Stefan
Member
Thanf for intersting contributions. But all of them seem to have implicit cube rotations (wide turns). Who can find "pure" generators? Will 3 generators suffice?
Was Tim's answer not sufficient?
Thanf for intersting contributions. But all of them seem to have implicit cube rotations (wide turns). Who can find "pure" generators? Will 3 generators suffice?
Thanf for intersting contributions. But all of them seem to have implicit cube rotations (wide turns). Who can find "pure" generators? Will 3 generators suffice?
Was Tim's answer not sufficient?
The six face turns answer???? That's trivial.
The six face turns answer???? That's trivial.
If it's oh so trivial, then why did you ask in the first place?
The six face turns answer???? That's trivial.
If it's oh so trivial, then why did you ask in the first place?
I wanted to know if something better is possible (fewer generators). Duh!
Per
So it must be
(a,b)(c,7)(e,8). The a must be from 1..3 and the b must be
from 4..6, else you couldn't move pieces between these orbits,
so call it (1,4)(c,7)(e,8). There's only a few possibilities left. Who
can finish this off without enumerating them explicitly?
The six face turns answer???? That's trivial.
If it's oh so trivial, then why did you ask in the first place?
I wanted to know if something better is possible (fewer generators). Duh!
Per
What Tim said is a proof that fewer than 6 generators is not possible for the supergroup.
Let me explain it slightly more than Tim did. Consider the effect of your generators on just the face centres. In other words, take the cube apart, and only use the spider. Your generators generate the supergroup, and so must also generate all 4^6 possible positions of the centres alone. Each generator has at most order 4 when considered as acting only on the centres. The order you apply them also doesn't matter (they form an Abelian group). Therefore you can get at most 4^g positions with g generators, and so you need at least 6. This is why {U,D,R,L,F,B} is minimal.