The Korean king's magic square: a brilliant algorithm in a k-drama (plus geomagic squares)

[@thicknavyrain]

You know it's going to be a great video when you're only one minute in and you've already seen three excellent results.

[@landsgevaer]

You know that it is going to be a great video when it is Mathologer, you have just started the first second of the video, and you've already seen three comments pointing out how exceptional the video is. I am confident that I am going to agree with your assessment.

[@craigdupree1687]

You know it's going to be a great video when it's posted by Mathologer.

[@neonblack211]

you knoiw its going to be a great video when its mathologer

[@gracenc]

You know it's going to be a good video when the intellectuals watching the channel make idempotent statements just to emphasise how great the video is. No judgment, @neonblack211 :)

[@kesotone]

KZbin needs a super like button for this level of content. Bravo Mathologer!

[@jonpopelka]

There’s a reason that the all-time mathematical greats like Euler, Ramanujan, Laplace, and Fermat were fascinated by magic squares and other patterns! It’s the knack and habit of recognizing those underlying structures which led them to some of the greatest insights and advances in mathematical history. THIS is why it’s so important to teach our kids more than rote memorization of numbers, facts, tables, and theories; we need to teach them how to see them as patterns which lead to other patterns within even further patterns. People with the gift of innate intuition about patterns are the people who change the world.

[@Mathologer]

What's also interesting is that a LOT of people who are into maths look down on things like magic squares not realising how important these little things are in the grand scheme of things :)

[@islandbuoy4]

my gift to the world is pointing out the OBVIOUS associations between the proof the fake Pitagoras used for the right triangle theorem, the 3x3 Lo Shu magic square, the 5x5 Rotas Sator palindrome and the 12,000+ year old 2D CHIRAL swastika ... but the ignorant do not want to acknowledge this insight due to WW2 crimes committed against humanity ... the truth is genius, hidden in plain sight and a bitter pill to swallow ....

[@zoomlifecoach]

@@Mathologerhi mathologer I love your videos thank you for making such great content

[@user-yc3fw6vq5n]

This is why 바둑/圍棋/囲碁 is important to teach to al children.

[@mlmimichaellucasmontereyin6765]

Absolutely rite on! Indeed, if we just taught kids such integral fundamentals + doing "Japanese" addition (etc.) with chopsticks (etc.) then we'd have no deficit of STEM whiz kids over here in the [not] greatest nation of all. Sigh... 😞

[@ytrichardsenior]

Dear Mathologer. I can see how much work goes into these video's, but please never stop doing them. When I was a child I remember asking myself why we didn't have a TV channel that just showed educational programs. You (and a handful of others) make youtube what it can and should be. Thank you.

[@KNemo1999]

Oh, that's what an empty lunch box means. I just thought my mom was a little absent-minded. But this makes so much more sense.

[@TheShadowOfMars]

Empty lunchbox = you don't need any food = you're a dead man

[@frankharr9466]

I hope she wasn't too disappointed and decided better of it.

[@nhuthuynhminh2603]

💀

[@freewheelburning8834]

either absent-minded or just a socialist either way I think Karl Marx would approve

[@Axman6]

Oh no…

[@DarknessGu1deMe]

18:28 mentions the general rule of filling odd magic square we were taught in China in primary school. The rules was written as so: 一居上行正中央，依次斜填切莫忘 上出框时向下放，右出框时向左放 排重便在下格填，右上排重一个样 Translation: 1. put 1 in the middle of the first row 2. fill next consecutive numbers diagonally 3. when going out on the top, put into the bottom row 4. when going out on the right, put into the left column 5. if the square is occupied, put into the square below 6. if going diagonally on the top right, the same rules applies

[@Mathologer]

Very nice. The Chinese text looks like it may rhyme. Does it?

[@DarknessGu1deMe]

@@Mathologer It does rhyme! and it is deliberately written in the form of seven-word-poem (much like the solution poem to the Chinese remainder theorem). On this same topic, there's another, much more ancient (and famous) "poem" on just the 3x3 magic square: 九宫之义，法以灵龟，二四为肩，六八为足，左三右七，戴九履一，五居中央 Translation: """ The way to fill a 9-square palace, is to imagine a turtle (back): 2 and 4 as the shoulder, 6 and 8 as feet, 3 on the left, 7 on the right, 9 as hat, 1 as shoe, and 5 in the middle """ This text came from an ancient manuscript, that says these numbers/pattern comes on a turtle, and it's a sign of miracle. I guess that's part of what you said in the video: some people think magic square is truly magical.

[@Mathologer]

@@DarknessGu1deMe That's great. Thank you very much for sharing this with me :)

[@howareyou4400]

@@DarknessGu1deMe There is a much simpler version by 杨辉(1127~1279) 九子斜排，上下对易，左右相更，四维挺出。 Which translates roughly to: 1. arrange the 9 numbers "along the diagonal direction" (3 X 3 tilted square) 2. switch top and bottom 3. switch left and right 4. "stick out" the 4 corners.

[@nidalapisme]

@@DarknessGu1deMe I think the poem ever showed up in 1994 chinese tv series "Legend of the Condor Heroes" when the main character was trying to solve the 3x3 magic square puzzle to open up a door to a secret place. The female character solved the puzzle whilst citing the poem after she saw a turtle nearby. I remember she put 5 in the middle square as her last move to complete the puzzle. cmiiw.

[@nanamacapagal8342]

8:24 The magic constant is sum of all numbers in the square divided by the number of rows. For those looking for a concise formula: it should be (n²(n²+1)/2)/n or simply n(n²+1)/2. Plugging in 33 will give us the magic constant of 17985. 18:48 Go up-right one space. If you exit the board, wrap around to the other side. If you run into a space already filled in, drop straight down one space instead.

[@Mathologer]

That's it :)

[@karmageddon9136]

You're right on. You know the sum of each row, but how do you figure out the individual numbers? This video showed an interesting technique.

[@dan-florinchereches4892]

I was using another way of reaching the solution. Is there a reason to add every single number together? Since we have an odd number of pieces intuitively the middle number multiplied by the number of rows is going to be the only answer to the question pertaining the sum. This is immediately obvious because if a number larger than this was the answer then there necessarily exists another column/row which is lacking. Average value is (n^2+1)/2 and we can multiply by n, reaching the same conclusion. I love your algebra though.

[@nidalapisme]

I think it's called the Siamese method for solving magic square in that fashion. cmiiw.

[@riccardosarti3234]

Conjecture: the number at the center of an odd magic square is always the middle number in the list, i.e. the average number in each cell i.e. (n^2+1)/2 (which is an integer as n is odd). Can you prove it? If this is false, can you find a counter-example? Also, given n odd, how many magic squares do exist (with the equivalence relation given by the trivial horizontal/vertical reflections and 90° rotations)?

[@jeff__w]

I can't believe you mentioned _Tree With Deep Roots._ Knowing next to nothing about Korean or Korean, I stumbled onto that drama around 2011 and was intrigued by the story of Hangul woven into it. I learned to read it and eventually became somewhat proficient in Korean, all because of a random K-drama! Funny how those things work.

[@Mathologer]

I liked the drama but not the way it ended :)

[@jeff__w]

@@Mathologer oh, yeah, I forgot the ending. By around the midpoint or so it began to get a bit ridiculous, I thought, but I did watch it till the end. (And it _did_ get me into Hangul and Korean quite unexpectedly.) _Edit:_ The original comment should read “Korea or Korean.” Oops!

[@johnchessant3012]

Neat proof of an amazing result! I'll have to watch that show because it sounds like the best use of math in a film / tv show ever. Also, nice pi shirt!

[@ahcuah9526]

It's not a T-shirt, it's a Π-shirt! (Is there something encoded in blue vs white dots?)

[@jonathanrichards593]

@@ahcuah9526 It looks as if every third digit is in white; I'm unaware of any significance in that sequence of numbers, but with Mathologer I just *never* know!

[@MisterMajister]

One of your finest videos in my opinion, I really enjoyed it!

[@Mathologer]

Something different. Glad you liked it :)

[@luxinvictus9018]

this channel is absolutely fantastic!

[@WonSikShim]

여기서 한국 드라마를 보게 될 줄은 꿈에도 몰랐네요. Never expected to see K-drama in this channel!!

[@Mathologer]

There are a couple of other K-dramas that would be worth covering mathswise. In particular, Melancholia has got some good stuff :) en.wikipedia.org/wiki/Melancholia_(TV_series)

[@박부성-e5j]

I never expected I could see a Korean drama in your channel. :-)

[@richardschreier3866]

Thanks for crafting such a lovely video. The hours spent in photoshop yielded a beautiful result, and the delightful subject matter made the video a joy to watch.

[@kk-wl2yg]

The level of video and animation is amazing. This is a huge and talented work! СПАСИБО БОЛЬШОЕ 👌

[@jacoboribilik3253]

The proof sketch was brillianty displayed. That "aha" moment you experience when it finally sinks in is priceless, almost addictive.

[@FrankHarwald]

This is honestly the best video on magic squares on KZbin & the best comprehensible video on the topic I've ever seen. SUPER! B)

[@Carlos-qz7ul]

I'm blown away by the amount of work necessary to build such a video, besides the knowledge and the insight needed 🙉

[@cparks1000000]

It's always exciting when these come out!

[@JCOpUntukIndonesia]

And turns out, there is a 3D equivalent for magic squares called a magic cube. Going beyond 3D, we have magic hypercube. How can I never hear of this before? I love that the well-known classic problem has a lot of different variations to tinker about. Especially when each has a unique approach to the problem. The wonder of math always amazes me.

[@FLScrabbler]

Reminds me of wordsquares that are cubed. For instance "CUBE" can be squared like this: CUBE UGLY BLUE EYES But then each of those other three words can be made into a wordsquare as well, which is then considered to have cubed the word "cube": UGLY GLUE LULL YELP BLUE L### U### E### etc.! (I'm sorry, I have forgotten the rest...)

[@SgtSupaman]

@@FLScrabbler , I've tried working out the rest of this cube from what you've provided, but I am unsatisfied with the set up. The cube properties applied to the used words duplicates "lull" as stemming from the L in "ugly" and the L in "blue", and then duplicates "yelp" as stemming from the Y in "ugly" and the Y in "eyes" (I don't know if it is possible, but I would think a cube with unique words at every step would be more interesting). Then that sticks you with a 'ul' to start a word with, which is a tough fill. Here is the final result I came to: CUBE UGLY BLUE EYES UGLY GLUE LULL YELP BLUE LULL ULTS ELSE EYES YELP ELSE SPED

[@FLScrabbler]

@@SgtSupaman Very nice! Well done..!

[@MichaelDarrow-tr1mn]

because if you try to make a magic cube it doesn't work.

[@PC_Simo]

@@SgtSupaman Very nice. Also; if you allow proper names, then one possibility would be to replace ”ULTS” with ”ULAM”, and ”ELSE” with ”ELMO”. 😌👍🏻

[@yomrcheng]

Your animations were amazing, I would have never been able to visualize this without your excellent animations. I was truly impressed, I learned a lot thank you!

[@johnchessant3012]

My favorite fact about the Dürer magic square: it was completed in the year 1514, and in the middle of the bottom row you can see it says 15 14. The only similar example I know of is the Basel problem, posed in 1644 whose solution is pi^2/6 = 1.644... However, sources differ on whether it was first posed in 1644 or 1650.

[@Mathologer]

Actually the Durer square has a couple of other magic properties. In particular the four 2x2 blocks in the corners and the one in the middle also all add to the magic constant. Sadly these extra properties are not replicated by the geomagic counterpart :)

[@FLScrabbler]

@@Mathologer This property is the reason why Dürer was able to put the year in middle of the bottom row after generating the MS by inverting the diagonals: The two central columns were swapped without affecting the magic constant...

[@thek3nger]

I followed a lot of math youtuber for years. Mathologer is really the only one that consistently blows my mind.

[@Mathologer]

Mission accomplished :)

[@SmartHobbies]

I love seeing Magic Squares used in Sudoku variants. Thanks for sharing.

[@wegwang7795]

It really makes me glad to see more China-related or Eastern Asia-related videos here!

[@NoobsDeSroobs]

This might be the most mindblowing piece of math I have ever seen. I had problems focusing on the rest of the video because my mind was reeling from the extreme and simplistic beauty of this structure!

[@alwysrite]

you are a genius and such a good teacher who is willing to share to the world, thankyou

[@agranero6]

Ok. Im am a huge fan of k-dramas. And now you gave me a double reason to watch this one.

[@Mathologer]

Maybe also watch Melancholia. That one has a lot of nice math(s), too :)

[@agranero6]

@@Mathologer Thanks for the tip.

[@kmjohnny]

So this is why sudoku named one of their optional rule as magic square.

[@omrizemer6323]

I was shaking my head with disbelief half the time while watching this video. Amazing

[@_abdul]

Love the Pi Tshirt, Nice Visualisation of Pi.

[@Quazgaa]

I was proud of myself for decoding your shirt 🧐 Not everyone can be a genius

[@78Mathius]

Great, now I have so many questions I want answered. Shapes: All shapes must be constructed from an integer number of squares connecting to each other across full edges. N=side length C=the magic number. 1. For what N is there a constant shape that is a square? 2. Is there an N with a set of shapes that only add to a square? 3. If a square can be built, it must have a C with a square root equal to or larger than the triangular number of N. Is there an N where all squares are possible once the minimum square size is reached? 4. Earliest N with a set of consecutive integers that can build a square? 5. So many more.

[@alokaggarwal6859]

The diagonal construction of magic squares and the geo magic squares were both superbly presented... Really interesting!

[@Mathologer]

Thank you very much, glad you liked it :)

[@alexpotts6520]

In case you're wondering, the dots on his shirt are the digits of pi

[@moin85]

Q: What’s the only thing better than a Mathologer video? A: Another Mathologer video

[@ZER0--]

Brilliant. I will never forget how to make magic squares. Love it. I don't know if it will come in handy but I love it.

[@bloodypommelstudios7144]

I discovered a similar trick to the king's magic square by taking an x-Sudoku and adding 9 * (n-1) from a 3x3 magic square to each cell in the corresponding region. It works because Sudoku grids are Latin squares and Latin squares are just magic squares with repeating numbers so just like your example it's taking another pattern with consistent sums and adding them together to make each number unique. You can use a similar technique to iterate magic squares creating any square of a length of power 3 (or any length multiplied by another) and it even works for magic cubes, I've checked up to length 125 by iterating a 5x5x5 twice. This video got me thinking you could construct a 4x4 latin square using playing cards and add 0, 4, 8 or 12 to each suit to create a magic square. It works.

[@Piffsnow]

Mathologer never disapoints. I'm so glad to know this channel.

[@cameronmyron5776]

What is interesting is that you can also turn any magic square into a new one by adding a non-zero integer constant to every square (the summations will change by the side length times the non-zero integer constant). Additionally there is probably some way to generate a new magic square by taking the modulus of every square (need to be careful about creating repeating numbers with certain values for the modulus. Edit: This actually might not be possible. I don’t have an example that it would work without causing a duplicate value).

[@Mathologer]

Yep. There are a couple of other transformations that turn magic squares into new magic squares. Have a look here for a summary: en.wikipedia.org/wiki/Magic_square#Transformations_that_preserve_the_magic_property

[@cameronmyron5776]

@@Mathologer Since I made the comment I was investigating the group structure of flipping the square across the different axes and until I saw the wiki article you provided I completely missed that the group was isomorphic to D8 and could be simplified from 4 operations (a flip across the 4 different axes) to 2 (90 degree turn clockwise and a flip across one of the axes). Thank you for sharing the wiki article.

[@CalvinsWorldNews]

I saw this done as a trick by a teacher, who generated magic squares to a set numbers on demand, in reality they were just doing some maths tricks in their head. Very impressive today but as a child it was like genuine sorcery.

[@SaturnCanuck]

Thanks again for anther great video. As I said i watch these on a nice Sunday afternoon to relax. :)

[@Peccomment]

Your videos make may days… everytime… thank you.

[@tom7]

Exceptional storytelling!

[@davidalbrecht9117]

Great video. I really loved learning about Bachet's algorithm and geomagic squares. I was also reminded of orthogonal latin squares when you described Bachlet's algorithm.

[@Mathologer]

I was actually thinking of mentioning orthogonal Latin squares but in the end decided against it :)

[@jeanconstant]

Fascinating. Thank you for the clarity of the demonstration & the inspiration that comes with it!

[@Mathologer]

Glad you enjoyed it!

[@stevereckamp3152]

Wanted to generalize the formula for the magic sum. for an n x n magic cube, I noticed that the sum of the "columns" are just Σk (k=1..n) and the sum of the "rows" is nΣk (k =1..n) - n² (the equivalent of nΣk (k =0..n-1) . That means the sum of all of those unique values (and the magic sum of the square) = Σk (k=1..n) + nΣk (k =1..n) - n² = (n + 1) nΣk (k =1..n) - n² substituting n/2(n+1) for Σk (k =1..n) = n/2 (n+1)² - n² And some math autopilot = n/2 (n² + 2n + 1) - n² = n³/2 + n² + n/2 - n² magic sum of n order magic square = *n/2 (n² + 1)* S₃₃ = 33 * (33² + 1)/2 = 33 * (1089 + 1)/2 = 33 * 1090/2 = 33 * 545 = *1795* _Also note it will always be an integer because n is odd so n² is odd and (n² + 1) is even and divisible by 2._

[@daviddilaura4614]

As always: beautifully and clearly presented.

[@n3cr0he11]

You here use some magic to put this video together so nicely! 😍

[@garybartnik1509]

Lovely video on the magic square by a gentleman who is NEVER SQUARE!!! Gary in dreamland. Have a nice dream!!🙂☁️☁️☁️⛅💫🌟🌟🌟🌟🌟👉☁️⏰☁️👈

[@PC_Simo]

Of course, a nice feature of the 3*5 magic rectangle is the fact that it quite well approximates the golden rectangle, considering the size of its denominator.

[@wmafyouni]

Marvelous content! The hidden theme of balancing the numbers around the mean itched my mind, but once I saw the solution I was in awe of the elegance of Bachet's algorithm. Can't wait to implement it in python.

[@Mathologer]

Definitely let me know when you are done :)

[@grandomart]

or try magic(n) in matlab and you are done :)

[@wmafyouni]

​@@grandomart As awesome as it is that someone decided to include a base function for magic squares in MATLAB, it is not open-source

[@wmafyouni]

@@Mathologer I added it in a reply, but it didn't post. Is it because I included a link? The URL leads to a github repo that contains the source code. Not sure if theres a comment filter for posts with links.

[@torbjrnwikestad28]

I obsessed a while ago about completing a program for random search for magic squares. I am not naturally inclined to think in terms of math, so the solution, albeit working, was highly inefficient. I were fascinated by the huge possibility space of the combinatorics behind the problem. I think magic squares of size 7 and beyond has been statistically explored, but not fully mapped, because number of solutions (but even more so, the number of non-magic arrangements) so quickly explodes into inexhaustible spaces. Seeing this video, I appreciate more math as a tool to express the logical patterns behind the magic squares. I think there must be mathematical expressions still to be discoved that generalises the rules and constraints for all the possible solutions of higher order magic squares.

[@sungpackhong4144]

I was not expecting Sejong and k drama from this channel

[@Mathologer]

There are actually quite a few k dramas with interesting maths built into them. I probably will cover Melancholia at some point: en.wikipedia.org/wiki/Melancholia_(TV_series)

[@mlmimichaellucasmontereyin6765]

Bravissimo! Thanks again guys. I love the way that your exploration of truly basic fundamentals of geometric-numeric logic inspire new ideas about efficient coding, data structures, etc. Cheers etc. ~ M

[@NaHBrO733]

The magic square proof is brilliant! To generalize it, i turned the tiles' numbers into coordinates for a n*n square, T=x+(y-1)*n, 1(1,1), 2->(2,1) etc. arrange the tiles into the big tilted square as the video shows, for every tile T, the only tiles that share a common x or y coordinate are the ones that share the same diagonal with T Thus, every tile on the same horizontal/vertical with T must have different x and y coordinate (that is the first part) Now we define horizontal and vertical distance, which is the number of horizontal and vertical steps required to move a tile to another square (empty or another tile) As the video says, when creating the magic square, all tiles outside moves exactly n steps to it's destination, which gives us a distance (horizontal or vertical) of n However, as there are only n squares on a diagonal, the maximum (horizontal or vertical) distance between T and any tiles that share the same x or y coordinate

[@bob-ym3gk]

loshu is never old!(just inscribed on a turtle shell🐢).Love this channel!

[@miguelarribas9990]

"Since this will probably be my only ever Mathologer video on magic squares" ... I am already waiting for the next video on magic squares (or maybe magic cubes?) 🤩🤣

[@neokart2660]

My fav YT Channel. Amazing Video.

[@pesilaratnayake162]

Great video. Magic squares are classic puzzles. The insights I help my students understand for 3x3 magic squares is the sum is triple the central number, and all lines through the centre are arithmetic sequences. This can be seen in the general solution to the 3x3 magic square (with a, b and c), which we derive if the student is up for it. This makes solving 3x3 easy with very little information required - I think any 3 values can be used except if they are one vertex and the opposite two edges, or all three values in a line through the centre (assuming any solutions exist). In those cases, the problem is underdetermined.

[@avoirdupois1]

I loved playing with the cube puzzles like those when I was a child. The manipulation of the pieces, not just in hand, but also in my mind lead me to a greater understanding of mechanics, physics, and engineering. Thank you for this great video!

[@cybersandoval]

visual thinker to the point of mathematical difficulties, so these proofs scratch an elusive itch, so satisfying to watch those shapes fit

[@VaradMahashabde]

Homeworks : 1) Considering a general nxn square, it is simplest to add the main diagonal, whose points stay in that diagonal. It becomes the middlemost row post transformation. Their points are (i,n-i) (1 at (0,0)) and the values are n+(n-1)*i for i from 0 to n-1. Adding, we get n^2 + n(n-1)^2 /2 = (n^3 +n)/2. For a 33x33, we get 17985.

[@quantumgaming9180]

I got 17688 ;-;

[@jursamaj]

The way I went: there are 33*33=1089 tiles on the board. The sum of all tiles 1 to 1089 is the 1089th triangular number: 1089*1090/2=593,505. Divide *that* by the 33 tiles in any row, and you get 17,985.

[@Nikolas_Davis]

@@jursamaj That's how I did it also. Very simple and straightforward 🙂

[@TimothyHilgenberg]

Another great video to share with my DP HL students. Thank you!

[@far2ez]

I'm a natural-study at Math. Have been my whole life. I could derive and integrate common polynomial and other typical calc1/2 problems mentally when I was like 14. School couldn't keep up, though, and the internet wasn't meaningfully around yet. As a result, I found it boring as I got older, and I moved into comp sci instead, working at a FAANG company living the easy life. But you really bring out the romance in math out of me. I genuinely never found any appeal in any other mathematician on KZbin (even though I respect them and what they know, they just don't resonate with me). But your work is really incredible -- had you been doing your thing 20 years ago, I would've gone into math for sure. Keep up these videos. They are really so good.

[@worstwordmonger]

The magic square construction at 18:45 was the method I was shown many years ago. You start with one in that location, then go up and to the right one space for the next number. You pretend the grid wraps on itself, the top going to the bottom and the right to the left extremes, and if the square it is headed to one that is already occupied you go down one. This construction works for all odd grids as far as I know

[@Mathologer]

That's it. The problem I have with most texts on magic squares is that they hardly ever bother to prove anything :(

[@WildStar2002]

Fantastic video - loved it! 😍 I learned a method of constructing odd order magic squares when I was in Jr High School. It was much later that I became intrigued with even order magic squares. I finally managed to crack that by dividing the problem into even-odd (2, 6, 10, 14, etc) and even-even (4, 8, 12, 16, etc) cases with a different technique for each.

[@erank348]

the magic method in 18:47 is simply drawing diagonals but using modular arithmatic or if you perfer imaging the board goes on forever and then collapsing in back to a single board so using (row,column) notation : we have (5,3) draw a diagonal thats (6,4)->(1,4) then (2,5) diagonal (3,6) ->(3,1) then (4,2) . once the first 5 diagonal has ended go down 1 and draw the second 5 diagonal thats (3,2) then (4,3) MARTY : you've made a mistake no (4,3) green dot! (5,4) , (6,5)->(1,5) (2,6) ->(2,1) go down (1,1) , (2,2) and so on this ensures every column has all (x+5*y) sequences (x being the sequence in the 1-5 drawn by the y'th diagonal)

[@SuperDreamliner787]

I was showing the first part with the 3x3 grid to my 5th graders yesterday to teach them a technique to create their own 3x3 magic square. Their mind was blown. Thanks! Greetings from Leipzig 😄

[@Mathologer]

That's great :)

[@crigsbe]

Your videos are just beautiful ! ❤ CONGRATULATIONS ! 🎉 🎉🎉

[@nemesisurvivorleon]

I love how Taoism and the brilliant mathematical patterns therein are just casually a core part of Korean culture. South Korea's flag is literally the Taijitu and Ba Gua. It's great.

[@yinq5384]

Great video as always! 8:24 The sum of all tiles from an order-n magic square is S = 1+2+...+n^2 = n^2(n^2+1)/2. Thus the magic number should be S/n = n(n^2+1)/2. It's 17985 when n=33. 18:50 Extend the square to the whole plane by translating horizontally and vertically (similar to the flat torus). Start from the middle tile in the top row, go towards top-right diagonally 4 times, then go down 1 tile. Repeat. P.S: There are 5 different starting places of tile 1 for this method to work (to make sure that 11 through 15 are the five numbers on the "/" diagonal). Actually, it's better to start from tile 11 - we just need it to be on the "/" diagonal. BTW, are there any sum-and-product double magic squares? I vaguely remember having read about it before.

[@gackerman99]

Man to see my all-time favorite obscure K-drama come up in a Mathologer video apropos of nothing is going to mess with my head for a while.

[@Mathologer]

I liked the drama but in terms of historical k dramas not among my top 10. I really hated the gratuitous killing of two of the three main characters at the end of the drama :(

[@gackerman99]

I thought the lead tension was done (come on... TAM-AH!!! got you and you know it did!) well and most of all I liked the way the politics were handled. The "bad guys" were well-motivated. There was some well-handled anachronistic philosophy. You don't see that often. And it wouldn't be a proper kdrama if they didn't completely bungle the ending!

[@ProfessorBeautiful]

Like many others in Mathologer-land, this video has helped me with some elementary school students I tutor. They need exercise in simple arithmetic, and need even more a window into how exciting and powerful math can be. As for me, I am now 4 episodes into Tree With Deep Roots. I got goosebumps watching the king solve the 33x33 square. Now I'm reading all this 15th century Korean political history. What a gas!

[@sea34101]

Do you know that you are my favourite youtuber? This is incredible.

[@Mathologer]

I did not know but glad to find out that I am :)

[@bachirblackers7299]

Hhh iknow i have not seen the video yet but iwanna say welcome back Dr ❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤

[@bachirblackers7299]

MVP : Most valuable price = Dr price . Only thing to be said : wawww

[@jasontomlinson9697]

slight error at 18:48, there is no green circle on the 2nd row third column. As for the second method of creating the magic squares is to go up and to the left by 1 square (looping when going off the square) and if that square is already occupied going down 1 instead.

[@Mathologer]

Well spotted. Also have a look here en.wikipedia.org/wiki/Magic_square#A_method_for_constructing_a_magic_square_of_odd_order

[@EspritBerlin]

Vielen Dank für das tolle Video!

[@Norm_React_View.]

(IMO) your vids are just high quality the effort is the most to pick for a idea, Remember: Its just My own Opinion on the suggestion, Advice; "Try getting used to making a opinion on a topic youre interested in works for *me works for anybody".

[@eliyasne9695]

I have a different perspective on "the king's method". When I went about proving the method, I looked at the form the columbs take when we unwind the magic square back to the lunch box formation. Under that perspective each columb corresponds to two diagonals with a combined leangth equal to that of the columb. Than the key to proving the method is to show that when shifting from a columb to its' adjacent, the sum over one of the corresponding diagonals increase by exactly as much as the sum over the other will decrease.

[@Mathologer]

Nice :)

[@eliyasne9695]

@@gregoryford2532 Sorry, I'm not a native English speaker...

[@natalieeuley1734]

It was cool seeing you today at the Cook Museum at the electricity exhibit :)

[@Mathologer]

You should have talked to me :)

[@natalieeuley1734]

@@Mathologer as a volunteer, I was instructed to act like we didn't know who the KZbinrs were, but believe me, I was freaking out on the inside. And if I had said anything it would have come out in my voice 😅

[@Mathologer]

@@natalieeuley1734 Ah, that explains it. I guess some of these KZbinrs would have asked for that. Pity though :(

[@ZedaZ80]

*edit:* this was such a cool video For an n-by-n square, the sum is n(n²+1)/2 The entire square is 1+2+...+n*n = n²(n²+1)/2, then divide that up into n "slices."

[@Mathologer]

That's it :)

[@pablozumaran3997]

My 3½-year-old son is fascinated by the Numberblocks series. I can’t wait to see what he’ll make of this video when he’s older.

[@agranero6]

Incredible: this reduces the problem to a double mutually constrained Sudoku.

[@wkelly4963]

Very good you explained it in a way other people can understand. 👍

[@darrennew8211]

How delightful! I'm gonna 3D print one of these for sure.

[@ozaman-buzaman9300]

Thank you a lot professor Burkard, you are truly enlightening my life

[@bozhidarmihaylov]

Happy to see this 😊 Around the same time was wrapping a small pass-time book on the topic..nvm Great video, Thanks :)

[@ianhay527]

6:24 Sometimes you have to think inside the box

[@MuffinsAPlenty]

Interesting! I was actually informed of the Benjamin Franklin magic square (and magic circle) just a day before this video released! Now I'm thinking about things like geometric magic circles.

[@mijaelmarcelovillarroelchu6513]

Hello, I am a last year high school student in Bolivia, I am 17 years old, I decided to study mathematics permanently, but blindly because I don't know how much I like mathematics, or if I have an abstract mind for research, but still my love for math predominated despite the fact that I had already been deciding to study medicine but it was only because lately I have been losing myself in medical series and I had the illusion, in short, what depressed me about being a mathematician or even a physicist is that both sciences I feel that it has lost its essence, that is, it has been prostituted as an instrument of dissemination in order to make physics seem like child's play and mathematics like a spectacular adventure, in addition to the fact that many of their famous characters from both are so acquaintances that make you see the lives of these characters more than their work and even I have been very sympathetic with them since I value myself so little that I prefer to compare myself with people who had the same personality as me and triumph in life, so I sold myself that illusion. Another thing is that many famous mathematicians or physicists were born intelligent and gifted or else they became obsessed with matter from childhood and they investigated on their own and saw that they have talent, I have none of that, I chose study the degree as an epiphany and because if I was interested in those subjects, but I can't find a solid relationship with mathematics or physics, I can't and my worst fear is not knowing my talent or what I am suitable for, I perceive myself as someone insignificant who sometimes I think about death or the past, and as having been born in a country like Bolivia and being a student in a Catholic school in which I am an average student, even though I was recognized as the 3rd best, I am not able to assess This honor because it is from a school that I do not consider intellectually demanding or stimulating, I tried to join Cheenta's group (Indian youtube song about mathematics) but they rejected me or rather they eliminated me because at that moment I had given up being a mathematician, Now I'm looking for a mentor or tutor to help me strengthen my math skills. I hope someone cares to read this comment.

[@kohwenxu]

18:16 SPOILERS FOR ANSWER: (The method used) Start by placing 1 on the middle column and in the top row. Move: Place the next tile diagonally up 1 space and right 1 space. If the ‘move’ makes you go off the board, cycle back to the bottom row or to the left most column. (If you hit the corner do both) If the ‘move’ makes you end up on a space occupied by a tile placed previously, place the next tile the space directly below the previous one instead. (I.e. move 1 space down) This will ensure you fill the square with numbers in such a way that you will end up with a magic square.

[@kohwenxu]

In fact the 3x3 square shown at 1:11 is also a variation on this, basically once you’re done just turn the entire square upside down. (And if on paper just rewrite the numbers such that they are the right way up)

[@orisphera]

I think you can generalise this approach to get a different way. For example, you can start just below the centre and move down and left or, if that space is already occupied, two tiles down (once again, wrapping around in both cases). Here's the general construct: Let's say you want to make an nxn magic square and n is odd. (You probably can make a version for even n's, too.) Let's say (x, y) means moving x spaces right and y spaces down, wrapping around the edges, and x and y are computed modulo n, (x1, y1)+(x2, y2)=(x1+x2, y1+y2). Pick four numbers r1, r2, dx1, dx2 (r1≠r2) smaller than n and coprime with it; you can repeat the numbers, but r1 and r2 must be different. Let's say d1=(dx1, dx1*r1), d2=(dx2, dx2*r2), d3=d1+d2. Start at the centre and, without writing there, move once by a move computed as follows: each part is a half of (-x for even x and n-x for odd x), where x is the corresponding part of d3. If, for both d1 and d2, the sum of its parts is coprime with n, you can move by (1, 1) any number of times, and if, for both d1 and d2, the difference of the parts is coprime with n, you can move by (1, -1) any number of times. As a consequence, if both, you can skip the previous paragraph and just pick any starting location. Now, for i from 1 to n², put i in the current square and move by d1; if that space is already occupied, move by d3 instead or, equivalently, move by d2 without cancelling/undoing the move by d1. (The square after the first attempt is already occupied iff i is divisible by n, i.e., after every n moves; the square after the second attempts is only already occupied in the end.)

[@jacobbaer785]

The starting square in this method always winds up being the one directly below the center. So, to restate the "king's method" in this procedural way: Start at one square below the center with 1. Move down-left one, add the next number, and so on. Wrap around as needed. If you reach a square that's already occupied, move down two squares from your current position to add the next number.

[@kohwenxu]

@@jacobbaer785 That’s an interesting way to look at it!

[@kohwenxu]

@@orisphera That could be correct.

[@four-wingedpaperplanes6829]

I think a part 2 would be great. Geomagic squares. Self tiling, where all of the pieces combine to make larger versions of a single pieces and rearrange the pieces and get any of the other pieces. And....... Geomagic squares that are fractal. With any row combining to make the target shape but also all of the pieces combning to make the same shape but bigger.

[@Mathologer]

Yes, a lot more variations are possible. Maybe have a look at Lee Sallows's book first, or his online gallery of geomagic squares (link in the description)

[@four-wingedpaperplanes6829]

@@Mathologer The book is already on my list to get. The online gallery is amazing

[@davicorosello1588]

Magic, as always! 👍🏼👌🏼👏🏼

[@Sam.Lord.Cambodia]

This may be the one Mathologer Video to approach being "age appropriate for all". Thanks.

[@hillelfinder428]

In reference to the problem at (8:24), the magic constant of the King's 33x33 magic square is 17,985. In general, the Magic Constant for the King's Magic Square is n*(n^2+1)/2 where n is odd. As can be seen at (11:46), the magic constant is given by a sum natural numbers, and a sum of multiples. Specifically, it is the sum of the natural numbers up to and including n [1+2+3...+n], and the sum of the first n multiples of n [0n+1n+2n+3n...+(n-1)n]. We can be sure that this will always be the case, since when labeling the tiles, we always begin by filling a diagonal (as seen at 6:50) and then the squareness of the setup guarantees that the other diagonal will always also have the same number of tiles. Important to this proof, we know that 1+2+3...+n = n(n+1)/2. So, the King's magic constant (M) is: M = (1+2+3...+n) + (0n+1n+2n+3n...+(n-1)n) M = (1+2+3...+n) + n*(1+2+3...+(n-1)) M = n(n+1)/2 + n*n(n-1)/2 M = n*(n^2+1)/2

[@jiminurminen1332]

That last geomagic square has some nice structure build in. Number one is of course just a little square, but from that point on you just build towers of three next to each other as the corresponding number increases. To me that seems so elegant - the Mother of all geomagic squares almost!

[@Mathologer]

Good point :)

[@saxbend]

Lesson learnt. There's always more to magic squares. Was not expecting that!

[@VikasKumar_iitb]

Just mind blowing

[@Paul71H]

28:10 - This is the exact set of shapes that is used for the pieces in the board game "Blokus," produced in the US by Mattel. (But there is a mistake in the video. The stairstep 5-square piece has an extra square, making it a 6-square piece.)

[@addyraptor3675]

Gonna be another banger!

[@Tehom1]

The column sum of the 33 square is 17985. It's just the number of squares (1089) into Gauss' sum formula n*(n+1)/2 over the number of columns (or equivalently rows).

[@franepoljak9605]

Using that logic, the column sum for 3x3 square would be 36 (it's 15)

[@franepoljak9605]

Actually I just checked and result reallly is 17985. I'm still confused about the formula though.

[@franepoljak9605]

Ah, I thought it said n*(n-1). Got it!

[@Tehom1]

@@franepoljak9605 No, because you didn't do the math right. 3x3 => 9 squares n*(n+1)/2 => 9*10/2 => 90/2 => 45 Divide by the number of columns => 45/3 => 15

[@larsprins3200]

If we look at the finished magic square, we can imagine that it has been filled using the following process: Put the average number 13 in the center. Fill the anti-diagonal using consecutive numbers with 13 in the center, that is, with 10+1, 10+2, 10+3, 10+4 and 10+5. These sum up to 65 as was explained in the video. Now we fill the main diagonal with the numbers 3+0, 3+5, 3+10, 3+15 and 3+20, again with 13 in the center, also summing up to 65. Now start at any remaining number on the main diagonal, e.g., 3+5, and move diagonally down/left counting up and diagonally up/right counting down until all numbers of the respective group, i.e., 1+5, 2+5, 3+5, 4+5, 5+5 (or 1+k.5 to 5+k.5 in general) have been filled in. Do this for each number on the main diagonal. When you reach the edge of the magic square, cycle around! Just imagine a second square adjacent to where you reached the edge and see which field of the adjacent square the next diagonal step would end up in. Filling up the square diagonally this way ensures that all numbers of a group 1+k.5 to 5+k.5 are all in different rows and columns. All numbers of the form k+0.5 to k+4.5 are also all in different rows and columns, because that is true for the main diagonal and the filling up process starts from there. This ensures that all rows and columns add up to 65. *The cycling around when you reach the edge of the board is equivalent to the royal shifting method.* This is not quite rigorous but you get the idea.