The ultimate tower of Hanoi algorithm

There must be millions of people who have heard of the Tower of Hanoi puzzle and the simple algorithm that generates the simplest solution. But what happens when you are playing the game not with three pegs, as in the original puzzle, but with 4, 5, 6 etc. pegs? Hardly anybody seems to know that there are also really really beautiful solutions which are believed to be optimal but whose optimality has only been proved for four pegs. Even less people know that you can boil down all these optimal solutions into simple no-brainer recipes that allow you to effortless execute these solutions from scratch. Clearly a job for the Mathologer. Get ready to dazzle your computer science friends :)
I also talk about 466/885, the Power of Hanoi constant and a number of other Hanoi facts off the beaten track. And the whole thing has a Dr Who hook which is also very cute.
00:00 Intro
01:58 Chapter 1: The doctor vs. the toymaker
14:27 Chapter 2: Hanoi constant
21:21 Chapter 3: The Reve's puzzle
28:04 A beautiful shortest solution for 10 discs and 4 pegs (discs and super-disks)
30:23 Chapter 4: Unprovable algorithm
35:43 A beautiful shortest solution for 10 discs and 5 pegs (discs, super-discs and super-super-discs)
37:17 Supporters
Here are some references for you to check out:
Andreas M. Hinz et al. - The Tower of Hanoi - Myths and Maths, 2nd edition (2018, Birkhäuser Basel) That's the book I mentioned in the video.
The Dr Who episode (well the part that's not been lost) www.dailymotion.com/video/x11...
The 3Blue1Brown video that I mention • Binary, Hanoi and Sier...
Thierry Bousch, La quatrième tour de Hanoi, tinyurl.com/4p3fudu7
That's the paper that pins down things for four pegs.
Andreas M. Hinz, Dudeney and Frame-Stewart Numbers,
A nice paper explaining the connection between Dudeneys's work and Frame-Stewart. Also worth reading for the historical details. tinyurl.com/t8xb2e5t
A. van de Liefvoort, An Iterative Algorithm for the Reve's Puzzle, tinyurl.com/h5cxfy5u
I found this one useful.
Paul K. Stockmeyer, www.cs.wm.edu/~pkstoc/toh.html A couple of very nice papers including a huge bibliography.
Ben Houston & Hassan Masum, Explorations in 4-peg Tower of Hanoi, tinyurl.com/mw95tnek This paper has some pictures of state graphs for the 4-peg puzzle.
towersofhanoi.info/Animate.aspx Very fancy animation of mulit-peg tower of Hanoi. Sadly, it just comes across as a mess of moves for more than three pegs. Programmers, you really should rise to my challenge to animate the 4-peg algorithm the way I present it in this video.
Here is the link to the wiki page for the Celestial toymaker Dr Who episode en.wikipedia.org/wiki/The_Cel...
Makes very interesting reading. Especially the fact that most of this episode has been lost I find pretty amazing. That's also why I only show a still image from the relevant part of the episode and play some audio snippet.
Music: Fresh Fallen Snow and Morning Mandolin both by Chris Haugen, Mumbai effect. All from the free KZbin audio library.
Enjoy!
Burkard

[@Nikolas_Davis]

"If you're ever captured by an evil toymaker, you'll be ready" And they say math has no real-world applications.

[@farrankhawaja9856]

Underrated comment :)

[@warrenbosch3581]

Very cool!

[@Shadow-xb2ce]

I much prefer an evil toymaker to the "motorcycle mathematicians" my sarcastic (as I was told) Calc II teacher used to love throwing us.

[@orstorzsok6708]

:D :D :D

[@titaniumtomato7247]

@@Shadow-xb2ce ?

[@outputcoupler7819]

That's it, I'm stealing that. I'm no longer a Software Engineer, I'm an Algorithmologist.

[@Mathologer]

Let's hope this word catches on :)

[@citolero]

17:40 "This recipe also solves another Hanoish puzzle." I think the correct adjective is "Hannoying".

[@Mathologer]

:)

[@StefanReich]

Hanoisy

[@nickbradley8354]

So Hanoid! Saigonna say that. Halong ago you see this? Gotta be quicker next Thai.

[@leif1075]

@@Mathologer Did you really learn the towers of Hanoi in primary school?? I hope you can PLEASE respond.

[@NStripleseven]

Hahaha

[@henridelagardere264]

40 minutes of Professor Polster + links to 400 hours of reading = 1 helluva education Every single time, thank you for all you're doing, Mathologer!

[@Mathologer]

Enjoy :)

[@andrewmartin3671]

Hear!

[@SaveSoilSaveSoil]

I know right? I LOVE the references in the description of Mathologer videos

[@l00d3r]

I guess it is a BYOW (bring your own wine) party

[@ajhokie130]

Don't let other KZbinrs doing the same topics stop you. Please! I like to see different takes on the same topic from my favorite KZbinrs.

[@Mathologer]

There must be millions of people who have heard of the Tower of Hanoi puzzle and the simple algorithm that generates the simplest solution. But what happens when you are playing the game not with three pegs, as in the original puzzle, but with 4, 5, 6 etc. pegs? Hardly anybody seems to know that there are also really really beautiful solutions which are believed to be optimal but whose optimality has only been proved for four pegs. Even less people know that you can boil down all these optimal solutions into simple no-brainer recipes that allow you to effortless execute these solutions from scratch. Clearly a job for the Mathologer. Enjoy :)

[@superman00001]

Hello Mathologer - there is a nice (possibly related, to an extent) puzzle that asks how many different unique ways a spider can put on his shoes and socks, given that each foot must receive a sock followed by a shoe. Can you solve it? (I did, and it’s a nice solution!).

[@tcadityaa]

Good evening Mr Bukard, I have something to share with you. Our Biology teacher(yes! Our Biology teacher!) told us about an approximation relating Pythagorean triplets. I ran it on Python and its accuracy(in terms of how good the approximation is, or in other words how consistenly low the error percentage is) is very good, though its accuracy rate decreases as we crank up the size. Should we discuss it further here sir? Or how shall i contact you. Because Sir, I assure you, this is not some attention seeking stunt. I trued so hard to figure it out but the 100% accuracy numbers just pop up randomly...I have no idea....

[@ChefSalad]

You should get yourself a :CueCat as a technological mascot to go with your website. The best part is that it's cat-shaped. It might even still work, if you can find a computer with a PS/2 keyboard port to use it with.

[@johnchessant3012]

Is it just me or is it very easy to read this paragraph in Mathologer's voice? :)

[@Mathologer]

@@superman00001 Ah, that's a nice one. Had not heard of this one before :)

[@notahotshot]

The best thing about Reve's puzzle; "We're going to have some cheese."

[@timbeaton5045]

But don't do it with camembert on a hot day. You'll end up with one super puddle of cheese.

[@heaslyben]

Did you wonder if the words "cheeses", "stools", and "moves" were chosen for (juvenile) humor? Perhaps by asking I've revealed too much about my own sense of humor.

[@rayhanmansoor2951]

So happy to see a new video. I was really upset when you didn't upload in February.

[@Mathologer]

Yes, sadly insanely busy at uni at the moment. COVID is still messing things up in a big way for us and so hardly any time for Mathologer :(

[@niladriroy7650]

@@Mathologer 😢

[@nonachyourbusiness1164]

@@Mathologer How is it affecting you guys? My uni is completely remote now and emphasis has been put on self-study

[@Mathologer]

@@nonachyourbusiness1164 Well in Australia we seem to have COVID pretty much under control with no new cases out in the open for a month. This means that life is pretty much back to normal, except it isn't. My uni tries to be as much as possible back to face-to-face. Sounds good but is incredibly messy with the government still requiring social distancing which results in all venues only being at half or less capacity. and also things changing all the time in a seismic way as soon as there is the slightest indication that the virus has managed to escape again from hotel quarantine :(

[@johnchessant3012]

13:27 In 1,023 moves, moving the smallest disc every other move means we're moving it 512 times, which is 2 (mod 3). So we need to move it counter-clockwise (A -> C -> B -> A) to make sure it ends at B. In general, 2^(n-1) is 1 (mod 3) if n is odd and 2 (mod 3) if n is even, so we should move the smallest disc clockwise if n is odd and counter-clockwise if n is even.

[@zanti4132]

As the idea of clockwise vs counterclockwise isn't clear if the three pegs are in a straight line, I think it's easier to remember (a) if n is odd, first move the small disk to the peg that will hold the ending stack (b) if n is even, first move the small disk to the peg that will not hold the ending stack. Then build the smaller towers by alternating between the two pegs until you can move the largest disk to the peg that will hold the ending stack. This essentially reduces the puzzle to solving the tower with n-1 disks. Keep going like this until you are done.

[@johannesh7610]

I thought about it with the recursion: upper ten to B requires upper nine to C, which requires upper eight to B, 7 to C, 6 B, 5 C,...

[@Paul_Hanson]

@@zanti4132 In that case, instead of clockwise vs. counter-clockwise it's left and right. And here we hark back to the old video games where if you moved off the right side of the screen, you reappeared on the left side of the screen.

[@DerKiesch]

@@zanti4132 I personally would find it easier to remember: 1. Odd and even n need a different starting move to end up in the right spot. 2. You check where you end up for the smallest odd stack (1 --> which is only one move). 3. If you have an odd n, it's the same move as in 2, if even n, it's the "opposite" move to 2. 4. After that you just continue moving the small piece in the same "direction" (clockwise / counterclockwise or left/right). Thats an easy way to convince yourself that you have the right direction

[@PC_Simo]

@@DerKiesch I agree about that being the easiest way to remember this 👍🏻.

[@peter_p_r_zhang]

The toy maker himself wouldn't expect this video to come out one day during that time

[@trueriver1950]

I think he was a rogue time lord himself. Not only that but both the doctor and the toy maker had probably time travelled to 2021 and seen the Mathologer video...

[@jacoblojewski8729]

If the tower has an odd number of discs, first move the smallest to the place you want the tower to end up. If even, move to the space you don't want the tower to end. Then just consistently follow the direction you moved the smaller disc to (clockwise/counterclockwise if arranged as a triangle, to the left/right with wraparound if arranged in a line).

[@deffinatalee7699]

23:33 there’s another lovely pattern that’s easy to miss in this table; look at the antidiagonals. See anything familiar?

[@RaquelFoster]

In 1983 I got a Commodore 64 for my 8th birthday. My dad got some 5-1/4" floppy disks full of public domain BASIC games and one of them was Tower of Hanoi. That was the last time I heard of Tower of Hanoi. 🙂

[@Mathologer]

Pretty sure that was about the same time that I got my Commodore 64. I still remember animating the basic algorithm on this computer. Ah yes, the good old days :)

[@RaquelFoster]

@@Mathologer Back when animating anything was impressive! I read the Programmer's Reference Guide and was making sprites with a bunch of POKE statements and moving them around the screen. I thought I was going to get hired by Atari as soon as I got the word out! But anything was more advanced than those public domain games. I think they came out of user groups where people just mailed floppies around. I remember a reactor simulator called Nuke 64 which I wanted to like but it was ridiculously simple. But there was a Monopoly game called MONOPOLE-64 that I played a lot. There was some good-looking stuff that only used text mode graphics.

[@HansMaurer.]

@@Mathologer Did the same with the 80286 PCs that the school had bought for the first ever Grundkurs Informatik (using the recursive algorithm, learned about the circle patterns only later in university).

[@Nikolas_Davis]

The Tower of Hanoi was the first non-trivial recursion problem I came in contact with, as an example in an introductory textbook of CS in my freshman year at university. I was instantly hooked! :-) None of that boring factorial stuff; now, this was a *real*, powerful algorithm!

[@RaquelFoster]

@@Nikolas_Davis That's awesome. I don't know if I ever even saw the word recursion in school. I never had any interesting classes. In 1989 I begged my parents for a Turbo Pascal (With Objects! LOL!) compiler for my 14th birthday because I saw it in a store. Then in 1991 I got into a computer camp that had an application with several questions but only one that involved writing code. All you needed to do was find numbers that were both prime numbers and part of the Fibonacci series. I thought it was so easy I needed to make my answer clever to stand out (only 14 kids got into the camp at Ohio Supercomputer Center) so I made a super-small recursive function (not that I knew it was called recursion) that found Fibonacci numbers up to a limit then checked if they were prime as it fell back out. When I got to the camp I asked how they graded it and they said I was the only kid who even got it right LOL. But I got to play with a Cray Y-MP that summer! I hated school so much I kept dropping out because I'd do all my CS work in the first week of class then I'd just have classes I couldn't pay attention in. But one time I kept going until I got an internship before I dropped out. 🙂

[@m4riel]

After checking a feel cases, I think that to get the tower from A to B, you start by moving the smallest disc to: • C, if there are 2n discs; • B, if there are 2n+1 discs.

[@inyobill]

Ooooooo, nice thought, I'll have to verify. Well played mate.

[@blackmephistopheles2273]

You can see that this has to be the case, due to how you move the first (0+1) disk.

[@rmsgrey]

If you pay attention, you may notice that odd-numbered discs always move one way, while even numbered discs move the other way. Since moving the entire tower involves moving the largest disc once, you want it to move in the right direction. If it has the same parity as the smallest disc, you move them in the same direction; if it's the other parity, they move in opposite directions.

[@EebstertheGreat]

@@blackmephistopheles2273 You can also work it out by feel as the OP suggests. There are only two possibilities, and they should depend only on the parity of the number of disks, because they depend only on the parity of some permutation in the computation. It all comes down to one choice, and it's deterministic, so either left first means left at the end or left first means right at the end. So you actually only need to check a single case. When there is n=1 disk, you move it onto the target peg. So if there are n=2k+1 disks, make your first move toward the target peg, and if n=2k, toward the other peg.

[@ivanskopin7723]

When you don't have the corners but you have the pegs, I would phrase it as "if it's an even number, move the smallest disc to your final destination, if it's an odd number, move it to the peg that you don't want the tower to end up on".

[@AngryArmadillo]

These are some of your most visually satisfying animations to date, which is really saying something!

[@eshh183]

I remember watching 3B1B's video on Hanoi and feeling such awe, that a seemingly simple puzzle could weave recursion and binary counting together. I never thought anything could replicate that awe again! But damn you! Thank you soo much for making me experience that awe again, and this time much muchh more! I just can't wait to get on my computer and try to animate all the animations that you showed today (and perhaps even a generalised version for small n using Frame-Stewart algorithm?). But seriously lot's of love and respect man for making me fall in love with Maths over and over again!

[@Mathologer]

That's great. Cannot wait for the first animations to pop up :)

[@MatthiasGorgens]

Well, binary counting is recursive just by itself already.

[@hybmnzz2658]

Honestly 3B1Bs most underrated video

[@AAAAAA-gj2di]

Don't cuss on a Great Professor who's probably much older than you

[@UCFc1XDsWoHaZmXom2KVxvuA]

Increbile, the amount of astractions one's mind can demarcate around a given subject. Just wow, maths is beautiful

[@NatePrawdzik]

The Toymaker is Hanoi'ing me again.

[@conovan5081]

I checked your channel three times this week, I was expecting this, LET'S GOOOOO

[@Mathologer]

Yes, sadly insanely busy at work and hardly any time for Mathologer :)

[@sidimohamedbenelmalih7133]

Yes mathologer video when i'm bored is the best of the best

[@Gravitraxer_AangCZ]

It is indeed!

[@inyobill]

31:05 I believe I read, was mentioned in one of my Logic classes, or ???: For every Provable True Theorem, there are an infinite number of True Unprovable theorems. "There are more things in heaven and earth, Horatio, than are dreamt of in your philosophy.".

[@istariknight1]

As others have shared, I also really enjoy Mathologer's take on these great topics even if other math channels I follow like 3B1B cover them! Every minute spent on this production by the team was worth its weight in gold (or cheese disks) I always get this sense like it's Christmas morning when I see a new video from you guys, thanks for pumping up my weekend!

[@Mathologer]

:)

[@wompastompa3692]

For n disks, the shorthand trick I remember is that odd heights start by moving the small disk to the destination peg and even starts by moving it to the intermediate peg.

[@KillianDefaoite]

Me every time a Mathologer video comes out: *visible excitement*

[@moonblaze2713]

Bioware put this in KotOR, and I had to memorize this thankfully simple recipe because I did multiple playthroughs. Years later, they put it in Mass Effect 1 and I still remembered it. And years later I saw this video and still remembered it. Stuff you do repeatedly as a kid really sticks with you, huh?

[@pisquaredoversix]

I have seen a lot of comments before saying that they needed the video before, but I didn't expect this one to be one of those. This was the topic of my research paper for my gifted class last year, specifically the smallest amount of moves to win a k-poled Hanoi Tower, and I plan to carry on this year. I haven't finished the video, but thank you for this wonderful video anyway. Cheers from Korea! Edit: Never thought about 4-peg Hanoi Towers with the concept of superdisks. Thanks again.

[@Mathologer]

Well, have fun with your research paper :)

[@DoReMeDesign]

2nd problem is 3^n because there are n discs, each of which has 3 'choices' of pegs - same as the number of vertices on the sierpinski triangle 3rd problem can be done by comparing coefficients of 12/59 and 18sqrt(17)/1003, upon which irrational parts drop out - not sure I want to do this.. -btw, thanks Mathologer!

[@HansMaurer.]

3^n-1 actually (for the path length)

[@Mathologer]

@@HansMaurer. That's it :)

[@keshavgoel2310]

For the second problem, recursion is T(n) = 3T(n-1) +2 as largest disc moves twice and the n-1 disc pile has to be moved from one extreme to another thrice. Solving the recursion gives the promised beautiful answer - 3^n-1

[@taibilimunduan]

less one, because the initial position does not count as a move

[@Lightn0x]

I used to solve towers of hanoi on fights because I had no internet. Solving the 10-peg version with 1023 moves must've been one of the proudest moments in my life :D Edit: huh, I didn't know the "smallest disk" trick explained at 12:18. I just tried to follow the recursive algorithm in my head. Way to trivialize my life's achievment, Mathologer! /s

[@Mathologer]

:)

[@10gbo_pizza]

Sir can I say, despite the time, your videos are ABSOLUTELY AMAZING! KEEP UP THE GOOD WORK! :)

[@danielmassart]

for the record, T. Bousch also has a strong contender in the "best title for a math paper, ever" category : Le Poisson n'a pas d'arête (Annales de l'Institut Henri Poincaré, 1999)

[@NoNameAtAll2]

Translate?

[@somebodyhere3160]

@@NoNameAtAll2 google translate tells me this:the Fish has no bones

[@Person1873]

The trick I always used for the tower of Hanoi is based on odd or even quantity of discs. If you have an odd number of discs then the first move will be to the target tower. If you have an even number, then your first move will be to the other empty tower.

[@TrimutiusToo]

I figured out best solution for tower of hanoi when i was 6... My dad gave to me tower of hanoi puzzle with 8 peaces and so that i don't bother him said that i should find best solution and left me like that, and then was very surprised that 3 hours later i came to him and said "255", he already even forgot that he gave me the puzzle, but back in those days i figured out the recursive solution to tower of hanoi which impressed my father quite a bit... I was always good with recursive algorithms...

[@Mathologer]

Cool :)

[@trueriver1950]

The iterative algorithm is harder to spot, and would have impressed him even more (I won't give it here, as several other commented already have)

[@kitrit8445]

Круто

[@maxsch.6555]

Oh, what could be nicer than a weekend with a mathologer video :)

[@hamiltonianpathondodecahed5236]

"Oh , I know about this puzzle" "how tf I never thought about multiple pegs"

[@tonyalanmarchant7330]

I love the little giggles. Could be a bady in a bond movie

[@Mathologer]

If they ever offer me a role as a mathematical supervillain I'd definitely go for it :)

[@tonyalanmarchant7330]

@@Mathologer you most defo should.an if the role comes along? I assume you know all about percentages 😁

[@xevira]

13:50 if n is odd: clockwise if n is even: counter-clockwise or another way to phrase it: if n is odd: first move is to the target position if n is even: first move is to the OTHER position

[@Hahahahaaahaahaa]

I think, because of how common the Tower puzzle is, and how little I've thought about it before, this is one of the most beautiful videos you've done. Despite many years of studying more complex topics, making "simple" problems elegant often gives me the greatest appreciation of mathematics. Thank you as always!

[@AlexanderQ689]

Excellent video, I love the visuals & extending the tower of hanoi puzzle beyond what's typically talked about

[@stromboli183]

Easily one of the best channels on all of youtube. Thank you once again for a very interesting, entertaining, learnful, and inspiring video, professor Burkard! 🙌

[@AmitSingh-sf5qp]

My interest in maths came back after my exam .

[@HAL-oj4jb]

I didn't even notice that the movie and tv references were gone, nice to see a revival!

[@trueriver1950]

Try opening the pod bay doors...

[@danpost5651]

14:00 For odd number of disks, move clockwise; for even, move counter-clockwise. Easy to see. if one disk (or odd), move directly to ending peg; otherwise, not.

[@stulora3172]

15:52 I love the Michael Ende quote

[@stulora3172]

unter uns.... war das Absicht?

[@buxeessingh2571]

For those who don't know: the reason you only saw a still and heard an audio is that only photos from the TV broadcast and a duplicate audio of the telecast are what BBC have in the archives of the first three episodes. Addendum: there was a proposed sequel to "The Celestial Toymaker" called "Nightmare Fair". Rumour had it that an early draft of the script included the Doctor solving the minimal path problem discussed here for the Trilogic game.

[@Mathologer]

Had not heard about this before :)

[@alvarezjulio3800]

Once time again a Wonderful Video from Mathologer !!! Thank you.

[@denny141196]

Decided to implement in python before watching. An interesting point I came across was that when recursing, the midpoint and destination swap for subtower move (so the biggest disc can move to the destination unimpeded). As the algorithm recurses down to the base case of 1 (not 0), there are n-1 such midpoint/destination swaps for a tower of size n. Hence, for towers with an odd number of discs, there are an even number of mid/dst swaps, which all cancel out -> first move should be the first disc on to the intended overall destination. Vice versa for even number of discs.

[@MatejCrhak]

I remember finding formula for the smallest number of moves required for n disks and an algorithm to solve it, when I were 12 years old. Those were great times.

[@abrvalg321]

0:56 that puzzle becomes really hard really quick if allow to input not only number disks but sticks as well.

[@namantenguriya]

•Always loving your videos. 🥰🥰🥰 Love from India 🇮🇳🇮🇳🇮🇳. •By the way, if you plug in 3 for n, you get 《946/243》(=3.893) •Please make videos on ☆☆☆ Unsolved problems in Mathematics ☆☆☆.

[@cybersecuritytutorials8762]

Amazing!!! Enjoy your storytelling style to explain the mathematical concepts and algorithms.

[@sldl]

While I never faced it, I was considering the programming interview question: "Solve the Tower of Hanoi puzzle using a recursive function." Here's my problem with this interview question: It doesn't actually *need* a recursive function. I'm of the opinion that if you can craft more supportable code, however less "elegant," it is, you will be doing right by your client. Here's how I would structure it (using English rules for pseudocode): Assume N disks (Labeled Disk 1 through Disk N) Assume 3 rods (Labeled 1, 2, and 3) Assume all disks are first placed on Rod 1. All moves follow the rules (The only disk that can ever be moved to either rod is Disk 1. All other exposed disks can only move to the rod not occupied by Disk 1.) Rule 0: (One time move) 　If N is even, move Disk 1 to Rod 2; 　　Else, move Disk 1 to Rod 3. LOOP ((N^2)-2)/2 times: 　Step 1: 　　Move the smallest disk that is not Disk 1. 　Step 2: 　　If Disk 1 is not on Disk 2, move Disk 1 to Disk 2; 　　　Else, 　　　If the two non-1 disks are both even or both odd 　　　　Move Disk 1 to cover the larger disk 　　　　Else, 　　　　Move Disk 1 to cover the even disk That's it. One single one-time move, and a loop that bounces back and forth between a non-decisive move, and a 3-option decision move. I believe this approach would be easier to read and maintain by someone who came along later. "Elegance" is great if it saves time, and is clear and easy to maintain. But it doesn't have client value in and of itself.

[@melekhine]

Every one of your videos is so amazing, Professor Burkard. Best on the internets

[@_majortom_]

beautifuly done as always. I have enjoyed every part of this video. thanx!

[@walrusman8691]

This kind of single move algorithms often show up in computer science and an often overlooked follow up question is if their were "multithreading" where if two subsequent movements occurred to and from different pegs they could be done simultaneously as 1 move. Obviously this would only apply to larger n but I would be curious if the solutions for that were actually superior given how ordered the solutions seem to be already.

[@heaslyben]

I love that question. It makes me think of context switching as well. Could any process (aka intelligent but forgetful prisoner) wake up, inspect the current state of the board, and make the next correct move? My instinct is "yes" but I'm not sure.

[@aeleron0577]

@@heaslyben probably. Of course it depends on the actual problem, but one could use a hamming code approach for that. Aka coding a failsafe

[@TheJoaovascorodrigue]

Your videos are mind-ticklers! Thank you

[@leonidassavalas745]

There was no Easter egg at the end, but I always watch the entire (always wonderful) video(s). Thank you, Mathologer!

[@Mathologer]

That's great :)

[@vivekpal1004]

Such a great explanation! Excellent!!!

[@sidazhou]

Already smashed the like button multiple times before Intro was finished. Such was the enthusiasm to learn.

[@Mathologer]

That's great :)

[@chiranjivipokharel1152]

Always so curious to see vedios , thanks a lot mythologer , love from Nepal

[@Jukka70]

So happy to see the video for this month. Wish they were more times then once a month, but am sure it takes a long time to make these, and happy that you make them

[@Mathologer]

Yes, sadly my uni job keeps me insanely busy these days. That and family leaves hardly any time for Mathologer. Anyway, I'll keep trying to put one out every month.... :)

[@PapaFlammy69]

Let's gooooooooooooooooooooooooo! :D

[@Mathologer]

Guten Abend Jens :)

[@meeeshroom]

Papa Flammy ist hier!

[@maxwellsequation4887]

Yaaaaaaaaaaaayyyyyyyyyy

[@adityaruplaha]

Hello papa

[@PapaFlammy69]

@@Mathologer Ich hoffe du hattest ein schönes Wochenende bis dato :3

[@notabotta3901]

Couldn’t be happier to see this video and try the challenges!

[@blackmephistopheles2273]

#1: Nine disks need forty-one (41) moves. My solution is the "minimum of (two times a smaller number of disks plus the minimum of the remaining disk moves) method. #2: An observation: if d (# of disks) < p (# of pegs), then you ALWAYS only need 2*d-1 moves. Basically, having more pegs than disks means you can move every disk to its own peg, including the base disk (to the designated ending peg), then move all of the other disks back on to the next bigger disk, which you should have already moved to that ending peg: (d - 1) + 1 + (d - 1) = 2d - 1. #3: Likewise, if d = p, then m = 2d + 1, since you will have to move whichever disk is on the ending peg (or move another disk, if the ending peg is the only one available to the biggest disk), move the biggie, move the disk you had to move beforehand, and then move all the other disks onto the biggie: (d - 1) + 1(clear off the ending peg) + 1(biggie) + 1(split the double stack up with the starting peg) + (d - 1) = 2d + 1.

[@laierr]

I'm so glad to see (literally) that your color scheme became more colorblind-friendly!

[@AlessandroDruetto]

18:16 (shortest solution with extra adjacent-peg rule): it is obvious that we have to visit all the nodes in the graph, and a path traversing x nodes exactly once requires x-1 arcs. So, the question boils down to evaluate the number of nodes in such a graph, that is essentially composed by all the possible valid configurations of disks on pegs. With 1 disk we have 3 valid configurations (the smallest triangle), with 2 disks we have 9 valid configurations (the 2nd-smallest triangle), with 3 disks we have 27 valid configurations (...), etc. Note that when we assign a "set" of disks to a peg, their placement is uniquely determined (since it must comply the no-big-over-small rule) by the total ordering of disk sizes. Then, the total number of configurations is the number of ways to assign n disks to 3 pegs; that is the question "with n available digits, how many numbers can I build that are 3-digits long" and the answer is 3^n. Having 2^n nodes in the graph, having to traverse all of them exactly once, the shortest "adjacent-peg" solution is of length 3^n-1 with 3 pegs; more generally with p pegs the shortest "adjacent-peg" solution is of length p^n-1 by the same reasoning.

[@yonatan1myers]

for the Q in 14:00: if mod(n,2) == 0 start by moving the smallest disk to the peg that is not your target, else start by moving it to the target

[@trtlphnx]

Love Your Mind, My Friend,,, Keep up the Incredible Work; You have a #1 Fan Here!!!

[@troemax]

Wonderful video. Thank you for making these!

[@keshavgoel2310]

For the second problem, recursion is T(n) = 3T(n-1) +2 as largest disc moves twice and the n-1 disc pile has to be moved from one extreme to another thrice. Solving the recursion gives the promised beautiful answer - 3^n-1

[@jorgevegarevolledo]

Todos tus videos son espectaculares. Gracias por tomarte el trabajo de hacerlos, son realmente inspiradores.

[@xCorvus7x]

13:27 The doctor has to move the tip counter-clockwise because B in that picture corresponds to the lower right corner of the triangle we had looked at before. In general, if you have a tower with n disks, you have to move the tip clockwise for odd n and counter-clockwise for even n for the tower to end up at B (according to this picture). Moving a tower with an even number of disks to B means that you first move a tower with one less disk and thus an odd number of disks to C and since a single-disk tower can be moved with one move only, the tip of a tower with an odd number of disks moves initially to the place where the whole tower will eventually end up. Therefore, if the tower has an even number of disks, the tip has to move to a different place than the final location.

[@stingrayx2182]

One of your best videos, thanks!

[@manorueda1432]

Funny and awesome, as always! I've enjoyed it. I'll think about the puzzle with for and five pegs!

[@maxreenoch1661]

Mathologer: *Explains mathematical induction in under a minute* Teachers: *take notes furiously*

[@StankyPickle1]

Beautiful visualizations!

[@casanova2829]

Incredible explanation!

[@ripe_aces]

14:43, this wicked sense of humor had me in stitches.

[@SaveSoilSaveSoil]

I find myself revisiting your hugely helpful references again. Kudos to Team Mathologer!

[@Mathologer]

Glad to hear it!

[@dctcristino]

Perfect, my friend! Thanks!

[@surfing_youtube]

Thank you for this excellent video. It was worth watching.

[@Joker22593]

Ah, Dynamic Programming! Excellent.

[@woodpecker8546]

What a fantastic video again! Thank you very much!

[@johnchessant3012]

Brilliant and very satisfying video!

[@PC_Simo]

18:20 The number of moves is just the number of possible states (3^n) minus 1, since the algorithm visits each state exactly once; and since the initial state requires 0 moves to reach from the initial state. The number of pegs is always 3; so, the number of moves, for n discs, is: (3^n)-1. Tada! 😎

[@d4django]

I am ready for the tomaker now! Thankyou Mr Mathologer

[@ryanbell3704]

18:10 I believe the answer is just 3^n-1, since that’s how many possibly positions there are (3 choices for each disc, so 3^n possibly positions, so 3^n-1 moves to get through all of them) really neat how that only-adjacent rule gets us from 2^n-1 to 3^n-1

[@paulklee5790]

I distinctly remember seeing this on our steam driven television the first time round... by then I’d graduated from watching from behind the sofa (for safety reasons...), it had a big impact on me ... I made a version of the game at school and started a craze for playing it that lasted two whole weeks! Nerd heaven.....

[@varunachar87]

To me, it is amazing that the state transition graph is (1) connected and (2) planar. The connectedness I can make sense of. But I can think of no obvious explanation for its being planar.

[@Muhahahahaz]

It’s because there’s only 3 pegs, so you just make a bunch of triangles But with 4 pegs, you might need tetrahedrons… 🤔

[@Veptis]

It's often used to teach recursion. And last semester our professor used Fibonacci sequence and towers of Hanoi as two examples to show different programming languages. Including postscript, which was quite fascinating

[@Tiessie]

Your videos are so interesting and inspiring

[@georget8008]

The answer to your question is counterclockwise. I came across the Hanoi Towers problem as a freshman in a course on algorithms and complexity. The original problem was set up with some 70+ discs. The textbook proved that if each move lasted for 1 second and we did no mistakes, then the time required to complete the puzzle would have been more than the time from the Big Bang until now.

[@prashantshukla6018]

Oh gosh I was craving such stuffs 😂. Love ur creations sir huge love and respect from India . Thank u sir!

[@technowey]

Thank you for another excellent video.

[@dcterr1]

By the way, Dr. Who was a great show! Great choice for introducing the Tower of Hanoi puzzle!

[@cameronlakeview7068]

I'm currently finding tower of Hanoi a good muse problem for getting my flat in order! (Long overdue tidying) So extra specially well pleased I watched this recently! (I'm picking things up off one surface, and putting them down on another temporarily.. Tower of Hanoi, I've learned (through this video) teaches that it's much less work with more permanent (rather than temporary - eg a table top or 2, rather than balancing pins on the tops of empty beer bottles while waiting to see if enough needless etc will emerge from the mess quantity to warrant a "sewing stuff box"...

[@josephcalzaretta8516]

This was so cool that I couldn't stop thinking about it. 30:14 "For my programmer friends..." I wrote a demo you can see at jcalz.github.io/hanoi-demo/ that (I think) implements and animates the Frame-Stewart algorithm. Last time I posted this I think it got sent to the spam graveyard, so hopefully his one will stick.

[@Muhahahahaz]

Awesome work! I immediately put in 5 pegs with 100 disks, and surprisingly it didn’t break lol (I set it to a speed of 0.01s so that I could see the whole thing in under a minute)

[@Muhahahahaz]

I also just tried 100 disks with 101 pegs (which is the minimal number of pegs required to minimize disk movement - it only requires 199 moves, which is 2 moves for every disk except the largest one) Much faster than 5 pegs! But obviously not very interesting lol

[@Mutual_Information]

Tower of Hanoi is a super popular interview question, so much so that I bet interviewers have to tweak the question so it's new to interviewees. I imagine asking about more pegs is a frequent extension. Now I feel prepared. Great vid

[@Mathologer]

A simple solution is to add more pegs before you start moving the discs. For 10 discs and 11 pegs the puzzle becomes trivial :)

[@pullingrabbitsouttaahat]

Another Gem From Mathologer. Hats Off !

[@matthewdodd1262]

13:55 Answer is pretty simple. If N is odd, you move it clockwise to end on B, and counterclockwise to end on C If N is even, you move it clockwise to end on C, qnd counterclockwise to end on B. It has to do with where disc 2 and disc 3 land, and proving that its to do with odd-even parity, and that it holds for all N

[@samuelwaller4924]

I just recently found a code golf where the challenge was to quine that output a unique program every iteration, without repeat, for however many iterations. I immediately thought of Gray code, and have been absorbed. This is great!