The Hanging Sign Puzzle in 59.93 sec

hi people, i made this video for the #VeritasiumContest (www.veritasium.com/contest) because they were looking for 60-sec videos and I was thinking "Man, I haven't made any short-form STEM videos in a while." So I fixed that today!
(for Veritasium's team) Here's my email: [e-mail]
This also might be my shortest video in like 5 years! Also, the video is 59.93 seconds long. Also, I uploaded it on August 30, and it's due on August 31.
Thanks to James Woma for introducing this puzzle to me in 2017:
/ @jameswoma1140
Thanks to "Objects in Motion" for encouraging me to join the Veritasium Science Communication Contest:
/ @objectsinmotion
Thanks to my dad for building the wooden pegboard!
Also, here's a cool paper about this puzzle from the State University of New York: arxiv.org/pdf/1203.3602.pdf
Music: "As Time Passes" from www.zapsplat.com
www.zapsplat.com/?s=as+time+p...

[@OrbitalRose_01]

finally, a clear explanation of pegging

[@darksentinel082]

I think this is the comment that will finally make me lose it

[@astrawby]

I'm not so sure, I'm still having trouble with my game of Peggle

[@Taib-Atte]

Good comment

[@arthurrandom2137]

Ah yes, today I've found gold

[@Samata94]

I checked the comments to see if someone already made the joke so I didn't have to. altough I was going with "Ah I do love a good pegging."

[@ShortHax]

This man explained it really well. I won’t be surprised if he speedruns this with 10 pegs next

[@lyonidus3073]

Mhm.

[@Nelo390]

Mhm.

[@penjamin1479]

Mhm.

[@ultrio325]

Mhm yeah.

[@ChurchOfThought]

Are there any examples of this kind of phenomenon with more complex terms like matrix multiplication instead? Pretty much any invertible group can be used - it shouldn't even need to be a commuting one, should it?

[@durdleduc8520]

I’ve seen this puzzle explained in other ways that I could never accurately grasp, but this 60-second video gave me the “OH” moment almost immediately. good luck in the contest!

[@danybeam]

IDK if this was it for you but I find this one has a very elegant? Minimalist? Both? Way of explaining it which makes it more accessible 🤔

[@carykh]

Thanks durdleduc! I appreciate it :)

[@rebucato3142]

I remember a guest video on Tom Scott’s channel about this problem but I couldn’t understand it Cary explained it in a minute and it’s so much easier to understand

[@Cyranek]

3 pegs speedrun would get me all the babes

[@DannySullivanMusic]

hahahaha pro comment

[@najhydro]

true

[@prabas8190]

3 pegs is super ez

[@HanzCastroyearsago]

this video is practically instructions for that

[@efperel]

I saw dozens of videos on this problem, and this had the cleanest explanation

[@Henrix1998]

kzbin.info/www/bejne/rmbLZKyKrcp2m9k Matt Parker

[@datarioplays]

@@Henrix1998 Edit your comment so that it doesn't look like something a spammer comments

[@real-jd5rc]

@@datarioplays eit my www.youtube.com

[@MCLooyverse]

It seems like he just gave the math and briefly explained it to those for whom it was not already obvious, whereas most people try to explain it even further to the layman, as if the math were not self-explanatory.

[@user-wr2uy9pj4m]

3 pegs? Sounds like it *sounds* complicated, but isn't actually complicated

[@phuonghuynhanh9879]

then imagine the middle peg to have C and C', and we will kinda work the same i guess??? hang on... that's why i have to cubing right now i guess

[@phuonghuynhanh9879]

so I think the answer may be ACBA'C'B'

[@drakeenor5951]

@@phuonghuynhanh9879 No, removing C doesn’t make it fall. Or any letter in that solution I think.

[@anarchosnowflakist786]

B’A'C'ABA'B'CBA I think it works if you remove the A : B’C'BB'CB B’C'CB B’B nothing if you remove the B : A'C'AA'CA A'C'CA A'A nothing if you remove the C : B’A'ABA'B'BA B’BA'A nothing edit : I wouldn't say it was that complicated but I did spend a few minutes to solve it

[@drakeenor5951]

@@anarchosnowflakist786 That is actually correct. The new pattern is just the same one as before (ABA’B’) but copied twice (where the second one begins with B) and separated by C: (ABA’B’)C[BAB’A’]C’

[@TheNatureThread]

When the rope goes over peg A rightward, let's call it A. When it goes over it leftward, let's call it A'. Cubers: Hey, I've seen this one before! It's a classic!

[@johndoes_art]

He is actually a speed cuber

[@melvintnh328]

yep

[@Rhys_1000]

wait is A and A' are musical notes???

[@TheNatureThread]

@@Rhys_1000 Cubing terms

[@Rhys_1000]

@@TheNatureThread oh

[@Anthony-vu8bl]

*Algebraic topology explanation:* this is because the fundamental group of the twice-punctured plane is the free group F2=, which is not abelian. Removing a puncture is considering the map → by a→a and b→1. Now, is abelian, so the sign falls.

[@pimcoenders-with-a-c1725]

Very interesting way to look at this; I know about topology (fundamental groups) and group theory, but I've never heard of free groups; generalisation should be pretty easy using these free groups

[@bettercalldelta]

🤓

[@scarcedude3353]

@@bettercalldelta When you're too ignorant to make the effort of understanding something so you hit them with the 🤓

[@georgecantu856]

I didn't know ablelian was related to topology! I just remember it blowing me away in my Modern Algebra class

[@CommentBanana]

this was such a short a sweet video that i was expecting a punchline at the end.

[@decb.7959]

It's part of a challenge to explain a puzzle in 1 minute.

[@PentameronSV]

Somewhat unexpectedly clear and easy to understand!

[@wilh3lmmusic]

WHYAREYOUHERE

[@YellowBunny]

That just sounds like group theory with practical applications.

[@mthf5839]

"Let's call that move A" -there's no way I am going to understand this without rewatching several times, eh? *proceeds to get explanation so clear, that mind implodes*

[@ahreuwu]

I've seen this being explained several times before and never understood it, your video made it really clear and now I finally understand it!

[@bennihtm]

Same

[@user-wr2uy9pj4m]

I remember that from Tom Scott's video lol

[@alexandertheok9610]

Me too! I'm honestly surprised Cary managed to compress this down to 60 seconds

[@user-wr2uy9pj4m]

@@alexandertheok9610 yup, and tbh I understood it better than in Tom's video (although it wasn't Tom who explained it)

[@IloveRumania]

xnopyt

[@erzar.1730]

That was epic Carych

[@dryued6874]

Карыч

[@DanHaiduc]

Algebra is overpowered! Amazing Cary!

[@TheSpacecraftX]

This legitimately is the best explanation I've seen of this. I'm familiar with this problem and have seen introductory knot theory content before but this is easily the one that left the strongest impression I understand what's happening.

When you suggested to try it myself, I paused, figured out a solution with basically just somewhat educated guesses and then correcting any leftover rotations after removing one rod by introducing additional rotations, then, when I had a solution, I spent longer to simplify it than to come up with it, because I confused myself, then I eventually ended up with the exact same solution that you had. A'B'AB, BAB'A' and B'A'BA would also be solutions (and lots of more complicated ones), but I got the exact same one. Then I saw the video and it was a lot easier to understand and to generalise than anything I had drawn, tried or thought in the minutes. Well done! (BTW, I used an actual thread and two pins stuck upwards through a cloth.) About the three rod version: This has an interesting twist, because you can't think about "A first, A' later", it doesn't work that way (you can't have A…B…A', because it would stay when C is removed and you can't have A…C…A', because it would stay when B is removed, so nothing could be between A and A', so you couldn't make anything). Instead, you absolutely need to use a cancellation of A'…A at least once, meaning you need at least two pairs of one letter. And once you get that idea, getting the solution is pretty easy, thanks to your method: ABA'CAB'A'C' (Edit: Or not. This "solution" doesn't actually work. Not sure what went wrong when I validated it, but this comment thread contains a proper general solution: kzbin.info/www/bejne/mJnQfGxqgaqkg7c&lc=UgzQ30xtnh4m5LLXXpJ4AaABAg ) (BTW, I used "A" and "a" in my notes, that's less confusing for me, because you can never accidentally delete only a letter and forget the apostrophe and it's all nicely monospace in Notepad++.)

[@alexanderbudianto7794]

Who'd knew that making a hanging sign to fall down has something to do with group theory (kinda)?

[@05degrees]

This is really legit group theory, something to do with commutators, but I don’t know about it more right now, just heard sometime ago.

[@alexanderbudianto7794]

@@05degrees Actually, now that I think about it, it really is group theory. For the case of 2 pegs, the group has infinitely many elements, and let's just say that the operator is *. Obviously, the operation is closed under operator *. It is obviously associative (e.g. (A*B)*A=A*(B*A)). It has the identity element, e, such that for all N in the set, N*e=e*N=N (e.g. A*e=e*A=A); in this case, e is the "not going around any peg" action. For all N in the set, there exists its inverse, N', such that N*N'=N'*N=e (e.g. A*A'=e). Therefore, it is in fact a group.

[@05degrees]

@@alexanderbudianto7794 yep

[@alexanderbudianto7794]

@@05degrees Actually, maybe it isn't. I'm pretty sure A*C does not exist, since you can't go from peg A to peg C without going through peg B, which means that it is not closed.

[@alexanderbudianto7794]

Actually, never mind. It is a group. I found out that it's better to define A as "going over peg A from left to right" and A' as "going over peg A from right to left" instead of using rotations. Please read Jim Gu's comment and its replies for more info.

[@PieMan12]

That notation reminds me a lot of the notation for Rubik's cubes lol

[@SpeedyCorky]

i just appreciate how quick and to the point this video is. no intro, no wasting time, just straight, quick, here is what you need to know

[@DGCubes]

This is so cool! I hadn't heard of this problem before but you helped me understand it really well, you're a great teacher :)

[@The_NSeven]

And easy to understand when you're a cuber :)

[@fryphillipj560]

That's an amazing example of abstraction. Great video 👍

[@collintmay]

Ok, 3-pegs: If 2 pegs can be solved with A B A' B', then we can reduce the problem using these moves as a grouping. When we remove either A or B, it cancels them both out together, so we can treat the two as a single peg. Thus, the 3-peg problem is really just the 2-peg problem with some nested 2-peg problems within it. If the algorithm for 2 pegs is A B A' B', then the 3-peg algorithm will take the same form. Let's construct a solution. A B A' B' Substitute C for A Substitute (A B A' B') for B Substitute (B A B' A') for B' Final solution: C (A B A' B') C' (B A B' A') Bonus: If you want to invert a group of moves, so as to generate the "prime" version of it, all you have to do is reverse the moves and switch every prime move to a normal one and every normal move to a prime one. Eg. A B A' B' -> B A B' A' Using this algorithm, you can derive solutions for an arbitrary number of pegs, however it becomes very cumbersome, and the optimal solution increase by a factor of 2n+2 for each peg you add. 4 pegs: D (C (A B A' B') C' (B A B' A')) D' ((A B A' B') C (B A B' A'))

[@socialist-oat]

The fact that before the video, I had to watch an ad that was even longer is embarrassing.

[@kfftfuftur]

N pegs are easy: First lets use upper and lower case letters to make things more compact and easier to read. If we have a sequence mirroring it will preserve whether it falls or not. So "ABab" -> "baBA" likewise we can substitute every letter by the same letter of the opposite case and also preserve the falling property. "ABab" -> "abAB" If we both mirror and flip the capitalisation of any sequenze wen can generate its inverse sequence that will cancel out with the original sequence. Trivially if we apply this to a sequence of length one this will only flip the capitalisation of the only element. Since this is compatible with the rules we used up until know lets modify the change in capitalisation to mean mirror and flip case. This way X can be a sequence and x would be the inverse sequence. Now lets prove N pegs: trivially the solution for one peg, lets call it A is "A". If we know that for N pegs the sequence X makes it fall then we can add a peg Y and use the sequence "XYxy" to still make it fall. Since there are 26 letters in the alphabet here are the solutions for up to 26 pegs: 1 peg: "A" 2 pegs: "ABab" 3 pegs: "ABabCBAbac" 4 pegs: "ABabCBAbacDCABabcBAbad" 5 pegs: "ABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbae" 6 pegs: "ABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbaf" 7 pegs: "ABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafGFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbag" 8 pegs: "ABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafGFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbagHGABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafgFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbah" 9 pegs: "ABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafGFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbagHGABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafgFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbahIHABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafGFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbaghGABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafgFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbai" 10 pegs: "ABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafGFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbagHGABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafgFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbahIHABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafGFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbaghGABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafgFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbaiJIABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafGFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbagHGABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafgFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbahiHABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafGFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbaghGABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafgFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbaj" 11 pegs: "ABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafGFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbagHGABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafgFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbahIHABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafGFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbaghGABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafgFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbaiJIABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafGFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbagHGABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafgFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbahiHABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafGFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbaghGABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafgFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbajKJABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafGFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbagHGABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafgFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbahIHABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafGFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbaghGABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafgFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbaijIABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafGFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbagHGABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafgFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbahiHABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafGFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbaghGABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafgFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbak" Nevermind: The solution for 26 pegs would be around 100 megabytes large. 11 pegs are the most I could fit while succesfully submitting the comment.

[@itsphoenixingtime]

Holy shit. Thats a lot of moves o_o. Is there a way to related n pegs by a mathematical function? It seems interesting to me the growth seems to be exponential in some way.

[@gordonriess4681]

i guess ABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafGFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbagHGABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafgFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbahIHABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafGFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbaghGABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafgFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbaiJIABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafGFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbagHGABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafgFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbahiHABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafGFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbaghGABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafgFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbajKJABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafGFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbagHGABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafgFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbahIHABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafGFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbaghGABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafgFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbaijIABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafGFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbagHGABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafgFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbahiHABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafGFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbaghGABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaeFEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbafgFABabCBAbacDCABabcBAbadEDABabCBAbacdCABabcBAbaefEABabCBAbacDCABabcBAbadeDABabCBAbacdCABabcBAbak is a word now

[@jimgu2578]

I’ve found a solution, not just for three pegs, but for any number of pegs! (Spoilers) For three pegs, ABA’B’CBAB’A’C’ is a solution. The idea is that since we already combined A and B in the video’s solution: ABA’B’, we denote this combination as A+B. A+B is a knot if both remains, and null if either is removed. The knot we want is A+B+C, which is equal to (A+B)+C, so (ABA’B’)C(ABA’B’)’C’ works, and it simplifies to the answer ABA’B’CBAB’A’C’. This way of thinking can be used for any number of pegs. For example, for four pegs, we only need to simplify (A+B+C)+D, or (A+B+C)D(A+B+C)’D’

[@alexanderbudianto7794]

Is that really the solution, though? I'm concerned with the last two actions, A'C'. How can you go from the first peg to the third peg without going through the second peg?

[@aslpuppy1026]

@@alexanderbudianto7794 When you remove A, ABA’B’ and BAB’A’ go away leaving CC’ which is nothing, this the signs falls

[@alexanderbudianto7794]

@@aslpuppy1026 You're not answering my question. I already figured out what you're saying. My problem is when the rope goes counterclockwise around peg A and then immediately goes counterclockwise around peg C. How is it going to do that without passing through peg B? It's either going to be A'BC' (from bottom of A going up to top of B then going down to bottom of C) or A'B'C' (staying at the bottom from A to C).

[@aslpuppy1026]

@@alexanderbudianto7794 Sorry, I misunderstood your question. I think the answer to your question is that the rope will pass below peg B but not to around it. Hope that helped.

[@alexanderbudianto7794]

@@aslpuppy1026 This will probably go nowhere unless we mathematically define what exactly it means for the rope to "go around the peg", such as the minimum angle of direction change. By the way, I just realized that when the rope starts at the left side of A, then goes over A, then below B, then over C and ending on the right side of C, if you take out peg C, the rope will only hold on to peg A. I wonder if that means that that position is AC and not AB'C, since if it were AB'C, then erasing C would yield AB', even though the rope won't hold on to peg B...

[@jblen]

That was a super clear but also very logical explanation. Abstracting a problem into a sequence of values really is the core of computing huh.

[@Dundereshock]

I want to see a large-scale simulation of this with 100 pegs or something, just out of curiosity

[@1996Pinocchio]

It does not scale linearly, so for 100 pegs you probably have to go around pegs 10k times or so

[@kfftfuftur]

@@1996Pinocchio well my recursive solution would need to go around the pegs exactly 1901475900342344102245054808062 times.

[@Nnm26]

@@kfftfuftur 17152 for optimal solution

[@pmj_studio4065]

You could also only use the first and the last peg and just put the rope on top of the remaining 98 pegs. That would definitely work 😁

[@xoxshielakixox]

Even though I missed the deadline for the coding competition and TWOW, I'll enter this Veritasium Contest!

[@0Smile0]

short but informative, great video!

[@cate01a]

Very good method of describing this, thanks!

[@thelordflygon6169]

God I’ve seen so many videos explaining this this with like, knot theory and tons of visual representations and stuff but none have been simpler than this one. Thank you so much lol

[@desertia]

Carych this was so cool your channel has great content haha

[@joelcrombie4902]

Finally someone answered my question about rope and pegs

[@stanleychan3212]

Nice simple application of group theory. Good job.

[@benwinstanleymusic]

Amazing explanation, nice work!

[@NithinJune]

this surprisingly was the clearest explanation

[@astralblob]

This is really easy to understand!

[@patag5092]

THIS is why I'm subscribed to you

[@johnweber4504]

ok but this is acually REALLY well made

[@lnx0007]

Any solution for two pegs has an inverse that would nullify it, just take the solution and reverse the order of the turns, and the handedness of the turns: ABA'B' has inverse solution BAB'A' where if you put both solutions on the pegs they would cancel each other out. So we can relabel them D and D' and treat them as a single peg. Then follow the same pattern for the two peg solution: DCD'C. Expanded, it looks like this: ABA'B'CBAB'A'C'.

[@misaalanshori]

Okay thats a really good explanation. I remember watching a video about this and didn't really understand how it worked in the end.

[@LockPickingCuber]

Really nicely explained!

[@diggoran]

I feel like your rubiks cube background helped with the explanation of this solution :)

[@BFBPSPstandwithukraine]

Wow! You are Master of Puzzles :D

[@Ascended_Spirit]

I have no idea what this is about, but I am here 11 minutes after it was posted

[@Squaduck]

This is a beautiful explanation! IMO you explained it better than that one Tom Scott video hosted by Up and Atom

[@IloveRumania]

xnopyt

[@Andyman620]

this solution was so intuitive, it feels obvious now. excellent job

[@DannySullivanMusic]

It makes me greatly joyful you realize what's up

[@chessie2003]

Good luck!

[@vinegarlegate24]

really great video, just fantastic.

[@masterofdesaster5367]

never thaught that would be so interesting

[@RedStinger_0]

Ooooo, intriguing!

[@bennihtm]

He explained it so good

[@makeup_by_jeki]

Hey you back!

[@Cmin7th]

I love your youtube channel :)

[@Folk_Rocket]

great explanation!!!

[@UltimateDuck]

i tried it and accidently created a rift in the universe

[@bodek]

this reminds me of sethbling when he listed bugs in early minecraft. not a second wasted

[@kyooboy]

That's amazing

[@ramizr]

this is so cool :)

[@stellacheri]

Nice real-life application of algebraic topology.

[@RamiSlicer]

Cool Cary Knight Holler uploaded!

[@j_quyatt]

good job!

[@BariumCobaltNitrog3n]

Only a mathematician would want a sign that falls down.

[@Observ45er]

This is more of a solution looking for a problem than a problem looking for a solution.

[@AminalCreacher]

this is such a good explanation what the heck

[@otesunki]

[x is shorthand for X'] the pattern is XYxy, where X is a lower solution and Y is a single peg 1 peg: A 2 pegs: (A)B(a)b 3 pegs: (ABab)C(BAba)c 4 pegs: (ABabCBAbac)D(CABabcBAba)d N pegs: (N-1 pegs)N(SGEP 1-n)n

[@pyromaniaxe3393]

Hey Cary if you take flowers petals off it makes your little stick man

[@Mozartenhimer]

This is what I'd call the right abstraction.

[@jonathanfuchs6262]

comutator logic like this is a lot of times found in rubik's cubes very interesting.

[@yinyangyt8749]

Interesting video

[@Ranzha_]

carykh: posts a commutator *BLDing and FMCing intensifies*

[@IloveRumania]

Cool!

[@Ludix147]

The quickest solution i found for 3 pegs: C ABA'B' C' BAB'A' Whichever peg you pull, the other moves collide with their inverses. Not sure if it's the simplest possible solution though.

[@nothingtoseemiano9895]

I blinked when i read "6 minutes ago"

[@niyate8331]

Great video. 🙂🙂 - Fellow participant

[@RGC_animation]

Well let's hope he wins the Veritasium contest

[@r3dp9]

One of those quirky situations where abstract math and the real world correlate in a semi intuitive way.

[@bernardoalbano1816]

Very cool method

[@user-jc2lz6jb2e]

Solution for 3: By using the solution for 2, we get (ABA'B')C(ABA'B')'C' This equals ABA'B'CBAB'A'C' (inverse of multiple moves reverses order of moves and inverts each move individually, i.e. (AB)' = B'A'. Inverse of "putting your socks then shoes" is "taking off your shoes then socks")

[@anarchosnowflakist786]

B’A'C'ABA'B'CBA works too

[@itsphoenixingtime]

What about n-pegs? It would be cool if there was a general solution for n amount of pegs

[@user-jc2lz6jb2e]

@@itsphoenixingtime inductively use the solution for the previous case. So if the solution was S for n-1, and say we use the letter X for the nth peg, then you can create the solution XSX'S'.

[@user-jc2lz6jb2e]

@@anarchosnowflakist786 there are infinitely-many solutions. For example you can take any solution and do it twice or three times or however many times you want. You could even switch around the letters, replacing every A with a C and every C with an A. In math jargon: You basically want the elements in the commutator subgroup of the free group on 3 generators that are some product of all three of the generators (and their inverses), which is itself a subgroup. All nontrivial subgroups of a free group are infinite, so there's no unique solution.

[@itsphoenixingtime]

@@user-jc2lz6jb2e So something like mathematical induction?

[@treestomped6954]

I'm not ready for three pegs Cary

[@noel975]

The intro was so short that I didn’t understand the math problem but the explanation was great.

[@GDError]

Flashback to the Mickey Mouse beanstalk

[@vanillawaffle7303]

I love how in the thumbnail it kinda looks like a guillotine

[@IloveRumania]

Robespierre: To the-

[@ZDNAnimators]

good experience!

[@The_NSeven]

Ah, so cubing can help solve random puzzles on the internet ;)

[@Greenfernfern]

We have waited a while month for another video, and this is what we get?

[@shivaprakash4717]

Nice

[@JMIDEV]

New vid yay

[@AsutaIsVkei]

Cary key hole posted!

[@ultrio325]

I remember watching a video about this but I can't remember

[@integrando1847]

yeah, very good, who will win this contest? i am excited, i am also participating

[@stArismynameee]

yes, you can weave the rope such that the sign stays up even if one of the three pegs are removed. you just only weave the rope with 2 pegs, and remove the other peg that isn’t being used. the sign will stay up

[@otesunki]

no, the question was how to weave it so that the sign _falls down_ if _any_ peg is removed

[@JMIDEV]

I got early the same day I subscribed XD

[@degariuslozak2169]

Saw this on a Tom Scott fan video

[@kapilbusawah7169]

So this is what my gf meant by "let's peg tonight". Cool!

[@macrorca]

this is cool

[@DannySullivanMusic]

pretty great you see reality for what it is

[@want-diversecontent3887]

Well that's simple

[@uy-ge3dm]

Didn't expect to see commutators from group theory to show up here!

[@DannySullivanMusic]

yep. _100%_ true