Thursday, 21 March 2019

Puzzle from CUBE

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?


Monday, 18 March 2019

Math at the Movies: x+y

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.
https://amzn.to/2T3mog9

Further reading links:

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

Friday, 15 March 2019

Answer to Fermat's Room puzzle

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:
  1. You are the truth-teller
  2. 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.




Tuesday, 5 March 2019

Maths at the Movies: CUBE

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


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. 
https://amzn.to/2TFJlT6

Further reading links:


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



Friday, 1 March 2019

Puzzle from Fermat's Room

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:
  1. You cannot ask questions like "Will it rain tomorrow?", because neither the truth teller, nor the liar can be sure.
  2. 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.
  3. You cannot ask something like "Will you answer No to this question?", because the truth-teller can't answer this question.
  4. All decisions must be based on the yes and/or no answers only.
  5. This puzzle is not about "how to find a way around the rules".

Thursday, 28 February 2019

Answer to the Good Will Hunting puzzle

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?


Tuesday, 19 February 2019

Maths at the Movies: Fermat's Room

Welcome to the strangely erotic episode of Maths at, where we watch the tense, psychological thriller, Fermat's Room (or La HabitaciĆ³n 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:

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

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