Follow-up: Birthday Magic Square singingbanana 2022-12-05 | This is my follow-up video to my birthday magic squares videoyoutu.be/hNn0j4Kay8g---Yeah, I know the audio is messed up. It's not the mic, it's because I recorded via OBS, and that messed up the audio. Sometimes it's difficult to get all the tech working. I tried.If the audio is really hurting your ears, I process the audio a bit and made an unlisted version of this video with different audio youtu.be/ah4WplJqaDAAt 2:20 I point at a group of four and say it adds to 43. That was a mistake.At 27:40 I said April but typed in 03. Whoops.---Here is my excel spreadsheet for creating birthday magic squaressingingbanana.com/birthdaymagicsquarescreator.xlsxHere is the page by Christian Lawson-Perfectsomethingorotherwhatever.com/birthday-magic-squareAnd I forgot to say that Christian is occasionally available to make things like this for commission.Wikipedia: en.wikipedia.org/wiki/Magic_squareEuler's method: en.wikipedia.org/wiki/Magic_square#Method_of_superpositionpandiagonal squares: en.wikipedia.org/wiki/Pandiagonal_magic_square And here are the videos for Further Maths Wales:youtube.com/playlist?list=PL0ThtFmyYmsslQn1M1KHmHwYBKBsEZG5e
How to make a Birthday Magic Square singingbanana 2022-11-20 | A birthday magic square is a special type of magic square that uses someone's birth date in the top row. Then every row, column, and main diagonal add up to the special birthday number - and a whole lot of other magic properties as well!-----Notes:Some people have realised it isn't always possible to make a birthday magic square that avoid negative values or repeats. In fact, if your total is less than 34, then you will either have repeats or negative values.Since a total less than 34 is unavoidable for some birth dates, I advise you don't worry about it.But if you embrace negative values, I think you can always avoid (additional) repeats.
Follow-up: Barbie electronic typewriter singingbanana 2022-08-01 | Here is a copy of the description from the Barbie video:----I first found this story on the crypto museum website, which has great information about the Barbie typewriter (and other cipher machines) cryptomuseum.com/crypto/mehano/barbie-------Thanks to Sarah Everett from Just My Typewriter, check out her channel here: youtube.com/c/JustMyTypewriterAnd Sarah's own Barbie video here: youtu.be/axgzXdN-3lo-------Here's the link for my cryptography course: singingbanana.com/codescourseAnd in case you have any problems with that link, here is the long link udemy.com/course/the-mathematics-of-cryptography-with-dr-james-grime/learn/lecture/28599464?referralCode=A9E03BB265345A92B6C4Make sure you use my link because it includes my referral code.------UPDATES FROM THE COMMENTSThe most important update comes from a comment by Rosie Fay, who spotted that cipher 1 is just a cycle on 90 symbols.What does that mean? We often write codes in two-row notation like thisPlaintext: ABCDECiphertext: DAEBCBut we can write that as two chains: A becomes D becomes B becomes A, and C becomes E becomes C.Or more compactly we write (ADB)(CE).Barbie cipher 1, is just one long chain, using 90 symbols, like this:mjvgxdlbcorneaithsfpuqkyzw;¢+%:,W.MZBFCA (right angle bracket) RSETHONILDUGYPQKJV (left angle bracket) X1234056789-¨§£€='_)(*#@?/^!&"YouTube doesn't allow the angled brackets in the description so I wrote them out, but that is meant to be one chain of 90 symbols.Rosie Fay then also spotted that cipher 2 is cipher 1 applied twice (we call that the second power).Cipher 3 is cipher 1 applied three times (third power).And cipher 4 is cipher 1 applied four times (fourth power).So there are some fun things we can discover from this.1. We can apply the ciphers in any order, and order does not matter (they are commutative).For example, cipher 2 followed by cipher 4 is the same as cipher 4 followed by cipher 2. In fact they are both equal to cipher 1 applied six times.2. If you applied cipher 1 90 times, you would get back to the original message.3. The decryption cipher for cipher 1 is cipher 1 to the 89th power.The decryption cipher for cipher 2 is cipher 1 to the 88th power.The decryption cipher for cipher 3 is cipher 1 to the 87th power.The decryption cipher for cipher 4 is cipher 1 to the 86th power.And you could think of these as eight different ciphers. But they work in pairs to code and decode.
The Barbie electronic typewriter - with Just My Typewriter singingbanana 2022-06-21 | I first found this story on the crypto museum website, which has great information about the Barbie typewriter (and other cipher machines) cryptomuseum.com/crypto/mehano/barbie-------Thanks to Sarah Everett from Just My Typewriter, check out her channel here: youtube.com/c/JustMyTypewriterAnd Sarah's own Barbie video here: youtu.be/axgzXdN-3loUPDATE 15th May 2023: I sent my Barbie typewriter to Sarah, who has made a follow-up video about that youtu.be/5uyyVbelSc0-------Here's the link for my cryptography course: singingbanana.com/codescourseAnd in case you have any problems with that link, here is the long link udemy.com/course/the-mathematics-of-cryptography-with-dr-james-grime/learn/lecture/28599464?referralCode=A9E03BB265345A92B6C4Make sure you use my link because it includes my referral code.------UPDATES FROM THE COMMENTSThe most important update comes from a comment by Rosie Fay, who spotted that cipher 1 is just a cycle on 90 symbols.What does that mean? We often write codes in two-row notation like thisPlaintext: ABCDECiphertext: DAEBCBut we can write that as two chains: A becomes D becomes B becomes A, and C becomes E becomes C.Or more compactly we write (ADB)(CE).Barbie cipher 1, is just one long chain, using 90 symbols, like this:mjvgxdlbcorneaithsfpuqkyzw;¢+%:,W.MZBFCA (right angle bracket) RSETHONILDUGYPQKJV (left angle bracket) X1234056789-¨§£€='_)(*#@?/^!&"YouTube doesn't allow the angled brackets in the description so I wrote them out, but that is meant to be one chain of 90 symbols.Rosie Fay then also spotted that cipher 2 is cipher 1 applied twice (we call that the second power).Cipher 3 is cipher 1 applied three times (third power).And cipher 4 is cipher 1 applied four times (fourth power).So there are some fun things we can discover from this.1. We can apply the ciphers in any order, and order does not matter (they are commutative).For example, cipher 2 followed by cipher 4 is the same as cipher 4 followed by cipher 2. In fact they are both equal to cipher 1 applied six times.2. If you applied cipher 1 90 times, you would get back to the original message.3. The decryption cipher for cipher 1 is cipher 1 to the 89th power.The decryption cipher for cipher 2 is cipher 1 to the 88th power.The decryption cipher for cipher 3 is cipher 1 to the 87th power.The decryption cipher for cipher 4 is cipher 1 to the 86th power.And you could think of these as eight different ciphers. But they work in pairs to code and decode.
Follow-Up: Finite Difference Method singingbanana 2022-06-13 | Original Video here: youtu.be/scQ51q_1nhwVideos mentioned:James Tanton youtu.be/_5vU48kf7NYMathologer youtu.be/4AuV93LOPcEMore on Gilbreath's conjecture here: https://primes.utm.edu/glossary/page.php?sort=GilbreathsConjectureHere is finite differences on wikipediaen.wikipedia.org/wiki/Finite_difference#Newton's_seriesI didn't mention:Max Peeters gave the following interesting comment:"John Conway and Richard Guy explain how this method can be further extended in their book 'The Book of Numbers'. They share 2 of them, the first works for simple exponential functions as well (such as 4^n - 3^n), and the second works for sequences where each term is a sum of previous terms (like Fibonnaci's sequence).For anyone interested: wolfram has info on both of them, they're called "Jackson's Difference Fan" and "Quotient-Difference Table"."mathworld.wolfram.com/JacksonsDifferenceFan.htmlmathworld.wolfram.com/Quotient-DifferenceTable.html
The Finite Difference Method singingbanana 2022-06-07 | Find a polynomial with the finite difference method. Take successive differences of a sequence to find the polynomial that made it.Let me try to anticipate some questions:1. What if we have a sequence that doesn't start at x = 0?There's a general form of Newton's Forward Difference Formula for a sequence that starts at x = a, with 1 unit steps. Here it is on wikipediaen.wikipedia.org/wiki/Finite_difference#Newton's_series2. What if the steps are not unit steps?There is an even more general form of Newton's Forward Difference Formula for a sequence that starts at x = a and has equally spaced steps of size h. You can see it on wikipedia at the end of the Newton series section here en.wikipedia.org/wiki/Finite_difference#Newton's_series3. The finite difference method leads to a whole branch of maths called finite difference calculus.In finite difference calculus, the difference operator (that I called D(x)) is analogous to differentiation. Then, Newton's Difference Formula is analogous to Taylor series, and there is a whole bunch of other formulas that are analogous to calculus. But it's all for polynomials rather than any analytic function.4. I want more YouTube videos about the finite difference method please!James Tanton did a great one here, very approachable description of the idea youtu.be/_5vU48kf7NYAnd Mathologer has gone into more of the details, especially the connection to calculus, here youtu.be/4AuV93LOPcE5. Are a few constants enough to know we have a polynomial?Unfortunately, you might have something that looks like a row a constants, but it is not guarantee that the procedure has finished. The formula you end up with will fit the data you currently have.So, if it is a mystery formula, then what you have is a good guess.For example, we could start with a sequence for x^2: 0, 1, 4, 9, 16, ...And the finite difference method will find the formula f(x) = x^2.But if we want to be naughty, we could add any random number to the end of that sequence:0, 1, 4, 9, 16, 73, ....And now we get the formula: f(x) = x + x(x-1) + (48/120)(x)(x-1)(x-2)(x-3)(x-4) = (48x)/5 - 19x^2 + 14x^3 - 4x^4 + (2x^5)/5.This agrees with x^2 for the first few values, and then gives the value 73.6. Max Peeters gave the following interesting comment:"John Conway and Richard Guy explain how this method can be further extended in their book 'The Book of Numbers'. They share 2 of them, the first works for simple exponential functions as well (such as 4^n - 3^n), and the second works for sequences where each term is a sum of previous terms (like Fibonnaci's sequence).For anyone interested: wolfram has info on both of them, they're called "Jackson's Difference Fan" and "Quotient-Difference Table"."mathworld.wolfram.com/JacksonsDifferenceFan.htmlmathworld.wolfram.com/Quotient-DifferenceTable.html7. I think I buried one of the most interesting things about this method, that this is how the Charles Babbage Difference Engine worked. See wikipedia for all its history, and a description of how it worked by differences (basically what I said in the video) en.wikipedia.org/wiki/Difference_engine#Charles_Babbage's_difference_engines8. I like the handwritten notes style, and I like that they are imperfect. But the formula at 3:49 looks a bit blobby. It says f(x) = (D^(2)(0)/2)x^2 + (D(0) - D^(2)(0)/2)x + D^(0)(0). And that can then be rearranged to say f(x) = D^(2)(0) (x(x-1)/2) + D(0)(x) + D^(0)(0).9. Oh and if you want to check your answer for hexagonal numbers, here it is:en.wikipedia.org/wiki/Hexagonal_number
Follow-up: British Flag Theorem singingbanana 2022-06-01 | UPDATE: I've had a think about those integer solutions rectangles. Here is one way to construct the rectangle.Take two pythagorean triples: (u, v, w) and (x, y, z);Then we can make four pythagorean triples that fit together, namely (ux, vx, wx), (vx, vy, vz), (uy, vy, wy), (uy, ux, uz).It turns out that's how I made my example, with (u, v, w) = (12, 35, 37) and (x, y, z) = (3, 4, 5).I don't know if there are other ways to do it.Richard Holmes wonders whether this is the only way to construct the rectangles, but found a counterexample: (25, 60, 65), (25, 312, 313), (91, 312, 325), (60, 91, 109)These are four genuinely different pythag triples. I don't know how to make other examples like this.Oh, and as supermarc45 pointed out in the comments, you could solve it using 8 copies of one pythagorean triple. I knew that, of course, but that would be boring.--Here is the original video youtu.be/scZMQEHEt1wAnd here is the description from that video, copied over to this one:1. Is there a 3d version of this?There is. First of all, the point can be above/below a rectangle, and if we connect the four corners to the point (now in 3d space), it is still true.But also, I've just checked for a cuboid and it's still true. If AB is a space diagonal (for example, from the bottom left corner of the cuboid to the top right corner of the opposite face), and CD the other space diagonal (for example, bottom right to top left of opposite face), then a^2 + b^2 = c^2 + d^2. You can prove that by splitting the height, width, and depth into u, v, w, x, y, z and doing 3d pythagoras on that. The two sides are equal to u^2 + v^2 + w^2 + x^2 + y^2 + z^2.2. Can this be done for a parallelograms?There is a version for parallelograms, although a^2 + b^2 does not equal c^2 + d^2. Instead, the two sides differ by a value that is independent of the choice of point. (I will leave that as a challenge for you, I might do the answer in a follow-up video).2a. ironpencil observes that if we place two copies of the rectangle, side-by-side, then the diagonals a, b, c, d form a quadrilateral with orthogonal diagonals. In that case a and b are opposite sides, c and d are opposite sides and a^2 + b^2 = c^2 + d^2.The diagonals would have length (w+x) and (y+z), and the area of the quadrilateral will be (w+x)(y+z)/2.3. Can we make w, x, y, z and diagonals a, b, c, d all integers?You can! If (w, z, a) are all integers, it is called a pythagorean triple. We need to find four pythagorean triples, (w, z, a), (x, y, b), (w, y, c) and (x, z, d) so they can fit together to make a rectangle. That's how I made my example, with pythagorean triples (280, 210, 350), (72, 96, 120), (280, 96, 296) and (72, 210, 222), making a 352 by 306 rectangle.UPDATE: I've had a think about those integer solutions rectangles. Here is one way to construct the rectangle.Take two pythagorean triples: (u, v, w) and (x, y, z);Then we can make four pythagorean triples that fit together, namely (ux, vx, wx), (vx, vy, vz), (uy, vy, wy), (uy, ux, uz).It turns out that's how I made my example, with (u, v, w) = (12, 35, 37) and (x, y, z) = (3, 4, 5).I don't know if there are other ways to do it.Oh, and as supermarc45 pointed out in the comments, you could solve it using 8 copies of one pythagorean triple. I knew that, of course, but that would be boring.4. What other flag theorems could we have?Here is the Dutch National Flag Problem: en.wikipedia.org/wiki/Dutch_national_flag_problemAnd someone suggested pythagoras is the Trinidad and Tobago flag theorem en.wikipedia.org/wiki/Flag_of_Trinidad_and_Tobago5. How can pythagoras be a special case of the British Flag Theorem, when you use pythag to prove the British Flag Theorem?We could prove the British Flag Theorem the same way we prove pythag, without using pythag itself. There are a few hundred ways to do that, take your pick. In other words, Pythag and BFT are equivalent theorems.6. Is there some sort of British celebration going on?There is. The Queen has been queen for 70 years. For the record, I'm not bothered about that, but there will be lots of flags about and I'm using that as an excuse to talk about maths.
The British Flag Theorem singingbanana 2022-05-24 | The British Flag Theorem connects any point to the corners of a rectangle to calculate distances.It turns out I don't know how to salute. The Royal Navy salute that way if that makes you feel better.I would appreciate it if you kept the comments light. Reading lots of hot takes isn't much fun for me.UPDATE: Follow-up video here: youtu.be/OaTnZXaFTBcSome good questions, and answers, from the comments:1. Is there a 3d version of this?There is. First of all, the point can be above/below a rectangle, and if we connect the four corners to the point (now in 3d space), it is still true.But also, I've just checked for a cuboid and it's still true. If AB is a space diagonal (for example, from the bottom left corner of the cuboid to the top right corner of the opposite face), and CD the other space diagonal (for example, bottom right to top left of opposite face), then a^2 + b^2 = c^2 + d^2. You can prove that by splitting the height, width, and depth into u, v, w, x, y, z and doing 3d pythagoras on that. The two sides are equal to u^2 + v^2 + w^2 + x^2 + y^2 + z^2.2. Can this be done for a parallelograms?There is a version for parallelograms, although a^2 + b^2 does not equal c^2 + d^2. Instead, the two sides differ by a value that is independent of the choice of point. (I will leave that as a challenge for you, I might do the answer in a follow-up video). UPDATE: I did youtu.be/OaTnZXaFTBc2a. ironpencil observes that if we place two copies of the rectangle, side-by-side, then the diagonals a, b, c, d form a quadrilateral with orthogonal diagonals. In that case a and b are opposite sides, c and d are opposite sides and a^2 + b^2 = c^2 + d^2.The diagonals would have length (w+x) and (y+z), and the area of the quadrilateral will be (w+x)(y+z)/2.3. Can we make w, x, y, z and diagonals a, b, c, d all integers?You can! If (w, z, a) are all integers, it is called a pythagorean triple. We need to find four pythagorean triples, (w, z, a), (x, y, b), (w, y, c) and (x, z, d) so they can fit together to make a rectangle. That's how I made my example, with pythagorean triples (280, 210, 350), (72, 96, 120), (280, 96, 296) and (72, 210, 222), making a 352 by 306 rectangle.UPDATE: I've had a think about those integer solutions rectangles. Here is one way to construct the rectangle.Take two pythagorean triples: (u, v, w) and (x, y, z);Then we can make four pythagorean triples that fit together, namely (ux, vx, wx), (vx, vy, vz), (uy, vy, wy), (uy, ux, uz).It turns out that's how I made my example, with (u, v, w) = (12, 35, 37) and (x, y, z) = (3, 4, 5).I don't know if there are other ways to do it.Richard Holmes wonders whether this is the only way to construct the rectangles, but found a counterexample: (25, 60, 65), (25, 312, 313), (91, 312, 325), (60, 91, 109)These are four genuinely different pythag triples. I don't know how to make other examples like this.Oh, and as supermarc45 pointed out in the comments, you could solve it using 8 copies of one pythagorean triple. I knew that, of course, but that would be boring.4. What other flag theorems could we have?Here is the Dutch National Flag Problem: en.wikipedia.org/wiki/Dutch_national_flag_problemAnd someone suggested pythagoras is the Trinidad and Tobago flag theorem en.wikipedia.org/wiki/Flag_of_Trinidad_and_Tobago5. How can pythagoras be a special case of the British Flag Theorem, when you use pythag to prove the British Flag Theorem?We could prove the British Flag Theorem the same way we prove pythag, without using pythag itself. There are a few hundred ways to do that, take your pick. In other words, Pythag and BFT are equivalent theorems.6. Is there some sort of British celebration going on?There is. The Queen has been queen for 70 years. For the record, I'm not bothered about that, but there will be lots of flags about and I'm using that as an excuse to talk about maths.
Mastermind with Steve Mould singingbanana 2022-03-20 | Thanks Steve! youtube.com/c/SteveMouldPlay online: codebreaker.eoin.coThanks Eoin.Word Mastermind: stressfle.nikolaus.ukThanks Alex.I recommend these apps for Android:Mastermind (but called Bulls and Cows) play.google.com/store/apps/details?id=org.xbasoft.master_mind&pcampaignid=web_shareBulls and Cows play.google.com/store/apps/details?id=com.vist.bullscows&pcampaignid=web_share----------------Let's go through some Mastermind algorithms. Here are two human methods:Randomly Consistent: Picking any remaining valid combination at random. (Swaszek 1999).FIRST GUESS = ANY, MAX = 10, AVG = 4.638.Here is an example of 10 consistent guesses: cut-the-knot.org/ctk/Rafler.shtmlRandomly Consistent can be improved with FIRST GUESS = 1123, then MAX = 9 and AVG = 4.58.Simple: This algorithm just goes through the combinations in numerical order from 1111 to 6666, and picking the next consistent combination. (Shapiro 1983).FIRST GUESS: 1111, MAX = 9, AVG = 5.765This method can be improved with a FIRST GUESS = 5466, then MAX = 7, AVG = 4.646.----------------Here is the Least Worst Case Scenario method:Least Worst Case Scenario (simple): Consider the table of responses at each step, and choose a combination (from the remaining combinations) with the least worse case scenario. (Norvig 1984).FIRST GUESS = 1122, MAX = 6, AVG = 4.478Least Worst Case Scenario (full): Consider the table of responses at each step, and choose the combination (from all combinations) with the least worse case scenario. (Donald Knuth 1977).FIRST GUESS = 1122, MAX = 5, AVG = 4.476.***CORRECTION: I said 4.478 not 4.476 in the video. I mixed up the two Worst Case Scenario methods***In Knuth's paper, he gives an example of a game that needs an inconsistent move to guarantee a solve in 5 steps.https://www.cs.uni.edu/~wallingf/teaching/cs3530/resources/knuth-mastermind.pdfUsing Least Worst Algorithm the most difficult codes are 1221, 2354, 3311, 4524, 5656, 6643. ----------------Here are some other methods:Expected Case: Consider the table of responses, and choose the combination with the smallest average scenario. (Irving 1979).FIRST GUESS 1123, MAX = 6, AVG = 4.395.Entropy: Using the table of responses, maximise entropy. (Neuwirth, 1982).FIRST GUESS 1234, MAX = 6, AVG = 4.415.Entropy is an idea from Information Theory and is quite technical. 3b1b goes into it in detail in his wordle video here: youtu.be/v68zYyaEmEAMost Parts: Using the table of responses, pick the choice with the most non-zero parts. (Kooi 2005).FIRST GUESS = 1123, MAX = 6, AVG = 4.374.The idea here is to divide the remaining possibilities into as many buckets as possible. This increases the probability of a lucky guess.Interestingly, this method performs well in the 4-digit, 6 colour mastermind, but not as well with other numbers of digits and colours.----------------And here is the Optimal method:Optimal: Deep search to create a look-up table that gives the lowest average. (Koyoma and Lai 1993).FIRST GUESS = 1123, MAX = 6, AVG = 4.340.Interesting fact, there is only one combination that takes 6 steps, (namely 4421).Adjusted Optimal: Deep search to find the lowest average, with a max of 5. (Koyoma and Lai 1993).FIRST GUESS = 1123, MAX = 5, AVG = 4.341.This is something that got edited out of the video for time. With a few adjustments, the optimal max is reduced to 5, with a tiny increase to the average.----------------Here is another humanly possible method, which I think is fun, but is very different:Static Mastermind: A set of 6 fixed combinations, that can completely determine any secret combination on the seventh step.GUESSES = 1212, 2263, 3344, 4554, 5316, 6156.Here is the article: cut-the-knot.org/ctk/Mastermind.shtml(My list is different to the article, because I tried to make it more memorable).----------------Bulls and Cows: The original game. Using pen and paper, a 4-digit number is chosen without repeats.Random: FIRST GUESS = ANY, MAX = 8, AVG = 5.445Least Worst Case: FIRST GUESS = 1234, MAX = 7, AVG = 5.380Optimal: FIRST GUESS = 1234, MAX = 7, AVG = 5.213----------------Generalisations: What about mastermind using n positions, and k colours? We have partial information about maximums and averages. There is a good summary here https://people.cs.clemson.edu/~goddard/papers/mastermindRevisited.pdf----------------Further information:I found good information and references here: mathworld.wolfram.com/Mastermind.htmlAnd here: http://www.serkangur.freeservers.comAnd wikipedia, obviously: en.wikipedia.org/wiki/Bulls_and_Cowsen.wikipedia.org/wiki/Mastermind_(board_game)I then got the book by Serkan Gur and I thought it was excellent: amazon.com/dp/B01M17PMIQSerkan kindly helped me out with some of the details. Thank you Serkan.
Introducing MathsCity Leeds singingbanana 2021-10-01 | Introducing MathsCity Leeds, the UK's first hands-on maths discovery centre. Step inside a giant bubble, explore our laser ring of fire, and solve our hands-on puzzle challenges.MathsCity Leeds can be found in the Trinity shopping centre, on the first floor, between Boots and Customer Services.Find out more and book tickets online at mathscity.co.uk
A MegaFavNumbers Thank You! singingbanana 2020-09-08 | See the full #MegaFavNumbers playlist here youtube.com/playlist?list=PLar4u0v66vIodqt3KSZPsYyuULD5meoAo
MegaFavNumbers - The Even Amicable Numbers Conjecture singingbanana 2020-08-19 | This video is part of the MegaFavNumbers project. Maths YouTubers have come together to make videos about their favourite numbers bigger than one million, which we are calling #MegaFavNumbers.We want *you*, the viewers, to join in! Make your own video about your favourite mega-number. You can think of a cool big number, or think of a cool topic first and hang a mega-number on it.Upload your videos to YouTube with the hashtag #MegaFavNumbers and with MegaFavNumbers in the title, and your video will be added to the megafavnumbers playlist.Submit your videos anytime before Wednesday 2nd September to be added to the MegaFavNumbers playlist!MegaFavNumbers Playlist: youtube.com/playlist?list=PLar4u0v66vIodqt3KSZPsYyuULD5meoAoAmicable Numbers on Wikipedia: en.wikipedia.org/wiki/Amicable_numbersHere is a list of amicable numbers:http://www.vaxasoftware.com/doc_eduen/mat/numamigos_eng.pdfA list of sums of amicable numbersoeis.org/A180164Remember, a test for divisibility by 9 is to add all the digits of the number together, and to do the same with the result, until you get a single digit. If the end result is 9 then the original number was a multiple of 9.Here is a list of even amicable numbers whose sum is not divisible by 9 oeis.org/A291550And here are those complicated mathematical conditionsams.org/journals/mcom/1969-23-107/S0025-5718-1969-0248074-6/S0025-5718-1969-0248074-6.pdf
Im making videos on MathsWorldUK and this is me telling you about it singingbanana 2020-06-12 | See the videos at youtube.com/user/MathsWorldUKAt the moment, we've made 12 and they will be released over the next few weeks.Here's more about MathsWorldUK http://mathsworlduk.com
Bayes Billiards with Tom Crawford singingbanana 2020-01-24 | Bayes' Theorem allows us to assign a probability to an unknown fact.Thomas Bayes himself described an experiment with a billiard table, which is brilliantly explained by Hannah Fry and Matt Parker here youtube.com/watch?v=7GgLSnQ48osBrian Cox and David Spiegelhalter did a 1-dimensional version similar to our experiment here youtube.com/watch?v=-e8wOcaascMAfter we filmed this video, 3blue1brown released his own Bayesian video youtube.com/watch?v=HZGCoVF3YvMFor more of Tom Crawford see his channel youtube.com/tomrocksmathsOur experiment failed pretty badly really. For some behind-the-scenes information, this was our third attempt at the experiment, the first two were a little slow. The previous attempts were a lot more accurate. Oh well.Why did we fail? Maybe because the balls were colliding they were not independent. Maybe Tom wasn't random enough. In which case, our assumption that each position is equally likely could be updated.For more information on this experiment see nature.com/articles/nbt0904-1177The main point, I believe, is that for limited data, the Bayesian approach (using the average estimate) is more accurate.Viewer, Penny Lane, has run a simulation of this experiment, which did show that Bayesian was slightly more accurate than Frequentist (so this real-life attempt was probably a toss up for who did best):"Here's the output of my script:simulated 14 balls 10000000 timesthe Bayesian approach won over the Frequentist one 50.44934% of the times6.65762% of the simulations were ties, so 42.89304% were Frequentist winsthe mean deviation from the real value for the Bayesian approach was: 0.0800043179873167the mean deviation from the real value for the Frequentist approach was: 0.08422584547321381"See the comment here http://youtube.com/watch?v=rwtDBhD6Mq0&lc=UgzUC4MfbeHpIPQsBSR4AaABAgAnd code here pastebin.com/yyCjtnEuAnother comment I liked talked about what would happen if all the balls had been to the right. In that case the frequentist approach would put the position on the extreme left, p=0. While the Bayesian approach would put the position at p = 1/16 = 0.0625, so a little way from the extreme. And that sounds sensible.People who read the description are the best people. If you have read this, I probably need cheering up after the failure of this experiment, so tell me a joke in the comments. Thanks.
Help us make a maths discovery centre in the UK singingbanana 2019-04-26 | I'm talking to Dr Katie Chicot CEO of Maths World UK ( http://mathsworlduk.com ) - an organisation dedicated to creating a maths discovery centre in the UK. Although maths discovery centres exist in other countries we have nothing like it in the UK, that is what Maths World UK want to change.We ask for your ideas for what you would like to see in a maths discovery centre and show off a few puzzles and pretty mathematical objects.At the moment Maths World UK has funding to creating a travelling exhibition that will be visiting science centres. Maths World UK are currently looking for funding to create a permanent home.I really want to see a maths discovery centre in UK, but at the moment we have no permanent home. So if you know any friendly millionaires, let them know. (That's not a joke, that's what we need).
A Pythagorean Theorem for Pentagons + Einsteins Proof singingbanana 2019-04-02 | Pythagoras's Theorem is the most famous theorem in mathematics, commonly stated as "the square on the hypotenuse of a right-angled triangle is equal to the sum of the squares on the other two sides."However, Pythagoras's Theorem is not just for squares. In fact it works for any shape. The proof relies on the fact that scaling a shape by c will scale the area by c^2. Then, if Pythagoras's Theorem is true then the area of the shape on the hypotenuse will be equal to the total areas of the similar shapes on the other two sides.More succinctly, if Pythagoras's Theorem is true then the areas will be equal.But we can prove Pythagoras's Theorem itself by running that argument in reverse - if the shapes have equal area then Pythagoras's Theorem is true.This is an argument an 11 year old Albert Einstein used to prove Pythagoras's Theorem for himself.There are a couple of things I wished I said clearer in the Einstein proof: The Einstein proof divides the triangle so we have three right-angled triangles (but I think that was clear from the picture)Secondly, the three triangles are scaled versions of a triangle with a hypotenuse of length 1 and area X, which then have areas scaled by a^2, b^2 and c^2. (I just said "some triangle").This topic has been done before by a couple of the big maths YouTube channels, which I didn't know at the time (or forgot).Numberphile did it in 2014 youtube.com/watch?v=ItiFO5y36kwAnd Mathologer did it in 2018 youtube.com/watch?v=p-0SOWbzUYIA little historical note, Pythagoras's Theorem appears twice in Euclid's Elements, the famous squares version appears in Book 1.47, and in Book 6.31 it is there again, this time for any shape.
Fermats Last Theorem for rational and irrational exponents singingbanana 2019-02-27 | Fermat's Last Theorem states the equation x^n + y^n = z^n has no integer solutions for positive integer exponents greater than 2. However, Fermat's Last Theorem says nothing about exponents that are not positive integers.Note: x, y and z are meant to be positive integers, which I should have said in the video. Whoops.This video introduces some results for rational and irrational exponents.For rational exponents, k/m, k must be equal to 1 or 2.If we allow complex roots, then we have strange solutions with x=y=z and m divisible by 6.For irrational exponents no general results exist but we know there are infinitely many integer solutions, in this video I give a couple of examples.Many of the examples in this video, as well as the proof for rational exponents, were taken from this paper by Frank Morgan (2010) maa.org/sites/default/files/pdf/cmj_ftp/CMJ/May%202010/3%20Articles/1%20Morgan/Morgan9_5_09.pdfHere is another description of the same proof, with a bit more detail https://www.math.leidenuniv.nl/~hwl/PUBLICATIONS/1992d/art.pdfAnother proof for rational exponents is here, as well as the result with complex roots, by Bennett, Glass, Székely (2004)https://digitalcommons.lmu.edu/cgi/viewcontent.cgi?referer=&httpsredir=1&article=1103&context=math_facThe result for rationals seems to have been first proven by R. Oblath. Quelques proprietes arithmetiques des radicaux (Hungarian). In Comptes Rendus du Premier Congres des Mathematiciens Hongrois, 27 Aout–2 Septembre 1950, pages 445–450. Akad´emiai Kiado,Budapest, 1952.
The Elo Rating System for Chess and Beyond singingbanana 2019-02-15 | The Elo Rating system is a method to rate players in chess and other competitive games. A new player starts with a rating of 1000. This rating will go up if they win games, and go down if they lose games. Over time a player's rating becomes a true reflection of their ability - relative to the population.My video was mostly based on A Comprehensive Guide to Chess Ratings by Prof Mark E Glickman http://www.glicko.net/research/acjpaper.pdfBelow are some of the things I wanted to talk about, but cut so the video wasn't too long!Some explanations of the Elo rating system say it is based on the normal distribution, which is not quite true. Elo's original idea did model each player's ability as a normal distribution. The difference between the two players strengths would then also be a normal distribution. However, the formula for a normal distribution is a bit messy so today it is preferred to model each player using an extreme value distribution. The difference between the two players strengths is then a logistic distribution. This has the property that if a player has a rating 400 points more than another player they are 10 times more likely to win, this makes the formula nicer to use. Practically, the difference between a logistic distribution and the normal distribution is small.Logistic distribution on Wikipedia en.wikipedia.org/wiki/Logistic_distribution?wprov=sfla1We replace e with base 10, s=400, mu=R_A - R_B and x=0 in the cdf.For the update formula I say that your rating can increase or decrease by a maximum of 32 points, and I said there was no special reason for that. This value is called the K-factor, and the higher the K-factor the more weight you give to the players tournament performance (and so less weight to their pre-tournament performance). For high level chess tournaments they use a K-factor of 16 as it is believed their pre-tournament rating is about right, so their rating will not fluctuate as much. Some tournaments use different K-factors.In the original Elo system, draws are not included, instead they are considered to be equivalent to half a win and half a loss. The paper by Mark Glickman above contains a formula that includes draws. Similarly the paper contains a formula that includes the advantage to white.Another criticism of Elo is the reliability of the rating. The rating of an infrequent player is a less reliable measure of that player's strength, so to address this problem Mark Glickman devised Glicko and Glicko2. See descriptions of these methods at http://www.glicko.net/glicko.htmlOn the plus side, the Elo system was leagues ahead of what it replaced, known as the Harkness system. I originally intended to explain the Harkness system as well, so here are the paragraphs I cut:"In the Harkness system an average was taken of everyone's rating, then at the end of the tournament if the percentage of games you won was 50% then your new rating was the average rating. If you did better or worse than 50% then 10 points was added or subtracted to the average rating for every percentage point above or below 50.This system was not the best and could produce some strange results. For example, it was possible for a player to lose every game and still gain points."This video was suggested by Outray Chess. The maths is a bit harder, but I liked the idea so I made a in-front-of-a-wall video.
The Infinite Game of Chess (with Outray Chess) singingbanana 2019-02-08 | An infinite game of chess with the Thue-Morse sequence.To avoid an infinite game of chess there was a rule that declared that a game would end if any sequence of moves were repeated three times in a row.However Dutch mathematician Max Euwe showed that the Thue-Morse sequence can define an infinite game since it contains no finite sequence that is repeated three times in a row. The Thue-Morse sequence is made from building blocks of 0110 and 1001, so we know it cannot contain short repetitions like 000, 111, 010101 or 101010.Double digits in the Thue-Morse sequence always appear in the odd positions (starting from position zero), which is not possible if a sequence of odd length is repeated. So the Thue-Morse sequence does not contain any finite sequence of odd length repeated three times in a row.If we remove every second digit of the Thue-Morse sequence we will still have the Thue-Morse sequence. If you apply this to any finite sequence of even length that is repeated three times in a row, you will get a sequence half the length that also repeats three times in a row. Repeat this process until you reach a sequence of odd length repeated three times or a short sequence repeated three times. Since we know this shorter repeated sequence is not contained in the Thue-Morse sequence it implies the original repeated sequence is not contained in the Thue-Morse sequence.The argument above is enough to show that the Thue-Morse sequence does not contain a finite sequence of any length repeated three times in a row.You can read a little more detail here https://homepages-fb.thm.de/boergens/english/problems/problem059englloe.htmThanks to Outray Chess for making this video.Outray Chess youtube.com/channel/UCVfSsCg38hOzrezIFvMz9oA/featuredHost: Rune FriborgCamera: Mathis EskjærEditing: Mathis Eskjær and Rune FriborgFinally, here is our video about Hugh Alexander, as promised youtu.be/im75EwDXEzg
Alan Turings lost radio broadcast rerecorded singingbanana 2017-12-24 | On the 15th of May 1951 the BBC broadcasted a short lecture on the radio by the mathematician Alan Turing. His lecture was titled “Can Digital Computers Think?” and was part of a series of lectures which featured other leading figures in computing at the time. Unfortunately, these recording no longer exist, along with all other recordings of Alan Turing. The following is a rerecording of Alan Turing’s lost broadcast from his original script. See Turing's script at the Turing Digital Archive from King's College, Cambridge: turingarchive.kings.cam.ac.uk/publications-lectures-and-talks-amtb/amt-b-5For more information on the radio series see: http://oro.open.ac.uk/5609/1/01299654.pdfPhoto: Alan Turing (right) at the console of Mark II computer, c. 1951, at the University of Manchester.
Wythoffs Game (Get Home) singingbanana 2017-08-18 | Wythoff's Game is played on a chessboard. Two players take it in turns to move a piece. That piece can move any number of square to the left, and number of squares down, or any number of squares on a down-left diagonal. The winner is the player who moves the piece to the bottom-left square. What are the losing squares?See my first video with Katie Steckles here youtu.be/pzlpi7lJi4k--If we call the bottom-left square (0,0) then the losing squares are (1,2), (3,5), (4,7) and their reflections that swap the coordinates.The losing squares can be generated one at a time using the following two conditions: First, for the nth losing square, the difference between its coordinates is n. And second, each positive integer appears once and only once as either the x or y coordinate of a losing square. These two conditions have the effect of putting all losing squares on different rows, columns and diagonals.In 1907, Willem Wythoff proved that the nth losing square has coordinates (n*phi, n*phi^2) where phi is the golden ratio (1.618), and the two coordinates are rounded down to the previous integer. He showed that the golden ratio is the only number that will work in this way, giving the desired two properties of losing squares.Play an interactive version of the nim version of Wythoff's Game (called Last Biscuit here) on nrich: nrich.maths.org/1186Wythoff's Game on Wikipedia:en.wikipedia.org/wiki/Wythoff%27s_gameWythoff's Game on Mathworld: http://mathworld.wolfram.com/WythoffsGame.htmlWythoff's Proof: archive.org/stream/nieuwarchiefvoo03genogoog#page/n219/mode/2upAn excellent series of blog posts by Zachary Abel, read in reverse order (Thanks to Daniel Kelsall for this link) http://blog.zacharyabel.com/tag/wythoffs-gameWillem Wythoff: en.wikipedia.org/wiki/Willem_Abraham_WythoffRufus Isaacs: en.wikipedia.org/wiki/Rufus_Isaacs_(game_theorist)
Game: Get Home singingbanana 2017-08-15 | A game with a winning strategy.Place a piece a grid (like a chessboard). Two players take it in turns to move the piece. You can move any number of squares to the left; and number of squares down; and any number of squares left-down (SW diagonal). No other moves are allowed. The winner is the player who moves the piece to the bottom-left square.What is the winning strategy? Is there a winning strategy for every square? Does it matter if you go first or second?Katie Steckles youtube.com/user/st3cks
Twin Primes Problem singingbanana 2017-06-04 | Prove that when you multiply a pair of twin primes you get a number that has remainder 8 after division by 9. With one exception.I'm already getting some good solutions. It's interesting to see people do it in slightly different ways.I'll kill some space here just in case people can see the beginning of the description if it gets reposted.OK, I think the most succinct answers go along these lines:Twin primes must be of the form 3n-1 and 3n+1. Multiplying these two primes gives us 9n^2 - 1 = 9(n^2 - 1) + 8. So we have a remainder of 8 after division by 9. The exception is 3 and 5.I've summarised answers there, but Alienturnedhuman was the first commenter who said something like that.Some people used the same argument using 6n-1 and 6n+1. It is true that all primes larger than 3 are of this form. That works as well, but it's not quite as neat as above.Some people used modular arithmetic. I can't assume all viewers know modular arithmetic, but one argument is:All primes are 1, 2, 4, 5, 7, 8 mod 9 The possible pairs are2x4 = 8 mod 95x7 = 35 = 8 mod 98x1 = 8 mod 9.That's actually how I did it when given the problem. It's not the neatest solution.
Cambridge has a new mathsy train station singingbanana 2017-05-26 | Cambridge North is the new train station in Cambridge which features a mathematical design. The architects said the design was "derived from John Conway's Game of Life". Except it's not the Game of Life. It is Stephen Wolfram's Rule 135.Find out more about Rule 135 (or Rule 30, which is the same thing with the colours swapped) en.wikipedia.org/wiki/Rule_30Here's is Stephen Wolfram's reaction on his blog http://blog.stephenwolfram.com/2017/06/oh-my-gosh-its-covered-in-rule-30sThanks to SRD who spotted this on twitter twitter.com/Quendus/status/866707269299339264Here is the architect's website about Cambridge North http://www.atkinsglobal.com/en-gb/projects/cambridge-north-stationThe architects respond in the comments under this article http://aperiodical.com/2017/05/right-answer-for-the-wrong-reason-cellular-automaton-on-the-new-cambridge-north-station"Let me assure you it is the correct answer. We turned the pattern through 45 degrees, distorted the pixels to a slightly elongated diamond and played about with the panel dimensions to ensure the maximum gathering of openings around eye level for the passengers using the station. What we liked most about rule 30 was it was as close as we could find to a “random” non repeating pattern.Quintin Doyle, Senior Architectural Designer, Atkins."The irony is, Stephen Wolfram went to Oxford.------------------------Corrections:I misspoke three times. Silly mistakes, but more than usual, and quite close together. I miss YouTube annotations, that would have sorted it out."John Conway was" - That was definitely a slip of the tongue. My mind was picturing 1970 so was speaking in past tense."Take that Oxford" - I thought that was quite funny. Apparently Oxford has two stations too. So that spoilt the joke. "Stephen Wolfram is American" - He was British, and is now an American citizen. People didn't like me calling him American.And a personal request from me. When you comment you are talking directly to me. Please be respectful. I make videos in my spare time and for fun. I'm just a guy.
A visit from Rafael Procopio (Matemática Rio) singingbanana 2016-09-29 | I was visited by Rafael Procopio from Matemática Rio.YouTube channel: youtube.com/user/matematicario/featuredOur video about the Walkie-Talkie building: youtube.com/watch?v=2of5DuORwj8An interview we did: youtube.com/watch?v=oPoJMJ2HGsgRafael on twitter twitter.com/MatematicaRio
Origami Soma Cube singingbanana 2016-09-23 | The soma cube is a famous puzzle among mathematicians. Seven tetris-like pieces fit together to make a 3x3 cube. I made one from post-it notes.My instructions for the sonobe units: youtube.com/watch?v=uF72h-7xe0gA blog with instructions if you don't like mine (sob) luckpaperscissors.wordpress.com/2013/05/09/make-a-soma-cube-out-of-sonobe-modulesMy friend Alison Kiddle twitter.com/ajk_44
Sum of Fibonacci Numbers Trick singingbanana 2016-09-18 | A little trick to sum Fibonacci numbers. Try it out.
The International Maths Salute with Dr James Tanton singingbanana 2016-09-08 | I caught James at the MATRIX conference in Leeds in September 2016. I'm a fan of his work so this was cool. We grabbed a moment to make a super quick video.Go see James' videos on his YouTube channel youtube.com/user/DrJamesTantonSome more videos with James via the MAA youtube.com/playlist?list=PLevtNOOa6SZXVJvtROAFCC0oYt0ySTSo4James' Twitter: twitter.com/jamestantonAnd websites: http://gdaymath.com http://www.jamestanton.com
Ramanujan Review singingbanana 2016-05-02 | A sort of review of The Man Who Knew Infinity. And a *review* of other videos that came out this week.standupmaths: Ramanujan, 1729, Fermat's Last Theorem youtube.com/watch?v=_o0cIpLQApkMathologer: Ramanujan. Making sense of -1/12 youtube.com/watch?v=jcKRGpMiVTwsingingbanana: The Riemann Hypothesis youtube.com/watch?v=rGo2hsoJSbosingingbanana: Ramanujan Summation youtube.com/watch?v=8hgeIDY7We4Numberphile: Partitions youtube.com/watch?v=NjCIq58rZ8IOther videos I found too late to recommend in the video:Tipping Point Math: The Man Who Knew Infinity youtube.com/watch?v=P0idBBhGNgUMindYourDecisions: Ramanujan's Radical Brainteaser youtube.com/watch?v=r5BGIi84arY
Ramanujan Summation singingbanana 2016-05-01 | The third video in a series about Ramanujan.This one is about Ramanujan Summation.Here's the wikipedia page for further reading: en.wikipedia.org/wiki/Ramanujan_summationEuler-Maclaurin Formula en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula---------Here is an example of divergent summation, that hopefully shows its usefulness, and shows that it is rigorous and consistent with traditional summation.Imagine a series c_n that can be split into two other series a_n and b_n as follows:sum c_n = sum (a_n + b_n) = sum a_n + sum b_nUsing this we can work out sum c_n from the values of sum a_n and sum b_n. This is standard stuff when working with finite series, or convergent series.It is also possible for a convergent series to be the sum of two divergent series. So in the above, c_n is convergent and a_n, b_n are divergent.In that case, we can still work out sum c_n from the values of sum a_n and sum b_n, but now you have to use divergent summation.This only works if the divergent summation method is regular (gives the same answer as convergent summation when applied to c_n) and linear (so sum (a_n + b_n) = sum a_n + sum b_n)A specific example is sum(n) = 1 + 2 + 3 + ... and sum((1-n^3)/n^2) = 0 - 7/4 - 26/9 - .... These are both divergent series but, by Dirichlet regularization, their divergent sums are sum(n) = -1/12 and sum((1-n^3)/n^2) = (1 + 2pi^2)/12. Finally sum(n) + sum((1-n^3)/n^2) = pi^2/6 = sum(1/n^2) as expected.Some series are are harder to sum than others. So there are levels of summation that you can use. The levels are something like this:Finite series: Can be added together, multiplied, rearranged, as expected.Convergent series: Has all the properties of finite series, except a sequence of partial sums does not end with the value of the series. Instead, the limit of the sequence is used as the sum of the series. Example: geometric series with decreasing terms 1 + 1/2 + 1/4 + ... = 2Conditionally convergent series: Has all the properties of convergent series, but if you rearrange the terms you get different answers. Example: 1 - 1/2 + 1/3 - 1/4 + .... = ln(2)Divergent series: Would go to infinity by definition of convergent series. Various other methods can be applied to give a value. Some divergent series are harder to give a value than others. See below. Any divergent summation methods needs to agree with the limit when applied to convergent series (i.e. regular).Divergent series Cesaro summation: Can still be added and multiplied like convergent series (i.e. linear). Example: Grandi's series 1 - 1 + 1 - 1 + ... = 1/2Divergent series Euler summation: A method of analytic continuation. Still can be added and multiplied as expected (linear). Example: geometric series with increasing terms 1 + 2 + 4 + 8 +... = -1Divergent series Borel summation: Can give a value to harder series but still agrees with previous methods. Loses a property known as stability, where removing a term from the series does not simply subtract its value from the total sum. Adding and multiplying (linearity) still exists.Divergent series Ramanujan summation: Still linear. Can be used on the most stubborn divergent series, but depends on your choice of a parameter. I called the parameter 'a' in the video. When a is taken as infinity this method agrees with convergent sums. Examples when a=0: 1+1+1+1+... = -1/2. 1+2+3+4+...=-1/12. Example when a=1: 1 + 1/2 + 1/3 + 1/4 + ... = 0.5772... the euler-mascheroni constant.Zeta Function continuation: A method of analytic continuation. You will continuously approach the value, and agrees with Ramanujan summation. Example: 1^-s + 2^-s + 3^-s + ... = B_(s+1)/(s+1)For s=0 we get 1+1+1+... = -1/2 and for s=-1 we get 1+2+3+4+... = -1/12.Some may remember the numberphile video where Tony Padilla manipulates series to write 1+2+3+... in terms of Grandi's series, and got the -1/12 answer. Strictly speaking, Tony was manipulating the Riemann zeta function, then as a last step you take s=-1. Tony explains that method here: http://www.nottingham.ac.uk/~ppzap4/response.htmlDirichlet regularization: This takes zeta continuation one step further and can be used for series of the form sum f(n)n^-s. This is how I got an answer sum((1-n^3)/n^2) = (1 + 2pi^2)/12 in the example sum(n) + sum((1-n^3)/n^2) = pi^2/6 = sum(1/n^2).As you can see, convergence isn't synonymous with sum. Convergence is just one method of summation out of many. But the idea is that these different methods fit together so it makes sense to call these methods sums.
Ramanujans Pi Formula singingbanana 2016-04-30 | The second video in a series about Ramanujan. Continuing the biography and a look at another of Ramanujan's formulas. This one involves Ramanujan's pi formula.
Ramanujan: Knowing The Man Who Knew Infinity singingbanana 2016-04-29 | The first video in a series about Ramanujan. A bit of biography then a look at one of Ramanujan's formulas. This one involves infinite nested radicals.Here's a more formal look at Ramanujan's nested radical http://ac.els-cdn.com/S0377042704001906/1-s2.0-S0377042704001906-main.pdf?_tid=e42b3094-0e17-11e6-ba9c-00000aab0f6c&acdnat=1461940825_1ca334466b1da553a2c87ab24086f93dThe proof is quite hard without observation. (Let me know if the link doesn't work.) [14-10-2020 THE LINK CURRENTLY DOESN'T WORK, I'LL TRY AND FIND A LINK THAT DOES]
Married Problem survey result singingbanana 2016-03-28 | Here is my solution video: youtu.be/pTpBcxIg-9AHere is the solution and results from the problem by Alex Bellos theguardian.com/science/2016/mar/28/did-you-solve-it-the-logic-question-almost-everyone-gets-wrongAnd check out more puzzles from Alex Bellos here theguardian.com/science/series/alex-bellos-monday-puzzle
Married Problem with solution (plus extra problem) singingbanana 2016-03-28 | Jack is looking Anne, but Anne is looking at George.Jack is married, but George is not.Is a married person looking at an unmarried person?A) Yes B) No C) Cannot be determinedHere is the solution and results from the survey on Alex Bellos' Puzzle blog:theguardian.com/science/2016/mar/28/did-you-solve-it-the-logic-question-almost-everyone-gets-wrongA short video of my reaction to the survey youtu.be/-iSxls2f0QoAnd check out more puzzles from Alex Bellos here:theguardian.com/science/series/alex-bellos-monday-puzzle-----Some interesting variations have been suggested. These include:1. A coin is flipped three times. The first is a head. The third is a tail. Is a head ever followed directly by a tail? This is the same problem but it seems easier to solve.2. Jack is sat next to Anne. Anne is sat next to George. Jack is married, George is not. Is a married person sat next to an unmarried person? This makes the relations bidirectional and seems easier to solve.3. We can also extend the married problem to as many people as we like. If M is married, and U is unmarried, then M sees ? sees ? sees ? sees .... sees ? sees U. Is a married person looking at an unmarried person? Well, at some point there is a transition from Married to Unmarried and so the answer is yes.4. The dictionary definition of unmarried is "not married". This includes everything that is not married, i.e. widowed, divorced and single. Whatever you definition of marriage this makes married and unmarried a binary choice. However, here is another suggestion using "alive" and "dead" "I found a picture from 50 years ago of my grandmother and grandfather on a ski trip together. In the picture, my grandfather is looking at the ski instructor, and the ski instructor is looking at my grandmother. My grandfather is still alive but my grandmother died last year. Is someone in the picture who is still alive looking at someone who is now dead?"(waits for people to argue the ski instructor might be a zombie)----Here are some other observations I want to add after seeing some questions in the comments:1. You can construct an explicit example to whether an irrational number to an irrational power can be rational if we know transcendental numbers exist. (That means it's not a solution to a polynomial with rational coefficients). Then we can take x transcendental and x^log_x(2) = 2. x is irrational, and log_x(2) is irrational, otherwise we have x^(p/q) = 2 and so x^p = 2^q contradicting the transcendentality of x. So e^ln(2) = 2 would do it.2. Can a rational number to an irrational power be rational? Indeed, for example 10^log(2) = 2. It is easy to show log(2) is irrational, otherwise we would have 10^(p/q) = 2 and so 10^p = 2^q which violates unique prime factorisation.----A final word to those who argue that Anne might not be a person (a dog etc). If that had been the answer then the question would have been a riddle rather than a logic problem. A riddle invite you to guess information not presented in the question, or hidden in the question. A logic problem asks you to make deductions from the information presented in the question only, and specifically not using information from outside the question. It is always possible to circumvent logic problems with left-field answers, but that isn't really in the spirit of the question and would miss the point of the exercise. So it's important to know that you are answering a logic problem, and how they work, before you start.Since Alex and I run maths blogs we will only present logic problems and never riddles.-----If you have read the description you are part of an exclusive club. Welcome. Your challenge is to work a Steven Spielberg movie title into a comment below.
Married Problem - Will 80% of people get this problem wrong? singingbanana 2016-03-28 | Jack is looking Anne, but Anne is looking at George.Jack is married, but George is not.Is a married person looking at an unmarried person?A) Yes B) No C) Cannot be determinedUnfortunately the survey is now closed. Here is the question from Alex Bellos theguardian.com/science/2016/mar/28/can-you-solve-it-the-logic-question-almost-everyone-gets-wrongMy solution video is here youtu.be/pTpBcxIg-9AMy reaction to the survey result is here youtu.be/-iSxls2f0QoSurvey results and answer from Alex Bellos is here theguardian.com/science/2016/mar/28/did-you-solve-it-the-logic-question-almost-everyone-gets-wrongAnd check out more puzzles from Alex Bellos here: https://www.theguardian.com/science/s...
Maths Puzzle: The self descriptive number solution singingbanana 2016-01-09 | Solution to the self descriptive number puzzle youtube.com/watch?v=K6Qc4oK_HqYHere is the original version of the proof presented in the video http://singingbanana.com/self%20describing%20proof.gif[I've added some notes to the proof I presented in the video below]Thanks to kshitij Ranganekar for the suggestion.Katie Steckles: youtube.com/user/st3cksMatt Parker: youtube.com/user/standupmathsThat puzzle again:I have a ten digit number. The first digit tells me how many zeros are in the number. The second digit tells me how many ones are in the number. The third digit tells me how many twos are in the number, and so on until the tenth digit which tells me how many nines are in the number. What is my number?Proof:We have an n-digit number a_0a_1a_2....a_n-1, where a_i is the tally of the value i in the number. Also note the sum of the a_i will be n.a_0 is greater than 0. It cannot equal 0 without being a contradiction.Let the rest contain p non-zero valuesSo the tally of non-zero values is a_1 + a_2 + a_3 + ... + a_n-1We have also said we have p+1 non-zero values.So a_1 + a_2 + a_3 + ... + a_n-1 = p+1In other words, we have p non-zero values summing to p. This means most of the digits are 1, 0 and a 2. Except for a_0.If a_0 is greater than 2, write j for a_0. Put 1 for a_j, which forces a_1=2 and a_2=1. Put in j zeros to make a j+4 digit number which had a sum of values which is also j+4. So we are done. No other values can be added without breaking it.In other words, these are numbers of the form j2100....1000So we have;3211000421010005210010006210001000721000010008210000010009210000001000If a_0 less than or equal to 2 then the length of the number will be short. This is because the sum of a_i is equal to n, and all values are less than 2 - meaning a_3, a_4, a_5 etc equal 0.This gives the sporadic cases:1210202021200That is a complete list of all self-describing numbers in base ten.
Maths Puzzle: The self descriptive number singingbanana 2016-01-05 | I have a ten digit number. The first digit tells me how many zeros are in the number. The second digit tells me how many ones are in the number. The third digit tells me how many twos are in the number, and so on until the tenth digit which tells me how many nines are in the number. What is my number?
2016 - The start of a new (dozenal) century singingbanana 2015-12-29 | 2016 is the beginning of a new century - if you count in twelves. If you do that then the year 2016 becomes the year 1200. Counting in twelves is known as the dozenal system, and a "century" in dozenal is 144 years, better referred to as a biquennium.To those who point out a new century, or biquennium, doesn't start till 1201; this was deliberately not included in the video because I think a lot of people already know that. Yet it is the rollover of the digits that is commonly referred to as the turn of a century and there is no better phrase for that. I remember the dreadful bores who said the same thing about the millennium not starting till 2001. I would not want to be one of those people.JAN 04: number of people making the comment about 1201: twenty-three. (Approximately 4%).For more information about dozenal visit dozenal.org
Cheryls Birthday Problem - It depends on your point of view singingbanana 2015-04-15 | Cheryl's Birthday problem was a question in a recent Singapore Maths Olympiad. Here is the problem:You have Albert, Bernard and Cheryl. Cheryl says "my birthday is one of these ten dates"May 15 May 16 May 19June 17 June 18July 14 July 16August 14 August 15 August 17She gives Albert the month of her birthday. And Bernard the number.Then the following conversation occurs:1) Albert: I don’t know when the birthday is, but I know Bernard doesn’t know too.2) Bernard: At first I don’t know when the birthday is, but now I know.3) Albert: Then I know the birthday too.From that information, work out Cheryl's birthday.[The English is the original, slightly dodgy, English of the original question].---Here is Alex Bellos' post on the problem: http://www.theguardian.com/science/alexs-adventures-in-numberland/2015/apr/13/can-you-solve-the-singapore-primary-maths-question-that-went-viralAnd Alex solution: http://www.theguardian.com/science/alexs-adventures-in-numberland/2015/apr/13/how-to-solve-albert-bernard-and-cheryls-birthday-maths-problemAnd here is my reply, concentrating on that alternative answer: http://www.theguardian.com/science/alexs-adventures-in-numberland/2015/apr/15/why-the-cheryl-birthday-problem-turned-into-the-maths-version-of-thatdress---Something from the article I wrote, which I forgot to say in the video, the alternative solution completely changes the nature of the problem, because the alternative answer can be worked out from the first two statements only. The expected answer does need all three statements.---A bit more about that third answer. I think the reader is treating it as just a mystery date (made up of a month and number) which they have to work out. This fails to take into account that Albert has the month only, and Bernard the number only.Oh, another thing I forgot to say: the third answer is also what you get if you mix up the roles of Albert and Bernard - in which case the first statement is information about the date.---More about the idea of "knowledge":Let p be a statement, it's either true or false. Logic is the maths of these statements. We can modify them, for example we can take not(p) which has the opposite truth value. We can add and multiply statements as well using "or" and "and". Another action is knowledge(p) which might be true or false, but is not the same as p, and the following deductions are different.So if we remove the story it becomes the difference between reading the first statement as "Bernard doesn't know" and reading it as "Albert knows Bernard doesn't know". And what you can then deduce from that.For definitions and properties of knowledge in logic, here is the wikipedia pagehttp://en.wikipedia.org/wiki/Epistemic_modal_logic---The other point I forgot to make, if this was a serious question in mathematical logic it would be written symbolically in formal logic - and there would be no ambiguity. However, this is not meant to be high level maths, it's a puzzle. So by making it accessible, and dressing it up as a puzzle, ambiguity was accidentally introduced. I sympathise with that! Writing puzzles is hard.Daniel Gjörwell in the comment made a very good point: "The intention of the author was KNOWS as REALIZE. If the first statement would have read: "I don't know but I realize that Bernard doesn't know too"; then I would have got [the correct answer]"---If you are reading this you are in a special group of highly intelligent people who read the description in the video. Well done you. Your mission, if you choose to accept it, is to work a fish related pun into a comment without other people noticing. Good luck!
The Imitation Game Reaction singingbanana 2014-11-18 | For more of my thoughts read The Imitation Game FAQs here http://aperiodical.com/2014/11/an-alan-turing-expert-answers-your-the-imitation-game-questions
Lorenz: Hitlers Unbreakable Cipher Machine singingbanana 2014-09-07 | Many people have heard of Enigma before, the code machine used by Nazi Germany to send secret coded messages. Yet, some very clever code breakers were able to break that code and read those messages!But there was another cipher machine used by the Germans in WWII called the Lorenz machine, and this machine was even more difficult than Enigma, and was used by the top level of the Nazi Party.However the code breakers at Bletchley Park broke this code too, and could read secret messages from people like Adolf Hitler himself!It was mathematician Bill Tutte who discovered the breakthrough that allowed the Lorenz code to be broken. On the 10th of September 2014 a new memorial to Bill Tutte is unveiled in his hometown of Newmarket. Have a look at the Bill Tutte Memorial website at http://billtuttememorial.org.uk
Cubic Curve Calculator singingbanana 2014-05-04 | Turn a cubic curve (with no x^2 term) into a calculator for sums, multiplications and square-roots.Simply draw a line across the curve so that it intersects at three points, a, b and c. We may now label the curve in a way such that a + b = c or ab = c. I used the cubic curve y = x^3 - 3x, print out the curves below:cccalculator for addition: http://singingbanana.com/cccalculatoraddition.pngcccalculator for multiplication: http://singingbanana.com/cccalculatormultiplication.png(Or, if you prefer the curves with the axes:http://singingbanana.com/cccalculatoradditionwithaxes.pnghttp://singingbanana.com/cccalculatormultiplicationwithaxes.png)Here's an alternative scale for multiplication so it's more like a slide rule. Magnitude just requires common sense http://singingbanana.com/cccalculatormultiplicationaltscale.pngThis was pointed out to me when my friend Colin Wright shared this blog post http://cp4space.wordpress.com/2012/08/29/elliptic-curve-calculatorBut I think the description in this video is simpler, which is based on this post from 1953 by Richard K. Guy http://singingbanana.com/a%20single%20scale%20nomogram.jpgThanks to Phil Ramsden for doing the curve in Mathematica for me, and for reminding me how roots work.
2048 Induction Extra singingbanana 2014-04-10 | An extra bit from our 2048 video on Steve Mould's channel.Full video here: 2048 strategy and maths http://youtu.be/OO4tA5i7X9gPlay 2048 here http://gabrielecirulli.github.io/2048----------------------------A little bit on the maximum achievable tile. I said here you need k free cells to achieve 2^k and the maximum achievable tile is therefore 2^16. This was assuming we only generate 2-tiles.I also show a formula to calculate the total score of making the 2^k tile which is (k-1)2^k.That means the maximum score is when you fill the board from 2^16 to 2, which is sum_i=1^16 (i-1)2^i = 1,835,012.If we include generating 4-tiles as well, you can actually go one step further and achieve the 2^17 tile. In that case the maximum score would be when we fill the board from 2^17 to 2^2. If I did that using 4-tiles only then that would double the maximum score from 1,835,012 to 3,670,024.If we use 2-tiles only, with a few exceptions, then I reckon the maximum score is sum_i=2^17 (i-1)2^i - 16*4 = 3,932,100.(You need to subtract 4 sixteen times because I need to generate sixteen 4-tile for free to fill the board from 2^17 to 4).
Building Houses Solution singingbanana 2014-03-28 | Problem video http://www.youtube.com/watch?v=YMkziQhJkmMImagine the line from 0 to 1. Write a sequence of points such that the first two points occupy different halves, the first three occupy different thirds, the first four occupy different fourths, and so on. What is the longest sequence you can write? Building Houses game by Christian Perfect: http://christianp.github.io/building-housesFollow Christian Perfect on twitter twitter.com/christianpTwo proofs of the irregularity of distributions (18 point problem) are listed below. But be warned, neither of them are particularly easy reads.First is the proof as demonstrated in this video:Warmus, M. "A Supplementary Note on the Irregularities of Distributions." J. Number Th. 8, 260-263, 1976.http://bit.ly/1j7vgcRSecond is a different proof which uses Farey sequences:Berlekamp, E. R. and Graham, R. L. "Irregularities in the Distributions of Finite Sequences." J. Number Th. 2, 152-161, 1970.http://www.sciencedirect.com/science/article/pii/0022314X70900156
Building Houses Problem singingbanana 2014-03-26 | Imagine the line from 0 to 1. Write a sequence of points such that the first two points occupy different halves, the first three occupy different thirds, the first four occupy different fourths, and so on. What is the longest sequence you can write?Building Houses game by Christian Perfect: http://christianp.github.io/building-housesFollow Christian Perfect on twitter twitter.com/christianpSolution video now at http://youtu.be/Y5q-MzS-lPE
New Wikipedia sized proof explained with a puzzle singingbanana 2014-02-24 | A new mathematical proof was in the news this week, which partially solves the Erdos Discrepancy Problem. The proof was described as "bigger than Wikipedia". I attempt to explain the problem using a puzzle which you can try at home.The puzzle was my idea to explain it to you - that's not really how the problem is stated.New Scientist: http://www.newscientist.com/article/dn25068-wikipediasize-maths-proof-too-big-for-humans-to-check.html#.UwdX6_l_uSpThe Independent: http://www.independent.co.uk/life-style/gadgets-and-tech/news/computer-cracks-erds-puzzle--but-no-human-brain-can-check-the-answer-9137097.htmlTim Gowers Blog: http://gowers.wordpress.com/2014/02/11/recent-news-concerning-the-erdos-discrepancy-problemPolyMath Project: http://michaelnielsen.org/polymath1/index.php?title=The_Erd%C5%91s_discrepancy_problemThe paper: http://cgi.csc.liv.ac.uk/~konev/SAT14/sat14.pdfBoris Konev's webpage: http://cgi.csc.liv.ac.uk/~konev/SAT14One page proof that a sequence of twelve has discrepency of 2 http://www.dpmms.cam.ac.uk/~ardm/erdoschu.pdf
There are always two opposite points on the Earth with the same temperature singingbanana 2014-01-26 | There are always two points on opposite sides of the Earth with the exact same temperature. And we can prove that.Temperature changes continuously. If a and b are on opposite sides of the equator and D(a) = T(a) - T(b) is positive, then D(b) = T(b) - T(a) is negative. That means there must be some point x on the equator where D(x) = 0. At that point the two opposite sides are the same temperature.Mathematicians call this the Intermediate Value Theorem which means if there is a continuous function that changes from of a positive value to a negative value (or the other way around) then it must, at some point, pass through zero.Here's a great simulation by mrFryd4y for how this might look http://youtu.be/JzV3ombeXxYInformation: The Intermediate Value Theorem: http://en.wikipedia.org/wiki/Intermediate_value_theoremThe general result, The Borsuk-Ulam Theorem: http://en.wikipedia.org/wiki/Borsuk%E2%80%93Ulam_theoremA lovely full explanation of why there is a point on Earth where temperature *and* pressure are the same http://mathforum.org/library/drmath/view/54746.htmlIn laymans terms, continuous means you can draw the graph without taking pen from paper. It can be a smooth or jagged line, as long as the line isn't a strict vertical jump from one value to another.Bless you, but some people are really over-thinking the definition of temperature. How is temperature measured by meteorologists? They stick a thermometer in the air http://en.wikipedia.org/wiki/Temperature_measurement#Surface_air_temperatureTemperature is regarded to be continuous due to the second law of thermodynamics which causes the transfer of heat between areas of different temperature http://en.wikipedia.org/wiki/Heat_transfer The temperature gradient might be steep, but not discontinuous. This is not an extraordinary claim.No one reads the description anyway. If you have you are part of an exclusive club. If you have read the description let me know by typing the secret word, "pineapple", in the comments. In return I will give you my respect, and maybe a thumbs up or something.Pineapple Count (07-09-16): 186 pineapples.
The Riemann Hypothesis singingbanana 2014-01-17 | The Riemann Hypothesis is one of the Millennium Prize Problems and has something to do with primes. What's that all about? Rather than another hand-wavy explanation, I've tried to put in some details here. Some grown-up maths follows.More information: http://www.claymath.org/publications/riemanns-1859-manuscript-----------------CORRECTION: The functional form of the zeta function is a reflection around the point 0.5, not the line x=0.5. But you can think of that as a reflection in the x-axis followed by a reflection in the line x=0.5. For example, if we start with zeta(x + iy) and reflect it in the x-axis we get zeta(x - iy). Reflect that in the x=1/2 line we get (junk)*zeta(1 - x - iy). And reflect that in the x-axis again we get (junk)*zeta(1 - x + iy). When zeta(x + iy) = 0 so are all the others. The zeros then look like this http://en.wikipedia.org/wiki/File:Zero-free_region_for_the_Riemann_zeta-function.svg That's four zeros for the price of one.The zeros are still symmetric around the x=1/2 line. The reflection of zeta(x + iy) is (junk)*zeta(1 - x + iy). In the video I mixed up zeta(1 - x + iy) with zeta(1 - x - iy). But the point is the same, the error in the Prime Number Theorem is minimised if all the zeros lie on the line x=1/2.---------------------Some people have asked about my final claim that you don't get the prize for a counterexample. Here is the rule from the Clay Institute:"if a counterexample is proposed, the SAB will consider this counterexample after publication and the same two-year waiting period as for a proposed solution will apply. If, in the opinion of the SAB, the counterexample effectively resolves the problem then the SAB may recommend the award of the Prize. If the counterexample shows that the original problem survives after reformulation or elimination of some special case, then the SAB may recommend that a small prize be awarded to the author. The money for this prize will not be taken from the Millennium Prize Problem fund, but from other CMI funds."In other words, there are no guarantees either way.
The Maths of Star Trek singingbanana 2013-06-08 | The science of Star Trek has, in the past, been discussed in great detail - but what about the maths of Star Trek?Star Trek: The Original Series contains a surprising amount of mathematics, including; the probability we are alone in universe; a paradox that upset 20th century mathematicians as well as 23rd century androids; and the most important question of all -- when on a dangerous away mission, does the colour of your shirt really affect your chances of survival?Mathematician James Grime is joined by the relatively normal Stuart Laws as they discuss new vs old Star Trek.More from Stuart Laws http://youtube.com/turtlecanyoncomedyhttp://twitter.com/thisstuartlawsMore from James Grimehttp://youtube.com/singingbananahttp://twitter.com/jamesgrimeFull article by James Grime here:http://aperiodical.com/2013/04/the-maths-of-star-trek-the-original-series-part-iOuttakes:Maths of Independence Day: http://www.youtube.com/watch?v=_61yR776Up0
