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This year, the front page of mscroggs.co.uk will once again feature an advent calendar, just like last year and the year before. Behind each door, there will be a puzzle with a three digit solution. The solution to each day's puzzle forms part of a logic puzzle:

It's nearly Christmas and something terrible has happened: Santa and his two elves have been cursed! The curse has led Santa to forget which present three children—Alex, Ben and Carol—want and where they live.

The elves can still remember everything about Alex, Ben and Carol, but the curse is causing them to lie. One of the elves will lie on even numbered days and tell the truth on odd numbered days; the other elf will lie on odd numbered days and tell the truth on even numbered days. As is common in elf culture, each elf wears the same coloured clothes every day.

Each child lives in a different place and wants a different present. (But a present may be equal to a home.) The homes and presents are each represented by a number from 1 to 9.

Santa has called on you to help him work out the details he has forgotten. Behind each day (except Christmas Day), there is a puzzle with a three-digit answer. Each of these answers forms part of a fact that one of the elves tells you. You must work out which combination of clothes each elf wears, which one lies on each day, then put all the clues together to work out which presents need delivering to Alex, Ben and Carol, and where to deliver them.

Ten randomly selected people who solve all the puzzles and submit their answers to the logic puzzle using the form behind the door on the 25th will win prizes! A selection of the prizes are shown below, and will be added to throughout December.

The ten winners will also will one of these winners' medals:

As you solve the puzzles, your answers will be stored. This year, there is a new feature allowing you to synchronise your answers between multiple computers: simply enter your email address below the calendar, and you will be emailed a magic link to visit on your other devices.

Behind the door on Christmas Day, there will be a form allowing you to submit your answers. The winner will be randomly chosen from all those who submit the correct answer before the end of 2017. Each day's puzzle (and the entry form on Christmas Day) will be available from 5:00am GMT. But as the winners will be selected randomly, there's no need to get up at 5am on Christmas Day to enter!

To win a prize, you must submit your entry before the end of 2017. Only one entry will be accepted per person. If you have any questions, ask them in the comments below or on Twitter.

So once December is here, get solving! Good luck and have a very merry Christmas!

A few weeks ago, I took the copy of MENACE that I built to Manchester Science Festival, where it played around 300 games against the public while learning to play Noughts and Crosses. The group of us operating MENACE for the weekend included Matt Parker, who made two videos about it. Special thanks go to Matt, plus Katie Steckles, Alison Clarke, Andrew Taylor, Ashley Frankland, David Williams, Paul Taylor, Sam Headleand, Trent Burton, and Zoe Griffiths for helping to operate MENACE for the weekend.

As my original post about MENACE explains in more detail, MENACE is a machine built from 304 matchboxes that learns to play Noughts and Crosses. Each box displays a possible position that the machine can face and contains coloured beads that correspond to the moves it could make. At the end of each game, beads are added or removed depending on the outcome to teach MENACE to play better.

On Saturday, MENACE was set up with 8 beads of each colour in the first move box; 3 of each colour in the second move boxes; 2 of each colour in third move boxes; and 1 of each colour in the fourth move boxes. I had only included one copy of moves that are the same due to symmetry.

The plot below shows the number of beads in MENACE's first box as the day progressed.

Originally, we were planning to let MENACE learn over the course of both days, but it learned more quickly than we had expected on Saturday, so we reset is on Sunday, but set it up slightly differently. On Sunday, MENACE was set up with 4 beads of each colour in the first move box; 3 of each colour in the second move boxes; 2 of each colour in third move boxes; and 1 of each colour in the fourth move boxes. This time, we left all the beads in the boxes and didn't remove any due to symmetry.

The plot below shows the number of beads in MENACE's first box as the day progressed.

You can download the full set of data that we collected over the weekend here. This includes the first two moves and outcomes of all the games over the two days, plus the number of beads in each box at the end of each day. If you do something interesting (or non-interesting) with the data, let me know!

As a child, I was a huge fan of Captain Scarlet and the Mysterons, Gerry Anderson's puppet-starring sci-fi series. As an adult, I am still a huge fan of Captain Scarlet and the Mysterons. Set in 2068, the series follows Captain Scarlet and the other members of Spectrum as they attempt to protect Earth from the Mysterons. One of my favourite episodes of the series is the third: Big Ben Strikes Again.

In this episode, the Mysterons threaten to destroy London. They do this by hijacking a vehicle carrying a nuclear device, and driving it to a car park. In the car park, the driver of the vehicle wakes up and turns the radio on. Then something weird happens: Big Ben strikes thirteen!

Following this, the driver is knocked out again and wakes up in a side road somewhere. After hearing his story, Captain Blue works out that the car park must be 1500 yards away from Big Ben. Using this information, Captains Blue and Scarlet manage to track down the nuclear device and save the day.

After rewatching the episode recently, I realised that it would be possible to recreate this scene and hear Big Ben striking thirteen.

At the end of the episode, Captain Blue explains to Captain Scarlet that the effect was due to light travelling faster than sound: as the driver had the radio on, he could hear Ben's bongs both from the tower and through the radio. As radio waves travel faster than sound, the bongs over the radio can be heard earlier than the sound waves travelling through the air. Further from the tower, the gap between when the two bongs are heard is longer; and at just the right distance, the second bong on the radio will be heard at the same time as the first bong from the tower. This leads to the appearance of thirteen bongs: the first bong is just from the radio, the next eleven are both radio and from the tower, and the final bong is only from the tower.

Big Ben's bongs are approximately 4.2s apart, sound travels at 343m/s, and light travels at 3×10⁸m/s (this is so fast that it could be assumed that the radio waves arrive instantly without changing the answer). Using these, we perform the following calculation:

This is close to Captain Blue's calculation of 1500 yards (and to be fair to the Captain, he had to calculate it in his head in a few seconds). Plotting a circle of this radius centred at Big Ben gives the points where it may be possible to hear 13 bongs.

Again, the makers of Captain Scarlet got this right: their circle shown earlier is a very similar size to this one. To demonstrate that this does work (and with a little help from TD and her camera), I made the following video yesterday near Vauxhall station. I recommend using earphones to watch it as the later bongs are quite faint.

Tomorrow, the new 12-sided one pound coin is released.

Although I'm excited about meeting this new coin, I am also a little sad, as its release ends the era in which all British coins are shapes of constant width.

A shape of constant width is a shape that is the same width in every direction, so these shapes can roll without changing height. The most obvious such shape is a circle. But there are others, including the shape of the seven-sided 50p coin.

As shown below, each side of a 50p is part of a circle centred around the opposite corner. As a 50p rolls, its height is always the distance between one of the corners and the side opposite, or in other words the radius of this circle. As these circles are all the same size, the 50p is a shape of constant width.

Shapes of constant width can be created from any regular polygon with an odd number of sides, by replacing the sides by parts of circles centred at the opposite corner. The first few are shown below.

It's also possible to create shapes of constant width from irregular polygons with an odd number, but it's not possible to create them from polygons with an even number of sides. Therefore, the new 12-sided pound coin will be the first non-constant width British coin since the (also 12-sided) threepenny bit was phased out in 1971.

Back in 2014, I wrote to my MP in an attempt to find out why the new coin was not of a constant width. He forwarded my letter to the Treasury, but I never heard back from them.

When cutting a pizza into equal shaped pieces, the usual approach is to cut along a few diameters to make triangles. There are other ways to fairly share pizza, including the following (that has appeared here before as an answer to this puzzle):

The slices in this solution are closely related to a triangle of constant width. Solutions can be made using other shapes of constant width, including the following, made using a constant width pentagon and heptagon (50p):

There are many more ways to cut a pizza into equal pieces. You can find them in Infinite families of monohedral disk tilings by Joel Haddley and Stephen Worsley [1].

You can't use the shape of a new pound coin to cut a pizza though.

Take a long strip of paper. Fold it in half in the same direction a few times. Unfold it and look at the shape the edge of the paper makes. If you folded the paper \(n\) times, then the edge will make an order \(n\) dragon curve, so called because it faintly resembles a dragon. Each of the curves shown on the cover of issue 05 of Chalkdust is an order 10 dragon curve.

The dragon curves on the cover show that it is possible to tile the entire plane with copies of dragon curves of the same order. If any readers are looking for an excellent way to tile a bathroom, I recommend getting some dragon curve-shaped tiles made.

An order \(n\) dragon curve can be made by joining two order \(n-1\) dragon curves with a 90° angle between their tails. Therefore, by taking the cover's tiling of the plane with order 10 dragon curves, we may join them into pairs to get a tiling with order 11 dragon curves. We could repeat this to get tilings with order 12, 13, and so on... If we were to repeat this ad infinitum we would arrive at the conclusion that an order \(\infty\) dragon curve will cover the entire plane without crossing itself. In other words, an order \(\infty\) dragon curve is a space-filling curve.

Like so many other interesting bits of recreational maths, dragon curves were popularised by Martin Gardner in one of his Mathematical Games columns in Scientific American. In this column, it was noted that the endpoints of dragon curves of different orders (all starting at the same point) lie on a logarithmic spiral. This can be seen in the diagram below.

Although many of their properties have been known for a long time and are well studied, dragon curves continue to appear in new and interesting places. At last year's Maths Jam conference, Paul Taylor gave a talk about my favourite surprise occurrence of a dragon.

Normally when we write numbers, we write them in base ten, with the digits in the number representing (from right to left) ones, tens, hundreds, thousands, etc. Many readers will be familiar with binary numbers (base two), where the powers of two are used in the place of powers of ten, so the digits represent ones, twos, fours, eights, etc.

In his talk, Paul suggested looking at numbers in base -1+i (where i is the square root of -1; you can find more adventures of i here) using the digits 0 and 1. From right to left, the columns of numbers in this base have values 1, -1+i, -2i, 2+2i, -4, etc. The first 11 numbers in this base are shown below.

Complex numbers are often drawn on an Argand diagram: the real part of the number is plotted on the horizontal axis and the imaginary part on the vertical axis. The diagram to the left shows the numbers of ten digits or less in base -1+i on an Argand diagram. The points form an order 10 dragon curve! In fact, plotting numbers of \(n\) digits or less will draw an order \(n\) dragon curve.

Brilliantly, we may now use known properties of dragon curves to discover properties of base -1+i. A level \(\infty\) dragon curve covers the entire plane without intersecting itself: therefore every Gaussian integer (a number of the form \(a+\text{i} b\) where \(a\) and \(b\) are integers) has a unique representation in base -1+i. The endpoints of dragon curves lie on a logarithmic spiral: therefore numbers of the form \((-1+\text{i})^n\), where \(n\) is an integer, lie on a logarithmic spiral in the complex plane.