I knew someone who had a ton of algorithms for very specific cases. So if he needed say an edge 3 cycle that did not disturb anything else he knew a ton of different edge 3 cycle for all kinds of different 3 edge combinations.
One day several of us were sitting around playing with our cubes, and he hit some position where none of his algorithms worked and was stuck while he tried to find a new algorithm for that position. We asked why he didn't just use a conjugate of one of his other algorithms.
That was how the rest of us did it. I for example knew one edge 3 cycle that didn't disturb anything else. Call it T. If I needed any other edge 3 cycle I'd just find some transformation g such that gTg' did what I wanted, where g' is the inverse of g. gTg' is called a conjugate of T, and the set of all gTg' is called the conjugacy class of T.
It's easy to find g, because it doesn't matter how much g messes up the rest of the cube. All that matters is that it moves the 3 edges I want to permute into the position they need to be for T. Whatever else g messes up will be restored by g'.
He was astounded when we showed him that approach. It had never even occurred to him that such a thing was possible.
The funny thing is he was a math major, and had taken (and passed with a good grade) an abstract algebra course. He definitely knew about conjugates of group elements. He had just never noticed that group theory was applicable to Rubik's Cube.
Maybe I'll just show my algebraic ineptitude, but how do you know that g' restores "everything else"? I get that gg' = I, of course, but gT*g'? Rubik cube algebra isn't commutative, is it?
The part that is important is T has a very specific effect, such as rotating three corner pieces, that doesn’t change most of the cube. Then you pick a g that doesn’t affect those pieces either (except in the way that you want) so that gTg’ does what you want.
It doesn’t work in general for any arbitrary T of course since it’s not commutative as you noted. This is more of a cubing thing than an algebra thing.
You are right that it is not really an algebra thing. It's a bijection between finite sets thing. If you apply a bijective mapping from set A onto set B, then apply some permutation to B, and then apply the inverse mapping, you get a permutation of A that has the same structure as the permutation you applied to B.
By "same structure" I mean that if written as a product of disjoint cycles has the same number and sizes of cycles.
Abstract algebra is probably the first place most people would encounter it though, in the context of conjugates in groups.
Let's step away from the cube and consider permutations on a set of N items, labeled 1, 2, 3, ..., N. In the example below N is at least 9. We start out with item 1 at position 1, item 2 at position 2, and so on.
Suppose we want to move what is at 4 => 8, 8 => 9, and 9 => 4, leaving everything else unchanged but the only permutation we know that moves 3 items in a cycle leaving everything else unchanged moves 1 => 2, 2 => 3, and 3 => 1. This latter permutation we'll call T.
Consider any permutation g that does 4 => 1, 8 => 2, and 9 => 3. It may or may not move other things. The inverse of g, g', does 1 => 4, 2 => 8, and 3 => 9.
Let's work out what happens if we do gTg'.
First let is just look at items 4, 8, and 9, which are at positions 4, 8, and 9, respectively. g takes those to positions 1, 2, and 3. Then T takes positions 1, 2, 3 to 2, 3, 1, so what we have in positions 1, 2, 3 is items 9, 4, and 8 in that order. Finally g' takes what is in positions 1, 2, and 3 to 4, 8, and 9, so we end up with item 9 at 4, item 4 at 8, and item 8 at 9.
So for positions 4, 8, and 9, gTg' does 4 => 8, 8 => 9, and 9 => 4.
g necessarily had to move whatever was in 1, 2, and 3 out of the way to make room for 4, 8, and 9, so we have to consider what happens to what was at 1, 2, 3. Let's just look at 1. It has to go somewhere. Call that position p. So g does 1 => p, where p is not 1, 2, or 3. g' does p => 1.
T only moves things in 1, 2, 3, so after g moves whatever was originally at 1 to p, T leaves it along. g' then does p => 1, putting it back where it came from. So we see that gTg' does not move 1. Same reasoning applies for 2 and 3.
If g moves any x else other than 1, 2, 3, 4, 8, 9 we can use the same argument. x => y for some y that is not 1, 2, or 3. T does not move y. g' does y => x, putting x back where it came from.
Thus we can conclude that gTg' only moves 4, 8, and 9.
Another way you can visualize why it works when applied to the cube is by cheating a bit. First do g legitimately by actually doing the moves on your cube. Then instead of actually doing the moves for T just repaint the faces on the cubies that T permutes so it looks like you did T. Then legitimately do g'.
Since you haven't actually done T the only actual moves you have done are gg' which as you've noted is of course I, so everything is back where it started but with some of the faces on some of the cubies repainted. The repainted faces are exactly those that would have ended up moved if you had actually done T. The faces not repainted are those that would be undisturbed by gTg'.
The key here is that all these mappings are finite, one to one, and onto (bijective if we want to get fancy). If you apply such a mapping from A to B, then permute k elements in B, and then apply the inverse mapping, you end up with k elements permuted in A.
Most of the best speedcubers from the late 2000s era when I used to follow the "scene" went on to work in FAANG or mathematics research or quant something similarly difficult and analytical. There's no real future in cubing, I saw it as a way for especially gifted teens to demonstrate to the world how excellent their minds were at computation and pattern recognition. Not dissimilar to track or football or spelling, some young people possess the individual burning flame to want to be the best and personally I love watching them do so.
> Returning to New Orleans in late 1859 at the age of 22, he retired from active chess competition to begin his law career. Morphy never established a successful law practice and ultimately lived a life of idleness, living on his family's fortune. Despite appeals from his admirers, Morphy never returned to the game, and died in 1884 from a stroke at the age of 47.
Same thing happened with Tetris. For a long while it was dominated by a few folks who could hyper tap, but then someone figured out rolling—which is something “anyone” can do + even faster than hyper tapping—so that seems to be what everyone is adopting at top tier. For rolling it started with one guy and then a few months later there were like half a dozen including a couple of the hyper tappers.
Technically 144 a week, but yeah. That's a stretch even with spaced repetition. On the other hand OP is doing 15 second solves, i.e. 240 solves an hour (1hr per day) plus a lot more on the learning day, so cycling through all the 1LLL algorithms fairly frequently.
The context of the video is to demonstrate to technical speedcubers that he's achieved the feat he claims. By showing his solves on video, speedcubers can tell that he's using the algorithms he claims and not more common techniques to obfuscate the achievement. The length of the video is to provide even further evidence of the claim.
Edit: The achievement is that he's been able to minimize the number of turns he needs to complete the cube by memorizing ~4000 edge cases and the specific turns needed to solve the cube from those configurations, as opposed to generalized algorithms that require memorizing less edge cases at the expense of more turns.
This has been a known possibility since 2011, but this is the first documentation of someone demonstrating mastery of it.
Hmm I see, thank you. How do we know that he hasn't just learned the specific cube he's solving in the video, though? Is there some verifiable source of randomness?
I'm not a member of the speedcubing community (only adjacent to it through acquaintances), but as far as I know, the website he's using is standard in the community and those in the know who have viewed the video acknowledge it as legitimate.
Technically not a skip, just finishing the last layer using one look (1LLL). A last layer skip in the CFOP method would be when the last layer ends up in the solved configuration as the first two layers (F2L) are finished.
That is properly, genuinely impressive. I spent quite a lot of time finding the “simple” algorithms on my own, but this is a whole other world. Remarkable!
This is too much for a human.
It may be that there was another computer screen not shown on video, displaying particular steps to perform. Not saying I think it's fake, but the supposed proof video does not actually prove anything.
Consider, from someone who knows speedcubing, that it’s not 3,000+ random algorithms, more like a decision tree of algorithms. Many are similar and related; many are rotated or mirrored. Long practice and pattern recognition make most algorithm selection second-nature. Yes, there’s a ton of memorization involved, but the limited problem domain, and therefore easy mental modeling of it, make this exactly the kind of task at which humans excel.
Professional chess and go players can replay many, sometimes all, of their past games from memory. Surely that’s a greater feat.
Agreed, it seems in the ballpark of what a well-prepared 2400 (which is to say, very good, but not close to enough to make a living from playing, never mind world class) chess player might know in terms of opening theory.
As long as you can get them into muscle memory, they're literally hard to forget. I stopped cubing for 4 years, every algorithm was exactly where I left it mentally.
Speedcubing is extremely communal. Once the 6-second barrier was broken, 3 other people did it the same day. It's a mind game, and I'll bet we'll see another person who does this within a year.
Interesting. Maybe it was a psychological barrier like you say. Certainly it's easier to motivate yourself to do something if someone else has provably done it!
This is why we can't have nice things internet friends.