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

[@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]

🍑🧮🧮🧮⚖️😞💀🌝🌝🌚🌚

[@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

[@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

[@ajiprnk4821]

@@TheGrayCuber i have a question can a 4d rubiks cube do mod 1000 ?

[@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

[@spicetea4060]

I haven't been in a group homomorphism since college

[@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?

[@Boxytablet]

@@cubingboxSo then he would be using mod 70

[@PMA_ReginaldBoscoG]

​@@Boxytablet He is using modular multiplication, not addition. You are right about 67 being the inverse of 3. However the underlying operation and the number we use to divide the sum or product differs.

[@Boxytablet]

@@PMA_ReginaldBoscoG bro I’m not talking about addition, it’s because 67 is 3 less than 70, so 67 in mod 70 could be -3

[@PMA_ReginaldBoscoG]

@@Boxytablet you really are talking about addition. In the case of the set of all integers, -3 is the additive inverse of 3, whereas in the case of the set of integers modulo 70, 67 is the additive inverse of 3.

[@clementdato6328]

“Can we doNO” 😂

[@msolec2000]

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

[@itzmetanjim]

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

[@Rb_Perm]

No its pretty Easy actually

[@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.

[@memetech-]

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

[@error_6o6]

Fr bro caught me

[@obsidianflight8065]

not only once, he made me think about it 3 times! I was thirsty sure, but now I'm REALLY thirsty and can't contain myself anymore, to the water I go!

[@PMA_ReginaldBoscoG]

Hi Thirsty! I'm your subscriber.

[@romywilliamson4981]

I love that you are teaching the essenece of group theory and elementary number theory to cube-enthusiasts, without ever saying that's what you are doing :)

[@Higgsinophysics]

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

[@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. ❤

[@philia08]

Another speedcuber here! Hope You aren't fan of Max Park or somebody random from new gen speedcubers ☺️

[@kyanilcauli9002]

When the obscure theorems in Dummit and Foote finally appear to you in a visual form. Bless you. You just made my day.

[@PMA_ReginaldBoscoG]

It's a pain in the ass because our professor beat the shit outta us!😂

[@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!

[@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.

[@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.

[@beansprugget2505]

I'm not a cuber so maybe I'm missing something obvious, but I don't understand your addition. Are you saying that to add, eg 3 + 4, you first do 3, and then you make the 7 permutation? Where does the 4 come in? Or is it that one set of moves goes from 1, 2 , 3 etc each iteration (like how repeating U adds one in mod 4).

[@tl_dragonstars2877]

I was brushing teeths while watching, this : => 1) I have to watch the video a second time => 2) Iwon't be able to sleep because of too intense curiosity

[@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.

[@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?

[@voliol8070]

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

[@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.

[@katherinek6166]

"Can I interest you in a video about group theory?" - "No." - "How about a Rubik's Cube one?"

[@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.

[@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"

[@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?

[@ajiprnk4821]

woudnt make you have to use multiple algorithms for one number?

[@wyattstevens8574]

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

[@error_6o6]

*multiple angry mathematicians staring at you* Edit: btw this reply was made before watching this video so I thought the comment said “can we do mod 1 though” (oh well too late to change it now)

[@RowanFortier]

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

[@qwerty_qwerty]

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

[@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.

[@HyperCubes]

one of the most impressive and informative videos I've seen, and bonus points for using a rubiks cube (I'm a speedcuber so it makes me so happy to see videos like these) great video!!

[@gwenhidwy]

As a lover of group theory and Rubick's cube I thoroughly enjoyed this! You should explore comutators in the same light

[@MeepMu]

Removing the stickers was really funny to me for some reason

[@jesusvera7941]

oh man i was having a terrible insomnia, but this video had me falling sleep like 7 times, thanks you for your help! now... i have to take advantage of this feeling.

[@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

[@applimu7992]

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

[@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

[@Fur0rem]

Wow that was such a great video, didn't know rubix cubes could go that deep! Now it makes me wonder if this, or a similar concept, could somehow be used by computers, with arithmetic modulo 256, 65536, (all the powers of 2), since right now they still do multiplication the old fashioned way

[@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.

[@kotmooson6459]

This video shows that pretty much anything with memory and operations over it can be computer of some sort if really want to

[@alejrandom6592]

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

[@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

[@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

[@Knighttwister]

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

[@szlanty]

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

[@BaranCemCesme]

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

[@cubingbox]

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

[@yeokonma]

2 of my favorite things in one video. thank you

[@genandnic]

it goes in the square hole

[@cosmicScience-p5m]

Just a a nuclear fussion reaction started in my brain. 🤣🤣🤣 I can power whole city by this video.

[@KLR-3]

I wonder if anyone has adapted this as an alternative to the current popular blindfold method.

[@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

[@jonathanshuman5859]

This is an amazing video, loved it!

[@whitefwr]

So you made a calculator from the Rubik's Cube? Congrats!

[@skmgeek]

this is a really well-made video!

[@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

[@abhijeetghodgaonkar]

Insane yo, good explanation!

[@d1tnhauxa0rau]

my act of drinking water predates the thirst reminder in the video by 3 seconds

[@roxashikari3725]

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

[@reyuki-i]

12:27 what subject I must learn to understand the reason behind the technicalities?

[@reyuki-i]

Ah, it seems about abstract algebra (just read the video description)

[@TheGrayCuber]

depends on which technicalities. Abstract algebra is needed to determine which cycles, and then familiarity with the Rubik's cube to determine what can fit on the cube. or just wait bit because the next few months I'll be posting videos about both topics!

[@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

[@reyuki-i]

9:10 are you just guessing each possibility and ended up with 7^4, or is there a systematic way to arrive at that expression?

[@TheGrayCuber]

This is largely just guessing

[@Sjoerd-gk3wr]

Great video can’t wait for the next one

[@titimathrosgui5109]

can you give a link or a list of every modular multiplication below a number, like 100 ?

[@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 )

[@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!

[@TheGrayCuber]

991 is the highest prime modulus possbile

[@Zufalligeule]

@TheGrayCuber wow, it's surprisingly large! I've expected it to be around 100-200.

[@TheGrayCuber]

@@Zufalligeule 1260 is the maximum cycle possible on a cube, but 1261 is not prime. 990 is the second highest cycle that fits on a 3x3, and then adding 1 we get 991 which is prime!

[@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.

[@Leonardo-cw1dd]

holy shit you earned a sub from me. the math behind this is insane, it broken my brain lmfao

[@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.

[@ConradoPeter-hl5ij]

really creative thinking like this

[@aviralsood8141]

This is beautiful

[@tepan]

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

[@rossthebesiegebuilder3563]

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

[@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!

[@onur-karakus]

you should use de esser for your audio records

[@pradeepmalar327]

"I'm a little thirsty" and "this answer is wrong" are the 2 things I thought.

[@אביבשקד-נ2ד]

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

[@StewartStewart]

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

[@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.

[@ptitbgdu8040]

If you combine edges and corners cycles, can't you do more than a 24-cycle on a 3x3x3 cube? I thought there were 120-cycles in this case but maybe it doesn't work for what you want to do idk.

[@TheGrayCuber]

You can do cycles larger than 24 on the cube - it's just specifically 25 that doesn't work. More generally, any power of a prime higher than 24 won't work, such as 25, 27, 29, 32, etc

[@Candlest1112-hb8tj]

Pretty nice video!!!!!!!

[@CosmicHase]

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

[@reyuki-i]

How the hell does he come up with this brilliant idea!? awesome.

[@miners_haven]

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

[@zetadroid]

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

[@lionelinx7]

Beautiful

[@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...

[@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

[@thetopnick32]

What is the largest modulus possible on 3x3?

[@TheGrayCuber]

533,520 is the largest that I have found

[@TheBookDoctor]

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

[@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

[@MitruMesre]

can you use multiple cubes for larger cycles? or 4d cubes

[@TheGrayCuber]

Multiple cubes would allow for more cycles, but not larger cycles. But yes 4d cubes would be able to fit larger cycles since there are more stickers

[@MitruMesre]

@@TheGrayCuber rubik's cubes have 54 stickers. surely i can make any cycle up to and including 54 by cycling stickers instead of making actual moves? though swapping "blue" with "blue" wouldn't be possible to notice

[@TheGrayCuber]

Sure, if you allowed peeling off and moving stickers as a legal move then you could go all the way up to a 54 cycle. But then it would no longer matter that you're using a cube since you'd get the exact same properties with any 54 objects that can be permuted

[@telotawa]

can you do powers of two? or something else? what numbers can be done?

[@TheGrayCuber]

Powers of 2 work up through 64, but 128 would require a 32 cycle and therefore would not fit onto a 3x3

[@lanevalhalla1225]

i actually derived a formula for optimizing soludtions using a similar concept to that of yours

[@Ensign_games]

Do the megaminx one as I would love to see it

[@Aa49Aa494aA]

12:55 106? That's 107

[@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

[@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

[@Anonymous-df8it]

@@TheGrayCuber Why?

[@TheGrayCuber]

i dont know how to answer the 'Why?' other than 'nothing bigger fits on the cube'. A sort of satisfying answer is to look at the prime factorization. 1260 = 4*9*5*7. This is about the max that even fits on the cube. You can push the power of 2 up to 16 but that would require dropping either 5 or 7, resulting in a lower cycle. You can also use 11, but again that would require dropping some combination of other cycles that would result in something lower. And then any other prime factorization is impossible. Any possible cycle must divide 16*9*5*7*11

[@V530-15ICR]

13:38 background noise

[@YEEEEEEEEEEET999]

Can we do mod 200?

[@TheGrayCuber]

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