Puzzlebomb is a monthly puzzle compendium. Issue 53 of Puzzlebomb, for May 2016, can be found here:

Puzzlebomb – Issue 53 – May 2016

The solutions to Issue 53 will be posted at the same time as Issue 52.

Previous issues of Puzzlebomb, and their solutions, can be found at Puzzlebomb.co.uk.

]]>]]>Dear colleagues,

There is much dissatisfaction with the current state of research

publication, but little information on community attitudes and priorities.

I have started a survey which I hope you will fill in (I estimate 10-15

minutes, and it is completely anonymous). The results will be made publicly

available later this year. I hope that it will help to focus our efforts as

a community by allowing us to work toward broadly agreed goals. I want to

get as representative and as large a sample of the world mathematical

community as possible. Please forward to your local colleagues.Please answer this survey if and only if you have been involved with a

mathematical journal as editor, reviewer/referee, author or reader in the

last 3 years. By “mathematical” we also mean to include theoretical

computer science and mathematical statistics journals, and disciplinary

journals used by applied mathematicians. Essentially, any journal covered

by Mathematical Reviews qualifies.

The Kickstarter has already racked up many multiples of the original funding goal with three weeks still to go, so it’s at the “effectively a pre-order” stage. The posters start at \$26.

**Kickstarter: **Mandelmap poster by Bill Tavis.

**Conjecture** Every planar graph without 4-cycles and 5-cycles is 3-colourable.

In a paper just uploaded to the arXiv, Vincent Cohen-Addad, Michael Hebdige, Daniel Kral, Zhentao Li and Esteban Salgado show the construction of a graph with no cycles of length 4 or 5 which isn’t 3-colourable: it isn’t possible to assign colours to its vertices so that no pair of adjacent vertices have the same colour, using only three different colours. This is a counterexample to a conjecture of Richard Steinberg from 1976.

The problem was listed in the Open Problem Garden as of “outstanding” importance.

**Read the paper:** Steinberg’s conjecture is false

*via Parcly Taxel on Twitter*

*Warning: you could make a very strong argument I’ve thought far too much about something inconsequential. If that makes your stomach turn, look away now.*

This morning in the shower, I had an idle thought about my towel. It was, as always, folded neatly on the toilet seat. A problem that’s been bugging me for a few days is how to pick up the towel by a section of the long edge, so when it unfolds it’s the right way round.

* quiet in the back

The problem is that the short edge and the long edge look the same, and once I’ve folded the towel over a couple of times and had a shower only a madman* would remember which is which. But my towel isn’t square, so it occurred to me that either the longer or the shorter edge, after folding, could be the edge I want. Since I never make a diagonal fold, the long edge is only ever folded on top of the long edge, and likewise for the short edge. I fold the towel until it fits comfortably on top of the toilet seat, and by the time I’ve finished my shower I can’t be relied upon to remember which sequence of folds I did.

Which got me thinking about the ratio between the width and height of my towel: if I know this ratio then, by looking at the towel and counting the number of folds, I can work out which folds I’ve done, and hence which of the sides will unfold to be the long edge.

So let’s call the longer edge $w$, and the shorter edge $h$. Because my towel is both hard to fold and doesn’t show up well in photographs, I’ll use a sheet of A4 paper to show you how the method works.

First I set a handy marker for the shorter length, $h$, to compare against. Ideally I’d do this comparison by eye, but once the two measurements start getting close to each other it’s hard to tell. Another way to tell which is bigger would be to do a temporary diagonal fold in the paper, putting the two lengths directly on top of each other.

Then, I show that $w \gt h$.

So I fold the paper in half, and see that $\frac{w}{2} \lt h$.

To get a known length between $\frac{w}{2}$ and $w$, I unfold the paper and make a new fold in the right half, giving me a width of $w \left(\frac{1}{2} + \frac{1}{4} \right) = \frac{3}{4}w$. You can see that $\frac{3}{4}w \gt h$.

So I fold that quarter in half to get a width of $w \left(\frac{1}{2} + \frac{1}{8} \right) = \frac{5}{8}w \lt h$.

Again, to get a bigger width I unfold the last fold and make a new fold in the right portion, giving me $w \left(\frac{1}{2} + \frac{1}{8} + \frac{1}{16} \right) = \frac{11}{16}w \lt h$.

One last step gives me $w \left(\frac{1}{2} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} \right) = \frac{23}{32}w \gt h$.

At this point, I had reached the limits of my ability to accurately fold very thin strips of paper, so I stopped. The last measurement to the left of the marker I had was $\frac{11}{16}w$, and the last one to the right was $\frac{23}{32}w$. So I’ve got upper and lower bounds for $h$:

\[ \frac{11}{16}w \lt h \lt \frac{23}{32}w \]

My best bet for the real value of $h$ is the midpoint of my two bounds, so $h \approx \frac{45}{64}w$, or $\frac{w}{h} \approx \frac{64}{45} = 1.4\dot{2}$. That’s pretty darn close to $\sqrt{2} = 1.414…$! If I didn’t already know that every sheet of A4 is a silver rectangle, this would be enough for me to make a good guess.

The method I used is a *binary search*: at each step, I changed my estimate by half as much as I did in the previous step. If the resulting measurement was less than $h$, I added that amount to my approximation. So, the calculation went like this:

\begin{align}

w &\gt h & &✗\\[0.5em]

\frac{1}{2}w &\lt h & &+\frac{1}{2} \\[0.5em]

\frac{3}{4}w &\gt h & &✗ \\[0.5em]

\frac{5}{8}w &\lt h & &+\frac{1}{8} \\[0.5em]

\frac{11}{16}w &\lt h & &+\frac{1}{16} \\[0.5em]

\frac{23}{32}w &\gt h & &✗

\end{align}

That’s much more concisely written in binary as $\frac{h}{w} \gt 0.10110_2$. And because the last measurement was bigger than $h$, I know that $\frac{h}{w} \lt 0.10111_2$.

This method works for any ratio, not just $\sqrt{2}$ – it’ll produce lower and upper bounds, in binary, for the true value of the ratio. And if the ratio is a dyadic rational – a fraction of the form $\frac{a}{2^b}$ – you’ll eventually get a measurement that lines up exactly and the process will terminate!

If I was just trying to get an approximation for $\sqrt{2}$, obviously this paper-folding method isn’t the best – apart from anything else, A4 paper is defined as 297mm by 210mm, which could only ever give me four decimal places of $\sqrt{2}$. However, this binary search method to find $1/\sqrt{2}$ works in algebra too:

\begin{align}

1^2 &\gt \frac{1}{2} & & ✗\\[0.5em]

\left(\frac{1}{2}\right)^2 = \frac{1}{4} &\lt \frac{1}{2} & &+\frac{1}{2} \\[0.5em]

\left(\frac{3}{4}\right)^2 = \frac{9}{16} & \gt \frac{1}{2} & &✗\\[0.5em]

\left(\frac{5}{8}\right)^2 = \frac{25}{64} & \lt \frac{1}{2} & &+\frac{1}{8} \\[0.5em]

\left(\frac{11}{16}\right)^2 = \frac{121}{256} & \lt \frac{1}{2} & &+\frac{1}{16} \\[0.5em]

\vdots

\end{align}

Here, I’ve used the fact that $\frac{a}{b} \lt \sqrt{2}$ if $a^2 \lt 2b^2$ to easily decide what to do at each step. Because we’re now working purely in algebra, we can keep going forever.

As for my towel, the folding method tells me that its aspect ratio is bang on 11:16. Or, more likely, it’s 2:3 and I made a measurement error. $\frac{2}{3}$ isn’t a dyadic number, after all.

]]>The Carnival rounds up maths blog posts from all over the internet, including some from our own Aperiodical. See our Carnival of Mathematics page for more information.

]]>*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.

*This video is extremely shonky. Blame my phone, which can’t bring itself to record for more than 250 seconds at a time.*

**More information** about the Correntator.

Previously unseen footage has been unearthed by The Aperiodical’s crack team of investigative journalists of Kevin Bacon and Paul Erdős writing a paper together, and a still from this is shown above. This has massive consequences for the important topics of Erdős numbers, Bacon numbers and Erdős-Bacon numbers.

At the cutting edge of mathematical research, Erdős numbers are assigned to people who collaborated on a paper with Erdős or with someone who has an Erdős number. Erdős has, by definition, an Erdős number of zero. People who collaborated with him have an Erdős number of 1, people who collaborate with them have an Erdős number of 2, and so on.

Bacon numbers work similarly for people who appeared on film with Kevin Bacon, and like Erdős, Kevin Bacon has been chosen as the centre of this network, due to his large number of film appearances. Kevin Bacon similarly has a Bacon number of zero.

Erdős-Bacon numbers are assigned to people who have both an Erdős number and a Bacon number, and are simply the sum of the two numbers.

The fact of them collaborating on a paper gives Bacon an Erdős number of 1, and the fact that this was filmed gives Erdős a Bacon number of 1 (previously, his Bacon number was 5, through a chain including Ronald Graham, Merce Cunningham, Dennis Hopper and Chris Penn). This means Erdős and Bacon both have an Erdős-Bacon number of 1, the lowest possible number which, before this footage was discovered, was a value thought to be impossible. This new property of Erdős-Bacon numbers has consequences in Yang–Mills theory and may even provide a hint at a new approach to solving the Riemann hypothesis.

The brief footage of the collaboration was found in unused material filmed for the 1993 documentary *N Is a Number: A Portrait of Paul Erdős*. Filmed over a four-year period in four countries between 1988 and 1991, the documentary explores his career, personal life and sense of humour.

Puzzlebomb is a monthly puzzle compendium. Issue 52 of Puzzlebomb, for April 2016, can be found here:

Puzzlebomb – Issue 52 – April 2016

The solutions to Issue 52 will be posted at the same time as Issue 51.

Previous issues of Puzzlebomb, and their solutions, can be found at Puzzlebomb.co.uk.

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