Видео

Yes, you can do that. :) Is there any setting that I need to change for that?

@@cuberaccoon 👍👍

@@cuberaccoon No, it's not necessary to change nothing 👍

I didn't know that channel :)

What do you mean?

@@davidgrin9998 Danke!

Hmm?

Ja. Ich habe mittlerweile da noch Punkte drauf gemacht um die zu unterscheiden. Und gelöst habe ich den Cube auch bereits! :)

Siehe hier: www.reddit.com/r/Cubers/s/KzRcOd76h9

Ich habe mir auch einen Sudoku-Würfel aus einem preisgünstigen Sengso gemacht. Es stellte sich heraus, dass es ein sehr schwieriges Rätsel war. Komplizierter als der „Molekül“-Würfel. Ich habe auch vor, mir einen 4x4-Sudoku-Würfel zu machen. Ich hoffe, dass mein Kopf beim Zusammenbau nicht in Stücke zerfällt =)))

@@wowkusw2649 sehr cool! Lass es mich bitte wissen wie der 4x4 aussieht wenn er fertig ist.

@@cuberaccoon Sudoku 4x4? Sie müssen Aufkleber von 1 bis 16 ausdrucken und sie entweder der Reihe nach (was viel einfacher ist) oder chaotisch auf jede Seite kleben, aber das wird sehr schwierig sein.

Yes. This is also the maximal order in the group of the 2x2 cube.

Thanks! This avoids a rotation I assume. I'm not good at finger tricks so I would probably still do some awkward things here ^^

Das sind Standardnotationen. F = Front R = Rechts U = Oben Die Drehungen macht man standardmäßig im Uhrzeigersinn, der Strich zeigt die Drehung entgegen den Uhrzeigersinn an. freshcuber.wordpress.com/2012/01/18/zauberwuerfel-notation/ Wenn du weitere Fragen hast, stelle sie gerne.

Thanks for the suggestion! I will think about it.

Nein. Das zeigt lediglich dass es natürliche Zahlen n < m gibt sodass A^n = A^m. Denn dann sind die Wiederholungen halt 1,A,...,A^(m-1). Was hier behauptet wird ist dass man n=0 erreichen kann. Dafür braucht man eben die Kürzungseigenschaft wie im Video besprochen.

Ich verstehe was du meinst, aber aus A^n=A^m mit m>n folgt 1=A^0=A^(m-n). Oder anders ausgedrückt, wenn du an zwei Stellen die gleiche würfelkonfiguration hast, bist du zu beiden jeweils mit den gleichen zufolge gekommen, also sind deren Vorgänger gleich, die davor und so fort. Wir haben eine endliche Gruppe, also haben alle Elemente endliche Ordnung.

Du hast mit der kürzungseigenschaft ja recht und danke für das schöne Video, versteh mich bitte nicht falsch.

@@vrzudq Ja schon klar. Ich habe das Video so gemacht dass man keine Gruppen kennen muss. :)

Okay, mal ganz ohne mathematische begriffe: Mach ich eine zugfolge, so kann ich sie durch die rückwärts-umgekehrte auch wieder rückgängig machen. Somit sind zwei Würfel die nach Anwendung der gleichen zugfolge im gleichen Zustand landen, auch vorher schon gleich. Wende ich nun bei einem Würfel eine zugfolge beliebig oft an, muss es aufgrund der Endlichkeit der Zustände eine dopplung geben. Unter allen dopllungen eine früheste. Besteht diese früheste nicht aus dem anfangszustand mit dopplung, so würde wäre der Vorgänger auch schon eine dopplung. Widerspruch. Also wiederholt sich der anfangszustand nach einer endlichen Anwendung der zugfolge.

The green orange edge and the green and orange centers just relax all the time. Don't bother at all to join this experiment.

And I almost did at 18:57 and 30:31 🚨

Thanks! What is the name of the video and the RUclipsr?

​@@cuberaccoon The video is "How to solve a Rubik's Cube in 1260 moves" by Tanner Gwaltney

Zum 3x3 habe ich auch ein Video auf meinem Kanal. :)

Thank you, but I would say those people are legends who actually found this sequence of moves and proved that 1260 is the maximal order. :)

@@cuberaccoon your welcome and honestly I can’t do that my self

@@cuberaccoon you are awsome

Thanks! I am afraid that many (or most) of my videos will be in German, though. Maybe I will change that, not decided yet. Would it help you if I flag English content somehow in the title? Or the thumbnail? I appreciate your feedback, as it will be helpful also for other viewers. :)

@@cuberaccoon yes, it would help greatly if you did either of those!

I haven't taken the time to figure this case out, but you can either use the method from this video, or just use a tool like CubeTwister which outputs the order for you. Why are you asking, by the way? www.randelshofer.ch/cubetwister/

Why? 😅

Sorry to hear that. RUclips often deletes comments with links. I cannot do anything about this, I am afraid. I posted some comments with links here, but tbh I was surprised that RUclips didn't delete them.

You are welcome to post a video explaining how this works. Let's stay constructive :). I am not really good with finger tricks, yet.

@@cuberaccoon im horrible as youtuber, but I may try to find a tutorial if you want

I don't understand what you are saying. Maybe you refer to a different time stamp. And maybe you can change your tone.

​@@cuberaccoonoh yeah, u right, i meant 1:15

The issue is that even though I can do a B move directly in the home position, I did not manage to do the D' B D' fluently. Maybe there is a way, I don't know.

@@cuberaccoon i think that regripping would be faster than rotation.

I had to so that everyone is convinced, even those who didn't understand the mathematical proof :D

Yes. You can find a proof here: math.stackexchange.com/questions/2392906/determine-the-highest-order-of-an-element-of-a-rubiks-cube-group Also here in Theorem 4.4.11: people.kth.se/~boij/kandexjobbVT11/Material/rubikscube.pdf But I also plan a video about this.

The length of the sequence does not really matter. RURURURURURURURURURURURURURURU has order 7 (since RU has order 105), just to give a simple example.

@@cuberaccoon Thanks for the reply! I think I understand it, but I'm shocked that there is no higher order sequence. This absolutely blows my mind because there was always a dimension of complexity I assume the puzzle had which this shows cannot exist. It's a very intuitive result that shows unintuitive behavior!

The maximal order of the elements of a group (in mathematical terms) does not tell much about the structure and complexity of a group. The Rubik's cube group is still very complex (even though it can be represented as a semidirect product of (semi)direct products of well-known much smaller groups, see Theorem 4.3.18 in the pdf).

I am not sure about this. But when you scroll down you find another comment where people discuss this.

That's interesting, but here it doesn't really play a role. Pondering about this belongs more to the field of numerology, I would say.

Hehe. I actually wrote a Python script that generates the subtitles file for me, which then could be pasted into my editor. You can find it here. gist.github.com/ScriptRaccoon/41a53790980ca7fa288935adaa4367c3 So it required some manual work (clicking every time the alg is done) but still was manageable.

I think someone already found the devil's algorithm...i mean maybe you can search it, but maybe it doesn't work but who knows lol

Sorry for the delay. These were indeed a lot of questions, and it is hard to understand for me what you are really asking, overall. > "does the length of the algorithm matter?" No. > "... it could reach every permutation. Is that theoretically possible?" No. The Rubik's Cube group is not cyclic ( = there is no "Devil's algorithm", but I do not recommend this vague terminology). Your reasoning is incomplete. There is no algorithm A such that 1,A,A^2,... reaches all allowed permutations. If A consists of face moves X,Y,... and you decompose the powers accordingly, this is a completely different sequence. I think you are mixing things up here. > " I’m also going to assume this is the shortest order 1260 algorithm because you’re showing it to us." Yes. > "So if this alg gives us a string of 7560 permutations that don’t repeat, ..." This is not necessarily true, see the other point above. It is only guaranteed that the 1260 permutations are pairwise distinct, each time after completing the algorithm. > "If you’re repeating an algorithm is it possible that some cube states shows up more than once even if the solved solution only shows up once in 1260 cycles?" No (if I understand your question correctly). If A is an element of a group of order n > 0, then the elements 1,A,....,A^{n-1} are pairwise distinct. Again, not sure if this answers all of your questions.

@@cuberaccoon Thanks, looking back I probably could’ve worded/organised them better haha