Personal Blogs

In the words of the prophet Isaiah: ’ To the captives, ”Come out,” and to those in darkness, ”Be free”. ’ 

This is from a speech made by George Bush in 2003 and it's notable because of the nested quotes. Bush was quoting Isiah and the latter was quoting what I assume to be the words of God.

Bush's words were quoted in a Guardian article by George Monbiot and perhaps you can see where this is heading. 

Another writer 'Mr Smallweed' (actually Davis McKie) picked up on this and recognised its potential. He wrote a humorous piece in which he quoted himself explaining to Mrs Smallweed what what George Monbiot had written.

So at this point Smallweed is quoting himself quoting what Monbiot had written, in which he quoted Bush who had quoted Isiah who had quoted the words of God. The paragraph ends with a flurry of punctuation 

” ’ ” ’ ”

I hope you are keeping up, but we are not finished yet. It seems Mrs Smallweed met a friend, and Mr Smallweed quotes what she said to her friend, quoting what Mr Smallweed had told her earler.  So now

Mr Smallweed is quoting Mrs Smallweed quoting him quoting Monbiot quoting Bush quoting Isiah quoting God

and the triumphant end to the piece is 

“To the captives, ‘come out’, and to those in darkness, ‘be free’. ” ’ ” ’ ”

You can read the full story in this Guardian piece by David Marsh.

Here's a probability question I found in the marvellous problem collection Cut the Knot. 

In Figure 1 (a) we see two islands in a river and five bridges joining the islands to each bank and to one another.

Unfortunately a recent storm has caused damaged and each bridge has a 0.5 probability of having been swept away. What is the probability that a traveller can still cross the river?

This was harder than I was expecting and I found it quite confusing to think about. It did remind me though of a celebrated problem, the seven bridges of Königsberg, which prompted to dispense with some of the details and draw Fig. 1 (b), which captures the structure, and then after a bit of head-scratch I reasoned as follows. 

Each bridge is either intact or has been swept away, with an equal 0.5 probability of each. So there are 2x2x2x2x2 = 32 equally likely possibilities, not too many. So I could systematically enumerate the ones that allowed a crossing, which I did, and then just count them and divide by 32. Here they are (Figure 2), where the thick lines represent bridges that have survived the storm. I hope I have included them all!

So the probability is . I looked up the answer and sure enough 0.5 is the probability. But the problem setter had a much cleverer way of finding the answer!

Suppose a steamboat is coming along the river.

The dotted lines show the possible channels the boat might follow, if the corresponding bridge is has been swept away. Moreover if we draw a diagram of these channels and their interconnection it turns to be structurally identical to Figure 1 (b), and the boat will be able to pass along each channel with probability 0.5, because if the probability a bridge is intact is 0.5, the probability it has been swept away is also 0.5.

So probability(boat can pass along the river) = probability(a traveller can cross over the river)

And now for the masterstroke! If the boat can pass there must be an unbroken path along the river and the traveller is cut off. If the traveller can cross there must be an unbroken path across the river and the boat is cut off. So the two probabilities above must add up to 1 and hence both are 0.5.

This is a very pretty solution. It seems to have originated in a book Bas ^[1], as an example where arguing from symmetry gives a correct probability, which in some cases it may not.

And never forget the old malaprop: Don't burn your bridges before they're hatched.

[1] Bas C. van Fraassen, Laws and Symmetry, Oxford University Press, 1989

The word currant was originally applied what we now call raisins, which are dried grapes (compare French raisin). There was a Greek raisin industry in Corinth and raisins imported from there were called in Anglo-French reisin de Corauntz, borrowed into English as raysyn of Curans.

Over time this became shortened to currants. The OED gives this quote for 1620:

The small Raisins of Corinth, which we commonly call Currants. ^[1]

Later the word currant became used mainly for small fruits from other plants, blackcurrants, redcurrants and so on, while dried grapes, the original "raisins of Cornith", came to be called simply "raisins".

The name of the Greek city Corinth, in Greek Korinthos, is interesting in itself because it is probably borrowed from the non Indo-European language spoken in Greece before the Greeks arrived. It is the cluster nth (νθ) also found in menthe, "mint", and labyrinthos, and plinthos, "brick", that suggest this. So perhaps the currants in your bun preserve a word from a long dead Mediterranean language.

[1] T. Venner, Via Recta vii. 122

On Stack Overflow a Python beginner asked for help in writing a program to find a 4-digit number ABCD such when multiplied by 4 it reverses its digits, that is 4 x ABCD is DCBA. The kindly folks on Stack Overflow duly provided some help and the unique answer is 2178: 4 x 2178 is 8712.

This was new to me; although searching on "2178" threw up ~52,000,000 hits, this puzzle must have passed me by, until now.

Seeing the solution made me wonder if it could be generalised. There is quite a wide field of possible variations, but the first I looked was whether there is a find a 5-digit number that reversed itself when you multiply it by 4.

And there is: it's 21978. That looked familiar somehow and then I searched for a 6-digit counterpart. 219978 was the answer.

You can probably spot the pattern: we are just adding 9s in the middle. Will it work however many of them we insert? Yes it will; here's why.

Imagine we are doing a multiplication, 219.. lots of 9s...978 x 4.

You can see that for each 9 in the middle we shall have carried 3 and we do 4x9 = 36, plus the carry is 39, so we put down 9 and carry 3 again. This will continue as long as the digit being multiplied by 4 is a 9. When reach the 1 we will still have a carry of 3, and we do 4x1 = 4 plus the carry is 7, and then finally 4x2 = 8. So we can have as many 9s as we like sandwiched between 21 an 78 and multiplying by 4 will always reverse the number's digits, which is really cool.

Python deals with very long numbers really well, so just for fun

n = int('21'+'9'*200 + '78')

n

219999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999978

4*n

879999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999912

This question was B1 in the December 2025 Putnam competition, which is the top competition for math undergrads at US universities. It's nice because no specialised mathematical knowledge whatsoever is needed to grasp the question and the solution; both just involve simple ideas and the solution needs just a short chain of reasoning.

I'll give the question first and then clarify it a bit.

Suppose that each point in the plane is colored either red or green, subject to the following condition: For every three noncollinear points A,B,C of the same color, the center of the circle 

I'm going to change the colour scheme to red and blue, which will work better for anyone red-green colourblind. Here are a couple of sketches to help me explain. 

The question is asking us to suppose we have found a scheme that assigns to each point in the plane a colour, either red or blue, in a way that meets the conditions illustrated in the sketches above, i.e.

• If three red points lie on a circle, the centre of that circle will always be red, we will never find three red points on a circle with a blue centre.

• If three blue points lie on a circle, the centre of that circle will always be blue, we will never find three blue points on a circle with a red centre.

You might say don't any three points lie of a circle, but no, they might all lie on the the same straight line, so we want to rule that out.)

We are asked to prove that these conditions cannot be met unless every point in the plane is the same colour as every other point; every point is red or else every point is blue.

So here is my proof. The idea to imagine trying to colour the points of the plane, always keeping to the conditions bulleted above, using both both red ands blue. We start with two points, one red and one blue, and then try to extend the configuration, still keeping to the rules. We shall find that we must inevitably reach a position for which it is impossible to extend the colouring further without breaking the rules. So no such colouring can exist.

Consider the following diagram, where I have supposed we have two points of opposite colours. It make no difference which is which, so I have made A red and B blue.

From there we can infer colours for points C, D, E, F, G, in alphabetical order, by applying our rules, as follows. To avoid a whole sequence of diagrams each little different from the last I have coloured them in advance but you need to to imagine the colours of the points are not known until we determine each.

1. First observe that C and D must be different colours; for if both were red C, A and D would be three red points on a circle whos centre B is blue. If both were blue C, B and D would be three blue points on a circle whose centre A is red. Which of C and D is red which blue makes no material difference so we can make C red and D blue as shown.
2. Now the circle through C, B, D whose centre A is red has two blue points B and D on it already and another blue point on that circle would break our rules. This forces E and F to be red as shown.
3.  Next the circle through C, A, D whose centre B is blue has two red points A and C on it already and another red point on that circle would break our rules. This forces G to be blue as shown.

Now consider the point H which lies on the intersection of a small circle whose centre C is red and a larger circle whose centre B is blue. What colour should H be? Well the small circle with a red centre already has two blue points on it, so a third blue point is ruled out. On the other hand the larger circle with a blue centre already has two red points on it, so a third red point is ruled out. So H cannot be blue and it cannot be red either and it is impossible to colour it in a way consistent with our rules.

So we have found that if any two points are different colours it is impossible to extend that colouring to the whole plane and this the only colouring that meets the conditions is if every point is coloured red or every point is coloured blue.

I'm trying to teach myself a bit of LaTeX, which if you don't know is a specialised language that allows a user to enter expressions as plain text and have them rendered as mathematical notation.

For a while I've felt not knowing LaTeX is a gap in my toolkit but the immediate prompt is that I'm writing accessibility guides for a couple of modules and I'm looking how someone who is blind or has low vision can read and write mathematical notation, and LaTeX is one route, especially for anyone planning to go on to study more mathematics. So as an exercise I decided to write a post about a neat mathematical trick I learned recently. 

My Exercise

Suppose we're asked to solve the following quadratic equation by factorisation

This is not too hard. We just look for a pair of numbers that multiply to give and add to . These are and add to . So the equation factorises to

and the solutions are and .

But what if the coefficient of is greater than ? For example we might have

Now we have to consider all the factors of and of and figure out which combination is the correct one and it all gets a bit messy.

Here's a neat trick that makes this equation as easy to solve as the first example. We say (first coefficient times last), set the first coefficient to and form a new equation (but remember that , we will need it later!)

This is easy to factorise and we get , so the solutions are and . Of course these are not solutions of the original equation but now we bring that back into play, and divide the two solutions we have just found by it (makes a kind of sense you see, first we multiplied by , now we divide by it).

and

and, surprise, surprise, these are the solutions to the original equation!

Why does this work?

Multiplying the original equation by will give

or .

Now replace by and we get the second equation and the roots of this must be times those of the original equation.

Suppose we choose 10 digits 0-9 at random, one after another. For example, we might get

3, 0, 5, 2, 1, 5, 3, 1, 9, 2

How many distinct digits appear?  In the example there were 6 –

0, 1, 2, 3, 5, 

– but the answer could be anything from 1 (all the digits are the same), up to 10 (they are different). 

Can we work out the expected number of distinct digits that appear; how many different digits show up on average?

Consider the probability that 0 does not appear. The probability of the first digit not being 0 is , because 9 out of 10 digits aren't 0.  The probability that the second digit is not 0 is the same, and so on, up to the tenth digit.

The probability the none of the digits is 0, i.e. 0 does not appear, is therefore .

What then is the probability that 0 does appear. Well, this is easy. It either appears or it doesn't, and the two probabilities must therefore add up to 1. Hence the probability of 0 being present in our sequence is 

This is equivalent to saying that the expected number of times 0 will come up is 0.651. But there was nothing special about 0, the calculation would be the same for the other digits.

Now imagine we keep a score of what distinct digits appear, counting 1 if a digit appears and 0 if it doesn't. The probability that a digit appears is 0.651, in which case it will contribute 1 to the total. and 1 - 0.651 = 0.349 that it does not turn up and contributes 0. So it will make a net contribute of 0.651 x 1 + 0.349 x 0 = 0.651.

Now here is a piece of utter magic. There is a theorem, "Linearity of Expectation" which says that to get the overall score across all the digits all we need to do is add up the individual scores! This is actually quite surprising but it's true.

So we can just work out 10 x 0.651 = 6.51 and that is the average number of distinct digits we expect to see.

I always like to simulate these answers, in case I have something horribly wrong, so here is a short Python program

digits = [0,1,2,3,4,5,6,7,8,9]
count = 0
trials = 1000000

for i in range(trials):
    count = count + len(set(random.choices(digits, k=10)))

print('mean number of different digits =', round(count/trials, 2))

And here is the output

mean number of different digits = 6.51

But there is more. I started thinking what if we did the same thing in base 16, with 16 digits 0 - 9, A, B, C, D, E, F? Or used the whole alphabet of 26 letters, or went beyond that?  Suppose we used a n-character alphabet?

This smelt like a situation where Euler's famous number e ≈ 2.71828 might crop and when I investigated, there it was, right on cue. The beautiful result is that with an n-character alphabet, for large n the expected number of distinct characters in a run of n random choices approaches 

I got this rather nice problem from Cut the Knot, and it was originally from the Leningrad Maths Olympiad.

Can we visit each of the 8 vertices of a cube exactly once and return to the starting vertex by following a path made up of 8 straight line segments, each connecting a vertex of the cube to another of the cube's vertices?

The answer is yes and below we see one possible solution, with the dotted path meeting the these conditions.

However you will notice that in this tour 3 of the segments coincide with edges of the cube. Is it possible to find such a path where none of the segments coincide with an edge?

Prove that the answer is no, no such path can exist.

Solution in comments.

Apple are now selling motor vehicles, starting with something called the "iVan" but critics are sayings it's "terrible".

This was one of the first jokes I ever heard, in my infant school days.

When is a door not a door?

When it's ajar!

Yesterday I suddenly wondered, don't know why, where "ajar" comes from and what other things , if any, can be ajar (apart from a  jar obviously).

The OED gives the definition "

Of a door, gate, or window: so as to be slightly or partially open.

It can apply to anything that closes off a space, so long as it's got hinges. The last bit is important, at least historically, because the original form was on char, where char means something like "turn". In Old English it was spelt in variety of ways, e.g. cyrre, chere, and referred at first to time "coming round". The meaning was extended to something physically turning; the OED gives this example from 1510

The dure on chare it stude.¶

So there we have our door. The meaning was then extended further to something like a "turn of work" — compare ways turn is used; take turns, it's your turn, etc. — and changed its spelling to chore, a piece of work that needs to be done. And so ajar and chore are related words, rather surprisingly.

But chare has left a fossil trace in the language, charwoman.

¶ Gavin Douglas, King Hart

The Language-Lover's Lexipedia, by Joshua Blackburn. I learned of this wonderful book via the podcast and YouTube channel Words Unravelled, and it is an absolute joy.

The breadth and depth of Mr Backburn's research and scholarship are simply stunning, but at the same time he manages to maintain a light and playful tone. To give a flavour of the book here's a potted version of just one of the 200+ entries, from Abracadabra to Zyzzyva.

We've all heard of the Bee's Knees and the Cat's Pyjamas I suppose, and perhaps also the Duck's Quack, although that's less familiar. But until now I never real stopped to wonder where they came from. Turns out they were Flapper slang.

When I think about Flappers I have a hazy vision of the Roaring Twenties and bobbed hair style and the Charleston and all that sort of thing. Merriam-Webster's more informed definition is " a young woman predominantly of the 1910s and 1920s who showed freedom from conventions (as in style of dress and conduct)".

And they also had their own Flapper-Speak which included a craze for phrases consisting of the name of an animal followed by something mildly surprising or incongruous (the Duck' Quack is an outlier here). Mr Blackburn has dug up a whole bunch of examples, forgotten now apart from the three I gave above. Here are some of my favourites:

The Bullfrog's Beard

The Clam's Garters

The Eel's Ankle

The Hen's Eyebrows

The Kipper's Knickers

 And if you fancy a Bee's Knees cocktail Mr Blackburn has found the recipe.

Picture credit: Copilot 27 November 2025

A phonestheme is when a large number of words that start with the same sound (most typically a consonatn pair) to have related meanings. The phenomenon was first noted by J. R. Firth in 1930. A classic example is fl-, which I remember being fascinated by when first reading about it. Here are some words beginning with this consonant pair fl-. 

flit, flash, flutter, flung, flicker, flee, flip, flap, flare, flurry, fly, flow

What all these have in common is some notion of movement, especially sudden or repetitive movement.

The Wikpedia article gives a number of other examples, such as cl-, e,g. clop, clap, cling, clasp, clutch, all to do with moving an object; and sn-, e.g. snort, snitch, snot, snout, snivel, snitch, sneeze, sniff, sneer, snore, snorkel, all nose-related.

So is there some real effect here, is there something inherently fluttery about fl-? If so, is there some deep psychological connect? Or it it mimetic; does "flutter sound like something fluttering? Or is it somehow to do with how we make a "fl" sound, do we flap our tongue when we pronounce it? (probably not). And then we know from historical evidence that words can influence other similar sounding words, so that could be at work too.

Is there even enough statistical evidence to support the idea that fl- carries a semantic load? Perhaps we are just (as people do) see in a pattern because we always look for them? 

We might be able to provide some evidence by doing some clever statistical analysis, but it's slippery (there's another example of a phonestheme, sl-; slip, sled. slop, slime slither, slobber, sludge). But what criteria do we use to decide of a word belongs in the list? Should slow be counted? What about slag? Slant? Sleet? Slick?

I tried some rough experiments with the help of Copilot. It says

About 8–10% of common English words starting with “fl” relate to repeated or sudden movement.

Is that evidence. Well maybe. We'd have to come up with some list of representative and comparable (how to define what these mean?) semantic fields and see if 8-10% was unusually high.

Another of the examples in the Wikipedia articles is gl-, which is connected with shining or gleaming etc. I discussed this in a previous post A Little "Eggymology" and it is very likely to be a real connection but one that is simple to explain. The words probably all share an etymology and are descended from a PIE root *ghel-, "shining". So no profound mystery. This time Copilot tells me 6-7%. If the proportion for fl- words sharing the same semantic field is higher than that for gl- where (we think) an effect is known to exist what does that tell us? (But I haven't checjed Copilots findings.)

Another example quoted is stand, stable, state, stasis, but from memory this too is a group of words thought to have a common ancestry, a PIE root *sta-, "stand" or something like that.

So is a phonestheme a real thing? It's an intriguing concept, but the jury is still out. There seems to be a school of thought that believes it is just an illusion, but it seems too tat there are also serious scholars and ongoing research in support of there being something real in the idea. Either way I find it fascinating.