The Rubik's Cube is a Calculator

[@EvilSandwich]

Man. Abstract Algebra is a hell of a drug.

[@mekaindo]

wait till calculus

[@minerscale]

@@mekaindo who's taking abstract algebra before calculus?

[@PFnove]

​@@minerscaleme later today cuz this sounds interesting

[@Lilly-Lilac]

@@minerscale I mean... you could, I suppose, but it is just a bit unorthodox

[@mekaindo]

@@minerscale i dont know i tried to be funny

[@excelmaster2496]

0×0 "Remove the stickers" *removes the stickers* "Remove the stickers" *starts panicking*

[@mr.duckie._.]

then just remove the cube wait what do you do in case of 0x0x0???

[@wj11jam78]

​@@mr.duckie._. Remove stickers Stickers existed on cube, so remove cube cube existed in your hands, so...

[@olegtarasovrodionov]

just fix the rule like this: "Remove the stickers if they are not removed yet"

[@katie-ampersand]

​@@mr.duckie._. hammer

[@cheeseburgermonkey7104]

@@katie-ampersand 0^4: remove the hammer

[@1.4142]

when you can't bring a calculator to a test

[@petersusilo9588]

However it looks really complicated and probably took a long time.

[@petersusilo9588]

Also, for one who aren't able to play rubik, that would be extremely hard. I think it is just better to use the usual method.

[@lollol-tt3fx]

he didnt say that its a good method he said its a method​@@petersusilo9588

[@PFnove]

​@@petersusilo9588no shit Sherlock

[@ynycu]

🍑🧮🧮🧮⚖️😞💀🌝🌝🌚🌚

[@danielleidulvstadpereda5481]

I've been speedcubing for quite a few years, and this is by far the coolest thing involving Rubik's cubes I've come across!

[@TheGrayCuber]

Thank you, I'm glad you enjoyed the video!

[@tymikaseawood2596]

I like it too! 10:51

[@DarkAlgae]

by tricking me into being entertained by modular arithmatic, you earn my subscription.

[@rarebeeph1783]

is this just a sneaky introduction to group homomorphisms?

[@TheGrayCuber]

It sure is!

[@mathgeniuszach]

But without any of the complex math terms that obscure away the beauty of math to the layman viewer.

[@wymarsane7305]

I KNEW IT

[@thegnugod2108]

My gosh doing 3*67 and watching the cube turn back to its starting position was amazingly satisfying

[@cubingbox]

Isn't 67 then the inverse of 3?

[@memetech-]

1:05 hey, I *AM* thirsty, I should drink water.

[@error_6o6]

Fr bro caught me

[@art-of-imagination]

I'm a cuber as well as a math student but I've never thought anything like this. I appreciate you for giving this kinda mind-boggling aspect to see or use the mod. ❤

[@vaughnp3913]

I've spent far too much of my life watching cubing videos on youtube, and this has to be one of my all-time favorites! Thank you for making this - it's excellently done :)

[@TheGrayCuber]

Thank you for watching, I'm glad you enjoyed it!

[@clementdato6328]

“Can we doNO” 😂

[@msolec2000]

"We can make a religion out of"NO, don't.

[@ProactiveYellow]

Working through this is an interesting exploration of the normal subgroups of the rubik's group. It must be a challenge to make sure the algorithms for the larger cyclic group elements end up commutative.

[@willlagergaming8089]

Then my chess board is a graphing calculator. Also most sane mathematician.

[@mekaindo]

if the chess is a graphin calculator, what is checkers???

[@Bangaudaala]

​@@mekaindo binary?

[@mekaindo]

@@Bangaudaala thats a good idea

[@willlagergaming8089]

The Chinese calculator thing. Can't remember it's name

[@brenatevi]

@@willlagergaming8089 Abacus?

[@SamuelLiJ]

Interesting video. You can actually do all four elementary operations (addition, subtraction, multiplication, division) mod 10 on the 3x3, where defined and invertible, for the dumb reason that the symmetric group S10 embeds into the Rubik's group. So you just perform the action corresponding to how the elements 0 through 9 permute. (Actually S12 fits as well.) Note that you can add or subtract any number, but only multiply and (modular) divide by units.

[@alejrandom6592]

Nice, can you elaborate?

[@louisrustenholz7642]

@@alejrandom6592 For S12, you can simply use the 12 edge cubes, ignore both edge orientation and all about corner cubes, and notice that you can perform swaps between any two edges. (These swaps also swap corner cubes, but you choose to ignore it.)

[@louisrustenholz7642]

Then, for the encodings, assign the names '0', '1', ..., '11' to each edge (arbitrarily). For each operation of the form '+4', '-3', etc., encode it as the corresponding permutation (e.g. for +4, you get 0->4, 1->5, ...), which can always be built out of simple swaps. For multiplication/division, do the same game, restricting yourself to invertibles mod 12.

[@louisrustenholz7642]

For mod 10, play the same game and just ignore two edges.

[@MooImABunny]

I'm relearning group theory right now and I'm delighted that this video came out right now 😁 Also, embedding an Abelian group within the Rubik's cube in order to multiply numbers mod n is stick is a wild idea. I know these (non-trivial) subgroups exist, I know how group isomorphism works, but there's a whole other set of steps I'd need to do to come to the idea "I'll use the Rubik's cube to compute ab mod n"

[@PaulFisher]

I’m not very skilled with the Rubik’s Cube so I couldn’t handle the complex algorithms for 10, but I have found one for modulo 2: 1: do nothing and the decoding process: • cube has been destroyed: 0 • cube exists: 1 (orientation unimportant)

[@catstone]

I have found one for modulo 1: 0: destroy cube and the decoding process: • cube has been destroyed: 0

[@R6nken]

​@@catstone i think it's more like: 0: cube Decoding: * cube: 0

[@rayzhao491]

@@R6nken idea for mod 0: you don't need the cube. the cube does not exist. the cube has never existed. multiplication does not exist. the universe does not exist. there is only an eternal void.

[@voliol8070]

Ah, the cliffhanger. Looking forwards to the next video!

[@amogus4868]

This is why Rubik's cubes are loved by many. They are more than some toys.

[@alansun70]

I had one in Louisiana. I never thought of it this way.

[@GhostShadow.0316]

this is literally what I had looking for for the past months! this is so smart, thank you so much to make this video

[@alejrandom6592]

U crazy bro. This math so good it seems forbidden.

[@wymarsane7305]

9:58 The reason for this is, of course, that cube algorithms aren't abelian. The irony here is that commutators are an extremely useful concept for solving twisty puzzles precisely because the piece movements of one algorithm messes with the pieces of another algorithm.

[@Phylaetra]

I love your encoding schema! That is a great way to map modular arithmetic onto a non-abelian group! Although I am very disappointed that R2D2 was not an alg...

[@artemisSystem]

The reason it works is that the subgroup is abelian. But if you want R2 and D2 to be valid algs, you can't do that, because they don't commute, and your subgroup is then not abelian. Though i suppose R2D2 could be a base alg in itself, perhaps. It has a period of 6 though, so not sure it can be used for this? It's not clear to me what determines what cycles you need for a given mod, how that's determined, and if you can have multiple different cycle sets. I guess i'll have to wait for the next video.

[@mr.vladislav5746]

​@@artemisSystem I guess you could just do ℤ/6ℤ or (ℤ/7ℤ)× with R2D2, as both groups have one cycle of length 6. To answer your question, first and foremost, yes, you can have multiple different cycle sets by the Chinese Remainder Theorem. So one 6-cycle is isomorphic to a 2-cycle and a 3-cycle because these two numbers are coprime. However, a 4-cycle is NOT the same as two 2-cycles. So essentially, if we break our cycles into "elementary cycles", they all have a length that is a power of a prime. These are sometimes called something along the lines of "elementary divisors." For a given n, to find which cycles you need (the elementary ones, i.e. powers of primes), you need to analyze the multiplicative group (ℤ/nℤ)× (which has φ(n) elements where φ is the Euler Totient function; in the video he calls these the units, e.g. φ(10) = 4 because the four units mod 10 are 1, 3, 7, 9). This, in turn, is easily Googlable, i.e. to find what product of cyclic groups (ℤ/nℤ)× is isomorphic to. But if you want to find it yourself, there is something called the Structure Theorem of Finitely Generated Abelian Groups (SToFGAG), which states that any finitely generated (thus also any finite) abelian group is a direct product of cyclic groups, i.e. that is what allows this entire exercise. If we take the example of mod 15, there are 8 elements in (ℤ/15ℤ)× (specifically, 1, 2, 4, 7, 8, 11, 13, 14). Then, SToFGAG clearly says it must be isomorphic to one of the following: ➡ ℤ/8ℤ ➡ ℤ/4ℤ × ℤ/2ℤ ➡ ℤ/2ℤ × ℤ/2ℤ × ℤ/2ℤ simply because if an abelian group (we know modular multiplication is abelian) with 8 = 2³ elements is a direct product of cyclic groups, there simply are no other ways. In other words, every abelian groups with 8 elements is isomorphic to one of the three above. Furthermore, since (ℤ/15ℤ)× has no element of order 8 (easily checkable) but it has an element of order 4 (for example 2*2*2*2 = 16 ≡ 1 mod 15), it must be the middle case, i.e. (ℤ/15ℤ)× ≅ ℤ/4ℤ × ℤ/2ℤ so we conclude that the "structure" of the multiplicative group mod 15 is a 4-cycle and a 2-cycle, which can be encoded using the ways discussed in the video (e.g. by setting 2 to be R, 11 to be L2, and then everything else is generated by 2 and 11). However, it's another question whether the Rubik's cube is "big enough" to contain so and so many different cycles.

[@BaranCemCesme]

Great video. My idea for multiplying by 0 was exploding the cube but removing the stickers is way better.

[@itzmetanjim]

0:15 setting it to its default position is not as easy as the other steps (depending on who you are)

[@roxashikari3725]

This was an immediate like and subscribe for me. I love it.

[@arisweedler4703]

If your stickers all had arrows on them, that would introduce additional state for your cube. I once had a Rubik’s cube (or… a derivative of one) that had directioned stickers. The 9 stickers formed a picture with the right orientation. With the wrong orientation it looked scrambled. It was harder for this reason.

[@error_6o6]

Actually, only the center stickers’ direction matter, but I think that should be enough to improve the highest number it can multiply by.

[@arisweedler4703]

@@error_6o6 that’s super cool! I never noticed that with my cube. From what you’re saying, there are only 6*4 additional legal states that are added by taking this type of “sticker rotation” into account? Well maybe it’s not 6*4… but it’s a number small enough to be entirely represented by the orientation of all the middle stickers. I can understand that. I gotta think abt it a bit 😁

[@error_6o6]

@@arisweedler4703 I’m pretty sure the amount of states are multiplied by 4^6 divided by 2 because of weird parity stuff, but that should total to a multiplier of 2048, or, in simpler terms, a lot.

[@have-bear]

If you need to construct a n cycle algorithm, it doesn't have to find n positions for stickers. For example, one can construct a 45 cycle algorithm that move an edge piece between 9 positions and move a corner piece between 5 positions. Unforntunately, a 25 cycle algorithm still can't be constructed with this technique.

[@BlueDog15391]

That's what was bugging me while I was watching. Thanks for answering my question before I asked it =)

[@TheGrayCuber]

Yes, this is a great point! 25 and 23 are problematic because they are only divisible by one prime, and therefore can't be constructed from smaller cycles.

[@aloi4]

Because Z45 = Z9 × Z5

[@Anonymous-df8it]

@@TheGrayCuber Couldn't you have two simultaneous five cycles for 25?

[@Higgsinophysics]

A math video that also reminds you to drinks water. This is just the summit of youtube. Loved the video

[@applimu7992]

The multiplicative group of units are one of my favorite constructions in ring theory!!!

[@yeokonma]

2 of my favorite things in one video. thank you

[@szlanty]

the Gray Cuber doing a video with Cubes mentioned? its more likely than you think!

[@MeepMu]

Removing the stickers was really funny to me for some reason

[@escthedark3709]

This was supremely disappointing when it turned out that you couldn't do 5x8, but then supremely interesting when it turned out that you could do all sorts of other stuff.

[@Salsmachev]

Wow it's like is a slide rule took 10 times as many moves to use and gave you the least significant digits instead of the most significant digits. How... useful?

[@HzyMkwii]

BUT ITS A RUBIX CUBE so it’s cool

[@Danitux11]

I know you won't read this, but this actually just made my day better. This is such a cool concept and I'm very thankful this appeared in my page. Subbed.

[@TheGrayCuber]

I did read this! Thanks for the positive comment, I'm very glad that you enjoyed the video

[@jonathanshuman5859]

This is an amazing video, loved it!

[@skmgeek]

this is a really well-made video!

[@Rhys_1000]

It is actually possible to input even numbers if you use 6 * 5 = 0 in modular arithmetic: Since 6 * 6 = 6, 6 would be peeling the stickers partially and not doing anything else. And the other numbers: 2 = 6 * 7 4 = 6 * 9 8 = 6 * 3 And any face (ignoring the colors) that has peeled the other stickers is 5 Hope this helps!

[@vytah]

You still need to figure out which stickers to remove so that a cube with partially removed stickers can still be unambiguously interpreted as the correct result.

[@Rhys_1000]

5:27 6 and 5 can be these two

[@aloi4]

​@@Rhys_1000 No, because 5×3=5×7=5×9=5 5 need to remove all stickers from the up layer (except the center)

[@Rhys_1000]

​@@aloi4Actually, it still only matters if that specific kind of stickers are peeled to be 5

[@wyattstevens8574]

"Can we do mod one thou-" "NO."

[@error_6o6]

*multiple angry mathematicians staring at you*

[@aviralsood8141]

This is beautiful

[@genandnic]

it goes in the square hole

[@CookieMage27]

bro i was literally glugging water when you said "hm, im a little thirsty"😂😂😂

[@ben_adel3437]

I love this because this year i was feeling so desperate that i wanted to cheat using more 5x5 i didnt because like memorazing a cheating method is harder than actually learning the topic i needed for the exam but it's cool knowing i could've done it

[@cubingbox]

You could maybe also use corner twists as moves, but I don't know if you would use that

[@Zufalligeule]

Really cool. makes me wonder, whether there is a largest prime modulus that can be represented on a cube.

[@TheGrayCuber]

This is a really interesting problem!

[@Knighttwister]

when you said "I'm a little thirsty" i was literally grabing for my water bottle

[@RowanFortier]

You could use a 1x2xN cuboid to do N/2-digit binary multiplication

[@qwerty_qwerty]

rowanfortier?!?! 0 likes 0 replies??!?!!?

[@Sjoerd-gk3wr]

Great video can’t wait for the next one

[@artemis_furrson]

Finally a use for the multiplicative group of integers modulo n

[@adityakhanna113]

Oh my gosh, this is brilliant. I'm definitely very jealous to not have thought of it , considering all of my years of cube experience and "it's a group" propaganda. Couldn't put 2 and 2 together to make a 4 xD Also, i believe you can only use units because the cube's moves are reversible (i.e. a group). I like your idea of 2*5 ≈ 0 mod 10, but just to hammer in the point that this means 2 doesn't have an inverse, which every move on the cube does.

[@vinesthemonkey]

I didn't watch it, but it's an application of the Chinese Remainder Theorem. For finite Abelian groups, there's a unique factorization analogous to the fundamental theorem of algebra. The Rubik's group isn't abelian (for example R U is not equivalent to U R) but the cyclic subgroup generated by a sequence of moves as one element is (for example )

[@juzbecoz]

Mathematical beauty

[@tcaDNAp]

This is really full circle for TheGrayCuber, or should I say... full cycle

[@Sean-Exists]

My brain is melting

[@adityakhanna113]

It might be possible to do for larger numbers by using multiple cubes and exploiting chinese remainder theorem right? To do modulo 1000, you could do 125 (if possible) and 8

[@TheGrayCuber]

You've got the right idea, the CRT does help breakdown the structure, but the units mod 125 still need a 4 cycle and a 25 cycle. There isn't really a way to do that 25 cycle on an nxn

[@adityakhanna113]

​@@TheGrayCuber oh that's so true. The cycles are given by factors and CRT requires the same factors, so they possibly inherit the impossiblities

[@StewartStewart]

I'll watch this later, but my big question going in is what abelian subgroups you used to make it commutative.

[@sk1ller_604]

I am kinda thirsty 😰💔

[@HoSza1]

OK, but can you run DOOM on it?!

[@hoperanker8395]

Yes, but it's very low resolution (3x3), and your hands are the cpu. Alg implementation is left as an exercise for the reader.

[@tepan]

Can I caculate the number by which I can multiply the cube into its starting position?

[@CosmicHase]

How about larger cubes/smaller cubes? Would they be more accurate/less accurate,?

[@Ensign_games]

Do the megaminx one as I would love to see it

[@Sw3d15h_F1s4]

dumb/random idea: since the center squares never change with respect to eachother, can you use the orientation of the cube itself to get larger cycles? mix in pitch, yaw, and roll of the cube, and say define white up with some color facing you as the default starting position?

[@TheGrayCuber]

Yes this would allow at least an additional 4-cycle! But then I think you'd also need to 'fix' the algs, like saying that U must always be the white face instead whatever is on top

[@miners_haven]

I wonder what could be done on a Rubik's Tesseract

[@GuzmanTierno]

Awesome idea!

[@MrConverse]

12:56, small error: the audio says 106 but the graphic shows 107. I’m fairly certain that 107 is correct. Hope it helps. Great video!

[@TheGrayCuber]

yes, thank you. 107 is correct

[@Deixa_cats]

Instructions unclear: My cube is stickerless, how can I multiply by 0

[@LiamHighducheck]

This is so cool

[@quentin611]

I love what you're doing! it's the beautifulest way to introduce someone to homomorphism group

[@idontknow1630]

I just asked my rubix cube the seventeenth root of 6.9 but it hasn't said anything yet, am I doing something wrong?

[@Kyoz]

🤍 Use complex numbers. The real part is a pieces diatance from its home. You can even use negative numbers to represent the same distance on the other side. The complex part is the rotational orientation of the piece. i^2 = -1 which corresponds to a 180 degree move. This is because 180 degree moves will move a piece to the negative side of the cube. We can define the negative side to be 1 tetrahedron, since the cube is made of 2 tetrahedrons. The worst case scenario is the superflip, so we can set that equal to 1 to force all other in-between states to be between 0 and 1. The complex numbers can be multiplied together to get a sort of system state. Then we can minimize this system state and try to approach 0 using these algs as numbers. The system state, because its a bunch of complex numbers multiplied together, can be represented as a single vector in the complex plane. We can then use an exponential decay to slowly reduce the magnitude and angle of the vector until it reaches zero. Gods number is 20, so this should only take t=20 steps. The units are the 6 complex numbers that represent each side and counter clockwise. We do so using the appropriate alg determined by the state after each move. Like having a pilot function in quantum mechanics. Guiding how the system evolves over time.

[@rossthebesiegebuilder3563]

What do you mean that you solve 4x4 parity by ignoring it? You still have to deal with it at some point...?

[@crumblinggolem6327]

Could you bypass the 24 cycle limit by tying two stickers together? like rather than consider just the edge or just the corner, 1 'position' would be (edge1 at pos1 x corner 1 at pos 1), then two would be (edge1 at pos1 x corner 1 at pos 2), etc... up to (edge1 at pos1 x corner 1 at pos 24) then (edge1 at pos2 x corner 1 at pos 1) which would allow for cycles up to 24².

[@TheGrayCuber]

This is a great point! You can get a cycle higher than 24, it's just that 25 and 29 don't work specifically because they're prime powers over 24.

[@PretzelBS]

Fun fact: if you only use prime number bases, you won’t have any “problem” numbers :D

[@creepinator4587]

Thoughts: The prime numbers seem really important for this, since they avoid the "multiply to 0" problem, and seem to be related to the unit cycles Would each prime number just have 1 cycle? If they do each have 1 cycle, than would 23 be the largest prime you can fit on a cube? And therefore are you able to fit any modulus that only has prime factors less than 23? I eagerly await the next videos in this series, since they seem like they'll answer some of these questions

[@creepinator4587]

Scratch that, the "prime numbers have 1 cycle" conjecture is easily disproved by 7 having 3 cycles 2-4-6-1, 3-6-1, and 5-3-1

[@TheGrayCuber]

Yes, primes are important for this and they do each only have 1 cycle. Good observations! It does turn out that you can get higher primes than 23 onto the cube though. 29 needs a 28 cycle, but you can achieve a 28 cycle by mixing a 4 cycle and a 7 cycle. So therefore the problematic primes are ones like 83, where p-1 is divisible by a prime > 24. 82 = 2*41

[@TheGrayCuber]

7 is really interesting! It can be represented as just one 6 cycle, or a 2 cycle and a 3 cycle. It's the smallest number that offers such a choice

[@vytah]

@@creepinator4587 If you have two cycles A and B, you can always combine them into a single cycle lcm(A,B) by doing them both at the same time. If A and B are coprime, this just means A×B.

[@Phylaetra]

Oooh! The megaminx (did I spell that/hear that correctly)? So - 60 - but could you work in base 60? I wonder... What's the largest cycle in base n for mod n, n^2, etc...

[@rodrigoqteixeira]

Signed integers hardcoded style (that one that in computers can lead to -0 beeing a thing). Nice

[@aloi4]

In mod 10 (not only unity) r = remove stickers from the middle layer and the down layer r' = remove stickers from the the up layer 1 = Id 2 = r U' 3 = U 4 = r 2U 5 = r' 6 = r 7 = U' 8 = r U 9 = 2U 10 = 0 = r r'

[@TheGrayCuber]

Beautiful

[@ojosshiroy8544]

This is gonna get virale

[@TheBookDoctor]

I hope you're doing this as part of a paper for some math journal.

[@אביבשקד-נ2ד]

Try using a prime number as the mod so no numbers will multiply to that, meaining you can use all numbers

[@SerKubos]

Thats amazing bro

[@durza4297]

5:49 Does that means we can't multiply 2 by 7 mod 10 ? Can someone explain it to me, because it feels quite unsatisfying...

[@RandomBurfness]

If you throw in a gigaminx or similar but bigger, can you do stuff a megaminx can't? Probably.

[@TheGrayCuber]

Yes! The gigaminx has way more pieces to work with. It's still constrained by the 60 sticker limit just like the megaminx, so it can't fit any cycles over 60. But it can fit more sub-60 cycles than a megaminx

[@zetadroid]

Is the 5-cycle one of the cycles that are used to solve the puppet v1?

[@UnderTheRated]

This is magic; how'd you know I was thirsty? Thanks for reminding me to drink water btw :]

[@JohnGisMe]

How would this work on a stickerless cube?

[@sinetangentsecant1102]

does the number of stickers really matter? you can make an alg on an nxn that has a period of 24 by composing an 8-cycle (of corners and edges together due to permutation restrictions) and a 3-cycle together, which would make for a valid 24-cycle right? edit just realised it’s a 25cycle which is 5x5 so you can’t do it by composing smaller primes nvm

[@Tech35_]

Awesome

[@鄿]

What is the largest modulus possible on 3x3?

[@TheGrayCuber]

533,520 is the largest that I have found

[@rodrigoqteixeira]

Possible definition for units those whose gcd with b is 1, those that are co-prime with b

[@TheGrayCuber]

This is a very good definition for the units!

[@Phylaetra]

Tres cool!

[@mrdraw2087]

I probably need to watch 1000 instruction videos first in order to make sense of this.

[@VantasiaGD]

I did not watch the fuII thing but even just the beggining teIIs me that this is way too high quaIity for so IittIe

[@fgvcosmic6752]

I have a question about mod 1000; cant we consider 2 stickers at a time? A 25-cycle that depends on the relative position of _2 stickers_ And consider more than just a single face?

[@vytah]

No. The Rubik's cube group has no elements of order 25. The Z1000 has an element of order 25 (41). Therefore Z1000 is not a subgroup of the Rubik's cube group.

[@artemisSystem]

the problem is that 25=5x5, so the only way would be to make two 5-cycles, but they're both 5-cycles and are therefore always aligned, so that doesn't work

[@PFnove]

I have no idea what I just watched but later I'll get mod 97 multiplication working on a 3x3

[@antoniusnies-komponistpian2172]

It would probably have been easier if you would have explained it for addition first and then explained isomorphisms 😂 Then you would have drastically expanded the potential watchers

[@YEEEEEEEEEEET999]

Can we do mod 200?

[@TheGrayCuber]

Yes! You'd just need to add an extra 2-cycle using a 101-alg

[@Memzys]

thank you thegaycuber for this awesome youtube video

[@TheGrayCuber]

You're welcome

[@tomkerruish2982]

Typo

[@SpotErrOne]

типо

[@alejrandom6592]

I thought this was gonna be about how you can do some matrix multiplications [SO(3)] by just rotating ur cube

[@cheeseburgermonkey7104]

12:26 Does someone know how to explain why this doesn't work

[@TheGrayCuber]

If there were a 23-cycle, then 23 of the stickers would move around to each other's locations. But then that 24th sticker must not move - otherwise it would hit one of the 23 and interfere with the 23-cycle. But if that 24th sticker can't move, then neither can the other stickers on the same piece, which also interferes with the 23-cycle

[@Anonymous-df8it]

@@TheGrayCuber What's the largest cycle that *_does_* work?

[@TheGrayCuber]

1260 is the largest possible