Having spoken at the MathsJam annual conference in November 2016 about my previous phone spreadsheet on multiples of nine, I was contacted by a member of the audience with another interesting number fact they’d used a phone spreadsheet to investigate: my use of `=MID()`

to pick out individual digits had inspired them, and I thought I’d share it here in another of these columns (LOL spreadsheet jokes).

Twin primes, which are pairs of prime numbers with a difference of two, are well-studied. It’s not known whether there are infinitely many such pairs (although we’ve made some progress on that), but it’s suspected. The gaps between pairs get larger as you go up the number line, and there’s a few interesting bits known about them, including Brun’s Theorem. Today I’m going to investigate a thing about their products.

We can start by populating a spreadsheet with a list of pairs of twin primes, starting with the classic \((3,\!5)\) and proceeding from there. This data was obtained from the amazing resource at primes.utm.edu, maintained by known prime-basher Chris Caldwell. Putting this list of primes in the first column, and populating the second column with the number two higher gives us a lovely list of twin prime pairs.

In the next column, we calculate the product of each pair:

And in the fourth column, we can use a similar `MID()`

/ `VALUE()`

based construction to find the sum of the digits in this number:

Giving the not-especially-interesting result:

Or is it? In the fifth column, we can take `=MOD(`

and see what happens:*this value*,9)

With the exception of the first row, and let’s be honest the first one’s always weird anyway, you’ll see that all of these numbers have the value $8$ modulo $9$.

For an explanation of this, done in a bar after a few pints of course, we can thank sometime Aperiodical contributor and general maths guy Colin Beveridge, who contributes the following; if you’d like to try and work it out yourself, we’ve made these bullet points appear one at a time when you click, so you can use as many as you need and work the rest out.

- All twin prime pairs other than $(3,5)$ are of the form $6n +1$ and $6n – 1$ (it says it on Wikipedia so it must be true; left as a pleasing exercise for the reader)
- The product will therefore be of the form $36n^2 – 1$ (classic difference of two squares)
- This means the first part of this number will be divisible by $9$, as $36$ is divisible by $9$
- The whole thing must therefore be congruent to $8$ modulo $9$, as it’s a multiple of $9$ minus one.
- This means the sum of the digits of the number will also be $8$ modulo $9$.

A fun diversion for everyone. Thanks to Simon Allen, who sent me the email, and to Colin for his neat explanation, relayed via Simon.

If you’ve seen any nice number facts I can investigate using a spreadsheet on my phone, please send them in!

]]>We’re all (hopefully) aware that a pleasing property of numbers that are divisible by nine is that the sum of their digits is also divisible by nine.

It’s actually more well known that this works with multiples of three, and an even more pleasing fact is that the reason three and nine work is because nine is one less than the number base (10), and anything that’s a factor of this will also work – so, in base 13, this should work for multiples of 12, 6, 4, 3 and 2. Proving this is a bit of fun.

Once when I was thinking about this fact, an interesting secondary question occurred.

When thinking about the multiples of nine, I started to wonder: exactly which multiple of nine will you get when you add the digits together? Obviously, for all the small multiples of nine (less than, say, 81) they all add to $1 \times 9$, but once you reach 99 you have a sum of $18 = 2 \times 9$; and obviously bigger multiples of nine like 248426 have larger sums. So, how does this increase, and is there a pattern?

Since this struck me while I was out at dinner, and not equipped with my usual laptop/MS Excel combo, phone spreadsheets came to the rescue! I launched Google Sheets and created a quick table of values. Putting in a few early multiples of nine, highlighting down and using Auto-Fill created a list of multiples:

I then used the second column to work out the digit sum; my preferred method for this is using the `MID()`

function, which will return a sub-string of a sequence of letters or numbers. It takes three inputs – the thing you want a substring of (in this case, the cell to the left), the starting point, and the number of characters. I whipped up a quick formula which would find the first digit, and add it to the second, third and fourth digits (even though Google Sheets only gives you 1000 rows in a spreadsheet, so I wouldn’t ever need this many).

I also needed to add in the `VALUE()`

function, since `MID()`

returns text strings which it doesn’t always recognise as things you can add together, so I needed to convert them back to numbers before adding. This then gave me the list of digit sums for the multiples of 9. A quick final modification to the formula to divide this result by 9 gave me the information I was after – mostly, a string of 1s with a blip 2 at 99, then back to 1s again. How does this pattern progress?

You can see that the pattern alternates between 1s and 2s, increasing the number of 2s until it’s all 2s, then there’s the one 3 and it returns to the pattern, but this time starting slightly later on; then two 3s, and repeat, and the number of 3s increases until it’s all 3s, and then I assume we’d get a 4 (but my spreadsheet doesn’t have enough rows to actually see this).

This information looks nice in a table, but could probably be better represented in a graph, so I created a graph to display this data and make it more visible. It was around this point that the bill arrived, and my fellow diners started to get annoyed that I was just playing on my phone (and even though I offered to show them the graph, they weren’t interested). Here’s the graph:

You can see this number varying between 1 and 2, and the increasing and decreasing width of the segments until there’s a spike, which then increases in width too and the pattern continues. A more powerful package (and probably a bigger screen) could undoubtedly bring more clarity, but this was an interesting starting point and satisfied my curiosity – there IS a pattern, it IS quite pretty, and numbers are cool.

]]>Another name for a Rubik’s cube is ‘the Magic Cube’ – and Dr James Grime wondered if you could make a Magic Cube which incorporates its 2D friend, the Magic Square.

Several designs were considered for such a puzzle cube – early ideas included a cube with the numbers 1-54, each face being a normal magic square using the numbers 1-9, with a different multiple of 9 added to give a range for each face. After some searching and testing James settled on one particular type of magic square to make it work.

The Grime Cube is based around the **perimeter magic square** – an arrangement of the numbers 1-8 such that the three numbers on each edge sum to the same total. For example:

The square above has all four edges adding to 12 (and of course any rotation or reflection of this arrangement will also have this property). One nice interesting maths fact about such constructions is that only six valid perimeter magic squares exist for the numbers 1-8 (obviously, you can just add the same thing to each value to get another one, but that’s not as elegant).

What James Grime discovered was that you could use a Rubik’s cube, where the centres are fixed, to implement a puzzle involving such squares – since the centre of each face is fixed and only the edges move, you could try to arrange the pieces so that they add up to a number given in the centre. There only being six makes this an ideal thing to do on the faces of a cube, and the six squares pair up with each other as complements – so if you overlay two and sum them, the squares in the same place all add to 9.

This nice complement property meant that the cube could be designed to have opposite individual cubies (single chunks of cube) have their faces complement each other – so where there’s a 4 on one face of the cube, the opposite square will be a 5. Dr Grime also worked out a way to arrange six colours on the cube, so that each pair of mini-faces not only sum to 9, but are the same colour – and only one instance of each number/colour (e.g. red 5) occurs anywhere on the cube.

The final coup de grace for James was discovering a way to arrange the faces on the cube so that these colours could also be solved as a normal Rubik’s cube – an alternative arrangement where each face contains a single colour. This makes it a double puzzle – it can be solved so that either the magic squares work, and the opposite pairs match up, (left) or so that the colours on each face are all the same and there’s still 1-8 on each face (right).

As part of the testing process, I’ve had several goes at solving the cube and found it actually pretty pleasing – with my knowledge of how to solve a Rubik’s cube, I know how to put a particular edge or corner piece in a particular place, so I’ve found that very useful. I could find a pair of edge cubies that sit opposite each other (because e.g. the red 4 is always opposite the red 5) and then try other pairs until I get an arrangement that doesn’t use anything disallowed and lets me arrange the rest of the face so the totals are all the same. Then I can figure out which face I’ve just made, arrange it around the right centre face, and start working on the other faces around it. Others may have their own methods but I like how much it makes me think – not just blindly solving the colours, but also having to do sums and deduction along the way.

There’s another outstanding question about the cube, which Grime doesn’t know a good answer to yet – from the solved-for-magic-squares state, there should be a sequence of moves which translates it to the solved-for-colours state (and the maths tells us that this can be done in fewer than 20 moves, aka God’s Number). So far, James has found a path which takes 70 moves, but who will be the first to find the optimum path?

**The Grime Cube is on sale at MathsGear.co.uk for £14.99, and limited stock is available. **The last ordering date for delivery before Christmas in the UK is **19th December**, and more stock should be available in the New Year.

]]>

Q for my maths tweeps – recommendations wanted for Maths Journals suitable for a bright and engaged Sixth Form student. Suggestions?

— Colin Wright (@ColinTheMathmo) November 24, 2016

It led to a flurry of interesting replies, and here’s some of them.

Run by a group of students based at UCL, Chalkdust comes out four times a year and has editorials, fun features, interesting articles, cartoons and a prize crossword. It’s also available in a print version if you ask nicely and pay for postage.

Based at the University of Cambridge’s Millennium Maths Project, Plus has been a free online maths magazine for a good while – almost 20 years, with articles dating back as far as 1997. Plus has articles on diverse related topics, news, reviews, interviews and puzzles.

MAA’s Mathematical Monthly and MAA’s Math Horizons

The Mathematical Association of America has two regular journals, both requiring MAA membership to read online, but there are discounted student membership rates.

The Mathematical Association is a UK maths teachers’ organisation, and its magazine SYMmetry Plus, part of its Society of Young Mathematicians, is aimed at 10-18 year olds, with issues coming out three times a year. SYM members get a free copy, and it’s also available by subscription, costing around £20 for three issues.

Run by publishers Springer, the Mathematical Intelligencer is a proper journal, and while some articles require a subscription they have a subset of them available as open access.

The Royal Statistical Society runs this monthly stats-focused mag with a subscription or RSS membership, and online articles appearing a year after publication.

]]>

Here’s our annual round-up of what’s happening in sums/thinking at this year’s Manchester Science Festival. If you’re local, or will be in the area around 20th-30th October, here’s our picks of the finest number-based shows, talks and events.

**Saturday 22nd October, 7pm-9pm, Manchester235 Casino**

** Tickets £6; 18+**

Dust off your tuxedos and cocktail dresses for a night at the casino… in the name of science. This cabaret-style show explores the different scientific aspects of gambling, like the probability of winning and the psychology of body language. Plus, what happens to your brain when you gamble?

Psychologist **Paul Seager** explores deception and bluffing within the game of Poker. He talks about the use of verbal and non-verbal behavioural cues (‘tells’, in Poker parlance) in figuring out whether or not your opponents are trying to pull the wool over your eyes and steal all your chips.

Mathematician **Katie Steckles** (that’s me!) reveals the probability of drawing particular cards and dice rolls, and how you can use statistics to your advantage.

Neuroscientist **Nicola Ray** explores why gambling (in its non-addictive form) is so much fun. She’ll talk about how important brain regions are “hijacked” by the games played during gambling: the same regions that are responsible for ensuring we eat, procreate and fall in love are also the ones that ensure we keep playing even when we’re losing.

Casino Royale-style black-tie dress is optional, but warmly encouraged.

**Monday 24th-Sunday 30th October, 10am-5pm, Museum of Science and Industry**

** Drop in any time; main activities over 29th-30th weekend**

Manchester MegaPixel is part of the 2016 Manchester Science Festival, during which **I **and maths ninja **Matt Parker** will be building a gigantic pixel image display by colouring individual pixels using red, green and blue pens. This will model the way computer LCD screens use red, green and blue light to display photographs and images, but on a much larger scale!

The finished pixels will be arranged inside a large window at the museum, and will be on display for people to see the completed image. The finished MegaPixel will be over 10 metres high, and consist of around 8000 individual pixels, each of which has 300 coloured segments.

We’ll be colouring and building the pixel from **Monday 24th October**, finishing on **Sunday 30th, **and will also have other activities going on at the Museum of Science and Industry during the week, so you can learn about how image displays work, and help create the MegaPixel.

**Wednesday 26th October, 7pm-10.30pm, Pub/Zoo**

** Tickets £5; 18+**

From the brains behind Bright Club and Science Showoff comes Engineering Showoff, a chance to hear the funny side of building and looking after the structures, technology and ideas that surround us. Engineers from the north west’s universities and businesses take to the stage as stand-up comedians, sharing jokes and anecdotes from their professional lives. The gig is hosted by comedian and self-professed nerd Steve Cross.

**Thursday 27th October, 7pm-10.30pm, Museum of Science and Industry**

** Tickets free; 18+**

The Festival celebrates its 10th birthday this year – which is a very fine excuse to throw a party. Grab a slice of cake and:

Find out the scientific (mathematical tho) way to cut a cake with the Guardian’s **Alex Bellos**. Decorate your own cake and learn how to avoid a soggy bottom with **MetMunch**. Discover the secret science (maths tho) behind magic tricks with magician and ex-atomic physicist **Matt Pritchard. **Explore the psychology of why we love or loathe clowns with **Ginny Smith. **Discover the maths of chocolate fountains with **Adam Townsend. **Punch a bowl of custard and play musical chairs with **Science Made Simple. **Blow up a giant DNA double helix made of balloons with the **Museum of Science and Industry’s Explainers. **Embrace your inner child with some science-inspired face painting, and get ready to bust some moves at the **#HookedOnMusic silent disco.**

Party food will be served from the Warehouse Restaurant and the bar will be open all night.

**Friday 28th October, 6.30pm-7.30pm, Portico Library**

** Tickets £5/6/7; concessions available**

**Dr Jonathan Swinton** talks about a 1949 seminar in which pioneering mathematician Alan Turing discussed artificial intelligence (AI). It was during this seminar that some of the world’s first scepticisms about AI were raised. Can a machine think? Can it love? You’ll also hear a rare recording from 1976 by Max Newman, which discusses Turing and his work.

The participants in this Mancunian conversation were a remarkable mixture of economic migrants, asylum seekers and local talent. What combinations of thought and love attracted these thinkers to the soot-black, war-weary city? And why was Turing’s tale for so long unwritten in Manchester’s own history?

**Saturday 29th October, 7pm-10pm, Museum of Science and Industry**

** Tickets £9.50; 18+**

Miss the fun bits of your school science lessons? Then you’ll be pleased to hear that *After School Science Club* is back. Join **that** **Katie Steckles** and some colourful science stars for an adults-only evening of demonstrations and interactive fun. Plus a bar. And no homework (hurrah!).

**That Matt Parker**, television’s **Andrea Sella**, BBC Naked Scientists’ **Ginny Smith** and atmospheric scientist **Sophie Haslett** will also be there to talk to you about the science of rainbows, the rainbows of science and the maths behind colour TV. There’ll also be competition prizes, a giant painting wall and live experiments. It will (100% guaranteed) be spectrum-tacular.

**Friday 28th October – Saturday 19th November, 7.30pm & 2.30pm matinees, Royal Exchange Theatre, tickets from £16.50**

**Lecture on Saturday 29th October 5pm, free, Royal Exchange Theatre**

Can machines think? Is it possible to build a machine that thinks for itself? This classic play by **Hugh Whitemore** is set in the leafy surroundings of Bletchley Park at the height of the Second World War, where a brilliant young mathematician named Alan Turing was creating a machine to secure victory for Britain.

In the aftermath of victory, Turing arrived in Manchester with an even bigger task in mind – the development of the modern computer. It would be a task he left unfinished, publicly humiliated and destroyed by the revelation of his sexuality and prosecution for indecency. Turing’s most heroic hour is intertwined with the story of his betrayal and neglect by the nation he had helped in its darkest hour. Sheffield Theatres new Artistic Director **Robert Hastie** directs BAFTA winner **Daniel Rigby** in this major revival.

`1`

) and no glitter (`0`

) to encode binary messages in your nail varnish. We also posted an accompanying puzzle, stated as:
**Suppose I want to paint my nails on one hand differently every day for a month – so I need to use all 31 combinations involving glitter. Assuming that a nail painted with plain varnish can have glitter added, but a nail with glitter needs to be nail-varnish-removed before it can become a plain nail again, what order do I apply the different combinations so that you minimise the amount of nail varnish remover I’ll need to use?**

Here’s our take on the solution.

First of all, there’s a slightly subtle point to note – there’s a similar question you can ask, to which there is an easy and well-known answer. If instead we were asking how to minimise **the number of times I need to change a nail**, either from glitter to no glitter or back again, that’s the same as asking how to minimise the number of flips from 0 to 1 or from 1 to 0. The answer to this is to use a *Gray code*.

A Gray code – sometimes known as reflected binary code, and named after its creator Frank Gray – is an ordering of numbers such that each successive pair of numbers differ in only one bit (binary digit). Patented in 1947, it was originally designed to minimise errors in mechanical switching systems – if more than one switch has to flip to change between two numbers, they might not flip at exactly the same time, so the output might temporarily read or print out a number unintentionally. In the conventional order, binary numbers sometimes involve flipping multiple bits – such as going from 3 (`011`

) to 4 (`100`

), in which all three bits flip; the Gray code is an alternative ordering that can be worked out for any set of numbers $0 \ldots 2^n-1$ to ensure each switch is only one digit – for example, the numbers 0 to 7 can be written in the following order with only single bit-flips:

0 | `000` |

1 | `001` |

3 | `011` |

2 | `010` |

6 | `110` |

7 | `111` |

5 | `101` |

4 | `101` |

In fact, the binary combinations above are referred to as the *Gray code* for the numbers 0 to 7, in that order, so Gray code for 2 is 011 (even though that’s 3 in binary).

A Gray code for an $n$-bit sequence can be constructed from the $(n-1)$-bit Gray code, in quite a pleasing way. To go from 2-bit to 3-bit, take the sequence for 2-bit:

`00`

, `01`

, `11`

, `10`

Add a reflected version of this same string to the end:

`00`

, `01`

, `11`

, `10`

, `10`

, `11`

, `01`

, `00`

Then prefix the first half with `0`

s and the second half with `1`

s:

`000, `

`001`

, `011`

, `010`

, `110`

, `111`

, `101`

, `100`

This process can be repeated to generate the 4-bit sequence, and so on.

Since its creation, the Gray Code has been used in error correction and digital logic, and it can also be used to solve the Towers of Hanoi puzzle. As far as I know, the present work is the first application to personal grooming.

So, if I were to use the Gray Code ordering for my nails, I’d get the following sequence:

Order | Gray Code | Decimal equivalent | Order | Gray Code | Decimal equivalent | Order | Gray Code | Decimal equivalent |
---|---|---|---|---|---|---|---|---|

0 | `00000` |
0 | 11 | `01110` |
14 | 22 | `11101` |
29 |

1 | `00001` |
1 | 12 | `01010` |
10 | 23 | `11100` |
28 |

2 | `00011` |
3 | 13 | `01011` |
11 | 24 | `10100` |
20 |

3 | `00010` |
2 | 14 | `01001` |
9 | 25 | `10101` |
21 |

4 | `00110` |
6 | 15 | `01000` |
8 | 26 | `10111` |
23 |

5 | `00111` |
7 | 16 | `11000` |
24 | 27 | `10110` |
22 |

6 | `00101` |
5 | 17 | `11001` |
25 | 28 | `10010` |
18 |

7 | `00100` |
4 | 18 | `11011` |
27 | 29 | `10011` |
19 |

8 | `01100` |
12 | 19 | `11010` |
26 | 30 | `10001` |
17 |

9 | `01101` |
13 | 20 | `11110` |
30 | 31 | `10000` |
16 |

10 | `01111` |
15 | 21 | `11111` |
31 |

Each time I change from one nail to the next, I’m changing exactly one nail – either adding glitter, or removing it, and this is undeniably the most efficient way. But the question we asked in our puzzle was slightly different – since it’s much more effort to remove nail glitter than to add it, can we minimise the number of glitter removals, i.e. changes from 1 to 0?

If you’d like to try and work it out for yourself, go and do that now.

The first question you might ask is, is Gray Code the answer here too? In the list given above, there are 15 instances of switching from a 1 to a 0. But it can be done in fewer!

Our first breakthrough here was to find a solution more efficient than the Gray Code ordering, and we did find an ordering which takes 14 removes (for the record, it was 0, 2, 3, 1, 5, 4, 6, 7, 11, 10, 8, 9, 13, 12, 14, 15, 23, 22, 20, 21, 17, 19, 18, 16, 24, 26, 27, 25, 29, 28, 30, 31). But can we do any better?

It’s actually possible to prove that the lower bound on this — the fewest removals any valid solution can have — is 13. Let’s assume that each move consists of removing or adding exactly one nail’s glitter, and note that there are 31 ‘steps’ to achieve all 32 positions. If you start with no nails glittered and finish with all five glittered, the difference between the number of moves adding and removing glitter must be at most five, and therefore the minimal solution in terms of removals must have 18 adds and 13 removes making up the 31 steps.

All we need now is to actually find a way of doing it with only 13 removals. Ooh, here’s one:

Order | Binary | Decimal Equivalent | Order | Binary | Decimal Equivalent | Order | Binary | Decimal Equivalent |
---|---|---|---|---|---|---|---|---|

0 | `00000` |
0 | 11 | `01000` |
8 | 22 | `10010` |
20 |

1 | `00001` |
1 | 12 | `01010` |
10 | 23 | `10011` |
22 |

2 | `00011` |
3 | 13 | `01011` |
11 | 24 | `10111` |
23 |

3 | `00010` |
2 | 14 | `01111` |
15 | 25 | `10101` |
21 |

4 | `00110` |
6 | 15 | `01110` |
14 | 26 | `10001` |
17 |

5 | `00111` |
7 | 16 | `11110` |
30 | 27 | `10011` |
19 |

6 | `00101` |
5 | 17 | `11100` |
28 | 28 | `11011` |
27 |

7 | `00100` |
4 | 18 | `11000` |
24 | 29 | `11001` |
25 |

8 | `01100` |
12 | 19 | `11010` |
26 | 30 | `11101` |
29 |

9 | `01101` |
13 | 20 | `10010` |
18 | 31 | `11111` |
31 |

10 | `01001` |
9 | 21 | `10000` |
16 |

One obvious way to find the most efficient solution is to crunch it all through a computer: generate all the possible orderings, and count how many removals you need in each case. Of course, this would take a huge amount of computer time – the number of possible orderings is 32!, which is 263130836933693530167218012160000000, and as you can see below in the results of a simulation of 126,000 randomly-chosen orderings in R, only a tiny proportion achieve fewer than thirty removals, let alone thirteen.

So instead of relying on luck, we used a bit of *maths thinking*. The standard Gray Code that will solve you the Tower of Hanoi looks like this

\[1213121412131215121312141213121\]

where each digit represents the nail you change at each step (or which disk you move to a new peg). The problem with the standard solution for our purposes is that, starting with five unglittered nails, it doesn’t end with all fingers glittered — it ends with only digit 5 glittered. Fortunately, it’s easy to modify the sequence to change this.

Consider the simpler case of a three-fingered mathematician. They can imagine their nail-painting journey as a trip round the edges of a cube visiting every corner, with forwards and backwards movemement in each of the three dimensions correpsonding to glittering and unglittering each of the three nails. The picture below shows how the standard Gray Code solution can be modified to create a minimal-removals, no-glitter to all-glitter, solution, which on the cube corresponds to a path starting and ending at opposite corners.

Switching 1s for 2s and vice versa in the second 121 subection has done the trick here, but for the five-nails case it’s a bit more complex. It’s hard to visualise a trip round a five-dimensional hypercube, so let’s return to the digit sequence. Starting and ending at opposite corners happens if and only if each digit appears in the sequence an odd number of times (an even number of changes puts you back where you started). In the longer five-digit sequence we can still switch 1s and 2s in a 121 subsequence, but we can also switch around the 1s, 2s and 3s in a larger 1213121 subsequence, or even the 1s, 2s, 3s and 4s in one of the two big 121312141213121 chunks. A brief bit of faffing around before too long delivered this sequence:

\[1213121413121315232423212324232\]

where we’ve switched 2s with 3s in the second 1213121 chunk, and changed the 1s to 2s, 2s to 3s, 3s to 4s and the 4 to a 1 in the second 121312141213121 chunk. This sequence of digit-changes contains each digit an odd number of times, and gives the full sequence of nails tabulated above.

Having found this one solution, any equivalent solution made by permuting the five columns of binary digits will also only involve 13 removals; plus there are other tweaks you can make to the ordering using the method above. So, this solution generates hundreds of potential other 13-remove solutions. We’ve not ascertained whether all solutions that take exactly 13 can be found in this way from this one, or whether there are multiple classes out there, but this is pretty satisfying.

As always, CL-P has done his usual trick and built an interactive version of this problem, so you can play around with it for yourself and see if you can find a different solution with 13 removals:

“Principia Mathematica”, published in three volumes in 1910, 1912 and 1913, was a major work by mathematician and philosopher Bertrand Russell, with help from Alfred North Whitehead. The book contains a proof, starting from very basic axioms, that 1+1=2 – which takes over **360 pages**! It might seem excessive, but they work from only the most basic assumptions, and have to define firstly what they mean by ‘1’, ‘2’, ‘plus’, and ‘equals’. It’s all done in formal logic, and must surely be one of the longest proofs relative to the length and complexity of the statement it’s proving.

**PROOF SIZE: 0.1148 tennis courts**

- Principia Mathematica, on Wikipedia
- Russell and Whitehead, on The Story of Mathematics
- Restatement of the proof in more modern notation, at MetaMath Proof Explorer
- The proof starts on page 362 of Principia Mathematica Vol. 1

Famously one of the first proofs that required so many fiddly cases to check that its authors resorted to a computer, the Four Colour Theorem simply states that any diagram (or equivalently, any graph) drawn on a flat piece of paper can be coloured using at most four colours, so that any two adjacent parts are different colours. The original problem was stated in 1852, and while purported proofs were published in the 1880s, they were later found to be incorrect.

The first real proof, given by Appel and Haken in 1977, roughly consisted of finding a set of 1,936 minimal reducible structures that any diagram/graph that might not be four-colourable could be made up from, then checking each one by computer. Checking the maps one by one took over **1000 computer hours**, and accompanied a hand-checked component of the proof on **400 pages of microfiche**.

A shorter proof, involving a mere 633 reducible structures, was produced in 1996, and that proof was formalised in 2005 using the COQ computer proof assistant, which means it’s less reliant on cases checked by computer.

**PROOF SIZE: 1.8 times the area of an olympic swimming pool plus 370.4 watches of the film Avatar**

- Four Colour Theorem, on Wikipedia
- The Four Colour Theorem, at NRICH
- Formal Proof – The Four-Color Theorem, an article explaining the shorter COQ proof by Georges Gonthier in the Notices of the AMS (PDF)

‘The Enormous Theorem’ is always given as an alternative name for the classification of finite simple groups, but of course nobody actually ever calls it this in their work.

The classification of the finite simple groups — otherwise known as ‘CFSG’ or simply **The Enormous Theorem** — is a bit different to the other massive proofs listed here. It’s not a single work from one person or team, but rather a joint effort among dozens of mathematicians through the (nineteen-) eighties and nineties, a patchwork of results published in separate articles which together constitute the overall theorem.

A *group* is roughly speaking the set of interlocking symmetries of some concrete or abstract object, and the simple ones are those that are not ‘made by’ mashing two smaller symmetry groups together. The CFSG states that any finite simple group must be either a member of one of a set of precisely-defined infinite families of groups, or one of 26 specific outliers: the sporadic simple groups.

Wikipedia reckons that the aggregated length of the papers contributing to the proof is around **10,000 pages**. Work on a consolidated ‘second-generation’ proof, led by Daniel Gorenstein, is ongoing and is expected to result in an eleven-volume proof a mere 5,000 pages long. In fairness to the competition, it should be pointed out that these page counts are for the full articles/books, including all exposition as well as the bare proofs.

**PROOF SIZE: 1.4 basketball courts**

- Classification of finite simple groups, on Wikipedia
- An enormous theorem: the classification of finite simple groups, by Richard Elwes at Plus Magazine
- Rewriting the enormous theorem, by Rachel Thomas at Plus Magazine
- The Status of the Classification of the Finite Simple Groups, by Michael Aschbacher (PDF)

Kepler Conjecture/FLYSPECK

The Kepler Conjecture, originally posited in the 17th century by Johannes Kepler, relates to the density of spheres packed in 3D space. Kepler conjectured that the ‘cubic close packing’ (the one where you put a hexagonal grid of spheres on a flat surface and then stack another one on top, but offset so the balls are over the gaps) is the most efficient way to pack spheres in 3D space – the one with the least empty space left in between.

While it was long suspected to be true, nobody managed to formally prove it until Thomas Hales in 1998. Hales’ proof involved around **300 pages of notes** and **3 gigabytes of computer programs, data and results**. It was a ‘proof by exhaustion’ which involved checking many individual cases. Referees on the proof said they were ‘99% certain’ the proof was correct, and hence that this was the most efficient 3D packing.

But that wasn’t enough for Thomas Hales – as we covered here when it was completed, the FLYSPECK project (named as it is a Formal Proof of the Kepler conjecture, and ‘flyspeck’ is a word that contains all of those letters in that order, but it’s not as good as the word ‘flapjack’) was a further project undertaken to formalise and check the previous proof. It took from its start in 2003 until September 2014 to complete, and used proof assistants Isabelle and HOL Light.

**PROOF SIZE: 1.43 downloads of the film Titanic in HD, plus about 0.94 Harry Potter and the Prisoner of Azkabans**

- The Flyspeck project is complete: we know how to stack balls!
- Git Repo for FLYSPECK
- The Kepler Conjecture, on Wikipedia

Another recent bit of maths that’s made headlines by having a massive proof was the Erdős Discrepancy problem – back in February 2014, a proof using an SAT solver by Boris Konev and Alexei Lisitsa of the University of Liverpool hit the headlines because it was ‘the size of Wikipedia’ (around 13 gigabytes).

The problem asks whether it’s possible to come up with an infinite sequence of +1s and -1s and a ‘target’ number in such a way that you can never get past the target by adding together regularly-spaced terms from the sequence. (Going lower than minus-the-target-number also counts, in case you thought you had trumped the proof with your clever sequence of just -1s.) James Grime has explained the problem in a video with snakes and a cliff. The proof showed that in fact you can’t always make the required sequence.

Luckily, Terence Tao came to the rescue in September 2015, with a smaller hand-crafted proof developed in collaboration with the Polymath project.

**PROOF SIZE: around 3,250 holiday snaps taken on a 10 megapixel camera**

- A SAT Attack on the Erdős Discrepancy Conjecture by Boris Konev and Alexei Lisitsa
- Erdős’s discrepancy problem now less of a problem
- New Wikipedia sized proof explained with a puzzle – James Grime on YouTube
- Terence Tao has solved the Erdős discrepancy problem!

Claiming to be the ‘largest proof ever’, the Boolean Pythagorean Triples theorem relates to the question of whether it’s possible to split all numbers into two groups, neither of which contains a complete Pythagorean triple. For example, 3, 4 and 5 form a triple, and to find a valid split they would have to not all be in the same half – but then 5 couldn’t also be in the same half as 12 and 13, and so on.

It’s much better explained by Evelyn Lamb in her post in Nature, but a team of researchers have shown that not only is it not possible to do this, it’s not even possible to split the numbers 1 to 8000 in this way without getting stuck. It might not sound like ground-breaking mathematical knowledge we need right now, but it ties in to Ramsey Theory and other combinatorial questions. Proving it took 2 days for a computer running 800 parallel processors, and generated **200 terabytes of data**.

**PROOF SIZE: Amount of data generated by CERN every 2.92 days**

- Solving and Verifying the boolean Pythagorean Triples problem via Cube-and-Conquer by Marijn J. H. Heule, Oliver Kullmann and Victor W. Marek
- Two-hundred Terabyte maths proof is largest ever, by Evelyn Lamb at Nature
- Boolean Pythagorean Triples theorem on Wikipedia

It’s Eurovision time again! A chance for everyone to enjoy musical performances that are either good or so bad they’re good, ridiculous staging, and hilarious costumes, all sprinkled with a gently sarcastic Irish voiceover (if you’re lucky enough to be watching in the UK).

BUT WHAT’S THIS? They’ve changed the voting system? Don’t worry – some mathematicians are here to straighten it out for you.

A hangover from an old attempt at a unified European TV channel, the **Eurovision Song Contest** takes place each year in the country that won it the previous year. That country, along with five other countries who put a lot of cash in (France, Germany, Spain, Italy and the United Kingdom) get automatic entry to the contest, and semi-finals are held in the week running up to decide which countries will make up the rest of the 26 entrants, whittled down from 40.

Under the old Eurovision voting system, each country, including eliminated semi-finalists, held a telephone vote on the night of the show, with the option to vote for any of the 25 or 26 other countries (no, you can’t vote for yourself). The results of that phone poll, alongside the votes of a jury of musical experts, were condensed down to a set of points given by each country – 12, 10 and 8 down to 1, ranking their top 10 countries in order. The points awarded by each country are announced by one if its minor celebrities, one country at a time over a patience-stretching hour-and-a-half or so, slowly revealing the final ranking.

Previously, we’ve written about the way the phone and jury information is combined to give the final points total, which is somewhat mathematically arbitrary, and never clearly explained or even really acknowledged.

Clearly, the Eurovision producers are avid Aperiodical readers, as they’ve rejigged this part of the system for the 2016 contest.

Under the new scoring system, jury and televoting results are kept separate, and each create their own ranking and set of 1, 2, 3, 4, 5, 6, 7, 8, 10 and 12 points; so twice as many points are awarded overall. The soporific 90-minute results drip-feed will only cover the results from the juries. After that, the results of the popular vote will just all be added on at the end in one go (the word ‘thwumpf’ has been used). The idea is that this keeps the results announcement more interesting: in past years it’s become mathematically impossible for any but the currently-leading country to win well before the end of the show. Since fully half the points are kept back until the end now, in theory any country could take 1st place at the last minute – if the callers at home are sufficiently contrarian relative to the expert juries.

This new system raises a couple of important questions. Is it likely to produce different results to the old system? And that big exciting announcement of the last half of the points – how much of a shake-up is it likely to create in the table?

Luckily, the kind bods at Eurovision provide the full jury and televote results for the last couple of years on their website, so we can simulate what would have happened if the new system had been in place. It’s time for SOME MATHS.

Here’s a table showing the top part of last year’s results:

Country | Score | Position |
---|---|---|

Sweden | 365 | 1 |

Russia | 303 | 2 |

Italy | 292 | 3 |

Belgium | 217 | 4 |

Australia | 196 | 5 |

Latvia | 186 | 6 |

Estonia | 106 | 7 |

Norway | 102 | 8 |

Israel | 97 | 9 |

Serbia | 53 | 10 |

Under the new system, each score would be split into a jury score and a televoting score, and the countries would be ranked in order firstly using the jury scores, then again after adding in the televoting results. Here’s how that would look:

For anyone wondering why Australia appears in this table, they were allowed to enter as a one-off in 2015 as part of the 60th anniversary celebrations of Eurovision. This was a total one-off, and won’t happen again. Except in 2016, when it’s happening again. But that’s probably it.

Country | Jury Score | Position based on jury score only | Televoting score | Total score from jury and televoting | Final position | Position under old system |
---|---|---|---|---|---|---|

Sweden | 353 | 1 | 272 | 625 | 1 | 1 |

Italy | 171 | 6 | 356 | 527 | 2 | 3 |

Russia | 234 | 3 | 286 | 520 | 3 | 2 |

Belgium | 186 | 5 | 190 | 376 | 4 | 4 |

Australia | 224 | 4 | 124 | 348 | 5 | 5 |

Latvia | 249 | 2 | 88 | 337 | 6 | 6 |

Norway | 163 | 7 | 37 | 200 | 7 | 8 |

Estonia | 53 | 11 | 144 | 197 | 8 | 7 |

Israel | 77 | 8 | 102 | 179 | 9 | 9 |

Georgia | 62 | 10 | 51 | 113 | 10 | 11 |

Serbia | 12 | 23 | 86 | 98 | 11 | 10 |

Armenia | 15 | 22 | 77 | 92 | 12 | 16 |

Romania | 21 | 21 | 69 | 90 | 13 | 15 |

Azerbaijan | 40 | 13 | 48 | 88 | 14 | 12 |

Montenegro | 44 | 12 | 34 | 78 | 15 | 13 |

Lithuania | 31 | 16 | 44 | 75 | 16 | 17 |

Cyprus | 63 | 9 | 8 | 71 | 17 | 21 |

Slovenia | 36 | 15 | 27 | 63 | 18 | 14 |

Greece | 29 | 17 | 24 | 53 | 19 | 18 |

Poland | 2 | 26 | 47 | 49 | 20 | 22 |

Hungary | 29 | 17 | 17 | 46 | 21 | 19 |

Austria | 40 | 13 | 0 | 40 | 22 | 25 |

Spain | 6 | 25 | 26 | 32 | 23 | 20 |

Germany | 24 | 19 | 5 | 29 | 24 | 25 |

France | 24 | 19 | 3 | 27 | 25 | 24 |

United Kingdom | 12 | 23 | 4 | 16 | 26 | 23 |

The most obvious fact here is that there’s not much difference between the rankings under the new system: we have the same winner, the same top 3 and top 5: in fact nobody moves by more than four positions. There would have been little change in 2014 either: the top 5 is place-for-place identical under the two systems.

So, we can rest easy that this switch probably won’t affect the overall results too much. But surely the big phone-vote reveal at the end is going to be total ranking chaos, rendering the entire jury reveal a pointless exercise?

In fact the data suggests not. There’s a lot of movement outside the top of the table – Serbia shoot up from 23rd to 11th and Cyprus crash from 9th to 17th – but the winner is unchanged. In 2014 the top 3 would all remain static, with Poland rising from 23rd to 6th the biggest change. In general we might suppose that ‘novelty’ acts are likely to shift up the table after failing to amuse the po-faced juries, but going down a treat with the more accepting/drunk public.

If you’d like to perform your own statistical analysis of the results of Eurovisions past, the data is all available on their website, and Excel spreadsheets can be downloaded for 2014 and 2015. Or, just watch it and cheer when the people are silly.

]]>Alongside this video, I also have an associated puzzle for you to think about.

At the very end of the video, there’s a clip showing my hand with every possible combination of varnish from 0 to 31 – shown in this gif:

The puzzle is:

Suppose I want to paint my nails on one hand differently every day for a month – so I need to use all 31 combinations involving glitter. Assuming that a nail painted with plain varnish can have glitter added, but a nail with glitter needs to be nail-varnish-removed before it can become a plain nail again, what order do I apply the different combinations so that you minimise the amount of nail varnish remover I’ll need to use?

For example, you could paint on 1 (00001) for the first day, then quite easily change it to a 3 (00011) for day 2 by glittering one nail, but to do 2 (00010) would require removing glitter. In taking the photos to make the video clip, I came up with a quick ordering on the back of an envelope, which did save me some time and effort from doing it in number order, but I’m pretty sure it’s not optimal. Can you find a better one?

We’ll be posting the answer, with the interesting maths behind all this, in the next week or so. In the meantime, no spoilers in the comments (here or on the video!)

]]>

*In this series of posts, Katie will be going on about some of her favourite board games and card games, and some of the interesting mathematics to be found there. If you’d like the chance to play a mathematical board game, why not find or start a Maths Arcade at university, or join your local MathsJam.*

6 Nimmt! (German: *Take 6!*) is a card-based game which involves a hand of numbered cards, each also containing a number of cow heads. The cards are played in rounds, and during each round everyone chooses a card to play, they’re played in order, and you may find yourself having to take cards. The aim of the game is to end with the fewest cow heads.

I was given a set of 6 Nimmt! cards for my birthday a few years ago, and even though the instructions were in German, the (German) friend who’d given it to me had kindly included a printout of the rules in English. Thankfully, the rules aren’t that complicated anyway, and it’s a great game.

The cards are numbered from 1 up to 104, and each contains one or more cow heads. In the German, it’s the word ‘hornochsen’, which Google Translate tells me means ‘horn oxen’ (thanks), and Wiktionary seems to think is an offensive term for an idiot, roughly ‘bullhead’. My English translation of the rules refer to them as ‘cattle heads’. I’m told Hornochse is also the name of a burger chain in Germany. This is all probably not relevant to gameplay. Anyway, the number of cow heads on each card follows a set pattern. By comparing different cards, we established it as the following:

- all the cards have at least one cow head on them
- if the number is a multiple of five, it will have two cow heads
- if the number is a multiple of ten, it will have three cow heads
- if the number is a multiple of eleven, it will have FIVE COW HEADS. This means that while numbers like 11, 22 etc have 5 cow heads, since 55 is both a multiple of 5 and 11, it has a massive 7 cow heads.

It’s not clear why the cow heads follow this pattern, although it does give a roughly even distribution throughout the deck, and it being a fixed pattern also means you can predict which cards are going to score big in the game and try to avoid them, a little bit.

The game starts with four cards on the table, each at the start of its own row. Each player is dealt ten cards, and gameplay proceeds as follows: on each turn, everyone picks a card to play. In practise, we tend to put our chosen card on the table and keep a hand on it, so it’s clear when everyone’s picked, and then we all turn over at the same time. Whoever has the lowest number goes first, then the next lowest, and so on until everyone has played.

When playing your card, you have to place it next to one of the gradually emerging rows of cards on the table – one which ends in a card lower than the one you’re playing, but specifically, it has to also be the closest. For example, if I’m playing 26 and the rows end in 14, 25, 60 and 85, it’s larger than both the 14 and the 25, but since 25 is closer to 26, I would have to play it there. This is referred to as the ‘lowest difference’ (‘niedrigste differenz’) rule. Each player plays their card in order until everyone has played.

If your card is lower than all the cards on the ends of the four rows, you have to take a whole row. It can be any row of your choice (although the smart money is on taking the one with the fewest cards and/or cow heads) and once you take the cards, they don’t go back into your regular hand, but into a separate ‘keep’ pile, which you use to score up at the end of the game.

Many discussions have taken place about why the game is called ‘Take 6!’, if you only actually take 5 cards; I wildly speculate that the direct translation is ‘6 – take!’, which means, when there’s 6, you take, although I guess the best way to translate that into English is to call it ‘Take 5!’, which makes more sense but angers Dave Brubeck.

The other situation in which you have to take cards into your ‘keep’ pile results from the final rule, which is that no row of cards can have more than 5 cards in it – so if your card would make it six, you have to take that whole row of cards. Your sixth card is left as the new start of that row, and you take five cards. At the end of the game, when everyone has played all 10 cards from their hands, the player with the fewest cow heads in their ‘keep’ pile is the winner.

Those are all the rules of the game – each hand of 10 cards takes 10 rounds to play, so the games are fairly short; the rules suggest that you continue playing hands until someone has reached a total of 66 cow heads, although we often stop after one hand. The box lists it as being for 2-10 players (and with 104 cards, you can deal exactly 10 hands of 10 cards and leave four on the table), but with fewer than 10 there will be plenty of cards left in the box and not dealt; playing with 10 people means you can in theory keep track of which cards are still out there and try to guess who’s got them, but who has time for that?

In terms of strategy, my friends and I have yet to work out anything that’s consistently good; if there’s a card on the end of a row that’s one less than a card you’re holding, then that could be a good card to play, as nobody will be able to beat you to playing there (if they have a lower card, which means they’d play first, it will be played on a different row); the only problem might be if someone can’t play higher than any of the cards there, and swipes that whole row. This fact also means that if you have a card that’s two or three above the end card, it’s slightly riskier – if someone has the card between, they’re more likely to play it, and there’s always a chance you’ll end up being forced to play the sixth card in that row and have to take 5.

Despite being entirely number-based, there’s very little mathematical calculation to do in the game, although trying to work out your strategy is a nice challenge; I’ve played it happily with groups of mathematicians and non-mathematicians, and everyone seems to enjoy it. There’s an element of chance, as your initial hand and the cards on the table are random, and there often seems to be one unfortunate player who keeps getting forced to take loads of cards, but it doesn’t always seem to be me, so that’s fine.

It’s also very possible to make your own version of this, since all you need is a set of 104 numbered cards, and to mark the cow heads (or a penalty concept of your choice) on the appropriate cards. I’d be interested to see how distributing the cow heads differently affects the gameplay.

The game (in a yellow box, shown above) is available to buy in nerdy game shops for less than £10 (I’ve definitely seen it in Travelling Man), and it’s also on Amazon. There appear to be German and English versions available, although they’re called the same thing. I’ve also seen a Hungarian version called ‘Vigyáz(z) 6!’, a hilariously bracketed z presumably necessitated by gendered verb endings, and here the name appears to make ‘watch out for 6!’, which makes a bit more sense. More accurate Hungarian language facts welcome.

There is also a commercially available variant of the game themed – and this is not a joke – around the TV series *The Walking Dead*, in which the cow heads are replaced by bullets, and all the cards have pictures of zombies on. It’s sold as ‘The Walking Dead Card Game‘, but the box acknowledges it’s the same game as 6 Nimmt!, by the same designer. There’s also some additional thematic rules, and instead of playing the usual game (known as ‘Survival Mode’) you can choose to play ‘Hero Mode’, in which the aim is to get as many bullets as possible, and you have character cards which you can play once per game giving you special powers, like getting to go first even if you haven’t got the lowest number.

The idea of Hero Mode interests me, since it’s easy to forget that the cards you start off with won’t be the ones you end up scoring, so if you have a high-value card it might be interesting to see if you can play it to increase your own chance of getting it back, since that has become a desirable outcome.

**You can play 6 Nimmt! online **at Board Game Arena.