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## Dragon Curves II

This post appeared in issue 05 of

*Chalkdust*. I strongly recommend reading the rest of*Chalkdust*.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.

Number in base -1+i | Complex number |

0 | 0 |

1 | 1 |

10 | -1+i |

11 | (-1+i)+(1)=i |

100 | -2i |

101 | (-2i)+(1)=1-2i |

110 | (-2i)+(-1+i)=-1-i |

111 | (-2i)+(-1+i)+(1)=-i |

1000 | 2+2i |

1001 | (2+2i)+(1)=3+2i |

1010 | (2+2i)+(-1+i)=1+3i |

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.

If you'd like to play with some dragon curves, you can download the Python code used
to make the pictures here.

### Similar Posts

Dragon Curves | The Mathematical Games of Martin Gardner | MENACE | Is MEDUSA the New BODMAS? |

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**2017-01-13 04:32:41**

## Is MEDUSA the New BODMAS?

I wrote this post with, and after much discussion with Adam Townsend. It also appeared on the Chalkdust Magazine blog.

Recently, Colin "IceCol" Beveridge blogged about something that's been irking him for a while: those annoying social media posts that tell you to work out a sum, such as \(3-3\times6+2\), and state that only $n$% of people will get it right (where \(n\) is quite small). Or as he calls it "fake maths".

This got me thinking about everyone's least favourite primary school acronym: BODMAS (sometimes known as BIDMAS, or PEMDAS if you're American). As I'm sure you've been trying to forget, BODMAS stands for "

**B**rackets, (to the power)**O**f,**D**ivision,**M**ultiplication,**A**ddition,**S**ubtraction" and tells you in which order the operations should be performed.Now, I agree that we all need to do operations in the same order (just imagine trying to explain your working out to someone who uses

*BADSOM*!) but BODMAS isn't the order mathematicians use. It's simply wrong. Take the sum \(4-3+1\) as an example. Anyone can tell you that the answer is 2. But BODMAS begs to differ: addition comes first, giving 0!The problem here is that in reality, we treat addition and subtraction as equally important, so sums involving just these two operations are calculated from left-to-right. This caveat is quite a lot more to remember on top of BODMAS, but there's actually no need: Doing all the subtractions before additions will always give you the same answer as going from left-to-right. The same applies to division and multiplication, but luckily these two are in the correct order already in BODMAS (but no luck if you're using PEMDAS).

So instead of BODMAS, we should be using

*BODMSA*. But that's unpronounceable, so instead we suggest that from now on you use**MEDUSA**. That's right,**MEDUSA**:**M**abano (*brackets*in Swahili)**E**xponentiation**D**ivision**U**kubuyabuyelela (*multiplication*in Zulu)**S**ubtraction**A**ddition

This is big news. MEDUSA vs BODMAS could be this year's pi vs tau... Although it's not actually the biggest issue when considering sums like \(3-3\times6+2\).

The real problem with \(3-3\times6+2\) is that it is written in a purposefully confusing and ambiguous order. Compare the following sums:

$$3-3\times6+2$$ $$3+2-3\times6$$ $$3+2-(3\times6)$$
In the latter two, it is much harder to make a mistake in the order of operations, because the correct order is much closer to normal left-to-right reading order, helping the reader to avoid common mistakes. Good mathematics is about good communication, not tricking people. This is why questions like this are "fake maths": real mathematicians would never ask them. If we take the time to write clearly, then I bet more than \(n\)% of people will be able get the correct answer.

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**2016-03-15 06:36:52**

## The Mathematical Games of Martin Gardner

This article first appeared in
issue 03 of

*Chalkdust*. I highly recommend reading the rest of the magazine (and trying to solve the crossnumber I wrote for the issue).It all began in December 1956, when an article about hexaflexagons was published in

*Scientific American*. A hexaflexagon is a hexagonal paper toy which can be folded and then opened out to reveal hidden faces. If you have never made a hexaflexagon, then you should stop reading and make one right now. Once you've done so, you will understand why the article led to a craze in New York; you will probably even create your own mini-craze because you will just*need*to show it to everyone you know.The author of the article was, of course, Martin Gardner.

Martin Gardner was born in 1914 and grew up in Tulsa, Oklahoma. He earned a bachelor's degree in philosophy from the University of Chicago and
after four years serving in the US Navy during the Second World War, he returned to Chicago and began writing. After a few years working on
children's magazines and the occasional article for adults, Gardner was introduced to John Tukey, one of the students who had been involved in
the creation of hexaflexagons.

Soon after the impact of the hexaflexagons article became clear, Gardner was asked if he had enough material to maintain a monthly column.
This column,

*Mathematical Games*, was written by Gardner every month from January 1956 for 26 years until December 1981. Throughout its run, the column introduced the world to a great number of mathematical ideas, including Penrose tiling, the Game of Life, public key encryption, the art of MC Escher, polyominoes and a matchbox machine learning robot called MENACE.### Life

Gardner regularly received topics for the column directly from their inventors. His collaborators included Roger Penrose, Raymond Smullyan,
Douglas Hofstadter, John Conway and many, many others. His closeness to researchers allowed him to write about ideas that
the general public were previously unaware of and share newly researched ideas with the world.

In 1970, for example, John Conway invented the Game of Life, often simply referred to as Life. A few weeks later, Conway showed the game to Gardner, allowing
him to write the first ever article about the now-popular game.

In Life, cells on a square lattice are either alive (black) or dead (white). The status of the cells in the next generation of the game is given by the following
three rules:

- Any live cell with one or no live neighbours dies of loneliness;
- Any live cell with four or more live neighbours dies of overcrowding;
- Any dead cell with exactly three live neighbours becomes alive.

For example, here is a starting configuration and its next two generations:

The collection of blocks on the right of this game is called a

*glider*, as it will glide to the right and upwards as the generations advance. If we start Life with a single glider, then the glider will glide across the board forever, always covering five squares: this starting position will not lead to the sad ending where everything is dead. It is not obvious, however, whether there is a starting configuration that will lead the number of occupied squares to increase without bound.Originally, Conway and Gardner thought that this was impossible, but after the article was published, a reader and mathematician called Bill Gosper
discovered the glider gun: a starting arrangement in Life that fires a glider every 30 generations. As each of these gliders will go on to live
forever, this starting configuration results in the number of live cells
perpetually increasing!

This discovery allowed Conway to prove that any Turing machine can be built within Life: starting
arrangements exist that can calculate the digits of pi, solve equations, or do any other calculation a computer is capable of (although very slowly)!

#### Encrypting with RSA

To encode the message \(809$, we will use the public key:

$$s=19\quad\text{and}\quad r=1769$$
The encoded message is the remainder when the message to the power of \(s\) is divided by \(r$:

$$809^{19}\equiv\mathbf{388}\mod1769$$
#### Decrypting with RSA

To decode the message, we need the two prime factors of \(r\) (\(29\) and \(61\)).
We multiply one less than each of these together:

\begin{align*}
a&=(29-1)\times(61-1)\\[-2pt]
&=1680.
\end{align*}
We now need to find a number \(t\) such that \(st\equiv1\mod a\). Or in other words:

$$19t\equiv1\mod 1680$$
One solution of this equation is \(t=619\) (calculated via the

*extended Euclidean algorithm*).Then we calculate the remainder when the encoded message to the power of \(t\) is divided by \(r\):

$$388^{619}\equiv\mathbf{809}\mod1769$$
### RSA

Another concept that made it into

*Mathematical Games*shortly after its discovery was public key cryptography. In mid-1977, mathematicians Ron Rivest, Adi Shamir and Leonard Adleman invented the method of encryption now known as RSA (the initials of their surnames). Here, messages are encoded using two publicly shared numbers, or keys. These numbers and the method used to encrypt messages can be publicly shared as knowing this information does not reveal how to decrypt the message. Rather, decryption of the message requires knowing the prime factors of one of the keys. If this key is the product of two very large prime numbers, then this is a very difficult task.### Something to think about

Gardner had no education in maths beyond high school, and at times had difficulty understanding the material he was writing about. He believed, however, that this was a strength and not a weakness: his struggle to understand led him to write in a way that other non-mathematicians could follow. This goes a long way to explaining the popularity of his column.

After Gardner finished working on the column, it was continued by Douglas Hofstadter and then AK Dewney before being passed down to Ian Stewart.

Gardner died in May 2010, leaving behind hundreds of books and articles. There could be no better way to end than with something for you to go
away and think about. These of course all come from Martin Gardner's

*Mathematical Games*:- Find a number base other than 10 in which 121 is a perfect square.
- Why do mirrors reverse left and right, but not up and down?
- Every square of a 5-by-5 chessboard is occupied by a knight.
- Is it possible for all 25 knights to move simultaneously in such a way that at the finish all cells are still occupied as before?

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**2015-10-21 14:27:38**

## How to Kick a Conversion

This post also appeared on the Chalkdust Magazine blog.

If you're like me, then you will be disappointed that all of the home nations have been knocked out of the Rugby World Cup. If you're

*really*like me, doing some maths related to rugby will cheer you up...The scoring system in rugby awards points in packets of 3, 5 and 7. This leads a number of interesting questions that you can find in my guest puzzle on Alex Bellos's Guardian blog. In this blog post, we will focus on another area of rugby: conversion kicking.

### Conversion Kicks

When a try is scored by putting the ball down behind the line, the scoring team gets to take a conversion kick. This kick must be taken in line with where the try was scored but it is up to the kicker how far away the kick should be taken. But how far back should the ball be taken to make the kick easiest?

One way to answer this question is to look to maximise the angle between the posts which the kicker will have to aim at: if the kick is taken too close to or too far from the goal line there will be a very thin angle to aim at. Somewhere between these extremes there will be a maximum angle to aim at.

When looking to maximise this angle, we can use one of the 'circle theorems' which have tormented many generations of GCSE maths students: 'angles subtended by the same arc at the circumference are equal'. This means that if a circle is drawn going through both posts, then the angle made at any point on this circle will be the same.

A larger circle drawn through the posts will give a smaller angle. If a vertical line is drawn which just touches the right of the circle, then the point at which it touches the circle will be the best place on this line to take a kick. This is because any other point on the line will be on a larger circle and so make a smaller angle.

Using this method for circles of different sizes leads to the following diagram, which shows where the kick should be taken for every position a try could be scored:

This, however, is not the best place to take the kick.

### Taking Account of Height

When a try is scored near the posts, the above method recommends a position from where the ball must be kicked at an impossibly steep angle to go over. To deal with this problem, we are going to have to look at the situation from the side.

When kicked, the ball will travel along a parabola (ignoring air resistance and wind as their effects will be small

^{[citation needed]}). Given a distance from the posts, there will be two angles which the ball can be kicked at and just make it over the bar. Kicking at any angle between these two will lead to a successful conversion. Again, we have an angle which we would like to maximise.However, the position where this angle is maximised is very unlikely to also maximise the angle we looked at earlier. To find the best place to kick from, we need to find a compromise point where both angles are quite big.

To do this, imagine that the kicker is standing inside a large sphere. For each point on the sphere, kicking the ball at the point will either lead to it going over or missing. We can draw a shape on the sphere so that aiming inside the shape will lead to scoring. Our sensible kicker will aim at the centre of this shape.

But our kicker will not be able to aim perfectly: there will be some random variation. We can predict that this variation will follow a Kent distribution, which is like a normal distribution but on the surface of a sphere. We can use this distribution to calculate the probability that our kicker will score. We would like to maximise this probability.

The Kent distribution can be adjusted to reflect the accuracy of the kicker. Below are the optimal kicking positions for an inaccurate, an average and a very accurate kicker.

As you might expect, the less accurate kicker should stand slightly further forwards to make it easier to aim. Perhaps surprisingly, the good kicker should stand further back when between the posts than when in line with the posts.

The model used to create these results could be further refined. Random variation in the speed of the kick could be introduced. Or the kick could be made to have more variation horizontally than vertically: there are parameters in the Kent distribution which allow this to be easily adjusted. In fact, data from players could be used to determine the best position for each player to kick from.

In addition to analysing conversions, this method could be used to determine the probability of scoring 3 points from any point on the pitch. This could be used in conjunction with the probability of scoring a try from a line-out to decide whether kicking a penalty for the posts or into touch is likely to lead to the most points.

Although estimating the probability of scoring from a line-out is a difficult task. Perhaps this will give you something to think about during the remaining matches of the tournament.

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**2015-10-08 04:38:58**

## How Much Will I Win on the New National Lottery?

This post also appeared on the Chalkdust Magazine blog. You can read the excellent second issue of Chalkdust here, including the £100 prize crossnumber which I set.

From today, the National Lottery's Lotto draw has 59 balls instead of 49. You may be thinking that this means there is now much less chance of winning. You would be right, except the prizes are also changing.

Camelot, who run the lottery, are saying that you are now "more likely to win a prize" and "more likely to become a millionaire". But what do these changes actually mean?

### The Changes

Until yesterday, Lotto had 49 balls. From today, there are 59 balls. Each ticket still has six numbers on it and six numbers, plus a bonus ball, are still chosen by the lottery machine. The old prizes were as follows:

Requirement | Estimated Prize |

Match all 6 normal balls | £2,000,000 |

Match 5 normal balls and the bonus ball | £50,000 |

Match 5 normal balls | £1,000 |

Match 4 normal balls | £100 |

Match 3 normal balls | £25 |

50 randomly picked tickets | £20,000 |

The prizes have changed to:

Requirement | Estimated Prize |

Match all 6 normal balls | £2,000,000 |

Match 5 normal balls and the bonus ball | £50,000 |

Match 5 normal balls | £1,000 |

Match 4 normal balls | £100 |

Match 3 normal balls | £25 |

Match 2 normal balls | Free lucky dip entry in next Lotto draw |

One randomly picked ticket | £1,000,000 |

20 other randomly picked tickets | £20,000 |

### Probability of Winning a Prize

The probability of winning each of these prizes can be calculated. For example, the probability of matching all 6 balls in the new lotto is $$\mathbb{P}(\mathrm{matching\ ball\ 1})\times \mathbb{P}(\mathrm{matching\ ball\ 2})\times...\times\mathbb{P}(\mathrm{matching\ ball\ 6})$$ $$=\frac{6}{59}\times\frac{5}{58}\times\frac{4}{57}\times\frac{3}{56}\times\frac{2}{55}\times\frac{1}{54}$$ $$=\frac{1}{45057474},$$ and the probability of matching 4 balls in the new lotto is $$(\mathrm{number\ of\ different\ ways\ of\ picking\ four\ balls\ out\ of\ six})\times\mathbb{P}(\mathrm{matching\ ball\ 1})\times\\...\times\mathbb{P}(\mathrm{matching\ ball\ 4})\times\mathbb{P}(\mathrm{not\ matching\ ball\ 5})\times\mathbb{P}(\mathrm{not\ matching\ ball\ 6})$$ $$=15\times\frac{6}{59}\times\frac{5}{58}\times\frac{4}{57}\times\frac{3}{56}\times\frac{53}{55}\times\frac{52}{54}$$ $$=\frac{3445}{7509579}.$$ In the second calculation, it is important to include the probabilities of not matching the other balls to prevent double counting the cases when more than 4 balls are matched.

Calculating a probability for every prize and then adding them up gives the probability of winning a prize. In the old draw, the probability of winning a prize was \(0.0186\). In the new draw, it is \(0.1083\). So Camelot are correct in claiming that you are now more likely to win a prize.

But not all prizes are equal: these probabilities do not take into account the values of the prizes. To analyse the actual winnings, we're going to have to look at the expected amount of money you will win. But first, let's look at Camelot's other claim: that under the new rules you are more likely to become a millionaire.

### Probability of Winning £1,000,000

In the old draw, the only way to win a million pounds was to match all six balls. The probability of this happening was \(0.00000007151\) or \(7.151\times 10^{-8}\).

In the new lottery, a million pounds can be won either by matching all six balls or by winning the millionaire raffle. This will lead to different probabilities of winning on Wednesdays and Saturdays due to different numbers of people buying tickets. Based on expected sales of 16.5 million tickets on Saturdays and 8.5 million tickets on Wednesdays, the chances of becoming a millionaire on a Wednesday or Saturday are \(0.0000001398\) (\(1.398\times 10^{-7}\)) and \(0.00000008280\) (\(8.280\times 10^{-8}\)) respectively.

These are both higher than the probability of winning a million in the old draw, so again Camelot are correct: you are now more likely to become a millionaire...

But the new chances of becoming a millionaire are actually even higher. The probabilities given above are the chances of winning a million in a given draw. But if two balls are matched, you win a lucky dip: you could win a million in the next draw without buying another ticket. We should include this in the probability calculated above, as you are still becoming a millionaire due to the original ticket you bought.

In order to count this, let \(A_W\) and \(A_S\) be the probabilities of winning a million in a given draw (as given above) on a Wednesday or a Saturday, let \(B_W\) and \(B_S\) be the probabilities of winning a million in this draw or due to future lucky dip tickets on a Wednesday or a Saturday (the values we want to find) and let \(p\) be the probability of matching two balls. We can write $$B_W=A_W+pB_S$$ and $$B_S=A_S+pB_W$$ since the probability of winning a million is the probability of winning in this draw (\(A\)) plus the probability of winning a lucky dip ticket and winning in the next draw (\(pB\)). Substituting and rearranging, we get $$B_W=\frac{A_W+pA_S}{1-p^2}$$ and $$B_W=\frac{A_S+pA_W}{1-p^2}.$$

Using this (and the values of \(A_S\) and \(A_W\) calculated earlier) gives us probabilities of \(0.0000001493\) (\(1.493\times 10^{-7}\)) and \(0.00000009736\) (\(9.736\times 10^{-8}\)) of becoming a millionaire on a Wednesday and a Saturday respectively. These are both significantly higher than the probability of becoming a millionaire in the old draw (\(7.151\times 10^{-8}\)).

Camelot's two claims—that you are more likely to win a prize and you are more likely to become a millionaire—are both correct. It sounds like the new lottery is a great deal, but so far we have not taken into account the size of the prizes you will win and have only shown that a very rare event will become slightly less rare. Probably the best way to measure how good a lottery is is by working out the amount of money you should expect to win, so let's now look at that.

### Expected Prize Money

To find the expected prize money, we must multiply the value of each prize by the probability of winning that prize and then add them up, or, in other words,

$$\sum_\mathrm{prizes}\mathrm{value\ of\ prize}\times\mathbb{P}(\mathrm{winning\ prize}).$$
Once this has been calculated, the chance of winning due to a free lucky dip entry must be taken into account as above.

In the old draw, after buying a ticket for £2, you could expect to win 78p or 83p on a Wednesday or Saturday respectively. In the new draw, the expected winnings have changed to 58p and 50p (Wednesday and Saturday respectively). Expressed in this way, it can be seen that although the headline changes look good, the overall value for money of the lottery has significantly decreased.

Looking on the bright side, this does mean that the lottery will make even more money that it can put towards charitable causes: the lottery remains an excellent way to donate your money to worthy charities!

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