yawning chasm just in front of them):

I watched a bit of Cornish cricket at the Lafrowda Festival in St Just, which seems to be much more fun than the activity called "cricket" that I have seen on TV:

I did quite a bit of moorland walking, but I was never quite sure whether I was inside or outside the areas of open moorland, as this gate in the middle of nowhere illustrates:

There appears to be only a loose correspondence between the moorland paths marked on the Ordnance Survey map and the actual paths on the ground (I double checked my position using my GPS locator), so I sometimes found myself wading through a sea of gorse and heather that was up to waist high in places. This scenery was very pretty to look at, but it tore my legs to shreds.

There is a move to enclose the moors and to graze cattle, which has caused uproar amongst some of the local inhabitants who have started a Save Penwith Moors campaign. My preference is for open moorland scenery unspoilt by fencing and cattle, and I hope yours is too.

Update (21 December 2008): I have just noticed that the 9 Maidens Common has had a reprieve from the cattle grazing plans (see here). Excellent news!

Butterflys in my bathroom

It has not been a particularly successful weekend for me plumbing-wise.

The butterfly flapped its wings.

My bathroom basin had a dripping tap, so I surmised that the tap needed a new washer. So, off I went to the local hardware store to buy said item, and returned to unscrew the top of the tap to fit the new washer. Unfortunately, the tap thread was completely frozen and no amount of fiddling about would shift it; I recall my plumber telling me that this tap had a problem several years ago. Never mind, I didn't like the taps anyway, so I decided to replace both taps on the basin, which would involve unscrewing the taps from below. So I applied some penetrating oil, and then went off to a large out-of-town hardware store where they had a good selection of taps, found what I wanted, and returned to fit them to my basin. I turned off the water at the main supply, and proceeded to use my under-the-sink wrench to loosen the nut connecting the water pipe to the tap, but rapidly discovered that the wrench's handle was far too short (and thin!) to apply sufficient torque to do the job, nor was it possible to attach anything to the handle to increase the torque. Off I went to the large hardware store again to buy a better under-the-sink wrench, which did the job once I inserted my hammer in its jaws to apply sufficient torque. Now all I needed to do was to detach the tap itself from the basin, so I applied the wrench again but found that the nut attaching the tap to the basin was frozen in place. To unfreeze it I reasoned that I could jiggle the tap backwards and forwards, using wrenches simultaneously above and below the basin. Jiggle, jiggle, wrench, curse, WRENCH ... crack!! The basin was now in several pieces, with cracks radiating from the tap that I had been working on. WTF happened? Oh well, the basin would now need to be replaced as well so I started to pull away the loose section behind the tap, and immediately cut my thumb by trapping it between the edges of two broken basin pieces. It was a deep cut about 2cm long and there was blood everywhere, so I had to retire for a while to mend my thumb. Later on, I returned to the basin to discover that the reason it had broken was that the part of the tap that passed through the basin had a square cross section, and the hole in the basin that it passed through was also square. Great!! Now I know that wrenching the tap around backwards and forwards was guaranteed to break the basin. Anyway, the tap was now free of the basin because I had smashed the basin, but there was enough of the basin left intact that I could still use it pending its replacement, so I put a new tap on the free end of the water pipe and gracefully dangled it over the broken edge of the basin. I turned the main water supply back on, so now I had a tap that didn't drip, but a basin that needed to be replaced.

The butterfly flapped its wings, and the puff of air developed into a gust of wind.

OK, so now I needed a new bashroom basin. I picked up a piece of the broken basin in order to use it as a colour swatch to match its nice pale blue colour to a new basin in the large out-of-town hardware store. Off I went to the store to find that all of their basins were white. Oh no! At first I assumed that they put the white basins on display and held a selection of coloured ones in the store room, but I rapidly discovered that they had only white. This was a large store, so this was as good as things would get for me. I realised with dawning horror that if I wanted a colour matched bathroom suite then I would need to replace the entire suite. I looked around nervously at the prices of whole suites, and realising that I would have to have it fitted professionally I mentally added in that cost as well. The total cost would not be less than £1000 in round figures. Perhaps I'll get used to having a broken bathroom basin! Maybe I could pass it off as an interesting new art form, and put a First Aid kit by the side of it for people to staunch their wounds when they cut themselves! No, that won't work. I'll have to spend loadsamoney on fixing this problem.

The butterfly flapped its wings, the puff of air developed into a gust of wind, and the gust of wind developed into a howling storm.

I should have got a plumber in to fix my dripping tap, but personal pride took over and made me attempt to do the job myself.

By the way, I have now found some specialist suppliers who sell discontinued coloured bathroom suites, so maybe I could replace only the basin, but I'll have a think about things to decide whether I might as well buy a whole suite anyway

NanoArt 2007 voting

I blogged a couple of times before about NanoArt 2007, see here (background information on the competition) and here (my 5-frame animation competition entry).

The NanoArt 2007 voting is now open here (I don't know why the year 2006 appears in this link!) until 31 March 2008.

Here are the voting steps (you can go straight to vote for my entry here):

1. Click on the album thumbnail to open the album.
2. Click on the image thumbnail to view the image.
3. Click on the number of stars you would like to rate this image.

My entry is different from the others because it is an animated GIF that you view here; this link is quoted beneath the static GIF that is displayed on the NanoArt 2007 competition web page. Unfortunately, the static GIF looks really boring, so I don't have much hope that many people will discover that my entry is actually an animation. Never mind, at least I tried.

Update (17 January 2008):

It is fascinating to watch the votes accumulate. At each stage you can see the number of votes cast thus far and their average rating, so if you take a peek often enough you can deduce each vote as it is cast, provided that the total number of votes is not so large that there are rounding errors in the average rating.

There is obviously someone who is trying to "spike" my entry by voting with a rating 0/5, whilst all the other votes that I have received have a high rating. How pathetic is that?!

Although I am curious to observe the pattern of voting, it is not that important to me what score I get because I have already achieved my goal which was to create an animated nanoartform. I would be interested to hear of any prior examples of this artform.

Interpreting mathematics

In the 5th January 2008 issue of New Scientist there is a short article by Mark Buchanan entitled "When it's time to sit back and think again", which discusses some of the results reported in an arXiv paper entitled "Symbolic manipulators affect mathematical mindsets" at http://arxiv.org/abs/0712.1187.

The phenomenon that is discussed in the paper is the tendency for people to switch off their brain when they use symbolic algebra programs (the paper specifically singles out Mathematica), but the problem is more widespread than this because it occurs with basic 4-function calculators (e.g. do I divide by 1.1 or multiply by 0.9?) or with advanced numerical software (e.g. why do the eigenvectors come out completely different for trivially different data?). This causes people to drop down into a calculational mode where they act merely as operators of the software/hardware, whilst not bothering to form a higher-level interpretation of what they are calculating (e.g. its physical interpretation).

This is like the difference between a worm's eye-view (e.g. low-level calculational mode) and a bird's eye-view (e.g. high-level physical interpretation mode). It is like the difference between having a local serialised view of each part of the problem that you are solving or a global parallelised view of the whole problem. It is like the difference between being a calculator or a visionary.

I know of people who are calculators but who can't see the grand picture, and who usually cannot communicate with anyone other than like-minded calculators. I know of people who are visionaries but who can't express their ideas in enough detail to carry them out, and who are highly articulate but whose apparent lack of rigour really annoys the ace calculators. I know of very few people who are both calculators and visionaries, but these people are really interesting to know.

The education system trains people to produce standard solutions to problems, so that everyone calculates using the same language. It is relatively easy to teach people to rote-learn standard procedures, and to then test them on this knowledge in exams. It is much less easy to teach people the skills that are needed to relate these calculations to the rest of the world, or so one would think.

My approach to counteracting the tendency to drop down into a calculational mode of thinking is to visualise what I am calculating; I try to avoid doing calculations that I can't visualise. When I draw a picture of what I want to calculate then the calculation itself follows almost automatically, and calculational subtleties (e.g. the epsilons and deltas) are easily resolved by referring back to the picture. I would go so far as to say that if I can't draw a picture then I don't understand what I am calculating.

I was amused to see that the principal example cited in the paper http://arxiv.org/abs/0712.1187 was one in which several students struggled to use Mathematica to evaluate the following integral (I have omitted various constants):

Integrate[x^2 Sin[x]^2, {x, -Infinity, Infinity}]

This is a clear example of students in calculational mode, who have adopted the worm's eye view of the problem as just being a calculation. They tried feeding the integral to Mathematica in various different ways, but without success. What they do not do was to diagnose their problem by simply visualising what they were calculating; it is not even necessary to know the physics behind this integral.

Using the same tool (i.e. Mathematica) that the students were using in their attempts to evaluate the integral, here is a plot of the integrand over a finite interval.

Normally, an integrand that is simple as this would not need to be plotted out explicitly because its behaviour is obvious from its structure, i.e. an x^2 factor that diverges times a Sin[x]^2 factor that oscillates between 0 and 1. Nevertheless, in this case I did plot it out as part of my ingrained habit of visualising calculations using Mathematica. The students should have been doing this as well, so I presume that they had not been very well tutored in their use of Mathematica. Had the students attempted even a rudimentary visualisation then they would have immediately realised that the limits of the integral they were trying to evaluate could not possibly be infinite.

As the visualisation habit becomes part of your way of working, you eventually reach a point where the solution of some problems comprises visualisation followed by calculation. There always remains a set of "difficult" problems for which your current set of visualisation techniques is inadequate, in which case you have to use pure calculation to get to a solution. But then you should be on the lookout for ways to capture the essence of your solution in a new visualisation technique.

Wouldn't it be nice if there was a standard set of visualisation techniques that you could use alongside the existing set of calculational techniques? If this set of visualisation techniques was carefully designed then it would be just as rigorous (and teachable) as standard calculational techniques. It would be a very interesting exercise to reformulate existing material using such a visual language; for instance, I made an attempt to do this sort of thing for the topology of the SO(3) rotation group here.

Nano flower

A while ago I blogged about NanoArt 2007 , which is an online competition to create nanoart. Well, it turns out that I had completely forgotten about this competition, so I have done nothing more than form a few vague ideas about what I planned to create as a submission. The closing date for submissions is 31 December 2007, which leaves very little free time to create a worthwhile submission, so I wondered how I could create something using the bits and pieces that I happen to have lying around already.

One of the 3 electron microscope pictures that are supplied for you to start from is this

which they fittingly call "Nano Flower", and you are allowed to submit up to 5 "artistic" images for the competition.

I think I will choose to submit a 5-frame animation generated using a variant of the Belousov-Zhabotinsky reaction simulation software that I described here, where I use the "Nano Flower" image as an external input to steer the the reaction dynamics. This means that I am simulating a modified version of the BZ reaction, where the surface on which the reaction occurs is weakly contaminated so that the BZ reaction dynamics vary across the surface.

Here is a 5-frame animation that I quickly created, which fortuitously turns out to be cyclic because the period of the BZ dynamics is 5 frames.

Well, that's one frame of the animation. I gave up trying to get the animation to upload successfully to blogger.com, so I have put it on my web site here.

I have called this work "Fizzix" for fairly obvious reasons. Now I will wait to see whether I have broken the submission rules for the NanoArt 2007 competition, which don't mention animation as being an acceptable artform, not even 5-frame animations.

Update (7 January 2008):

All of the NanoArt 2007 competition entries can be viewed here, and the album containing my single competition entry is here (with the animation hosted here). My competition entry is a lot less colourful than other peoples' entries, so it is unlikely to attract much attention. I have to hope that people click down to my animation to discover the true nature of my competition entry.

Molecular manufacturing

The Center for Responsible Nanotechnology has a nice overview of its current findings on molecular manufacturing here, which summarises the pros and cons of being able to manipulate matter in a controlled way at the molecular scale.

Molecular manufacturing is fundamentally different from the "heat and stir" approach to building things with molecules, because it aims to directly control the placement and interaction of individual molecules. This opens up lots of new possibilities for building things (beneficial or dangerous), and the world will not be the same when the transformation to molecular manufacturing has occurred.

This revolution in manufacturing will make the Industrial Revolution seem trivial in comparison, and it will happen within a small number of decades (CRN says 2 decades). This means that it is likely to have a big impact on the lives of most people alive today, which gives you an incentive to read all about it here.

Merry Yuletide

Here is a movie in which I have created an animation of a festive texture on the surface of a trefoil knot. Merry Yuletide!

How did I create this movie?

I have been playing around with the Belousov-Zhabotinsky reaction simulation that I described here.

One thing that you can do is to find limit cycles of the simulation, where the state of the array of cells returns to a state that it visited earlier in the simulation. The whole sequence of states between two such repeats (including one of the end points) is then a limit cycle of the BZ simulation. Such limit cycles must exist because the state space is finite in size, so it it inevitable that the simulation must eventually revisit states that it visited earlier, though starting from a random initial state the likelihood that this occurs in a given timescale decreases rapidly as the size of the array increases.

One example that I particularly like is a cycle of length 10 that I found on a toroidal 13 by 13 array (using the same parameter values and colour scheme as here), and I show 2 of these cycles in the movie below:

This cycle is unusual because it has a low symmetry, and because it is pretty to look at despite its short length.

Using this cyclic BZ solution on a torus it is relatively easy to create the movie at the start of this posting, by using it to texture the toroidal surface of a trefoil knot, and choosing a colour scheme that maximises its festive feel.

Nerd test

I happened upon Nerdiometer, which describes an on-line test for assessing how nerdy you are. You can find the test here.

I couldn't resist trying this one. I suspect that the results of the test are biassed by a selection effect where the nerdier you are the more likely you are to take the test in the first place.

Anyway, here is my result:

17% of people score higher than me, and 82% score lower. I was disappointed to score so highly, because as I filled in the multiple-choice questionnaire I could see that the extremely nerdy answers were off the scale compared to where I stood. Maybe that was a trick to let me make the choices that I did without being too embarrassed about them.

The questionnaire was extremely selective in the areas it covered, because it focussed on asking about nerdy things to the almost complete exclusion of asking you about anything else. So, if you spend only small fraction of your time being a nerd, you will register highly on the nerd-scale as defined by this test.

I wish there was a way to find out how each answer was weighted to produce the overall result. Oh no! Wishing for that must in itself be a nerdy thing to do! Aargh! This is renormalisation gone mad!

Update (12 December 2007):

The result I got when I tried version 2 of the test (see here) is

OK, so now I am a "cool nerd" which sounds pretty good to me, but I am annoyed that my Science/Math and Technology/Computer scores are deemed to be so low! Who are the people who designed this test anyway? I'm not playing any more, maybe...

Belousov-Zhabotinsky reaction

Whilst exploring Wikipedia I came across the Belousov-Zhabotinsky reaction, and the phenomenology of its reaction dynamics such as the development of propagating waves of reactants is described here. Although I previously knew about the BZ reaction I had never implemented a simulation of it. It turns out to be remarkably easy to implement in Mathematica, and the BZ reaction dynamics can readily be displayed as a very pretty movie.

Although the original purpose of this type of simulation was to model the BZ reaction dynamics, you could imagine using variants of this type of simulation for generating interesting dynamic art forms.

A slightly modified form of an algorithm to simulate the BZ reaction due to Professor A K Dewdney is listed here.

(i) Select an integer q in the range 2 through 255. Cells may be in any of the states 1 through q.
(ii) Select two integers k1 and k2 in the range 1 through 8 and an integer g in the range 0 through 100.
(iii) In the transition from one "step" to the next the state of each cell is changed once according to rules (iv) - (vii).
(iv) A cell in state q changes to state 1.
(v) A cell in state 1 changes to state a/k1 + b/k2 + 1 where a is the number of neighbors of the cell which are in states 2 through q-1 and b is the number of neighbors in state q.
(vi) A cell in any of states 2 through q-1 changes to S/(9 - c) + g, where S is the sum of the states of the cell and its neighbors and c is the number of neighbors in state 1.
(vii) If the application of rule (v) or rule (vi) would result in a cell having a state > q then the state of that cell becomes q.

After some experimentation to find the fastest (and clearest) implementation in Mathematica, I created a function for implementing a BZ update step with toroidal boundary conditions:

update[state_, q_, k1_, k2_, g_] :=
Module[
{ones, qs, kernel, total, totalones, totalqs},
ones = Map[If[#==1, 1, 0]&, state, {2}];
qs = Map[If[#==q, 1, 0]&, state, {2}];
kernel = Table[1, {3}, {3}];
total = ListCorrelate[kernel, state, {{2,2},{2,2}}];
totalones = ListCorrelate[kernel, ones, {{2,2},{2,2}}] - ones;
totalqs = ListCorrelate[kernel, qs, {{2,2},{2,2}}] - qs;
MapThread[If[#>q, q, #]&[Switch[#1, 1, Floor[#4/k1+#3/k2+1], q, 1, _, Floor[#5/(9-#2)+g]]]&, {state, totalones, totalqs, 8-totalones-totalqs, total}, 2]
];

I used this BZ update function to generate the above movie of the BZ reaction dynamics, using suitable parameter values that I copied from here.

statesequence =
With[
{n=64, q=200, k1=2, k2=3, g=70},
NestList[update[#, q, k1, k2, g]&, RandomInteger[{1,q}, {n,Floor[4/3n]}], 250]
];
movie = Map[ArrayPlot[#, ColorFunction->"Rainbow"]&, statesequence];
ListAnimate[movie]

The poor image quality in the movie is due to the high degree of image compression, rather than due to any limitations of the above algorithm. Also, each frame is separately scaled and mapped onto the colour lookup table, which is probably not the best way of creating the movie.

Enigmatic comments

This is a placeholder for comments on postings at Enigmatics.

Please remember to identify which Enigma problem you are commenting on.

Enigmatic movements

I have decided that my Mathematica solutions to the New Scientist Enigma problems are cluttering up this blog, so I will move them to a separate blog called Enigmatics which I will generate using the incredibly useful WorkLife FrameWork; believe me, using WLFW is like having a 24/7 personal assistant organising all of your Mathematica notebooks. This allows me to present my solutions in a form that is much closer to the original Mathematica, but it does not allow for comments on blog entries.

I have looked into the possibility of managing the comments myself, but the effort needed to counter the inevitable spam is too great. I will instead provide a link in each Enigmatics blog entry back to this blog, which will allow comments to be made using the Blogger commenting system.

NanoArt 2007

Nanoart is defined here as:

NanoArt is a new art discipline related to micro/nanosculptures created by artists/scientists through chemical/physical processes and/or natural micro/nanostructures that are visualized with powerful research tools like Scanning Electron Microscope and Atomic Force Microscope.

The process of creating nanoart is described here as:

I bring the small world in front of my audience through high resolution electron microscope scans of natural micro or nanostructures and sculptures I create at micro or nano scale by physical or/and chemical processing. I take further steps by mixing the realistic images of this structures with abstract colors, digitally painting and manipulating the monochromatic electron scans, and finally printing them with long-lasting inks on canvas or fine art paper (giclee prints). This way, the scientific images become artworks and could be showcased for a large audience to educate the public with creative images that are appealing and acceptable.

In a nutshell, the electron microscope provides the raw image, and the artist colours it in to produce an enhanced (i.e. artistic) result. This could range from a trivial colour tinting of the raw image, through to an artistic rendition that is based only very loosely on the raw image. Of course, I favour the more artistic style of nanoart, and I have some ideas on new ways of creating such artwork, but there isn't room in this tiny blog posting to tell you all about it!

Anyway, back to the title of this posting: NanoArt 2007. This is an online competition to create nanoart, and the submission deadline is 31 December 2007. For people without ready access to an electron microscope to create their own raw images there are 3 high resolution monochromatic electron scans provided here (nano-flower), here (micro & nano), and here (nano-crystals). I think it would be fun to enter this competition.

Dresden art gallery

Here is a nice photo of a very small part of the world famous Dresden Art Gallery:

© Ben Bartels and André Schulz
http://www.chessbase.com/images2/2004/dresden/dresden018.jpg

Here is a snapshot that I took of the same scene (with some more context) in Second Life:

The exact location of this duplicate of the Dresden Art Gallery in Second Life is http://slurl.com/secondlife/Dresden%20Gallery/77/121/26. Visitors to SL do not need to pay anything just to explore.

I won't show you inside the Second Life duplicate of the art gallery, so you will have to visit it yourself to enjoy it. You can have a look at the WIRED report here if you want more details. Set aside a spare half-day for this adventure in SL. You won't be disappointed.

Sand painting

I found this video of sand painting here at Backreaction:

I think it is a wonderful art form, and it has given me some ideas ...

Enigma 1469

Here is how Mathematica can be used to solve New Scientist Enigma number 1469 in the 17 November 2007 issue.

WARNING: See the comments for details. There is an error in the solution given below that I need to fix. In working on a fix I discovered a bug in the HamiltonianCycle function in Mathematica's Combinatorica package (this bug has been acknowledged by the author of the Combinatorica package) which destroyed any chance of my fix working fully correctly, although the correct solution to Enigma 1469 could be seen lurking in amongst the erroneous clutter. I will abandon my solution to Enigma 1469 because it has absorbed too much of my time already, and the discovery of the HamiltonianCycle bug can stand as a testimonial to the cleverness of Enigma 1469. I suppose that I ought to leave the rest of this posting intact otherwise the comments would be orphaned, and also there are some useful techniques illustrated in my erroneous solution.

Load the Combinatorica package.
Needs["Combinatorica`"];

Derive the formula for triangular numbers.
triangularformula=Sum[i,{i,n}]
1/2 n (1 + n)

Compute a list of all triangular numbers less than 1000.
triangularnumbers = Table[triangularformula, {n,44}]
{1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666, 703, 741, 780, 820, 861, 903, 946, 990}

Convert this to the corresponding list of lists of digits.
triangulardigits = IntegerDigits[triangularnumbers]
{{1}, {3}, {6}, {1, 0}, {1, 5}, {2, 1}, {2, 8}, {3, 6}, {4, 5}, {5, 5}, {6, 6}, {7, 8}, {9, 1}, {1, 0, 5}, {1, 2, 0}, {1, 3, 6}, {1, 5, 3}, {1, 7, 1}, {1, 9, 0}, {2, 1, 0}, {2, 3, 1}, {2, 5, 3}, {2, 7, 6}, {3, 0, 0}, {3, 2, 5}, {3, 5, 1}, {3, 7, 8}, {4, 0, 6}, {4, 3, 5}, {4, 6, 5}, {4, 9, 6}, {5, 2, 8}, {5, 6, 1}, {5, 9, 5}, {6, 3, 0}, {6, 6, 6}, {7, 0, 3}, {7, 4, 1}, {7, 8, 0}, {8, 2, 0}, {8, 6, 1}, {9, 0, 3}, {9, 4, 6}, {9, 9, 0}}

Construct the adjacency matrix of allowed adjacencies between triangular numbers.
adjacencies = Outer[(If[Last[#1]==First[#2], 1, 0])&, triangulardigits, triangulardigits, 1];

Convert the adjacency matrix to a set of rules that allow me to add a "tooltip" to each node of the corresponding the graph so that it can be explored by hovering the mouse over the graph: (1) extract the sparse array form of the adjacency matrix, (2) convert the sparse array into a list of rules for which graph nodes are linked by edges, (3) add a tooltip to each graph node.
adjacencies2 = SparseArray[adjacencies];
adjacencies3 = ((adjacencies2//ArrayRules//Most) /. HoldPattern[{i_,j_}->1] :> i->j);
adjacencies4 = adjacencies3 /. x_?IntegerQ :> Tooltip[x, triangularnumbers[[x]]];

Display the graph with tooltips.
GraphPlot[adjacencies4, DirectedEdges->True]
[Output omitted]

Convert the adjacency matrix to a graph. I need this representation in order to use graph manipulation algorithms.
g = FromAdjacencyMatrix[adjacencies, Type->Directed];

Extract the longest cycle in the graph, and extract the edges within this cycle.
longestcyclenodes = ExtractCycles[g]//Sort//Last//Most;
longestcycleedges = Partition[longestcyclenodes,2,1,{1,1}];

Display the graph with the longest cycle highlighted.
GraphPlot[adjacencies4, EdgeRenderingFunction -> (If[MemberQ[longestcycleedges,#2], {Red,Arrow[#1]}, {Opacity[0.3],LightGray,Arrow[#1]}]&)]

Convert the longest cycle to the corresponding a list of triangular numbers, and thence the sequence of digits encountered when going around the cycle.
longestcyclenumbers = triangularnumbers[[longestcyclenodes]];
longestcycledigits = longestcyclenumbers//IntegerDigits//Flatten
[Output omitted]

Compute the number of digits encountered when going around the longest cycle.
longestcycledigits//Length
[Solution omitted]