the individual v the commons

The twin prime conjecture, a famous open problem in mathematics, states that there are an infinite number of pairs of primes of the form \(p, p+2\).  Earlier this year, Yitang Zhang made a great breakthrough, publishing a proof that there are an infinite number of pairs of primes of the form \(p, p+H\), where \(H \leq 70,000,000\).

I've been watching what happened next with interest. Some people quickly seized on the result, and found better bounds. The encouraged Terrence Tao to propose a Polymath project, Polymath8,  to help coordinate efforts to understand Zhang's proof, and to reduce the bound on \(H\). Subsequent progress has reduced \(H\) considerably, with it currently standing at \(12,006\), and with as yet unconfirmed results of \(5,414\).

That's an amazing 4 orders of magnitude reduction, in just a couple of months.  To appreciate the progress, it's useful to look at a couple of graphs (based on that PolyMath wiki data). Here's how the best bound for \(H\) has fallen over the eight weeks since Zhang's paper was accepted:

best known value for \(H\), linear scale (diamond, confirmed result; +,  unconfirmed result)

best known value for \(H\), log scale (diamond, confirmed result; +,  unconfirmed result)

Here we see a fascinating synergy between individual and group efforts.  Zhang came up with the first, qualitative, breakthrough: a technique for providing a bound, and got a first (and now we see, rough) estimate.  Then the community gathered round, and through a process of cooperation (and presumably, a degree of competition, too), chipping away at the various definitions and terms, have quantitatively improved the technique.

So, to all those administrators trying to force us to work individually, or in groups, depending on the current fashion at headquarters -- the answer is clear: diversity works!  Some of the time progress is made by individuals, sometimes by groups, even on the same problem.  Don't assume one size fits all.

Progress here appears to have tapered off recently, with no new results reported in the last few days.  Is this due to the improvements having been pushed as far as possible (not likely, as there are several unconfirmed results yet), to enthusiasm having flagged (also unlikely, as the results are getting ever closer to the ultimate value of \(2\)), or due to it being vacation time? I'll continue watching with interest.

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the scientific process

There have been two well-publicised mathematical and scientific breakthroughs in the news recently, that have got the community all a-buzz, and that illustrate two quite different approaches to advancing understanding.

Yitang Zhang

In the first, a "virtually unknown" mathematician has made a breakthrough in the twin prime conjecture, a deep question in number theory.  A beautifully clear write-up from the Simons Foundation explains both the mathematical problem, and the process by which Zhang's contribution was handled by the mathematical community.  Zhang wrote a paper and submitted it to the Annals of Mathematics, a highly respected journal.  There is was refereed, and swiftly accepted, since, not only was it correct, it was "written with crystalline clarity and a total command of the topic’s current state of the art".  Being right is crucial, but being comprehensible to the community is also important: it is hard to check someone's work if they have written it in a language all their own.  So, kudos to Zhang for his achievement, and to the mathematical community for recognising it.

Eric Weinstein

The second well-publicised "breakthrough" is Eric Weinstein's "Geometric Unity", a contribution to particle physics.  Or is it?  Here we have another unknown worker, but this time his idea has been launched on the world through a public lecture and a newspaper article written by his sponsor and friend, mathematician Marcus du Soutoy. No publication, no journal submission, not even an unrefereed pre-print, and so no peer scrutiny is possible. Several commentators, including writers at New Scientist and Scientific American, and many science bloggers, have been highly critical of this process that amounts to "science via press conference".

Andrew Wiles

So, two announced breakthroughs by virtual unknowns, given two very different receptions.  Why the difference?  Well, because the conduct of science and mathematics is a process, not (just) a result. It's not some bizarre club, where you have to go through the right channels, jump through the community's hoops, in some arbitrary hazing process.  Those hoops are there for a reason.  It is so very easy to make a mistake, and the hoops are part of the process to help ensure that any mistakes are caught.  Remember the story of Andrew Wiles, slaving away in secrecy for seven years, proving Fermat's last theorem. Well, his first proof had a flaw. Fortunately, the error was correctable, after a lot of effort, and his proof stands. But other equally eminent mathematicians have thought they had made breakthroughs, only for peer review to determine otherwise.

Peer review is an essential part of the scientific process.  It's not there to keep out "outsiders". It's there to keep out errors.  Put up or shut up.  And show your working.

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Mathematical Models

Way back when I was at school, I came across a fascinating little book called Mathematical Models, by Cundy and Rollett.  It had instructions on how to build various mathematical objects, such as stellated polyhedra.  I liked the book so much I actually bought my own copy, new, for £2.95, which was a lot of money back then!

I made a few of the simpler models, but never got much further than the dodecahedron.  I certainly never got as far as making any of the fiddly stellated ones.

So today, when I came across a posting in Google+ about Anselm Levskaya's website polyHédronisme, I was taken right back to those days.  Playing with this interactive web-based systems is much easier than fiddling with card, glue, and scissors, though.  Type in a few commands, and a zoomable, rotatable polyhedron appears!

I've spent my afternoon playing around on this site, and reading up on Conway polyhedron notation that is used to define shapes, and now I can say I have at last "made" some of these polyhedra.

The small stellated dodecahedron is made by raising a pentagonal-based pyramid on every face of a regular dodecahedron. If the pentagons making up the dodecahedron have side length \(1\), then the height of each pyramid should be* \[ \frac{\sqrt{4\sqrt{5}-1}}{2} \approx 1.41\] The Conway notation command in polyHédronisme that achieves this is \(k(5,1.41)D\), which means: start with a dodecahedron \(D\), then raise a pyramid of height \(1.41\) on each \(5\)-sided face.

The great dodecahedron is made by making a pyramidal dimple in every face of a regular icosahedron. If the triangles making up the icosahedron have side length \(1\), then the depth of each pyramid should be* \[ \sqrt{\frac{1}{2}- \frac{\sqrt{5}}{6}} \approx 0.36\] The Conway notation command in polyHédronisme that achieves this is \(k(3,-0.36)I\), which means: start with an icosahedron \(I\), then indent a pyramid of height \(0.36\) on each \(3\)-sided face.

The great stellated dodecahedron is made by raising a pyramid on every face of a regular icosahedron. If the triangles making up the icosahedron have side length \(1\), then the height of each pyramid should be* \[ \sqrt{\frac{7+3\sqrt{5}}{6}} \approx1.51\] The Conway notation command in polyHédronisme that achieves this is \(k(3,1.51)I\), which means: start with an icosahedron \(I\), then raise a pyramid of height \(1.51\) on each \(3\)-sided face.

So much for standard polyhedra.  But polyHédronisme doesn't stop there.  I had great fun playing about with the notation language, producing weird and wonderful forms:

(i)  \(k(20,1)bk(3,2)I\)    (ii)  \(k(12,1)k(10,2)bk(5,1)D\)    (iii)  \(k(20,-0.3)k(6,0.3)bk(3,-0.3)I\)
(iv)  \(k(24,-0.5)k(6,0.2)k(20,-1)bk(3,-0.1)k(5,1)D\)

The results of play can only really be appreciated on the site itself, rotating the polyhedra, and getting a real feel for their shapes, with all their dips and bumps.  A marvellous site.

---

* The book Polyhedron Models has helpful stellation diagrams that allow these heights to be calculated, with a little trigonometry.

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ODE to a Petri net

Writing ordinary differential equations (ODEs) to model various natural world processes comes more readily to some than to others.  And once written, it can take some effort to pick apart their real-world meaning.  It would be nice to have a more visual form.  Petri nets are one such approach.

Epidemics

For example, consider the simple SIR model of epidemic infection.

• +++S+++ : those uninfected, but susceptible -- their number is reduced as they become infected, at a rate proportional to the number of susceptibles and the number of infected
• +++I+++ : those infected -- their number is increased by susceptibles who become infected;  it is also reduced as those infected recover, at a rate proportional to the number of infected
• +++R+++ : those recovered -- their number is increased by those who were infected recovering

For those happy with ODEs, it is straightforward to write down a set of coupled equations to model this:

$$\dot{S} = - i S I $$

$$\dot{I} = i S I - r I$$

$$\dot{R} = r I$$
In addition to natural language (the bullet list) and maths (the ODEs), there is another language useful to explain and understand models: diagrams.

For example, we could draw a simple state transition diagram to show the movement from susceptible to infected to recovered:

This captures some of the information, but not all of it.  A (continuous) Petri net can do better:

The circles are called "places", and represent the "things" involved: here the susceptibles, the infected, and the recovered.  The rectangles (the colours aren't significant) are called "transitions", and represent how the things in "input" places get transformed into things in "output" places.

• transition +++r+++: an infected comes in, and a recovered comes out
• transition +++i+++: a susceptible and an infected come in; two infected come out (the two infected outputs are the newly infected, and the original infecter)

This diagram has enough information in it to reproduce the original ODEs.  We have one ODE per place, with the terms being given by the transitions feeding that place.

• place +++R+++. +++R+++ has only one transition feeding it: transition +++r+++.  It is feeding into +++R+++ at a rate proportional to all the inputs to +++r+++: here just +++I+++.  Hence +++\dot{R} \propto I+++.  If we call the constant of proportionality (the rate constant) +++r+++, we get +++\dot{R} = r I+++
• place +++S+++. +++S+++ has only one transition, +++i+++, removing stuff from +++S+++.  It is removing at a rate proportional to all the inputs to +++i+++: here +++S+++ and +++I+++.  Calling the rate constant +++i+++, we get  +++\dot{S} = - i S I+++ (the minus sign is there because we a removing from +++S+++, and it is conventional to keep the rate constants positive)
• place +++I+++.  +++I+++ has two transitions: +++i+++ both feeding it (two input arrows) and removing from it (one output arrow), and +++r+++ removing from it.  We get +++\dot{I} = i S I - r I+++

Thus we have recovered the original equations.

We can write this focussing on the transitions, in a more algorithmic way (algorithms, or pseudo-code, are yet a further language we can use to explain, describe, and define processes).

• initialise the rates of change to each place to be +++0+++.  +++\dot{S}, \dot{I}, \dot{R} := 0+++
• transition +++r+++.  This is removing stuff from place +++I+++ and adding it to +++R+++ at a rate +++rI+++. Update the output place +++R+++ and input place +++I+++ appropriately: +++\dot{I} {-}{=}\ rI+++, +++\dot{R} {+}{=}\ rI+++
• transition +++i+++.  This is removing stuff from place +++I+++ and from place +++S+++ at a rate +++iSI+++ and adding it to +++I+++ at a rate +++2iSI+++ (from the two input arrows).  Hence there is a net input to place +++I+++ at a rate +++iSI+++.  Update the output place +++S+++ and net input place +++I+++ appropriately: +++\dot{I} {+}{=}\ iSI+++, +++\dot{S} {-}{=}\ iSI+++

We can write this as a general algorithm:

  for each place Pi
    Pi_dot := 0
  for each transition Ti
    let Pin = < Pin_1, ... , Pin_n > = list of n places,
                                       one for each input arrow of Ti;
        Pout = list of m places, one for each output arrow of Ti;
        t = Ti x Pin_1 x ... x Pin_n
    for each Pi in Pin
      Pi_dot -= t
    for each Pi in Pout
      Pi_dot += t

So now we have a diagrammatic form, and an ODE form, that are equivalent, and an algorithm to translate on to the other.  This is useful, because we can use them interchangeably, without risk of losing information.  In particular, notice how explanation accompanying the Petri net focusses on what is happening in the transitions, whilst that for the ODE form focusses on what is happening to the places.  Having different forms of explanation can be useful in different circumstances (modelling, communication, modification, validation, calculation, etc).

Catalysis

Although that all works well, the handling of the infecter in the +++i+++ transition seems a bit unnatural: infecter goes in, infecter comes out, resulting in an addition and subtraction of this rate.  The infecter is needed for the transition, and affects the rate of the transition, but is not itself changed by the transition.  In chemistry, this is called a catalyst, and there is some special Petri net syntax for it.  We can draw the SIR Petri net above equivalently as:

Here the dashed arrow means that +++I+++ is a catalyst: it is needed for the transition, but is not consumed by the transition. Hence there is now only one arrow out to +++I+++: since the catalyst wasn't consumed, it doesn't need to be replaced; the remaining single arrow represents the newly infected.

The algorithm needs a little  modification:

  for each place Pi
    Pi_dot := 0
  for each transition Ti
    let Pin = < Pin_1, ... , Pin_n > = list of n places, 
                                       one for each input arrow of Ti;
        Pcat (sublist of Pin) = list of catalytic input arrow of Ti;
        Pout = list of m places, one for each output arrow of Ti
    let t = Ti x Pin_1 x ... x Pin_n
    for each Pi in Pin \ Pcat
      Pi_dot -= t
    for each Pi in Pout
      Pi_dot += t

So the catalytic arrows still contribute to the functional form of the overall rate +++t+++, but not to the changes to the specific places.

Logistic equation

Possibly the simplest bounded growth model in biology is the logistic equation:
$$ \dot{N} = rN(1-N/K)$$where +++r+++ is the growth rate, and +++K+++ is the carrying capacity (so when +++K=N+++, +++\dot{N}=0+++).

This can be drawn as an equivalent Petri net:

• transition +++r+++ (birth): one in, two out
• transition +++rK+++ (competition): two in, one out

These two transitions can also be shown in a simpler catalytic form (if maybe not with the same intuition as before):

• transition +++r+++ (birth): one "catalyses" the birth of the other
• transition +++rK+++ (competition): one "catalyses" the death of the other

Lotka-Volterra predator-prey

The simple predator-prey model, usually cast as rabbits and foxes, has rabbits being born, predated on by foxes to produce more foxes, who then die.  A simplistic version of this might be:

• transition +++b+++ (birth): one rabbit in, two rabbits out
• transition +++d+++ (death): one fox dies
• transition +++p+++ (predation): one fox and one rabbit in, two foxes out

This however has a problem: it has a new fox produced every time a rabbit is eaten.  Real foxes need more food than this to reproduce.  We can't solve the problem by changing the rate +++p+++, as this affects the consumption of rabbits and production of foxes equally.  What we really need is for the consumption of a rabbit to produce a bit of a fox.  We can do this by adding a separate rate to the arrow:

• transition +++p+++ (predation): one fox and one rabbit in, one plus +++\epsilon+++ foxes out

The algorithm needs to be updated to multiply the rate by the weight of the arrow before adding/subtracting  as appropriate.  This then yields the familiar equations:

$$\dot{R} = R(b-pF)$$

$$\dot{F} = F(-d+p\epsilon R)$$

If we use the catalytic form, the diagram simplifies to:

• transition +++p+++ (predation): one fox "catalyses" the transformation of a rabbit into +++\epsilon+++ of a new fox

Lotka-Volterra competition

The simple competition model, usually cast as rabbits and sheep, has rabbits and sheep being born and dying following their own logistic equation, and also competing with each other for resources.

• transition +++rb,rs+++ (birth): one +++x+++ "catalyses" the birth of the next +++x+++
• transition +++rc,sc+++ (competition): one +++x+++ "catalyses" the death of another +++x+++
• transition +++rs+++ (rabbit/sheep competition): one sheep and one rabbit in, a proportion of each out

$$\dot{R} = R(rb-rcS) + (pr-1)rs RS$$

$$\dot{S} = S(sb-scS) + (ps-1)rs RS$$

Combining predator-prey and competition

The competition example has two logistic "subnets", showing how these diagrammatic forms can be readily combined.  So, for example, we could easily add some foxes to the brew:

$$\dot{S} = S(sb-scS) + (ps-1)rs RS$$

$$\dot{R} = R(rb-rcS) + (pr-1)rs RS -pRF$$

$$\dot{F} = F(-d+p\epsilon R)$$

If the foxes also worried the sheep, the diagram would get messier, but would still visually represent the relationships between the different components.

Diagrams v cartoons

Pictures can be very helpful at getting across ideas, but they have their problems if they are ambiguous, incomplete, or otherwise open to misinterpretation.

The (continuous) Petri nets shown here have a formal meaning: they can be translated into equivalent ODEs. They are not informal "cartoons", merely sketching some part of the meaning.  There is an algorithm from diagrams to equations, and it is possible to build a tools allowing the manipulation of diagrams and equations as two different "concrete syntaxes" of the same underlying model.  Which syntax to use depends on what you are doing: it truly is the best of both worlds.

Acknowledgments

• I first came across the formal link between continuous Petri nets and ODEs on Alexi Sharov's web site
• I was reminded of using Petri nets to model population dynamics on reading David Tanzer's guest post on the Azimuth site
• I drew the diagrams in graphviz
• The maths is formatted with LaTeX and displayed using MathJax

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a jewel of a probability puzzle

I've been spending way too much time reading John Baez' Azimuth blog. It's got lots of fascinating stuff, but one piece that really caught my eye is a probability puzzle:

Suppose I have a box of jewels. The average value of a jewel in the box is \$10. I randomly pull one out of the box. What’s the probability that its value is at least \$100?

First thought, of course, is: "I don't have enough information; the answer will depend on the distribution".  But in fact you can give quite a strong bound on the answer even without knowing the distribution, using Markov's inequality: the probability is +++\le 1/10+++.

At first, this looks impossible: how can this be independent of the distribution?  Surely I can have some strange collection of jewels that violates this bound?  But no, and in the comments section there are some great explanations that give an intuition about why this is so, including David Guild's:

As long as gems have non-negative value, then (probability of pulling a \$100 or better gem > 10%) implies that (average value > \$10). Since the average is exactly \$10, then the probability can’t be more than 10%.

A very clear proof is laid out by Greg Egan. I'll rework it here, for the general case of an average value of +++a+++ (rather than the specific \$10) and a big value of +++b+++ (rather than \$100), to get the general result.

Let +++A+++ be the set of all jewels, +++\#A+++ be the size of set +++A+++, and +++V(x)+++ be the value of jewel +++x+++. Let +++B+++ be the set of "big value" jewels
$$ B = \{x\in A | V(x) \ge b\}$$We know the average value of the jewels is +++a+++:
$$ a =  \frac{1}{\#A} \sum_{x \in A} V(x) $$If we replace the sum over all jewels +++A+++ with the sum over the smaller set of jewels +++B+++, we must get a smaller answer (assuming that all the values are non-negative: necessary in the general case, and implicit in the given example).  So:
$$ a \ge  \frac{1}{\#A} \sum_{x \in B} V(x) $$From the definition of +++B+++, we have +++\forall x \in B, V(x) \ge b+++.  This implies that
$$ \frac{1}{\#A} \sum_{x \in B} V(x) \ge  \frac{1}{\#A} \sum_{x \in B} b  $$We can do this last sum:
$$ \frac{1}{\#A} \sum_{x \in B} b =  \frac{\#B}{\#A} b$$Putting this all together, we have
$$ a \ge  \frac{1}{\#A} \sum_{x \in B} V(x)  \ge \frac{1}{\#A} \sum_{x \in B} b = \frac{\#B}{\#A} b$$So
$$ a \ge  \frac{\#B}{\#A} b$$Rearranging gives the result that the proportion of big value items to all items (the probability of drawing a big value item) is:
$$ \frac{\#B}{\#A} \le \frac{a}{b}$$For the example with +++a=10+++ and +++b = 100+++, this gives us +++1/10+++.

This derivation assumes that the chance of pulling out any jewel is the same.  But, as Baez explains, the result is independent of this.  If some jewels are more likely to be picked, and that likelihood is used to define the average value too, then the result stands. (I leave the proof as an exercise for the reader.)

So, from a problem that initially looked as if it doesn't have nearly enough information, we've moved to an intuition about why a result (albeit only a bound) can be given, and a simple proof of a general result for that bound: Markov's inequality.  The power of maths!

There's also a result for values that can go negative, which requires also knowing the standard deviation: Chebyshev's  inequality.  It's all on John Baez' blog.  Go there, and you to  might spend as much time reading around as I have!

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going to extremes

Last week I went to a talk by Jim Al-Kalili, called “9 Paradoxes” (which, not coincidentally, is the title of his latest book).

One style of paradox he talked about are the mathematical ones, where a plausible calculation is presented, but the conclusion is wrong.  He illustrated this with the Monty Hall problem, and the Missing Dollar puzzle.

I’ve written about the Monty Hall problem elsewhere, and how the resolution is much easier to see by taking a more extreme case.  The approach of changing the problem to an extreme case is not specific to the Monty Hall problem, but is a more widely applicable check.  In particular, is can be applied to help see through the Missing Dollar puzzle. 

The Missing Dollar puzzle is as follows:

Three friends book into a shared room at an hotel.  The rate is $30, so they pay $10 each.  Later, the clerk realises they have overpaid; the rate is actually $25.  He takes $5 from the till, and goes to give them their refund.  On the way he realises that he won’t be able to split $5 between the three, so gives them $1 each, and pockets the remaining $2. 

So they have each paid $10-$1=$9, which is a total of $27. With the $2 in the clerk’s pocket, that’s a total of $29.  The original payment was $30. What happened to the missing $1?

The answer is that this is the wrong calculation. They have paid $27.  Of this $2 is in the clerk’s pocket, and $25 is in the till to pay for the room.  The puzzle works because the two prices are so close, and so it isn't necessarily obvious on a fast telling of the puzzle that the $2 should be subtracted from the $27, rather than added to it.  Let’s use the same approach of taking it to extremes to make the problem more obvious.

Three friends book into a shared room at an hotel.  The rate is $3000, so they pay $1000 each.  Later, the clerk realises they have overpaid; the rate is actually $25.  He takes $2975 from the till, and goes to give them their refund.  On the way he realises that he won’t be able to split $2975 between the three, so gives them $991 each, and pockets the remaining $2. 

So they have each paid $1000-$991=$9, which is a total of $27. With the $2 in the clerk’s pocket, that’s a total of $29.  The original payment was $3000. What happened to the missing $2971?

It is much clearer now that is that this is the wrong calculation. They have paid $27.  Of this $2 is in the clerk’s pocket, and $25 is in the till to pay for the room.

Going to extremes doesn't work for everything, but it is quite a powerful argument sanity-checker.

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climbing up onto the shoulders of giants

If I have seen further it is by standing on the shoulders of Giants – Isaac Newton 

Newton was not first to say this (and it may or may not have been a jab at Hooke), but the idea is sound: we can get further because we don’t have to invent everything from scratch; we can build on what others have done before. So, if we need to solve a particular problem that needs calculus, we don’t have to invent calculus from scratch to do so, we can use what Newton (and Leibniz, of course) invented. Standing on their shoulders, we can see further.

But how do we get to stand on the giant’s shoulders? (I’ll keep the metaphor to a single giant, as standing on the shoulders of multiple giants sounds too much like a circus act. And I am focussing on the mathematical giant.) We aren’t born up there on the giant’s shoulders. While we don’t have to grow into the giant (invent calculus), we do have to climb up the giant (study calculus).

And the giant is getting ever bigger. On the one hand, this is good: being so much higher we can see so much further. On the other hand, what happens when we have to spend our entire lives climbing up the vast growing giant, and never reach the viewpoint on the ever-distant shoulders?

We need short cuts up the giant. Fortunately, other are building ropes and ladders and lifts: tools to climb the giant more easily. So, we now have computer algebra packages that can solve our differential and integral equations for us; we no long need to spend years studying and practising how to do this.

But wait! cry the purists. That is cheating.

there is no Royal Road to geometry – Euclid

Understanding an idea meant entangling it so thoroughly with all the other symbols in your mind that it changed the way you thought about everything. – Greg Egan

There are no shortcuts, the purists insist. Mathematics is not a “spectator sport”. You have to do it, be immersed in it, internalise it, in order to really understand it.  The youth of today, with their fancy calculators and computers, they don’t really understand arithmetic and algebra and calculus.  Get off my lawn!

There is nothing new under the sun when it comes to criticism of youth, of course. Plato, in Phaedrus, has Socrates rail against this new-fangled literacy:

 [writing] will introduce forgetfulness into the soul of those who learn it: they will not practice using their memory because they will put their trust in writing, which is external and depends on signs that belong to others, instead of trying to remember from the inside, completely on their own. You have not discovered a potion for remembering, but for reminding; you provide your students with the appearance of wisdom, not with its reality. Your invention will enable them to hear many things without being properly taught, and they will imagine that they have come to know much while for the most part they will know nothing. And they will be difficult to get along with, since they will merely appear to be wise instead of really being so. 

This sounds suspiciously similar to those modern complaints about using calculators rather than mental arithmetic, or using computer algebra programs rather than slogging through pages of pushing symbols around. These devices give only the “appearance of wisdom”.

I do have some sympathy with this view. There does seem to be a lot of blind trust in the output of calculators and computers. However, I’m not sure it is purely the fault of the calculators. There can be uncomprehending blind trust in symbol pushing, too. I remember, many years ago, being in a computer shop, buying four items. The shop assistant wrote down the prices, and laboriously added them up, with much crossing out. When they announced the total, I said “that’s wrong”. They got a bit huffy, but then I pointed out their total was too small: it was less than one of the items on the list! As well as their huffiness, I detected a faint feeling of puzzled wonder from the assistant: how had I known? Despite the hand calculation, the assistant had no feel for the numbers. Maybe Socrates would have said that they should have added the numbers in their head? (Notice here that I didn’t know what the right answer was, but I knew the suggested answer was wrong.)

Another example comes to mind, again from many years ago. We were buying some new pillows: four for £4.99 each. The shop assistant wrote down 4.99 four times in a list, and added them up. Meanwhile I was going “£4.99 is a penny less than £5, so that’s £20 minus 4p, or £19.96.” I had the right money ready by the time the assistant came up with the answer, and was again met with puzzled wonder. (I’m sure that’s the real reason supermarkets took the prices off their goods: to stop some customers freaking out the cashiers by having the right money ready!)

I recounted this pillow story to my mother, who, faster than I did my shortcut calculation, simply multiplied 4.99 by 4 in her head, and got the right answer. I was almost as much in awe of this feat as the shop assistant had been in mine. But which approach shows more understanding of numbers: my short cut or my mother’s brute force calculation? Is it possible that slogging through all those exercises merely enable us to do calculations quickly, without thinking? And if there is no thought, then what have we actually gained? After all, one can learn by rote and merely “parrot” remembered answers.

Back to climbing that giant. What we need is a way of taking short cuts up and of having the “feel” for the numbers. An approach that could work is critical thinking about the supplied results (whether supplied by computer, or by our own unthinking calculations). We can keep the feel by using even shorter short cuts and heuristics that give an approximate answer, as a sanity check. Those shorter cuts and heuristics supply the feel, and when they are done automatically, they are the feel.

So, education shouldn’t be focussed on getting students to wade through pages and pages of exercises, pushing symbols (be they numbers or letters) around (unless they enjoy that sort of thing, of course). It should be more focussed on training in the use of short-cut tools, education on where and how to apply the tools, and meta-training in critical thinking about the results those tools give. Then we can climb the ever-growing giant fast enough to get to the top in time to see something before we die, and confident that we’ll understand what we do see when we get there.

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arbitrary doubling

One reason for the news report about temperatures (amongst other things) falling is that it has been quite warm this last week. The average daytime temperature here in the UK for March is usually about 10°C, but it has been 20°C and more.

This warmth has been the source of some irritating newspaper headlines:

The Mirror has a headline with: “temperatures soar to twice the average for March”

BBC New reports that the Daily Telegraph (which possibly should know better) says : “a barbecue weekend is in prospect, with temperatures double the usual for March.” 

No, the temperature hasn’t doubled. If it had doubled, if would be nearer 300°C!

If something doubles in size, units don’t matter. Something can double in length from 1 metre to 2 metres, or equally from 100 cm to 200 cm, or even from 3.28 feet to 6.56 feet.

0°C, 0°F and 0K are all different temperatures
(yes, there is no degree symbol when
using the absolute kelvin units)

But let’s look at changing the temperature units, say to Fahrenheit (an obsolescent temperature scale where the freezing point of water is 32°F, and the boiling point is 212°F).  10°C is 50°F, so if the doubling was real, 20°C would be 100°F. But 20°C is only 68°F (a seemingly much more modest rise!), whereas 100°F is 38°C (or nearly quadruple that 10°C, by newspaper headline “logic”!).

Things are even weirder if we used the Delisle scale (an obsolete and peculiar temperature scale where the freezing point of water is 150°D, and the boiling point is 0°D, so higher numbers are colder temperatures). 10°C is 135°D; “doubling” this gives 270°D, which is -80°C, whereas 20°C is 120°D. (Although it may seem weird for numbers to go down as temperatures go up, Delisle was not alone: this is the direction the original Celsius scale went, with freezing being 100, and boiling being 0.)

So why does doubling work for lengths, but not for temperatures? What’s different about the temperatures is that they have an arbitrary zero point. 0 metres means no length, but 0°C doesn’t mean no heat. There’s nothing special, temperature-wise, about the freezing point of water; it’s an arbitrary zero point (as evidenced by the fact that the Fahrenheit scale chooses a completely different arbitrary zero point).

To be able to do the multiplication and get a meaningful doubling, we need a true zero point for temperature, the absolute temperature, measured on the Kelvin scale (or the Rankine scale if you prefer those good old small Fahrenheit-sized degrees). There is another condition: the scale needs to be linear (rather than say logarithmic, like acidity, sound loudness, or star brightness); temperature is a linear scale, so that’s okay.  (If you want to play around with temperature scales, there’s a nice temperature units converter on the web, although it says °K, when it should be just K).

Let’s look at that doubling calculation again, now using an absolute scale. 10°C is 283K (rounded to the nearest degree, since that 10°C isn’t supposed to be very precise). Then 283K x 2 = 566K – this time, a meaningful calculation. And 566K is 293°C, which is worryingly hot for March!

The same data, graphed with
an arbitrary zero and a true zero on the y axis

This trick of an arbitrary zero leading to nonsensical comparisons like “temperatures double” is used all the time in “presentation graphics” (often in newspapers!) with misleading y axes. If the y axis doesn’t show a true zero (or worse yet, is unlabelled), then beware!



Note that we can do this kind of multiplication if we are talking about temperature differences, no matter what the scale. A rise of 20°C is twice a rise of 10°C. (A rise of 10°C is a rise of 18°F; a rise of 20°C is a rise of 36°F ). Rises work this way because a rise of 0°C, which is also a rise of 0°F, is a true zero.

However, newspapers can even mangle this! Several decades ago (although I doubt things have improved today) I saw a newspaper report that included the immortal phrase: “a rise of 1°C (33°F)”.

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FractalLab

Okay, here's a compelling reason to upgrade my browser to one that supports WebGL:

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Coincidence!

I've only recently started this intermittent blog, and a few days ago I sent off an email to a friend, notifying him of its existence. We haven't communicated for a few months, so it was amusing to get back the response:

(interesting stuff about stuff ....)
Hey, I was in the middle of this when YOUR e-mail came in! Telepathy?

Now, I know he doesn't really think it's telepathy, but some people do get exercised about this sort of coincidence. But things like this happen all the time. And it's easy to see why.

It might seem like the question is: "What's the chance of me receiving an email from you just as I'm writing one to you, given we haven't emailed each other for ages?" Pretty small, I suspect. But actually, the question is really: "What's the chance of me receiving an email from you just as I'm writing one to you, given we haven't emailed each other for ages, and given that I've just received an email from you just as I'm writing you one?". That is, what is p(x|x) (the probability of x having happened, given that x has happened)? Well, it's one. You can't get less unlikely than that!

Okay, that seems a little unsatisfactory. It still seems somehow to be remarkably unlikely. What's the probability it will happen again? Very small. But what's the probability that some weird coincidence will happen again? Now that is rather high.

Let's assume that we think some event has a probability of one in a gazillion of happening (the probability prior to it actually having happened, that is). But there are equally gazillions of unlikely things that could happen. Say you get an email from a friend just as you were thinking of them. But they might have phoned you, or texted you, or visited you, or written to you. Or you might have seen them on TV, or read about them, or about someone with the same name. And you could have been thinking of any of your friends, or of anyone else, or of anything else.
There are oodles of possible unlikely coincidences. What are the odds that one of them happens?

The way probability works, it's easier to calculate the chance of none of them happening. Let's say the odds of each one of these things happening is one in N, where N is very large (one in a billion, one in a trillion, or more). So the probability is 1/N, and the probability of it not happening is 1-1/N (very nearly, but not quite, certain that it won't happen).

Now let's say the number of unlikely things that might happen is also this huge number N (it could be 10N, or N/10; the calculation is cleaner using N, but the overall flavour of the result still holds for other values).What is the probability that none of the N unlikely things happens? That is, what is the probability that the first thing doesn't happen, and the second thing doesn't happen, and ... all the way up to and the Nth thing doesn't happen?

We just multiply the individual probabilities together, so we get (1-1/N)^N. That's the probability of no coincidences, so the probability of at least one coincidence is p = 1-(1-1/N)^N. For N larger that about 100, p is about 63% (for 10N it's 99.99%, for N/10 it's 10%). That's a pretty good chance of a weird coincidence. And that's just today!

The moral is: when individual events are unlikely, but there are also a lot of events that could happen, something will almost certainly occur.

So, that email: unlikely coincidence? That coincidence, yes; some sort of coincidence happening, not really.

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Computer Science meets King's Cross

I haven't been to London for a while, but I was there last week to give a talk. As I got off my train at King's Cross, I thought I heard an announcement: "The train currently standing at platform zero is the...". What? Platform 0? I listened to check I wasn't hallucinating, and there it was again: "Platform zero for the ..."

Now, I know King's Cross quite well, I thought. It has eight platforms on the main concourse, and also platforms 9 to 11 hidden away in their own private little siding. (Forget finding platform 9-and-3/4s: I've know people fail to find the real but well-hidden platform nine!) But platform 0? Srsly? Only computer scientists start counting from 0!

So when I got onto the main concourse, I checked, and sure enough, there were signs to platform 0. When I got back in Google-range, I consulted that fount of all knowledge, wikipedia. Yes, there is now a platform 0, as part of the refurbishment scheme. On the one hand, it seems a shame they've gone with platform 0, since that means platform 9-and-3/4 is still a "mistake" (there are tracks, not platform, between 9 and 10). But on the other hand: platform 0. Cool.

So I was disappointed to discover that once the refurbishment is completed, the platforms will be renumbered, starting from the much more unimaginative 1. (Presumably resulting in hilarity for all concerned as the zombified commuters, with the current numbering scheme hardwired in their cortexes, start making off-by-one errors, and end up scattered all over the country.)

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