Maths at:

Maths at: The FUTURE!

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Thu, 1 Jun 2023 06:00:00 +0100

Back to the future. There's no maths in it. We just wanted to watch a good film. I suppose we should talk about it too.

In today's programme:

we travel through time at a rate of 1 second/second.
is Harry Potter better than Back to the Future?
how do the laws of physics change when we're moving?
all our knowledge is guided by the wonderful Naomi Wray.

Further reading:

book your tickets to the Back to the Future The Musical;
is time travel complicated? Take a look at this infographic and learn about 3 Theories of Time Travel;
have a glimpse at Hannah's programme partying with Steven Hawking;
watch Kip Thorne present on black holes at Cardiff University.

Subscribe via iTunes.
Follow us on twitter @PodcastMathsAt, as well as @ThomasEWoolley and @benmparker. Hannah has one of those new-fangled instagram s.

Maths at: Easter

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Sun, 9 Apr 2023 09:00:00 +0100

When is Easter? I literally have no clue. Today's Maths at: justifies this ignorance!

In today's programme:

we fail to do basic arithmetic;
should you use Salad Cream at Easter?
Thomas loses his mind.

Further reading:

So... when is Easter?
Check your ethical compass.

Subscribe via iTunes.
Follow us on twitter @PodcastMathsAt, as well as @ThomasEWoolley and @benmparker. Hannah has one of those new-fangled instagram s.

Maths at: Wordle

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Wed, 4 May 2022 16:33:00 +0100

It's like the last day of school, because we are playing games!

Wordle,

Guess who,

Hangman.

Never has maths been so much fun.

Further reading:

Subscribe via iTunes.
Follow us on twitter @PodcastMathsAt, as well as @ThomasEWoolley and @benmparker. Hannah has one of those new-fangled instagram s.

Maths at: The Number 23

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Sun, 23 May 2021 23:23:00 +0100

Did you know that:

the Earth tilts at 23 degrees on its axis? (Well it's actually closer to 23.5 and degrees are an arbitrary unit of measurement);
2 divided by 3 is 0.666 recurring? (Although if you round it properly it would be 0.667, which is the Devil's next door neighbour);
if you rearrange the letters in "The Number Twenty Three" you get "This is all Hogwash"?

Isn't numerology incredible?

This week we watched Joel Schumacher's 23rd film: "The Number 23", starring Jim Carrey. And we ask, perhaps, the most important question of all - was this film actually a secret sequel to Ace Ventura Pet Detective?

We're through the looking glass here people!

If you're interested in watching The Number 23 then please recondsider watching Paddington (one or two, they're both good.)

Your regulars, Thomas and Ben are joined by Hannah Veale and you can either tickle your eyeballs and/or your ear drums. We're so good to you.

Further reading:

How do you spot a conspiracy theory?
James Grime demonstrating Benford's law in an everyday paper.
???

Subscribe via iTunes.
Follow us on twitter @PodcastMathsAt, as well as @ThomasEWoolley and @benmparker. Hannah has one of those new-fangled instagram s.

Maths at: Bitcoin

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Fri, 9 Apr 2021 22:06:00 +0100

It is with a heavy heart that we have to inform all of our viewers and listeners that The Wonderful Liz, otherwise known as Lydia Menzies passed away in late 2020.

Lydia, by her own choice, wanted to be enigmatic. So although you gained glimpses into her past career as a school teacher you rarely got insight into the funny, passionate and ebullient life that she lived. Which is a great shame, because her impact on the lives around her was tremendous. So much so that she even received tributes in the house of commons from Angus MacNeil and Jacob Rees-Mogg.

A fundraising page for Breast Cancer Now was set up in her honour and we hope that you are able to donate whatever you can to this cause.

This podcast covers the maths of bitcoin and guest stars one of Lydia's oldest friends, Duncan Coutts, who is currently developing one of the bitcoin alternatives.

Did we learn what bitcoin is?

Or did we just start to question all money?

Join us in Lydia's final podcast. We hope you enjoy it.

Further reading:

What on earth is Haskell?
If you can't donate money to a cancer charity then perhaps you could donate your time through fundraising?

Subscribe via iTunes.
Follow us on twitter @PodcastMathsAt, as well as @ThomasEWoolley and @benmparker.

Maths at: Long-tailed tits & Lovelace

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Tue, 13 Oct 2020 09:00:00 +0100

TITS! Tits. TiTs... TIIIIIIITS,

(tits)

t...i...t...s

Yup, we we've found our level.

This week we're joined by the wonderful (soon to be doctor) Natasha Ellison (good luck, Natasha!), from the Unviersity of Sheffield and she takes us on a whirlwind tour of her thesis results of where you should look for tits in a forest and how the mathematics stems from understanding wolf pack dynamics.

Alongside our ecological leanings we also discuss Augusta Ada King, or Ada Lovelace to her friends. Tuesday 13th October marks Ada Lovelace day. As one of the first ever computer programmers to exist why not crack open your favourite programming language and have a go at saying:

HELLO WORLD!

Further reading:

Follow Natasha on twitter;
Learn more about Ada Lovelace day;
How to say Hello World in 28 different programming languages.

Subscribe via iTunes.
Follow us on twitter @PodcastMathsAt, as well as @ThomasEWoolley and @benmparker.

Maths at: Tidying & toilets

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Tue, 15 Sep 2020 09:33:00 +0100

It is a truth universally acknowledged, that noone likes tidying up. So, if you want a reason to not bother then give us a listen.

Once again your regulars, Ben, Liz and Thomas, seek answers to such questions as:

Why are people in London Freezing?
Do butterflies fart?
How many toilets do you need to make everyone happy?

All questions and no answers in today's Maths at:

Further reading:

House always messy? Just blame it on the second law of thermodynamics
No more queuing at the ladies' room
Do butterflies fart?

Subscribe via iTunes.
Follow us on twitter @PodcastMathsAt, as well as @ThomasEWoolley and @benmparker.

Maths at: Fantasy Football (Part 2)

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Fri, 14 Aug 2020 19:54:00 +0100

The first episode got us all up to speed as to what on earth Fantasy Football was all about. The Maths at: team is now ready to plunder the depths of Josh's strategies and Thomas has been editing in haste.

We can 100% guarantee that if you follow the strategy as specifed by Josh you will be in with a chance of winning and that's no Bull.

Once again your regulars, Ben and Thomas, are joined by the knowledgable Hannah Veale and the intriguing Josh Bull. Gather round as they seek answers to such questions as:

is it luck, or is it skill?
is it better to have consistent scorers, or one off star strikers?
why is everyone regressing to the mean?

Of course, the simple answer is that you get reddit to do the hard work for you! See how in today's podcast.

Further reading:

Josh Bull's Fantasy Football page, which contains all the history of his choices.
Why not have a go at your own Fantasy team through The official Fantasy Premier League?
For actual tips on Fantasy Football, Josh recommends Reddit.
Alongside being brilliant at everything Hannah Veale is also a photographer. Why not check her out?
Interested in the statistics that Ben showed? This paper quantifies the effect of having a crowd on your win ratio.

Subscribe via iTunes.
Follow us on twitter @PodcastMathsAt, as well as @ThomasEWoolley and @benmparker.

Maths at: Fantasy Football (Part 1)

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Thu, 13 Aug 2020 11:58:00 +0100

WELCOME TO A NEW DIMENSION OF MATHS AT:

We recorded our first ever video, so that we may delight your eyes, as well as your ear canals. However, for those of you still living in the 20th century we've stripped the video off. Of course, you'll miss the visual tom foolery, but we're sure that your imagination can fill in the blanks.

Fantasy football is the name of the game today and your regulars Ben and Thomas are joined by the knowledgable Hannah Veale and the intriguing Josh Bull. Josh recently topped the Fantasy Premier League and we though we'd take the time to interview him about his secrets.

At least, that was the plan.

Unfortunately, the idea of "fantasy" led Thomas astray. So if you're interested in knowing the answers to such questions as:

Why isn’t Gandalf available in fantasy football?
Would Gandalf would be a good striker?
Would Gandalf be a good head of a university?

Then we're the podcast for you!

Further reading:

Josh Bull's twitter account contains all the latest Fantasy Football tips and tricks (may actually just be mathematical musings).
Why not have a go at your own Fantasy team through The official Fantasy Premier League?
For actual tips on Fantasy Football, Josh recommends Reddit.
Alongside being brilliant at everything Hannah Veale is also a photographer. Why not check her out?

Subscribe via iTunes.
Follow us on twitter @PodcastMathsAt, as well as @ThomasEWoolley and @benmparker.

Maths at: The Election

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Tue, 10 Dec 2019 19:37:00 GMT

WE'RE BACK FOR A ONE OFF SPECIAL!

What could be special enough to bring us back? Well nothing more than a bloomin' election! That's right we're going to look at how maths infuses even our democratic duties.

Today's discussion points:

How are Thomas and Liz now that they're parents?
How do we deal with having no perfect voting systems?
Rude diagrams and asymptotes.

All this and more in this week's Maths at.

In you're interested in more political theory then the UK will be going to the polls on November 12th! Enough stats to make you go blind.

Further reading:

This YouTube Playlist provides a really wonderful overview of why all voting systems are bad.
Take a look at what your vote is really worth at voterpower.org.
How does tactical voting work?

Subscribe via iTunes.
Follow us on twitter @PodcastMathsAt, as well as @ThomasEWoolley and @benmparker.

Maths at: Season 2 Bloopers

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Wed, 22 May 2019 15:15:00 +0100

We can't be perfect all the time.

From everyone at maths at have a good year and we hope to be back at some point in the future.

Maths at: Magictown

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Tue, 21 May 2019 13:46:00 +0100

WE DID IT! Another series done!

Marriages, prenancies, black holes, none of these stopped us getting the best quality maths film chat from our mouths to your ears. The question is: after all that will there be a third series?

Today's discussion points:

Are opinion polls pointless?
Which way does the inequality point?
What instrument can you play?

All this and more in this weeks Maths at.

In you're interested in watching Travelling Salesman you don't need to buy it off Amazon, you can just watch it on YouTube! Look a James Stewart's pretty face below.

Further reading:

Have a look at the New York Times' scathing review of this charming film.
A Journalist’s Guide To Opinion Polls from the British polling council.
Four times when having an online poll was a bad idea.

Subscribe via iTunes.
Follow us on twitter @PodcastMathsAt, as well as @ThomasEWoolley and @benmparker.

Maths at: Travelling Salesman

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Mon, 6 May 2019 09:00:00 +0100

The government didn't want you to hear this podcast. They tried to pay us off! Just remember as you listen to it: we are not responsible for any repercussions if you decide to watch Travelling Salesman.

Today's discussion points:

How many tickets do you have to buy to win the lottery?
Squashed research!
How many actors can you have playing the same character?

All this and more in this weeks Maths at

In you're interested in watching Travelling Salesman you don't need to buy it off Amazon, you can just watch it on YouTube! They can't even give it away!

Further reading:

Should you feel guilty about watching the film on YouTube, you can give the creators your money, through their official website.
An easy description of the P vs NP problem.
Sorting algorithms through dance.
Color visualization of sorting algorithms.

Subscribe via iTunes.
Follow us on twitter @PodcastMathsAt, as well as @ThomasEWoolley and @benmparker.

Maths at: Contact

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Tue, 23 Apr 2019 10:58:00 +0100

Maths at... proudly* sponsored by Prof. Elizabeth Baldwin!

*we have no idea of how proud she is.

Care to sponsor an episode? It'll cost you 77p!

Sorry about the quality of Thomas' microphone, he had trouble recording. Won't happen again (until the next).

This week we watched Contact and are joined, once again, by

Dr William Taylor

His enthusiasm is so infectious! So whilst he tried to educate us, we try to divert his attention with such discussion points as:

Spice girls! A very strong start?
We didn't do the "you can't be Sirius Joke!" (Although I did, but no-one laughed so I cut it. But feel free to add it back in... you'll know when).
Jane Fonda, or Jodie Foster?

Thomas is tired, Liz is Pregnant and Ben can't find his mass, can you find it?

If you're interested in watching Contact you can follow the Amazon link below.

Further reading links:

Can you tell what the Arecibo Message is?
Hubble Image of Galaxy Cluster Converted Into Sound

Subscribe via iTunes.
Follow us on twitter @PodcastMathsAt, as well as @ThomasEWoolley and @benmparker.

Maths at the Movies: The Martian

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Tue, 9 Apr 2019 11:00:00 +0100

Another sexy, sexy episode this week as we are joined by the NAKED MATHEMATICIAN

Dr Thomas Crawford

Get ready to feel dull by comparison!

Today's discussion points include:

Everyone's links to Countdown;
Thomas tells us about the Greeks;
Can you tell the difference between Jeff Daniels and Jeff Bridges?

So put your feet up, get comfortable and get naked, as we talk about this week's movie The Martian.

If you're into watching The Martian, you can follow the link below.

Further reading links:

There is actually a website that maintains a fact file on every episode of Countdown. Why not have a look at how Ben did?
Recently the rover Opportunity made the news as it lasted 5,352 sols after being planned for only 90 sols, but what did it discover?

Fancy having a go at surviving through The Martian's maths problems?

Subscribe via iTunes.
Follow us on twitter @PodcastMathsAt, as well as @ThomasEWoolley and @benmparker.

Puzzle from CUBE

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Thu, 21 Mar 2019 11:39:00 GMT

Simple question this week:

Consider a clock with an hour hand and a minute hand. Starting from midnight, how many times do the hands cross each other in 24 hours.

Note, you don't count the starting point of midnight, with the hands overlapping as a crossing, but you do count the last moment, when the two hands overlap at midnight a day later.

Bonus Question:
Normally, clock hands travel clockwise around the clockface. Suppose now that the two hands are travelling in opposite directions. How many times do the hands cross in this case?

Math at the Movies: x+y

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Mon, 18 Mar 2019 21:10:00 GMT

Well, this was a pain in the backside to edit. The film is so tawdry and dull that we kept getting lost on tangents. Fear not though faithful listener, Thomas has edited the two hours of guff down to a single hour of solid... bronze.

Today's discussion points include:

How should you flip a mattress?
Does the culture you grow up in influence how you learn maths?
BUMFIT!

From our mouths to your ears, enjoy!

If you want to watch x+y, you can follow the link below.

Further reading links:

As per usual some artistic license was taken with this true story, this website provides the fact behind the fiction;
Fancy testing yourself with real IMO problems?
What is the difference between IMO problems and research mathematics?

Subscribe via iTunes.
Follow us on twitter @PodcastMathsAt, as well as @ThomasEWoolley and @benmparker.

Answer to Fermat's Room puzzle

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Fri, 15 Mar 2019 12:02:00 GMT

A very diffiicult puzzle this week!

Three people guard two doors. You know that:

one person always tells the truth;
one person always lies;
one person randomly decides whether to tell the truth or lie (assume lies and truth are equally likely);
the three know amongst themselves who they are.

You can ask two questions to the people. The answer to which must either be yes or no. What question do you ask and who do you ask?

This is an extension of the famous two person puzzle. Normally, you only have two guards, one tells the truth and one lies. You have to choose and open one of the doors, but you can only ask a single question to one of the guards.

What do you ask so you can pick the door to freedom?

In this case the solution is:
If I asked what door would lead to freedom, what door would the other guard point to?

This works by considering the two possible outcomes. Namely:

If you asked the truth-guard, the truth-guard would tell you that the liar-guard would point to the door that leads to death.
If you asked the liar-guard, the liar-guard would tell you that the truth-guard would point to the door that leads to death.

Therefore, no matter who you ask, the guards tell you which door leads to death, and therefore you can pick the other door.

This puzzle is so famous it's appeared many times in media

The inclusion of the trickster guard, however, changes the puzzle dramatically. Specifically, you questions have to work no matter who is being asked (truth-teller, liar, or trickster). Further, no matter what you ask, you always have to worry about the trickster screwing up your logic.

Thus, one strategy is to identify one person is NOT the trickster. We don't have to identify whether they are truth-teller, or liar.

Call the three gaurds A, B and C. You ask A:

"Is exactly one of these statements true:

You are the truth-teller
B is the trickster

If you get back the answer yes, then the possibilities are:

A is the truth-teller and B is the liar (1. true, 2. false, so one statement true, so answer is yes which truth-teller truthfully gives)
A is the trickster
A is the liar and B is the truth-teller (both statements false so answer is no which liar lies about)

In all three cases, B is not the trickster.

If you get back the answer no, then the possibilities are:

A is the truth-teller and B is the trickster (both statements true, so answer is no which truth-teller truthfully gives)
A is the trickster
A is the liar and B is the trickster (1. false, 2. true so one statement true so answer is yes which liar lies about)

In all three cases, C is not the trickster.

Once you have found a person who is not the trickster, just point to a door and ask the person:

"Would your exact opposite say this door leads to freedom?"

Thus, reducing the problem to the previous case.

Maths at the Movies: CUBE

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Tue, 5 Mar 2019 11:49:00 GMT

This week is a little different, and I've got to say perhaps one of the most interesting episodes we've ever done!

We are joined by the multi-talented

Dr Lyle Skains

who uses her research knowledge of evolutionary biology and digital literature to show us how to really write a good movie!

Highlights this week are:

Liz geeking out with Lyle, will they go on holiday together?
Ben misunderstanding publishing, will he ever get his cheese and wine?
Thomas reliving his childhood years, when did he stop being so nihilistic

If you're interested in watching CUBE you can follow the Amazon link below.

Further reading links:

Wondering what digital fiction looks like? Well try What football will look like in the future? and prepare to have your horizons expanded.
If you're interested in the future of literature then have a look at Lyle's recent book. It discusses such subjects as: indie publishers, hybrid authors, and fanfiction writers.
Lyle's also written plenty of digital fiction. Again, this is reading, but not as we know it.
If you're lazy and got be bothered to read then why not try Lyle's podcast Wonderbox publishing.
If you've seen CUBE, when not try CUBE ZERO and CUBE 2: Hypercube. They are equally as bonkers!
The geocities CUBE website that actually investigated the mathematics more than we did!

Subscribe via iTunes.
Follow us on twitter @PodcastMathsAt, as well as @ThomasEWoolley and @benmparker.

Puzzle from Fermat's Room

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Fri, 1 Mar 2019 14:34:00 GMT

Three people guard two doors. You know that:

one person always tells the truth;
one person always lies;
one person randomly decides whether to tell the truth or lie (assume lies and truth are equally likely);
the three know amongst themselves who they are.

You can ask two questions to the people. The answer to which must either be yes or no. What question do you ask and who do you ask?

Some further rules for the more pedantic:

You cannot ask questions like "Will it rain tomorrow?", because neither the truth teller, nor the liar can be sure.
You cannot ask questions like "What would you answer if I ask you blablabla?", because if you ask the random liar they don't what their next answer will be.
You cannot ask something like "Will you answer No to this question?", because the truth-teller can't answer this question.
All decisions must be based on the yes and/or no answers only.
This puzzle is not about "how to find a way around the rules".

Answer to the Good Will Hunting puzzle

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Thu, 28 Feb 2019 15:53:00 GMT

In our Good Will Hunting podcast we asked:

What is the highest number of eggs that you CAN'T make, when you have boxes of size 6, 9 and 20?

Turns out this is called the McNuggest number as McNuggets originally came in boxes of this size.

In this case, it turns out that 43 is the largest number you can't make, but how do you prove it?

Well we note that:
44 = 4x6+20
45 = 5x9
46 = 6+2x20
47 = 3x9+20
48 = 8x6
49 = 9+2x20

Since we have 6 consecutive numbers that can all be made from 6, 9 and 20, then every number there after can be made simply by adding an appropriate multiple of 6, e.g. 50 = 44+6, 51 = 45+6, etc.

Simple, no?

Maths at the Movies: Fermat's Room

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Tue, 19 Feb 2019 09:00:00 GMT

Welcome to the strangely erotic episode of Maths at, where we watch the tense, psychological thriller, Fermat's Room (or La Habitación de Fermat, for you Spanish speakers) and we ask the real questions of... WHAT HAPPENED ON THE BOAT?

As per usual, the time line is all wonky. This episode does follow on from A Beautiful Mind, but was recorded a long time after, so although we talk about our lives having changed dramatically, it's only bee two weeks for you and you already know what's happened if you've listened to our Christmas episode. It's so hard living in a linear timeline.

So if you want to know:

what Liz's ovaries sound like;
which superpower our hosts would rather have;
how Ben would overhaul examination procedures,

then join us in our latest episode of being distracted by pop corn makers.

If you're interested in watching Fermat's Room and want an easier time than we had in finding it, simply click the Amazon link below.

Further reading links:

videos of every puzzle from the film;
a complete writeup of questions and answers from the film
A duck popcorn maker, should you want such a thing.

Subscribe via iTunes.
Follow us on twitter @PodcastMathsAt, as well as @ThomasEWoolley and @benmparker.

Puzzle from Good Will Hunting

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Tue, 12 Feb 2019 12:13:00 GMT

Ben's local shop stocks eggs in boxes of capacity 6, 9, or 20 eggs. What is the highest number of eggs that you CAN'T make?

For example, you can make 29 with one 9 box and one 20 box, 29=9+20,

you can make 30 with a five 6 boxes, 30=5x6.

but you can't make 31.

For those wanting an extra puzzle, can you prove that your answer is correct. Namely, all numbers higher than your chosen integer can be written as a linear combination of 6, 9 or 12.

Answer to A Beautiful Mind puzzle

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Fri, 8 Feb 2019 15:00:00 GMT

In our podcast episode on A Beautiful Mind the following question was asked:

Two trains are on the same track. They start 100km apart and head towards each other at a speed of 50km/h.

Whilst these two trains are heading for their collision a fly starts out on the front of one train and zooms directly to the front of the other at a speed of 75km/h (see the animation above). Once the fly reaches the second train it immediately darts back to the front of the first train at the same speed and repeats this back and forth motion until the two trains collide and the fly is squashed on impact.

How far has the fly traveled, before it meets its demise?

One way to approach this problem is through infinite series. Namely, we find how far the fly during the first journey, the second journey, the third journey, etc. and add them all up. Thankfully, there is a fairly nice formula that provides this answer.

However, a much simpler way to calculate the distance is by realising that the changes in direction do not matter. Namely, all we are asking is how far can a fly travel in the hour it takes for the trains to hit each other? Clearly, this is simply 75 km. Sometimes, a moment's thought can save an hour's work!

As mentioned last time, John von Neumann was said to have immediately answered this problem, but when pressed on his solution method he said that he has used the infinite series method. Ah to have the mind of a genius!

This and other aspects of von Neumann's genius are discussed in Raymond Flood's excellent Gresham College talk, below (plus you get a bit of Alan Turing for free, which Thomas is always happy about).

Maths at the Movies: Good Will Hunting

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Tue, 5 Feb 2019 09:00:00 GMT

Last week we did A Beautiful Mind and now Good Will Hunting. We are really hitting all the well-known maths films at the moment aren't we?

More importantly joining us this week we have the wonderful

Dr James Grime

Philanthropist, playboy, billionaire... he is none of this things, but he may have identified the real Will Hunting!

This week we touch on such subjects as:

Is University a scam?
Good Will Hunting needs a prequel!
Will James and Liz ever write a paper about the maths of Dirty Dancing?

If you're interested in watching Good Will Hunting you can follow the Amazon link below.

Further reading links:

What is the Hadwiger-Nelson problem and who is Aubrey de Grey?
We also mention the fields of Combinatorics, Graph theory and Fourier theory.

Subscribe via iTunes.
Follow us on twitter @PodcastMathsAt, as well as @ThomasEWoolley and @benmparker.

Puzzle from A Beautiful Mind.

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Fri, 1 Feb 2019 16:30:00 GMT

A classic puzzle to start our second series. It appears in the background of A Beautiful Mind and it is said that the famous mathematician John von Neumann immediately answered with the correct result. But we'll talk about solutions later!

Animation illustrating the problem courtesy of MathWorld.

Two trains are on the same track. They start 100km apart and head towards each other at a speed of 50km/h.

Whilst these two trains are heading for their collision a fly starts out on the front of one train and zooms directly to the front of the other at a speed of 75km/h (see the animation above). Once the fly reaches the second train it immediately darts back to the front of the first train at the same speed and repeats this back and forth motion until the two trains collide and the fly is squashed on impact.

How far has the fly traveled, before it meets its demise?

If you think you have the answer comment below, tweet it to us @PodcastMathsAt, or email us at podcastmaths@gmail.com.

The answer will be posted next week.

Maths at the Movies: A Beautiful Mind

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Tue, 22 Jan 2019 09:00:00 GMT

Ok, so the "Maths at" timeline is a mess.

The Christmas episode was recorded in November.
The Christmas episode reveals secrets from later on in the series.
We tried to hide these secrets in the Mean Girls episode, which was recorded around seventh, but released first.
When we I say that we've had a complaint about Liz's language it from the pi day episode, not the Mean Girls episode.

Is that all clear? And this is all because the recording quality of this episode is a little dodgy!

Sorry about all that. Just pretend that Thomas, Ben and Liz are Time Lords.

Anyway, we had to get there eventually. Probably number one of many science film lists: A Beautiful Mind. The biopic of John Nash, a prodigy behind the field of game theory.

To help us discern our cooperators from our defectors we are joined by the wonderful

Dr Vince Knight

So, if you're wondering:

which queue to join;
which region in Risk to take;
or simply how to win at Monpoly,

then don't ask us, we're all about the theory!

If you're interested in watching A Beautiful Mind you can follow the Amazon link below.

Further reading links:

Subscribe via iTunes.
Follow us on twitter @PodcastMathsAt, as well as @ThomasEWoolley and @benmparker.

Maths at the Movies: Mean Girls

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Tue, 8 Jan 2019 09:00:00 GMT

Mean girls...

Mean girls?

MEAN GIRLS!

There is no font, size or punctuation mark that does full justice to Thomas' anger.

This week, for some inexplicable reason, we watched the movie: Mean Girls. Join us as:

Liz tries to convince Thomas that "Do you even go here?" is a funny line;
Thomas' blood pressure shoots through the roof;
Ben compares Mean Girls to the New Testament. and Liz compares it to a clockwork orange.

If you're interested in watching Mean Girls then please get help. Go and watch Marvelous Mrs Maisel instead.

However, if you are beyond help that you can follow the Amazon link below.

I mean come on! This meme doesn't even make sense.

Further reading links:

Subscribe via iTunes.
Follow us on twitter @PodcastMathsAt, as well as @ThomasEWoolley and @benmparker.

Maths at: Christmas 2018

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Tue, 25 Dec 2018 14:30:00 GMT

Nadolig Llawen to all our Listeners

We hope Sion Corn has visited you on this fine festive day. For indeed we are BACK! And this episode went a bit Welsh! Much to Ben's chagrin.

In this episode we find out:

the female version of phallic;
what eggnog actually is;
how NORAD tracks santa;
and how following stars probably won't lead you to the next messiah.

To ensure at least some expertise on the panel we are joined by the wonderful physicist, astronmer and raconteur:

Dr William Taylor

Further reading links:

download the free software Stellarium to track the stars positions throughout time and space;
track Santa using NORAD;
have a look at Quartz modelling of the Christmas lights turn on time, which produces the below graph that we talk about.

Subscribe via iTunes.

Follow us on twitter @PodcastMathsAt, as well as @ThomasEWoolley and @benmparker.

Bringing down the House

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Sun, 12 Aug 2018 02:25:00 +0100

Maths at the Library: How Six Students Took Vegas for Millions

Looking back since our first podcast, we've learnt a lot about the portrayal of mathematicians in movies, mostly as psychopaths or murderers; that most sports are played on rectangles (thanks, Donald Duck!); and that statisticians are mostly fat.

In our first ever podcast, we reviewed the movie "21". We generally found it a nifty and exciting movie documenting a little maths, and the dark forces of Las Vegas, and all enjoyed it. The basic plot involves a bunch of MIT students who use their memories and some team work, with some clever card-counting techniques to (perfectly legally) take the big casinos in Las Vegas for millions of dollars.

Recently, I read the book of the film, "Bringing down the House", which in many ways is even better than the film. The maths of card-counting, and how blackjack is a beatable game is summed up neatly in the book

Blackjack is the only game in the casino that is beatable over an extended period of time, because blackjack is subject to continuous probability. This simply means that what you see affects what you arre going to see. Blackjack is a game with a memory. If an ace comes out in the first round of a blackjack shoe, that means there is one less ace left in the rest of the deck. The odds of drawing another ace have gone down by a calculable fraction. In other words, the past has an effect on a the future....Blackjack is the only popular casino game where what you see affects what you are going to see.

We saw this again in Moneyball, where a statistician used the power of previous information on how baseball players had played to work out how they were going to play in the future.

Statistics, and it's related discipline Data Science, are everywhere these days. Websites use history about your previous buying habits to pepper you with adverts about what you are most likely to buy. Clinicians use previous results of drug trials to predict what the very best drug for any occasion will be. Self-learning cars are using tonnes of previous data to work out what to do in any dangerous situation.

Blackjack is not life- but the story about how, using elementary statistics and probability, as well as some acting skills, and some disguises, a bunch of students managed to win against one of the most sophisticated money-grabbing empires, is beautifully written, interesting, and exciting, and well worth a purchase for the mathematician in your life. See you in Vegas!

Friday factoid- Friday 13th occurs far too frequently.

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Fri, 13 Apr 2018 08:00:00 +0100

Friday 13th is an unlucky day in the UK. British people stay at home, don't make eye contact in public, and spend their day complaining, mostly about the weather.

Did you know, however, that the 13th is more likely to be a Friday than any other day? And also, coincidentally, Friday the 13th is the equally-most likely day number/day name combination to occur?

Why? We know the year for the Earth (the time it takes us to do one orbit of the Sun) is 365.242189 days. This is annoying for calendar makers, as we just can't put an extra 0.256 days into each year. So what do we do? Well, we try and even it out over a 400 year cycle. We fix the year to have 365 days normally (remember the old rubbish rhyme: "30 days has September, April, June and November, all the rest have 31, except February...")

February has 28 days, except in a leap year, where it has 29. A leap year is defined as

any year which is divisible by 4 except
years divisible by 100 are not leap years except
years divisible by 400 are leap years.

If we repeat this on a 400 year cycle, then this means that the average year length is 365.2425 days (or very close to it)

If we work out the number of days in 400 years, then this is 146097 days; by chance this is divisible by 7, so it means that if a day is a Monday, the same day 400 years later will be a Monday. In other words, the minimum period the calendar cycles over is 400 years. Note that 146097/400=365.2425, very close to Earth's year.

We can then run a simple computer script to count the number of times that each number/day of week combination occurs in a 400 year cycle, and graph the results

This shows us that the 13th is a Friday 688 days out of every 400 year cycle, more than any other day. Also, we can see that Friday 13th is the joint most common number/day of week combination (all the yellow squares in the diagram). I have left off the 29th,30th and 31st just because it's a nicer pattern this way.

So is this important? No! But next time you have bad luck on Friday 13th remember, this day will happen 688 days in the next 400 years so you'd better get used to it...

So what have we learned?

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Thu, 5 Apr 2018 09:36:00 +0100

Although we provided a fun marking scheme at the end of every episode we decided to actually plot our opinions in the following quality graphs. Horizontal axes provide a measure of quality of the film, whilst vertical axes provide a a measure of quality of the maths.

Our first graph illustrates all of our differing opinions. Each film has a different marker and the colour of each marker links to a particular person's tastes.

We can clearly see that Interstellar was a favourite all round, whilst The Oxford Murders took a battering from all sides.

Equally, although Thomas and Ben agreed that Pi wasn't a good Liz seemed to enjoy it more than them (still not good, just not as bad!). Proof, was perhaps the most divisive amongst the team.

A couple of interesting points is that there are very few, if any, points in the Bad Film/Good Maths category, or the Good Film/ Bad Maths. This is, of course easy to explain. The Bad Film/Good Maths section would probably be inhabited by videos of mathematical lectures. Whilst, (good) non-science films would inhabit the Good Film/Bad maths quadrant.

Below we plot the average values of the above data in order to visualise the trends better.

What is most striking about this representation is that most of the films tend to cluster around The Line of Equal Quality. This means that a film portraying good maths is also likely to be a good film overall, equally, a film presenting poor maths is likely to be poor overall.

Again, this makes sense, as we have been focusing on such mathematical films, the mathematical content will be central to its subject matter. Thus, a lot of the films quality will rest upon the mathematics representation.

Of course the outlier from this theory is The Imitation Game. A gripping, if slightly embellished, story with actually very little mathematics presented.

So, ten movies later and what have we learned?

A good maths film has to present both good maths and a good film.

Yes, mathematicians are often known for stating the obvious!

Answer to the Interstellar puzzle

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Mon, 2 Apr 2018 13:00:00 +0100

In our Interstellar podcast, Thomas posed the following puzzle:

A group of people with assorted eye colors (say 10 blue and 10 brown) live on an island.

Everyone can see the eye colour of everyone else, but they can't communicate to each other to tell each other their eye colour.

Every night at midnight, a ferry stops at the island. Any islanders who have figured out the colour of their own eyes can leave the island.

One day a sailor from the ferry gets off the boat and says:

"I can see someone who has blue eyes".

Everyone hears and understand the statement, but the sailor is immediately shot dead for communicating with the islanders and no-one ever speaks again. However, given this information some people are able to figure out their eye colour.

Who leaves the island, and on what night?

The simple answer is that all 10 blue-eyed people leave on the tenth day. However, the thinking behind this answer is probably more important than the answer itself. Critically, this is a difficult puzzle. You need to have an extended chain of thinking, but, before we make a chain, let's start with a single link.

Let's make the puzzle as simple as possible. Suppose on the island there is only one person with blue eyes and a load of people with other colour eyes (it doesn't matter how many, or what colour, as long as they're not blue). This blue-eyed person would immediately realize that they had blue eyes because they could see no one else with blue eyes. Therefore, that person would leave on the first night.

Suppose, now, there are two people with blue eyes. They would both leave on the second night, because they would each look at the other blue-eyed person on the second morning and realize that the only reason the other blue-eyed person wouldn't leave on the first night is because they see another person with blue eyes. Seeing no one else with blue eyes, each of these two people realize it must be them.

Carrying on this argument inductively we see that n blue-eyed people would leave on night n because on the n-1 previous night they cannot deduce that the other blue-eyed people are not leaving because of them.

This is a slightly hand-wavy proof, but can be made more rigorous, indeed Reddit has a very formal proof.

One part that still blows our mind is that everyone can always see everyone else's eyes. So in the case of 10 blue-eyed people on the island everyone knows that there are blue-eyed people on the island, so what information have they gained from the sailor?

The answer is common knowledge, but we'll let Wikipedia explain that.

Maths at: pi day

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Wed, 14 Mar 2018 09:00:00 GMT

Happy pi day everyone.

We figure that pi day is the like Hallowe'en for mathematicians. It's a day they get to cut loose and throw away their inhibitions.

As such Lorraine's intro pertaining to strong language is particularly pertinent for today's podcast. It starts out quite coarse.

So sit back, grab a slice of your favourite pie and enjoy as Ben sings you the song of his people.

Further reading links:

Why not listen to our review of the film pi;
The best web address known to man;
The song from pi's digits;
Search pi for meaningful numbers;
Weird ways of calculating pi;
Known digits of pi over time.

Subscribe via iTunes.

Follow us on twitter @PodcastMathsAt, as well as @ThomasEWoolley and @benmparker.

Puzzle from Interstellar

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Sat, 10 Mar 2018 11:14:00 GMT

In our Interstellar podcast, Thomas posed the following puzzle:

A group of people with assorted eye colors (say 10 blue and 10 brown) live on an island.

No one knows the color of their eyes as there are no mirrors on the island and the water is muddy so you can't use the reflection. For all each person knows, they could have green eyes!

However, everyone can see the eye colour of everyone else, but they can't communicate to each other to tell each other their eye colour.

Every night at midnight, a ferry stops at the island. Any islanders who have figured out the color of their own eyes can leave the island.

One day a sailor from the ferry gets off the boat and says:

"I can see someone who has blue eyes".

It's difficult, but possible.

If you think you have the answer comment below, tweet it to us @PodcastMathsAt, or email us at podcastmaths@gmail.com.

The answer will be posted next week.

Solution to The Oxford Murders puzzle

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Wed, 7 Mar 2018 17:27:00 GMT

In our Oxford Murders podcast, Ben posed the following puzzle:

Suppose you are going to play chess against two people: one person is really good, one person is quite bad. You are going to play three games and you always have to alternate your opponents. Namely, you can either choose to play the opponents in the order

Good, Bad, Good,

or you can play the opponents in the order

Bad, Good, Bad.

Which of these two play sequences gives you the optimal chance of winning two consecutive games?

You can approach this problem using probability and tree diagrams. However a little logic goes a long way.

Specifically, in order to win two consecutive games you have to win the middle game. Thus, it is best to put your weaker opponent in the middle. Thus, Good, Bad, Good is the best strategy.

An alternative way of also seeing this answer it that you're probably going to lose against the good player, so the Good, Bad, Good play order gives you two chances to win against the good player, rather than just one.

Simple no?

If you want a bit more rigor then Ben has created a YouTube video solution.

Alternatively, you could try three player chess and team up with the weak player to beat the good player. But that might be considered cheating...

Puzzle from The Oxford Murders

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Fri, 2 Mar 2018 15:58:00 GMT

In our Oxford Murders podcast, Ben posed the following puzzle:

Good, Bad, Good,

or you can play the opponents in the order

Bad, Good, Bad.

Which of these two play sequences gives you the optimal chance of winning two consecutive games?

If you think you have the answer comment below, tweet it to us @PodcastMathsAt, or email us at podcastmaths@gmail.com.

The answer will be posted next week, or you can listen to the answer in our Interstellar podcast

Flatland puzzle answer

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Thu, 1 Mar 2018 09:14:00 GMT

In our Flatland podcast Thomas asked the question:

What is six divided by two plus one times two?

You can hear the solution in our Oxford Murders podcast or read on below.

So what did you get?

4?
5?

Something else completely?

Well, whatever your answer is, you're completely correct (as long as you did the arithmetic correctly).

The problem with the phrase

What is six divided by two plus one times two?

is that it can be read in many different ways. For example, it could mean

6/2+1x2=5

(6/(2+1))x2=4

6/((2+1)x2)=1

or

6/(2+1x2)=3/2

The original statement is ambiguous and so providing a specific answer is difficult.

Your host, Thomas, was once interviewed on radio about a similar ambiguous statement. Have a listen below.

This sort of problem makes the rounds on the internet a lot, below is a Japanese version,

which shows that it is not just simply a problem with language. Mathematical operators can be written ambiguously as well.

Thus, if you arrive at such a question, the answer is not to angry, but try and understand how other people read the question.

Question from Flatland

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Fri, 23 Feb 2018 11:38:00 GMT

In our Flatland podcast Thomas asked the question:

What is six divided by two plus one times two?

Sounds easy right? Can't possibly get it wrong?

If you think you have the answer comment below, tweet it to us @PodcastMathsAt, or email us at podcastmaths@gmail.com.

Maths at: Valentine's day

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Wed, 14 Feb 2018 08:30:00 GMT

Although we've finished our first series, we couldn't miss St Valentine's day.

What could be more romantic than scientifically analysing love and romance using mathematics?

Amongst other things we discuss:

Why should you dump the first 37% of your partners?
How many snogs away from Prince Charles are you?
Are you a validator or an avoider?

Join your hosts as they show you their romantic side and be surprised that any of them actually have partners!

Further reading links:

Ben's bunkum length of love formula;
Drawing hearts graphically;
Jim Murray's Maths of Marriage book or online lecture;
Mobius strip earrings, for you significant other.

Subscribe via iTunes.

Follow us on twitter @PodcastMathsAt, as well as @ThomasEWoolley, @benmparker.

Maths at the Movies: Interstellar plus SPECIAL GUEST

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Tue, 30 Jan 2018 23:46:00 GMT

This week we watch Interstellar.

We are very excited to be joined by the expertise of

Cat Harris

Whereas Thomas and Ben are just about good enough understand their own mathematical fields Cat has a wide range of disciplines under her belt spanning both physics and movie effects. We couldn't ask for a better guest!

If you're interested in watching Interstellar you can follow the Amazon below.

Further reading links:

The hour long Interstellar documentary;
The Interstellar book;
All you could ever want to know about Sex in Space;
Dissolving the dead with alkaline hydrolysis.

Subscribe via iTunes.

Follow us on twitter @PodcastMathsAt, as well as @ThomasEWoolley, @benmparker and @cat_harris_fx

Friday Factoid: Once in a blue moon is one tenth of donkey's years!

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Fri, 26 Jan 2018 11:30:00 GMT

Our Friday Factoid: Once in a blue moon is one tenth of donkey's years!

This month sees a blue supermoon occurring at the same time as a lunar eclipse. But how rare are these things compared to others? The Maths at (mathsat.co.uk team investigate)

We all know the phrase “Once in a blue moon”, but how long is this period? We compare this periods with other well known periods to enable better mathematical precision on these matters.

A blue moon
Traditionally, a blue moon occurs when there are 13 moons in a year or, in a more recent definition when there are 2 moons in a calendar month…. Whichever moon we designate as the blue moon, this occurs once ever 2.7145 years, or once every 991.47 days.

Donkey’s years

Once in Donkey’s years is a common expresison, from the fact that donkey’s ears are very long thing. The Guineess world record for the oldest donkey ever is 54 years, but 25-30 years is more common. Thus "Once in Donkey’s years" is around once in 27.5 years, or once every 10044 days.

A super moon

This means that a full moon is closer in its orbit than at other times. Not that rare at all- and occurs about 25% of all moons. This term is not well defined, so this happens around once every 109 days.

A lifetime

A "once in a lifetime" experience clearly depends on how long someone lives, but from official UK statistics a newborn boy will expect to live 79.2 years and a girl 82.9 years. Thus a man will have the advantage of experiencing a once in a lifetime event slightly more often at once every 28927.8 days, and a woman once every 30279.23 days. You may be lucky and experience it more often!

Total Lunar Eclipses

Eclipses occur very frequently, but total eclipses only rarely. Of course, they’re not visible everywhere on Earth, but somewhere. Encyclopedia Brittanica lists these as 66 per century, or once every 1.512 years or every 553 days.

A fortnight

A fortnight is something Brits use to confuse Americans, like rubbers and irony, and is defined in British law as the frequency of rural buses, and is once every 14 days.

Super Blue Total Lunar Eclipses

OK, for all these to occur together, these are quite rare. Forbes estimates this as once every 265 years, or around once every 10 Donkey’s Years, or once every 96791.25 days. The last one was on December 30, 1982. However, just by chance, there’s another one coming along next year. Like buses- none come for a fortnight, then two come at once.

Super Blue Total Donkey Lunar Eclipses

Now you’re just being silly. This won't happen in a month of Sundays. This is mathematics- are you paying attention, as it will be on the test?

A month of Sundays.
This is a way of saying never, colloquially. Every ∞ days.

Biannually

This means once every six months, or once ever 182.75 days.

Biennially

Who invents these things? This means once every two years, not to be confused with the above, or once every 731 days.

Time immemorial

Defined in English Law to be since before the reign of King Richard 1st, 1189, or longer ago than anyone can remember. There’s over 828 years or around 302,600 days.

Conclusion: "Once in a blue moon" is a well defined phrase. A blue moon happens 10 times in donkey years, or about once every 71 fortnights, or about 30 times in a lifetime. It never happens in a month of Sundays, but has happened around 305 times since time immemorial.

For these and other interesting mathematical facts, listen to Dr Thomas Woolley and Dr Ben Parker, as well as the mysterious Liz, on the popular Maths podcast “Maths at:”, available at mathsat.co.uk, iTunes, and wherever you get your podcasts.

Watch along with MATM: Interstellar and SPECIAL GUEST!

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Fri, 26 Jan 2018 09:43:00 GMT

Next Friday we will be critiquing the maths and physics of Interstellar.

What's more, because this is the 10th episode (and because Ben and Thomas aren't physicists) we will be joined by a SPECIAL GUEST.

Not only is she a Cambridge physics graduate, but she now works in film digital effects. Who could ask for more?

Why not watch along with us? You can buy a digital version of the film through Amazon by clicking on the image below.

Maths at the Movies: The Oxford Murders

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Fri, 19 Jan 2018 11:59:00 GMT

This week we discuss we discuss The Oxford Murders:

Thomas laughs at the word bra;
Liz wants a prime number named after her;
Ben just wants his coffee bringing to him;
And we all think about mathematically defining pasta shapes.

Yup, it's a case of a bad movie, with very little to talk about. At least it made for a fun recording!

If you're interested in watch The Oxford Murders, then you're weird, but you can buy the DVD from Amazon below.

Further reading links:

Subscribe via iTunes.
Follow us on twitter @PodcastMathsAt, as well as @ThomasEWoolley and @benmparker.

Watch along with MATM: The Oxford Murders

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Fri, 12 Jan 2018 16:46:00 GMT

Don't do it.

The Oxford Murders is terrible.

(Thomas is very sorry for choosing it).

However, should you want to waste 104 minutes of your life you can buy the DVD from Amazon by following the link below.

Such a waste of talented people!
Such a waste of mathematical logic!

We didn't read it... but the book might be better?

Answer to the Moneyball problem

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Wed, 10 Jan 2018 16:24:00 GMT

Listen here!

During Moneyball Ben asked:

What is the probability that the next person you meet will have an above average number of ears?

Although most people have two ears, there are those with only one. Thus, the average number of ears is just below two.

However, since people with only one ear are quite rare this means you are most likely to meet a person with two ears next, meaning that the probability is nearly 1! We can't say it is exactly 1, because the next person you meet, although unlikely, may have a damaged ear.

Of course, if you live with a person with only one ear then your probability is significantly lower, because you'll likely bump into that person next.

It's all about the assumptions!

Listen to our Flatland podcast to hear more about this problem and learn why clarifying the averages we talk about have important consequences regarding the interpretation of the question.

Maths at the Movies: Flatland

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Fri, 5 Jan 2018 15:22:00 GMT

This week we take a walk in the lower dimensions as we talk about Flatland: The Movie.

How much is your life worth?
Could 2D animals exist?
Did George Orwell rip this story off by adding a third dimension?

All of these questions and more are interrupted in our Flatland: The Movie podcast.

Join us in our search for the third dimension and beyond!

If you're interested in watching Flatland: The Movie you can follow the Amazon link below.

Further reading links:

galleries of beautiful fractal images;
learn more about maths in the courtroom;
the the company behind the film have made a number of educational resources and other films;
why not read the book the film is based on?
This links takes you to one of Martin Gardener's review of Planiverse, Alexander Dewdney's fully thought out and scientifically accurate version of Flatland.

Subscribe via iTunes.
Follow us on twitter @PodcastMathsAt, as well as @ThomasEWoolley and @benmparker.

Puzzle from Moneyball

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Tue, 2 Jan 2018 09:00:00 GMT

Ben provided the following puzzle during our Moneyball podcast.

What is the probability that the next person you meet will have an above average number of ears?

Don't worry he gave a set of multiple choice answers, so, is the answer:

0?
nearly 0?
1/2?
nearly 1?
1?
π?

As a hint it isn't π.

Join us next time for the answer, when we discuss Flatland.

One the first day of Christmas my true love gave to me...

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Mon, 1 Jan 2018 13:01:00 GMT

Yes, Christmas maybe over, but the puzzles still remain! During our Christmas episode Thomas provided the following puzzle.

In the song "12 days of Christmas" how many presents do you get over all?

Note the lyrics are given below, but the main idea is that you are adding up consecutive triangular numbers. Namely, on the first day you get 1 present. On the second day you get 3 = 2+1 more presents, making 4 overall. On the third day you get 6 = 3 + 2 + 1 more presents, meaning 10 overall. Thus, in total, how many presents do you get?

You can find the answer during Maths At: Christmas.

We also gave you a bonus question that built on this ideas.

In the song "12 days of Christmas" how many legs are there in total?

As we discussed in the podcast there are many possible solutions to this question based on different assumptions. Do the milk maids have stools? Has a swan lost a leg and swimming in a circle?

Many possibilities! We will reveal our solution next time.

For those who want a consistent set of lyrics we used the following list:

First day:

A Partridge in a Pear Tree

Second day:
2 Turtle Doves
and a Partridge in a Pear Tree

Third day:
3 French Hens
2 Turtle Doves
and a Partridge in a Pear Tree

Fourth day:
4 Calling Birds
3 French Hens
2 Turtle Doves
and a Partridge in a Pear Tree

Fifth day:
5 Golden Rings
4 Calling Birds
3 French Hens
2 Turtle Doves
and a Partridge in a Pear Tree

Sixth day:
6 Geese a Laying
5 Golden Rings
4 Calling Birds
3 French Hens
2 Turtle Doves
and a Partridge in a Pear Tree

Seventh day:
7 Swans a Swimming
6 Geese a Laying
5 Golden Rings
4 Calling Birds
3 French Hens
2 Turtle Doves
and a Partridge in a Pear Tree

Eighth day:
8 Maids a Milking
7 Swans a Swimming
6 Geese a Laying
5 Golden Rings
4 Calling Birds
3 French Hens
2 Turtle Doves
and a Partridge in a Pear Tree

Ninth day:
9 Ladies Dancing
8 Maids a Milking
7 Swans a Swimming
6 Geese a Laying
5 Golden Rings
4 Calling Birds
3 French Hens
2 Turtle Doves
and a Partridge in a Pear Tree

Tenth day:
10 Lords a Leaping
9 Ladies Dancing
8 Maids a Milking
7 Swans a Swimming
6 Geese a Laying
5 Golden Rings
4 Calling Birds
3 French Hens
2 Turtle Doves
and a Partridge in a Pear Tree

Eleventh day:
11 Pipers Piping
10 Lords a Leaping
9 Ladies Dancing
8 Maids a Milking
7 Swans a Swimming
6 Geese a Laying
5 Golden Rings
4 Calling Birds
3 French Hens
2 Turtle Doves
and a Partridge in a Pear Tree

Twelfth day:
12 Drummers Drumming
11 Pipers Piping
10 Lords a Leaping
9 Ladies Dancing
8 Maids a Milking
7 Swans a Swimming
6 Geese a Laying
5 Golden Rings
4 Calling Birds
3 French Hens
2 Turtle Doves
and a Partridge in a Pear Tree

Watch along with MATM: Flatland

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Fri, 29 Dec 2017 16:11:00 GMT

It's been a long time coming, but we're finally back on track. Next Friday will be our 8th episode where we discuss and watch Flatland: The Movie.

Once again we have a star-studded cast of voice actors including Martin Sheen, Kristen Bell and Michael York.

You can buy a DVD copy from the company behind the film, or Amazon does stock imported US versions. However, they are quite expensive and will take a little while to arrive.

For those of you who do not want to seek an illegal online streaming copy (which we do not condone here at Maths At) there is always the possibility of reading the book by Edwin A. Abbott, which sticks pretty close to the movie, except for one or two plot points. The book also goes into much more detail about the Flatland universe.

Be there or be square.

Maths at: Christmas

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Mon, 25 Dec 2017 08:44:00 GMT

It's the most wonderful time... of the year.

Merry Christmas and happy holidays to all of our listeners.

Join us as we discuss (amongst other things):

our favourite Christmas films (not really mathematical);
the differences between a mathematicians office Christmas party and an actual Christmas party (slightly mathematical);
the number of presents your true is giving and how many legs they'll have (quite mathematical).

So fill your boots with Turkey and in your after dinner stupor enjoy our CHRISTMATHS! We're better than the Queen's speech.

GUEST INTRO AND OUTRO: Ma Woolley.

As festive as Ben can get!

Further reading links:

It's a Wonderful Life (Ben's favourite Christmas movie), Muppet's Christmas Carol (Liz' favourite Christmas movie) and Batman Returns (Thomas' favourite Christmas movie);
Sheffield University's Maths Society's formulas for a perfect Christmas tree;
A nice discussion on the Travelling Santa Problem.

Subscribe via iTunes.

A very merry christmas to all our listeners!

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Fri, 15 Dec 2017 16:20:00 GMT

A very merry Christmas to all our listeners. We're hoping to record one more seasonal themed podcast before Santa comes, so let us know your Christmas mathematical questions in the comments below.

Christmas card with apologies to Hilbert, all major churches, and humour in general.

MATM Appendix: Boy born on a Tuesday

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Fri, 15 Dec 2017 14:35:00 GMT

During Maths at the Movies: The Imitation Game Ben introduced the following puzzle:

I have two children. One of them is a boy.

What is the probability that both children are boys?

Although counter-intuitive, it can be shown that the solution is 1/3.

He also posed a similar, but different question:

I have two children. One of them is a boy who was born on a Tuesday.

What is the probability that both children are boys?

Although very similar, the answer turns out to be 13/27, which is quite different. However, Ben got confused when giving the answer during our pi podcast and paid penance by making a video

However, there is still some debate about the answer.

Join us this week as we record all from the same location and try and tease apart the complexity of weird conundrum.

Yup, we really are altogether!

Maths at the Movies: Moneyball

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Fri, 1 Dec 2017 12:56:00 GMT

In this (Hallowe'en themed?) episode we watch the movie Moneyball.

The story of a young mathematician, played by Jonah Hill, struggling to be heard amongst the angry men that make up baseball. And I think Brad Pitt was in there somewhere as well.

Can Jonah produce a winning team by redefining how statistics are used?
Why does Thomas keep saying the phrase "dick swinging"?
Where did Ben get a hold of those baseball sounds?

All of these questions and more are not answered in our Moneyball podcast.

If you're interested in watching Moneyball you can follow the Amazon link below.

Further reading links:

a really interesting sabermetrics talk at google;
the Baader-Meinhof Phenomenon;
pi after shave.

Subscribe via iTunes.
Follow us on twitter @PodcastMathsA, as well as @ThomasEWoolley and @benmparker.

Puzzle from pi: Crossing a river

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Mon, 27 Nov 2017 14:00:00 GMT

Although there was no Dev Patel, or tigers in this pi, we still managed to squeeze a boat into this week's puzzle.

This puzzle is a variant on an old favourite. We don't have chickens, corn, or foxes, but we still want to cross a river. Here's the question:

A husband and wife come to the edge of a river, where they find two children with a small boat. The boat can hold either one child, two children, or one adult. How do you get everyone across the river in the minimum possible number of crossings?

Post your answers below, and we'll give you the answer in the next podcast!

Watch along with MATM: Moneyball

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Sun, 26 Nov 2017 10:52:00 GMT

Brad Pitt is a sexy man. He's done action, comedy, drama and more. A true acting force to be reckoned with.

Join us as we watch him sit around in chairs and talk to men:

Young men;
Old men;
Sporty men;
Angry men;
Mathematical men.

All the men you could ever want in a film are right here for your viewing pleasure.

Yup, it's Moneyball: baseball, men and apparently some maths? We've really got to stop Ben choosing the films!

If you want to watch along with us then can buy a digital version of the film through Amazon by clicking on the image below.

Maths at the Movies: pi

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Fri, 17 Nov 2017 13:49:00 GMT

This week we watched pi.

Sadly there were no tigers, boats or Dev Patel in this movie.

Nope, this film was an avant garde, mood piece, seething with questioning the meaning of truth and its place in the lives of humans and the universe.

In other words pretentious.

However, none Darren Aronofsky's nonsense matters. What you should be listening for is Ben's rendition of "Euclid's people" a song of his own creation sung to the tune of "Common people" by pulp.

You've got to hear it to believe it.

If you're interested in watching pi you can follow the Amazon link below.

Further reading links:

play go online;
find your birthday in pi;
more pi facts than you can shake a stick at;
weird ways of calculating pi;
known digits of pi over time.

Subscribe via iTunes.

Answer to The Imitation Game

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Thu, 16 Nov 2017 18:24:00 GMT

As you'll hear in the next podcast Ben cocks up the explanation of his original questions from Maths at: the Movies, The Imitation Game:

I have two children. One of them is a boy and they were born on a Tuesday.

What is the probability that both children are boys?

This is a hard question, and Ben ****ed up the explanation when he tried to do it live. So, as penance, we made him sit down and explain it as a video.

Here's a simpler question written out much nicer:

I have two children. One of them is a boy.
What is the probability that both children are boys?

Now you may think the probability is 50%, but that is not so (note that we are assuming that boy and girl births are equally likely). The reason is because we have more information about the children.

Suppose we denote a boy by "b" and a girl by "g". Further, we capitalise the letter to denote the elder child. In this way we could have the following combinations of children:

However, we know we have at least one boy, so we can't have Gg. Out of the possibilities that are left, namely Bb, Gb and Bg, there is only one way to get two boys, the chance is 1/3! Counter-intuitive no?

Note that if we had posed the problem as I have two children and my eldest is a boy then (using the above argument) the probability of have a second boy is then 1/2.

Probability can be a tricksy animal. Even for a Cambridge educated lecturer!

Answer to The Man Who Knew Infinity

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Wed, 15 Nov 2017 16:13:00 GMT

At the end of Maths at: the Movies, The Man Who Knew Infinity Thomas posed two teasers to you.

A simple one to start you off.
I buy a bottle and a cork for £1.10. The bottle costs £1 more than the cork.

How much does the cork cost?

A moments thought should show that the bottle costs £1.05, whilst the cork costs only 5p. If you go it right first time well done! The answer most people tend to give if they don't pause for a second is 10p.

Now, for the more difficult question:

I live on a street with more than one house. All the houses on this street are numbered consecutively, 1, 2, 3,..., etc. Amazingly, I live in the house such that if you add up all the house numbers below me and all the house numbers above me then they come to exactly the same answer.

What is the minimum number of houses on this street and what is my house number?

The smallest answer, excluding the one house case is 8 houses on the street and I live at number 6, thus, 1+2+3+4+5=15=7+8.

There are actually an infinite number of increasing solutions to this problem. Although the solution can be found using basic algebra and a knowledge of continued fractions the details can get a bit hairy. Thus, I direct the interested reader to the following two wonderful expositions on the matter:

http://www.johnderbyshire.com/Opinions/Diaries/Puzzles/2009-06.html
http://www.angelfire.com/ak/ashoksandhya/winners2.html#PUZZANS4

Answer to Donald Duck in Mathmagic Land

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Tue, 14 Nov 2017 16:44:00 GMT

During Maths at: the Movies, Donald Duck in Mathmagic Land Ben presented the following conundrum.

Ben's greengrocer uses a balance scale, like the one seen above, and only has a 40kg weight. However, the greengrocer fortuitously broke the weight into four pieces of integer weight that will allow them to measure out every integer of kilograms from 1kg to 40kg. What are the four weights?

Normally, for a problem like this, you'd think of the binary sequence 1, 2, 4, ..., because, as shown in the gold chain problem, you can construct any number using combinations of these numbers. However, with only four weights, we would have 1, 2, 4, 8, from which we could produce a maximum of 15, falling far short of the 40kg total.

The crux of the problem is that in binary we can only add or not add a weight. In this problem, because we are using a set of balance scales we have three possibilities:
(a) Not add the weight, denoted 0;
(b) Add the weight to the left side, denoted L;
(c) Add the weight to the right side, dented R.

Because we have three possibilities, instead of two, we might think about using the numbers based around powers of 3 (the trinary system), rather than those based around powers of 2 (the binary system). Thus, our weights would be 1, 3, 9, 27. Adding these together does indeed give 40kg, but how would we use them to weigh out 2kg?

Put the 1kg on the left and the 3kg on the right. This produces a deficit of 2kg in the left pan, so we add apples to the left pan until it balances and, voila, we know we have two kilograms of apples.
Thus, 2 is represented as LR00 in our system.

What about 5kg? Similar to the above put the 1 and 3kg weights in the left and the 9 in the right. This produces a deficit of 5kgs in the left pan. Thus, 5kg is represented by LLR0.

Using this ideas we can produce the following table

Weight to be measured	Trinary encoding	Effective calculation
1	R000	1=1
2	LR00	-1+3=2
3	0R00	3=3
4	RR00	1+3=4
5	LLR0	-1-3+9=5
6	0LR0	-3+9=6
7	RLR0	1-3+9=7
8	L0R0	-1+9=8
9	00R0	9=9
10	R0R0	1+9=10
11	LRR0	-1+3+9=11
12	0RR0	3+9=12
13	RRR0	1+3+9=13
14	LLLR	-1-3-9+27=14
15	0LLR	-3-9+27=15
16	RLLR	1-3-9+27=16
17	L0LR	-1-9+27=17
18	00LR	-9+27=18
19	R0LR	1-9+27=19
20	LRLR	-1+3-9+27=20
21	0RLR	3-9+27=21
22	RRLR	1+3-9+27=22
23	LL0R	-1-3+27=23
24	0L0R	-3+27=24
25	RL0R	1-3+27=25
26	L00R	-1+27=26
27	000R	27=27
28	R00R	1+27=28
29	LR0R	-1+3+27=29
30	0R0L	3+27=30
31	RR0R	1+3+27=31
32	LLRR	-1-3+9+27=32
33	0LRR	-3+9+27=33
34	RLRR	1-3+9+27
35	L0RR	-1+9+27=35
36	00RR	9+27=36
37	R0RR	1+9+27=37
38	LRRR	-1+3+9+27=38
39	0RRR	3+9+27=39
40	RRRR	1+3+9+29=40

Watch along with MATM: pi

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Sat, 11 Nov 2017 07:50:00 GMT

Apple?

Cherry?

Raspberry?

Unfortunately, none of these.

This Friday we'll be releasing our pi podcast. Darren Aronofsky's surreal, disturbing, art house, pretentious, feature debut about the descent into insanity of the Jewish Mathematician Max Cohen.

One of the reasons Thomas started this podcast was to talk about this film and to ensure that everyone hated it as much as him.

We don't recommend it, but if you want to watch along with us then can buy a digital version of the film through Amazon by clicking on the image below.

Puzzle from The Imitation Game

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Fri, 10 Nov 2017 13:23:00 GMT

At the end of Maths at: the Movies, The Imitation Game Ben presented the following question

I have two children. One of them is a boy and they were born on a Tuesday.

What is the probability that both children are boys?

Now, unfortunately, Ben royally screws up the explanation of this answer in the next podcast. However, we are recording an appendix episode to put the world to rights. In the mean time though we can solve the slightly simpler puzzle:

I have two children. One of them is a boy.

What is the probability that both children are boys?

The answer will appear in the next podcast.

Puzzle from The Man Who Knew Infinity

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Thu, 9 Nov 2017 09:03:00 GMT

And a more difficult one for you to chew on that was apparently given to Ramanujan himself.

I live on a street with more than one house. All the houses on this street are numbered consecutively, 1, 2, 3,..., etc. Amazingly, I live in the house such that if you add up all the house numbers below me and all the house numbers above me then they come to exactly the same answer.

What is the minimum number of houses on this street and what is my house number?

Post your comments or answers below.

If you want to know the answer you can either wait until next week or listen to the next podcast, Maths at: the Movies, The Imitation Game.

Puzzle from Donald Duck in Mathmagic Land

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Wed, 8 Nov 2017 15:14:00 GMT

During Maths at: the Movies, Donald Duck in Mathmagic Land Ben presented the following conundrum.

Until recently my greengrocer sold apples in multiples of 40kg. The apples are weighed out using an old set of balance scales and a 40kg weight (see the above picture). However, the greengrocer dropped their weight and it broke into four pieces weighing four different integer values. Just before the greengrocer threw the pieces away I stopped them and showed that the set of four smaller weights could be used to measure out every integer of kilograms from 1kg to 40kg.

What were the values of the four smaller weights?

Post your comments or answers below.

If you want to know the answer you can either wait until next week or listen to the next podcast, Maths at: the Movies, The Man Who Knew Infinity.

Maths at the Movies: The Imitation Game plus SPECIAL GUEST

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Fri, 3 Nov 2017 08:28:00 GMT

In this episode we watch the movie The Imitation Game.

Alongside your regular team of Thomas, Ben and Liz there was only one mathematician with the expertise who could take us through this movie with grace, wit and wisdom. And that mathematician wasn't available so we got

Dr James Grimes

instead.

Join us for episode five of Maths at: The Movies as we separate fact from fiction about the life of Alan Turing.

If you're interested in watching The Imitation Game you can follow the Amazon link below.

Further reading links:

more than you would ever want to know about Alan Turing;
Turing's biography by Andrew Hodges;
more than you would ever want to know about James Grime;
finding the truth behind the movies;
comparing actors and their real-life counter parts.
play with an online enigma machine;
play with an online Turing machine.

Subscribe via iTunes.

Watch along with MATM: The Imitation Game and SPECIAL GUEST!

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Mon, 30 Oct 2017 11:27:00 GMT

This Friday, we will be watching Banister Crumblebench starring in The Imitation Game. The film follows the life of monumental mathematician Alan Turing and his work on cracking the Enigma code during World War II.

The mathematical genius Alan Turing and marvellous actor Bumblesnuff Crimpysnitch.

Joining your regular team of Liz, Ben and Thomas will be a SPECIAL GUEST mathematician.

Who could it be?

Is it Bodysnatch Cummerbund?
Is it Buffalo Custardbath?
Is it Bundleup Catchyoudeath?

Will Thomas ever get tired of mispronouncing Benedict Cumberbatch's name?

None of these answers and more will be provided on Friday.

Why not watch along with us? You can buy a digital version of the film through Amazon by clicking on the image below.

Friday Factoid: There are 451 unique ways to make up 50p.

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Fri, 27 Oct 2017 00:35:00 +0100

During The Man who knew Infinity I mentioned an interesting problem:

How many ways were there of making coins add up to the value of 50p?

In the UK, we have coins of value 1,2,5,10,20,50,100, and 200. To get them to add up to 50, we can have three 10p pieces plus one 20p. Or we could have 25 tuppences (2p pieces). Or many other combinations.

I got this problem via a phone call from a friend of mine who said that their grandchild had been sent home from school with a challenge. I was in the pub with two statistics professors, and we all scratched our heads.

This problem turned out to be quite fun. I thought the answer would be quite large, but I was surprised that there are 451 ways. No wonder running the economy is so difficult!

I eventually came up with a graph which manages to show how we generalise this to any monetary amount.

See how quickly the graph rises with the amount of money to be made up in change? This seems a very difficult problem for a small person, so I suspect the teacher set it as a way to keep a bright child quiet for a few hours!

(Edited to add US Coins: Look how much difference there is for UK and US coins, mostly because USA does not have a 2c coin.)

I looked for a sensible way to do it, but the only real way I can explain it is to come up with some logical way to write these out. I cannot see a clever mathematical trick, so I used a recursive algorithm, which means that we keep using the algorithm within itself until we get the answer.

Either you use one 50p piece (count 1 way)

or you don't use any 50p piece (Call this X).

Now we need X:

Either you use two 20p pieces (call this X1)

or you use one 20p piece (Call this X2)

or you use no 20p pieces (Call this X3)

So X=X1+X2+X3

Now we find X1- we have to find ways to make 10p with no coin bigger than a 10p.

Either we use one 10p piece (count 1)

or we use none (Call this Y1)...

All this algebra produces a complicated sum. Although you can reuse bits of it, it is laborious, and probably very hard to get right.

I got bored in the end, so I cheated. I wrote this short piece of code in MatLab which will take any amount of money, any collection of change, and print out all possible combinations of change to get the required sum. The code can be found below.

The function takes in three arguments, "amount", "possCoins" and "strIn".

amount is the amount of money we are partitioning;
possCoins is the set of coins we are partitioning the amount into. For example, in Britain we use coins of denomination of 1p, 2p, 5p, 10p, 20p, 50p, £1 and £2. In America, they have a 25c coin, rather than a 20c coin.
strIn is a string which is the list of all the coin combinations we have already found if we are part way through the recursion.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

function numWays = coins(amount,possCoins,strIn)
    if ((isempty(possCoins)) | (amount<=0))
            numWays=0;
    else
        if (amount==possCoins(1))
           strOut=[strIn,' ',num2str(possCoins(1))];
           disp(strOut)
           numWays=1+coins(amount,possCoins(2:length(possCoins)),strIn);
         else
           strOut=[strIn,' ',num2str(possCoins(1))];

numWays=coins(amount-possCoins(1),possCoins,strOut)+...

           coins(amount,possCoins(2:length(possCoins)),strIn);
        end
    end

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

The above code does essentially the same process as described. Using it for the 50p case, we first define the possible coins we can use:

>> possCoins=[50,20,10,5,2,1]
possCoins =

50 20 10 5 2 1

Then we define strIn to be empty (we start at the beginning so we haven't started the recursion) :

>> strIn=' '

Finally, we call the function above, which gives the following output, in a logical order:

>> coins(amount,possCoins,strIn)

50
20 20 10
20 20 5 5
20 20 5 2 2 1
20 20 5 2 1 1 1
20 20 5 1 1 1 1 1
20 20 2 2 2 2 2
20 20 2 2 2 2 1 1
20 20 2 2 2 1 1 1 1
20 20 2 2 1 1 1 1 1 1
20 20 2 1 1 1 1 1 1 1 1
20 20 1 1 1 1 1 1 1 1 1 1
20 10 10 10
20 10 10 5 5
20 10 10 5 2 2 1
20 10 10 5 2 1 1 1
20 10 10 5 1 1 1 1 1
20 10 10 2 2 2 2 2
20 10 10 2 2 2 2 1 1
20 10 10 2 2 2 1 1 1 1
20 10 10 2 2 1 1 1 1 1 1
20 10 10 2 1 1 1 1 1 1 1 1
20 10 10 1 1 1 1 1 1 1 1 1 1
20 10 5 5 5 5
20 10 5 5 5 2 2 1
20 10 5 5 5 2 1 1 1
20 10 5 5 5 1 1 1 1 1
20 10 5 5 2 2 2 2 2
20 10 5 5 2 2 2 2 1 1
20 10 5 5 2 2 2 1 1 1 1
20 10 5 5 2 2 1 1 1 1 1 1
20 10 5 5 2 1 1 1 1 1 1 1 1
20 10 5 5 1 1 1 1 1 1 1 1 1 1
20 10 5 2 2 2 2 2 2 2 1
20 10 5 2 2 2 2 2 2 1 1 1
20 10 5 2 2 2 2 2 1 1 1 1 1
20 10 5 2 2 2 2 1 1 1 1 1 1 1
20 10 5 2 2 2 1 1 1 1 1 1 1 1 1
20 10 5 2 2 1 1 1 1 1 1 1 1 1 1 1
20 10 5 2 1 1 1 1 1 1 1 1 1 1 1 1 1
20 10 5 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
20 10 2 2 2 2 2 2 2 2 2 2
20 10 2 2 2 2 2 2 2 2 2 1 1
20 10 2 2 2 2 2 2 2 2 1 1 1 1
20 10 2 2 2 2 2 2 2 1 1 1 1 1 1
20 10 2 2 2 2 2 2 1 1 1 1 1 1 1 1
20 10 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1
20 10 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1
20 10 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1
20 10 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
20 10 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
20 10 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
20 5 5 5 5 5 5
20 5 5 5 5 5 2 2 1
20 5 5 5 5 5 2 1 1 1
20 5 5 5 5 5 1 1 1 1 1
20 5 5 5 5 2 2 2 2 2
20 5 5 5 5 2 2 2 2 1 1
20 5 5 5 5 2 2 2 1 1 1 1
20 5 5 5 5 2 2 1 1 1 1 1 1
20 5 5 5 5 2 1 1 1 1 1 1 1 1
20 5 5 5 5 1 1 1 1 1 1 1 1 1 1
20 5 5 5 2 2 2 2 2 2 2 1
20 5 5 5 2 2 2 2 2 2 1 1 1
20 5 5 5 2 2 2 2 2 1 1 1 1 1
20 5 5 5 2 2 2 2 1 1 1 1 1 1 1
20 5 5 5 2 2 2 1 1 1 1 1 1 1 1 1
20 5 5 5 2 2 1 1 1 1 1 1 1 1 1 1 1
20 5 5 5 2 1 1 1 1 1 1 1 1 1 1 1 1 1
20 5 5 5 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
20 5 5 2 2 2 2 2 2 2 2 2 2
20 5 5 2 2 2 2 2 2 2 2 2 1 1
20 5 5 2 2 2 2 2 2 2 2 1 1 1 1
20 5 5 2 2 2 2 2 2 2 1 1 1 1 1 1
20 5 5 2 2 2 2 2 2 1 1 1 1 1 1 1 1
20 5 5 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1
20 5 5 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1
20 5 5 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1
20 5 5 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
20 5 5 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
20 5 5 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
20 5 2 2 2 2 2 2 2 2 2 2 2 2 1
20 5 2 2 2 2 2 2 2 2 2 2 2 1 1 1
20 5 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1
20 5 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1
20 5 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1
20 5 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1
20 5 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1
20 5 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
20 5 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
20 5 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
20 5 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
20 5 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
20 5 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
20 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
20 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1
20 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1
20 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1
20 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1
20 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1
20 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1
20 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1
20 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
20 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
20 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
20 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
20 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
20 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
20 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
20 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
10 10 10 10 10
10 10 10 10 5 5
10 10 10 10 5 2 2 1
10 10 10 10 5 2 1 1 1
10 10 10 10 5 1 1 1 1 1
10 10 10 10 2 2 2 2 2
10 10 10 10 2 2 2 2 1 1
10 10 10 10 2 2 2 1 1 1 1
10 10 10 10 2 2 1 1 1 1 1 1
10 10 10 10 2 1 1 1 1 1 1 1 1
10 10 10 10 1 1 1 1 1 1 1 1 1 1
10 10 10 5 5 5 5
10 10 10 5 5 5 2 2 1
10 10 10 5 5 5 2 1 1 1
10 10 10 5 5 5 1 1 1 1 1
10 10 10 5 5 2 2 2 2 2
10 10 10 5 5 2 2 2 2 1 1
10 10 10 5 5 2 2 2 1 1 1 1
10 10 10 5 5 2 2 1 1 1 1 1 1
10 10 10 5 5 2 1 1 1 1 1 1 1 1
10 10 10 5 5 1 1 1 1 1 1 1 1 1 1
10 10 10 5 2 2 2 2 2 2 2 1
10 10 10 5 2 2 2 2 2 2 1 1 1
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2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
2 2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
2 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

ans =

451

Loop de loop patterns

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Mon, 23 Oct 2017 11:37:00 +0100

During The Man Who Knew Infinity, Thomas produced a picture for Liz based on a random set of numbers taken from Donald Duck in Mathemagic Land.

The picture was a Loop de Loop pattern.

The idea is that you consider a moving pointer, known as a "turtle". The turtle is given instructions to move around and the picture is made up of the turtle's track history. The set of rules that the turtle follows is:

the turtle moves along in a straight line a given distance. Here, the random numbers where used as the distances;
after moving each distance the turtle is turned 90 degrees to the right;
iterate the pattern until you get back to the beginning, or until you want to stop.

There are many websites that will allow you to produce such a picture using a programming language known as Logo.

For a simple example consider the list of numbers {50, 100, 150}. A single iteration of the Loop de Loop algorithm is given by

fd 50 rt 90 fd 100 rt 90 fd 150,

where "fd" means "go forward" the given amount and "rt" means "turn right" the given number of degrees. Following this set of instruction would produce a "J" type pattern, as shown below. Iterating the pattern can be done simply through

repeat 4[fd 50 rt 90 fd 100 rt 90 fd 150 rt 90].

This code will repeat the instructions 4 times and produce a "pin wheel" type pattern, shown below.

Iterations of the code using the set of numbers {50, 100, 150}. Top left - one iteration, top right - two iterations, bottom left - three iterations, bottom right - four iterations.

The numbers Thomas used (although apparently not the numbers that Liz gave) were

{7, 9, 4, 8, 5, 7, 4, 3, 9, 2, 5, 7, 3, 8, 4, 9, 7, 5, 2, 6, 3}.

Four iterations take you back to the beginning point; the four iterations are illustrated below. Once, you have made your pattern, you can colour it in and make it funky.

The four iterations of the Loop de Loop pattern, using the Donald Duck random number list.

Thomas generated his pictures using MatLab. The code is given below.

Why not produce your own Loop de Loop pattern and tweet it to us @PodcastMathsAt?

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% Clear all information

clear

close all

clc

N=[7 9 4 8 5 7 4 3 9 2 5 7 3 8 4 9 7 5 2 6 3]; %The list of numbers.

figure('position',[0 0 1 1]) %Set the figure space to fill the screen.

% Iteration 1
M=N;
theta=0;
x(1)=0;

for i=1:length(M)

x(i+1)=x(i)+M(i)*exp(theta*1i);
theta=theta+pi/2;

end
subplot(2,2,1)
plot(real(x),imag(x))
axis equal
axis off

% Iteration 2

M=[N N]
theta=0;
x(1)=0;
for i=1:length(M)

x(i+1)=x(i)+M(i)*exp(theta*1i);
theta=theta+pi/2;

end
subplot(2,2,2)
plot(real(x),imag(x))
axis equal
axis off

% Iteration 3
M=[N N N];
theta=0;
x(1)=0;

for i=1:length(M)

x(i+1)=x(i)+M(i)*exp(theta*1i);
theta=theta+pi/2;

end
subplot(2,2,3)
plot(real(x),imag(x))
axis equal
axis off

% Iteration 4
M=[N N N N];
theta=0;
x(1)=0;

for i=1:length(M)

    x(i+1)=x(i)+M(i)*exp(theta*1i);
    theta=theta+pi/2;

end
subplot(2,2,4)
plot(real(x),imag(x))
axis equal
axis off

Maths at the Movies: The Man Who Knew Infinity

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Fri, 20 Oct 2017 22:09:00 +0100

In this episode we watch the movie The Man Who Knew Infinity.

And in this episode we learn that we shouldn't be mean to Dev Patel.

You can admire Jeremy Iron's subtle performance of introversion;
You can wonder if maths is like Jazz;
You can even question whether Ramanujan really was as brilliant as he was claimed to be.

You can do all of this... as long as you're not mean to Dev Patel.

Join us for episode four of Maths at: The Movies as we delve deep into the biopic of Srinivasa Ramanujan.

If you're interested in watching The Man Who Knew Infinity you can follow the Amazon link below.

Further reading links:

more than you would ever want to know about Ramanujan;
Hardy's famous apology;
puzzles based on Ramanujan's work;
buy the book behind the film;
Chinese multiplication;
Generating -1/12 from adding the integers.

Subscribe via iTunes.

Watch along with MATM

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Tue, 17 Oct 2017 20:22:00 +0100

Coming this Friday Maths at: The Movies reviews The Man Who Knew Infinity, the biopic of the incredible genius of Srinivasa Ramanujan.

As you will not doubt have noticed, maths films bring in the acting powerhouses: Kevin Spacey, Gwyneth Paltrow,... Donald Duck!

This film is no different. Between Dev Patel's puppy dog eyes and Jeremy Irons' introversion this film has some interesting, sad and poignant moments.

Why not watch along with us? You can buy a digital version of the film by clicking on the image below.

Answer to "Seven Gold Rings"

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Mon, 16 Oct 2017 11:23:00 +0100

In our "Proof" podcast, we set the following puzzle:

Tom employs a labourer, who is going to work for him for seven days.

Every day the labourer wants to be paid one gold ring for each day they have worked. So at the end of the week they should leave with seven gold rings.

Tom has a chain with seven gold links, but cannot simply give the laborourer all seven links, because the labourer wouldn't return. Equally, the labourer wants to ensure that they leave every day with as many gold links as days that they've worked, otherwise, they, again, won't return.

Tom could cut each link and pay the labourer one link per day, using seven cuts. But this wastes a lot of gold.

What is the minimum number of links that Tom must cut in order to make sure the labourer can always leave with the correct number of gold links?

What was your answer?

We can do it in one cut! Here's how:

Cut only the fifth ring.

and rearrange to form the binary sequence 1, 2 and 4 links.

Using these links we can pay the labourer the exact amount everyday, on the understanding that we can take back the links we've already given them.

Explicitly,

Day 1: give the 1 link piece;

Day 2: take back the 1 link piece, and give the 2 link piece;

Day 3: give the 1 link piece (the labourer already has the two, so they have three links altogether);

Day 4: take back the 1 and 2 link pieces and give the 4 link piece;

Day 5: give the 1 link piece;

Day 6: take back the 1 link piece, and give the 2 link piece;

Day 7: give the 1 link piece.

Thus, at the end of the seven days the labourer has all seven links and was never short-changed!

This puzzle relies on the fact that all the numbers between 1 and 7 can be written in binary, using three bits:

1=001,

2=010,

3=011,

4=100,

5=101,

6=110,

7=111.

Consider the extension to this problem. Tom wants the labourer to work for 15 days and has a chain of 15 gold links. What is the minimum number of cuts would he need to do now?

Think you know the answer? Comment below, tweet at us or email us at podcastmaths@gmail.com

Maths at the Movies: Donald Duck in Mathmagic Land

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Fri, 6 Oct 2017 14:25:00 +0100

In this episode we watch the movie Donald Duck in Mathmagic Land.

Well, this was just weird.

Although the animation is beautiful, can you trust the company that told you that lemmings were suicidal to teach you mathematics?

Also we cover Donald Duck's terrible gun control and the billiards game that goes on forever!

Plus bees that make jam.

If you're interested in watching Donald Duck in Mathmagic Land, you can follow the Amazon link below.

Further reading links:

coloured honey from French bees;
the billiards diamond system;
Le Corbusier's mathematical architecture;
there was a book too?

Subscribe via iTunes.

So, how old is Ben?

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Wed, 4 Oct 2017 12:55:00 +0100

If you remember, at the end of Maths at: the Movies, 21, we asked you how old is Ben on his next Birthday, if he is currently 50 years, 50 Months, 50 Weeks, 50 Days, and 50 Hours old?

The best way to answer this question is to start with the hours and convert them to days,

i.e. 50 hours = 2 days 2 hours.

Then convert the 52 days into weeks,

52 days = 7 weeks 3 days,

and so on.

Following this process you should discover that Ben is currently 55 years old and, thus, will be 56 on his next birthday.

Doesn't Ben look good for his age?

Of course the easiest way is to get a computer to do it. Wolfram alpha may not be good for much, but unit conversions are where it excels.

Not too difficult to start us off with, but did you get it right? Tweet at us, @podcastmaths, or find us on facebook, to receive a shout-out on a future podcast and a warm fuzzy glow.

Next up - Maths with Donald Duck!

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Mon, 2 Oct 2017 21:00:00 +0100

There can be few more celebrated, more acclaimed, more accomplished mathematicians than Donald Duck.

For our next podcast "Maths at: The Movies" we're going to Mathmagic land with everybody's favourite Disney duck.... and you can watch this for free on YouTube.

Our podcast will be out later in the week, but tell us what you think in the comments below!

Quack quack!

Puzzle from Proof: Seven Gold Rings

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Tue, 26 Sep 2017 13:03:00 +0100

In our latest podcast about the movie "Proof", we set the following puzzle:

Tom employs a labourer, who is going to work for him for seven days.

Every day he wants to pay the labourer one gold ring for each day they have worked.

Unfortunately Tom is a strange person who keeps his gold in the form of a chain with seven gold links, and a miser who doesn't like wasting gold. What is the minimum number of links that Tom must cut in order to make sure the labourer can always leave with the correct number of gold links?

Post your comments or answers below, and we'll give you the answer at the end of our next podcast.

Maths at the Movies: Proof

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Fri, 22 Sep 2017 13:47:00 +0100

In this episode we watch the movie Proof.

Did Gwyneth Paltrow really prove the theorem? What theorem was she trying to prove? How many vaginas does a Time Lord have?

All these questions and more are discussed in this weeks Maths at the Movies.

If you're interested in watching Proof you can follow the Amazon link below.

Further reading links:

more about the amazing life of Sophie Germain;
who was Erdos?;
calculate your Erdos number;
what are prime numbers good for?;
the ideas behind the Four Colour Theorem;
buy the book of the play.

Subscribe via iTunes.

MATM Appendix: Who the hell are we?

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Wed, 20 Sep 2017 15:10:00 +0100

You've listened to us talk, but what are our credentials? What do we know? What are our backgrounds? When should you take our "facts" at face value and when should you call bullshit?

Join us in this appendix to better understand your hosts.

Further reading links:

you can see this earlier blog post about us;
Thomas E. Woolley's home page;
Ben M. Parkers' profile;
more about Liz.

Subscribe via ITunes.

Want to join in on the fun?

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Tue, 19 Sep 2017 10:46:00 +0100

Our next "Maths at: The Movies" podcast will be appearing soon, and this week we chat about the movie "Proof", starring Gwyneth Paltrow and Anthony Hopkins. The podcast talks about the concept of proof, certainty, and how to behave at a funeral.

We had great fun watching and discussing it, if you want to watch along with us, you can rent or download the movie now by clicking the image below.

Puzzle from 21: How old will Ben be on his next birthday?

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Mon, 18 Sep 2017 15:56:00 +0100

Here's the puzzle from the first episode of the podcast, Maths at: the Movies, 21, where we learnt that mathematicians like Fibonacci Birthday cakes.

However, here's the question for this week. Comment below, or send your answer to podcastmaths@gmail.com, tweet us @podcastmaths, find us on facebook, to receive a shout-out on a future podcast and a warm fuzzy glow.

"Ben is currently 50 years, 50 Months, 50 Weeks, 50 Days, and 50 Hours old. How old will he be on his next birthday?"

(Ben would like to point out he is nowhere near that old, despite Tom's insinuations)

Maths at the Movies: 21

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Tue, 12 Sep 2017 14:45:00 +0100

In our first ever episode we watch the movie 21.

This film has it all: racism, sexism, disableism... and a smattering of mathematics.

Join us as we peel back the layers of gambling and card counting to reveal why you should never keep your money in your ceiling and why you'll never be as good a mathematician as Kevin Spacey.

If you're interested in watching 21 you can follow the Amazon link below.

Further reading links:

more than you could ever want to know about the Fibonacci sequence;
a basic strategy guide to blackjack;
learn how to count cards from wikiHow or Scam School;
a technical guide to the Newton-Raphson technique.

Subscribe via iTunes.

Who are we?

podcastmaths@gmail.com (Thomas E. Woolley, Ben M. Parker, The Wonderful Liz) — Mon, 11 Sep 2017 14:47:00 +0100

Thomas E. Woolley

is a Lecturer in Applied Mathematics at Cardiff University. He was an undergraduate, graduate, doctorate and post-doctorate at the University of Oxford, St John's College since 2004 and thought it was about time to leave in 2017.

He specialises in mathematical biology and focuses on the pattern formation behind fish spots and zebra stripes. He also researches mathematical models of stem cells movement. The hope is that by understanding how stem cells move we can influence them and, thus, speed up the healing process.

When not doing mathematics he is a keen participant in mathematical outreach workshops and has given a variety of popular maths lectures nationally and internationally. He has previously worked for the BBC, illustrated Marcus du Sautoy’s book, "The Number Mysteries"; set questions for the popular maths show “Dara O’Briains school of hard sums” and helped redesign the the London Science Museum's Mathematics Gallery.

Ben M. Parker

is a Lecturer in Statistics, in Mathematical Sciences, and the Southampton Statistical Sciences Research Institute. He received a MA in Mathematics and Diploma in Computer Science from University of Cambridge, Pembroke College.

Previous employments include:

working on an EPSRC grant at Queen Mary, University of London, entitled “Design of Experiments for Complex, Large Scale Networks".
PhD from Queen Mary, University of London, thesis: “Design of Experiments for Packet Communication Networks”.
MSc in Applied Statistics and Operational Research from Birkbeck College, University of London.

The Wonderful Liz

is a [REDACTED] who can [REDACTED] with her [REDACTED]. Incredibly, she once [REDACTED] a [REDACTED], which is truly a motivating example of the human spirit that everyone can learn from.

In her spare time she likes nothing more than [REDACTED] with her [REDACTED], or [REDACTED] whilst sipping on a [REDACTED].

She wishes she could say more about her adventures in [REDACTED], but unfortunately, they are far too amazing and no one would believe her.