"I'm just an average cuber who decided to make a video about one of his favourite toys." You just nailed the magnificence of SoME there. This video wouldn't have been created if you hadn't felt inspired to make a contest entry, and the world is a better place for it existing. Rubik's Cube happens to be one of *my* favourite toys too, so maybe I'm a tad biased here. I own so many of them, including one that's identically reflective on all sides, but sides vary in **size**, so a scrambled cube is actually non-cubic! That's fun to put in front of people :)

A detail: the group does not consist of the cube, or configurations of the cube. It consists of all the actions you can perform on a cube.

@ethanbottomley-mason8447

8 ай бұрын

That is not really the case. The rubiks cube group really does correspond to the legal configurations of the cube. Two configurations are composed by following the moves required to do the first, then following the moves required to get the second. This then realizes the rubiks cube group as a subgroup of S_48. There are lots of groups that act on the rubiks cube though, such as Z^*5, the free group on 5 elements, which has a transitive action given by turning 5 of the faces. Note that the last face may be obtained as a combination of the other 5. This action is not faithful.

@BryanLu0

8 ай бұрын

​@@ethanbottomley-mason8447The state of the cube can be thought of as the result of a series of actions, but a group refers to actions performed on an object, not the actual state of the object.

@theodoreastor3443

6 ай бұрын

Building off of what the other person said, if group of actions you can perform on the cube would be the free group on 6 elements. You need some relations on the actions that distinct sets of moves represent the same thing. e.g. R^4=U^4=e

@ethanbottomley-mason8447

6 ай бұрын

@@theodoreastor3443 If you add relations to your free group to encode all of the relations of the rubiks cube, then you end up just putting a group structure on the collection of legal configurations of the rubiks cube. In that case you have a transitive faithful action. If you have a group G acting transitively and faithfully on a set S, then you can put a group structure on S as follows: Fix some element A of S, then for any two elements B and C of S, since G acts transitively, then there are elements x and y of G such that xA=B and yA=C, then define the group operation of S to be BC=(xy)A. Note that since G acts faithfully, then x and y are unique, so the operation is well-defined. You can check that this indeed makes S a group, the identity element will be A. Furthermore the group structure on S is isomorphic to G. So if you take a free group on 6 generators, all the face rotations, then impose the relations of the rubiks cube on that free group, then that free group yields a group structure on the set of legal cube configurations which is just the rubiks cube group and so the free group with those relations is exactly the rubiks cube group.

OMG. I love the use of pigeonhole principle, It so useful yet not many people aware of it

Your English was great! Really interesting video and made groups seem a lot more relevant and useful/interesting than when I first learned about them!

it is such a shame that you have so few subscribers and so few views for such a fun and informative video. I'm both a speedcuber and someone who really enjoys maths , so getting to relearn about groups and group theory in the form of a rubik's cube was really fun! Don't let the views discourage you and continue on making amazing content my friend.

@HereToSin

8 ай бұрын

I second this comment!

@samueljehanno

8 ай бұрын

+1

great video! Loved rubiks cube since middle school and now am a math major as a senior in college

I explored something like this when first learning how to solve a rubiks cube. And I decided to take it a step further for fun. What if I apply a rotation like R U and then reposition the cube (I forget the notation)by taking the bottom face and making it the right face for example, and continuing on like that. It was always a whole number multiple of rotations compared to the same thing without the 'repositioning' of the cube added into the moves which I thought was pretty cool

@kinexkid

10 ай бұрын

It got even more fun when I got my first 5x5x5 and could take the same pattern and apply it. I ended up with some crazy patterns that I knew would always go back to being normal no matter how many rides I had to repeat it. It was like a fidget spinner, before fidget spinners existed lol

@imtoxiiq423

10 ай бұрын

Really interesting to hear! I've left out cube rotations on purpose since they easily confuse non-cubers and it gets hard to keep track of which layer is which. I was able to do that because your algorithm (R U z' [z' being the cube rotation]) can be (like every other algorithm) described with the fundamental moves R U L D F B. It would be (R U D R L D U L), but that is just very inconvenient to write down or execute on the cube in real life, so we introduce more notations that are closer to how cubers handle the cube while solving. That is where all the different turn metrics like HalfTurnMetric, SliceTurnMetric etc. come from :)

This is great! Very impressive for such a small channel :)

Oh yeah, this guy cubes

I'm just an average cuber that decided to watch! Fascinating subject!

A very high-quality video. I wonder if there are any other principals from group theory that show interesting properties of the Rubik's cube that might even be considered counter intuitive?

Great video my dude, well done.

Very well done as an intro. Maybe in the next part you could show how some actions leave parts of teh cube invaraint, discuss commutators and normal subgroups and show a logical way of constructing an algorithm that solves the cube.

A very slick production and your English is excellent. I do hope this is the start of a series on the Rubik Cube and group theory. You could go on to commutators next.

This is among the more interesting cube videos I've seen! 🤓 🖖

Amazing content

A nice thing about Rubik’s cubes is that, if a sequence leaves the cube the same, then it’ll leave every other valid cube the same. You addressed this in the video, but this isn’t with any group. If you’ve got an arbitrary group acting on some mathematical object, then if an element g leaves the object the same, then taking another element h, applying that to the object, and then applying g will not be the same as only applying h. I feel that, if there’s a 1 to 1 mapping between states and elements in a group, that group is like the “natural” group for that object, if that makes sense. Quality video!

@-minushyphen1two379

5 ай бұрын

How would that be possible? Isn’t g in the kernel of the homomorphism into the symmetric group on that object, so that it acts as the identity on the object?

@enpeacemusic192

5 ай бұрын

@@-minushyphen1two379 the kernel of the group action is the intersection of the stabilizers, not just the stabilizer itself. In this particular case it is the stabilizer, but that’s rare. It means that the quotient group G/G^x acts freely on the set of Rubik’s cubes

I spent an afternoon looking into the Rubik's cube group theory - see wikipedia, for instance. There's a lot more there.

Nice video. Don't have anything to say really, just want to boost your video by making this comment because you really deserve it

I tried to solve the rubiks cube using group theory as well though not successful as of now. some nice insights though on a 2x2 cube: define an equivalence class that says 2 configurations (or elements in a general group case) are equivalent if there is a rotation that moves the whole cube mapping between them. it is clearly a normal subgroup (physics says that rotating an object shouldn't change its inner composition) and we can consider the quotient group. this reduces the cube by 4*6=24 as implied by the orbit stabilizer theorem (stab one side orb all sides). the quotient group is generated by 3 elements a,b,c , the orthogonal rotations in 3d space. they all generate a C_4 group. using lagranges theorem and trying random composition using only a,b shows that the group generated is above 500 elements. in particular it is divisible by 2,3,5,7,13,17 2 and 3 at least twice(unless I miscounted which would be a little embarrassing) the group is not abelian but conjugation almost always acts like an identity in the sense that the group generated by conjugation is isomorphism to that is C_4 the rubik's cube sounds a lot like free groups which is a very general way of constructing larger groups. good luck finding a group theory solution, apparently 3x3 rubik's cube is very hard and there are relatively recent papers are them. 2x2 should be much more doable and studying the group generated by 2 orthogonal rotation is a good start :) '

Great video, but I'd dispute the title, as the pointer clock is no more than Z_12.

Groups are descriptions of symmetry

14:15 isn't g^0 the neutral element? This proof shows that g^n=e for any positive n, but to say generally "for some n", it seems trivial.

@imtoxiiq423

8 ай бұрын

Yeah, the wording is a bit vague at that part. It should be "positive n" rather than just "n"

Which animation engine did you use for this video? OpenGL or Cairo? Because the cube looks realistic.😮

@imtoxiiq423

8 ай бұрын

My friend made and animated the cube by himself in Blender. I was blown away too when I saw it for the first time :) If I remember correctly, it was modeled after the Moyu RS3 M 2020

@PMA_ReginaldBoscoG

8 ай бұрын

@@imtoxiiq423 That's cool man. Kudos to both of you for bringing this masterpiece out in the Internet.

Sorry for the second comment but this is not applicable for an edit: In a finite group G and an element g, the set “generated” by g (which you called S in your video) is something called a subgroup of G. It’s basically a group embedded in G. i.e, S is a subset of G which adheres to the group rules, or axioms. Due to something called langrange’s theorem, which states that the order (size, or number of elements) of a subgroup must divide the order of the “parent” group. The proof is not too hard but does require more group theory knowledge than is presented here. Since S is a subgroup, the order of S must divide the order of G, and since n is the order of S (since after g^(n-1) it just wraps around to e, g, g^2, etc.), n must divide the order of G. Now it’s not very useful here, since G is absolutely monstrous, but it’s still an important fact, and shows that n isn’t just some arbitrary number that happens to work

youtube says this video is in 1080p but it looks to be 1080i

Random annoyance - why aren't speed records done as averages? The record is just whoever got a lucky starting configuration

@imtoxiiq423

8 ай бұрын

They are actually done as averages to reduce the amount of luck involved! The official standard for competitions is five solves. The best and worst solve do not count and then you take the arithmetic mean of the other three. The record for this average is currently 4.48 seconds by Yiheng Wang (www.worldcubeassociation.org/persons/2019WANY36 ). Also competition scrambles are checked to not be trivial, so that for example a one move solve will not happen in the future. Max Park's 3.13 is certainly lucky, but not many could achieve a similar time with the same scramble. Scrambles may sometimes be luckier than others, but the skill to recognize and execute a good scramble is still needed.

Minor correction: the individual pieces of the cube are referred to as "cubies", not "cubits".

@imtoxiiq423

8 ай бұрын

Yeah I've been made aware already. I probably misread it during my research.

I don't get it doesn't that group get formed by _transitions_ from cube state to cube state? why do you say you count group size when you count cube state number? one is, like, square of another!

@-minushyphen1two379

5 ай бұрын

Yes, the group is actually formed by sequences of moves, it’s just that there are the same number of states as sequences of moves, so it’s easier to count them. It’s also not true that one is a square of the other. You might think that you have to multiply the number of starting states by the number of ending states, but actually you overcount.

they are called cubies not cubits

@imtoxiiq423

9 ай бұрын

Oh... my bad, I probably misread it during my research for the video.

@Yugemostsuj

3 ай бұрын

I heard it as cube-ette like the French diminutive. Whether or not it was intentional that's what I'm calling them from now on