Math Behind the Cube

[AvGalen]

Short summary:
Use (and understand) the mighty alg above that calculates all possibile positions
Adjust the mighty alg to accomodate your specific situation (for cross: 4 edges already permuted and oriented correctly)
Finally: use smart symmetry reduction (hint: cube explorer or your own programming) to reduce the remaining cases by eliminating equivalent cases. I use the term equivalent for a reason because one (Johannes ) could argue that there are less then 21 (14) unsolved PLL's

 

[Johannes91]

Finally: use smart symmetry reduction (hint: cube explorer or your own programming) to reduce the remaining cases by eliminating equivalent cases.

So there's no easy way to do that on paper, without a program?

Ryan Heise mentions Burnside's Lemma here.

 

[teller]

About the OT:

There are 12 different group orbits on the cube, only one of which can actually "solve." So if you're just counting *all* possible cubie combinations, can't you just divide by 12 to get the correct number of "possible" combinations? Or am I missing something?

 

[Sin-H]

About the OT:

There are 12 different group orbits on the cube, only one of which can actually "solve." So if you're just counting *all* possible cubie combinations, can't you just divide by 12 to get the correct number of "possible" combinations? Or am I missing something?

That's exactly what you do when you divide by (2*2*3) because 2*2*3 = 12

but then you still get a lot of cases which would be equivalent for us, like a U Perm, then a Permutation which is a U Perm, but a different U adjustment, etc...

 

[guusrs]

1212 U-algs. explaining is easy, calculating definitely not!

Some algs results in many U-positions, they can be Front-color-dependent, U-turn dependent, mirror dependent and reverse-alg depentent. Take a sune, it covers 4x4x2+2=64 U-positions:

R U R' U R U2 R and
R U R' U R U2 R U and
R U R' U R U2 R U2 and
R U R' U R U2 R U' and
R U2 R' U' R U' R and
R U2 R' U' R U' R U and
R U2 R' U' R U' R U' and
R U2 R' U' R U' R U2 and
R' U2 R U R' U R and
R' U2 R U R' U R U and
etc...

So most algs should be multiplied with 64. Or reversed logic: There are at least 62208/64 U-algs

But some U-positions are not Front-color dependant (ie. PLL-X-Perm), U-turn dependent (ie. PLL-Z-Perm), mirror dependent (ie PLL-L-Perm) and revserse-alg (ie PLL-L-Perm) depentent. If you can calculate these exceptions (which is pretty hard, they'll overlap) you will notice it'll yields 1212 algs.

BTW. in mid 80's we didn't calculate, we wrote out all positions, covered it by algs and by counting we discovered it yielded 1211 algs.

Gus

 

[Stefan]

Ok, now that it's out there anyway, here's a 1985 explanation:
http://www.geocities.com/jaapsch/puzzles/cubic7.htm#p8

Gus, do you know whether that was the first time it was calculated? Explanation is the same as yours.

Btw I found it like this:
http://www.google.com/search?num=20...ls=en&hs=XHu&q="cube+lovers"+1212&btnG=Search
("cube lovers" was the name of the old mailing list, I was hoping to find it there)

 

[guusrs]

Ok, now that it's out there anyway, here's a 1985 explanation:
http://www.geocities.com/jaapsch/puzzles/cubic7.htm#p8

Gus, do you know whether that was the first time it was calculated? Explanation is the same as yours.

Btw I found it like this:
http://www.google.com/search?num=20...ls=en&hs=XHu&q="cube+lovers"+1212&btnG=Search
("cube lovers" was the name of the old mailing list, I was hoping to find it there)

Stefan,

John F. Jarvis' list was dated on august 15, 1983. He is probably the first who calculated the correct number of distinct U-algs: 1212. Without knowing this Anton Hanegraaf published a table in CFF 8 (Netherlands, oct 14, 1984) showing there are 1211 distinct U-algs. He counted, not calculated.

Also in early 1983 a group of dutch cubers completed a U-table with a maximum of 19 moves. We got it down to 16 moves in 1986.

All history now....

Gus

 

[AvGalen]

Except for the typo of 2984 this "group of dutch cubers", counting and manually finding sequences sounds so "old-school" right now.

I bet most people on this forums first reaction would be "why didn't they just used a computer and googled it"

 

[eastamazonantidote]

Yardwork kept me busy during the day, and then I realized I had to do some homework, and I'm at school right now (its over though, thankfully), so I haven't been on for a while. But I have had it figured out for a little while, so don't worry about the spoilers. Now, I spent a while trying to contemplate exactly waht you guys have told me in later posts: the lack of need for the division by 2 * 2 * 3. I figured out (later realizing Wikipedia and my own work had it all along) that those cases were already accounted for and that I was being really, really dumb and stubborn thinking I needed to eliminate even more cases.

Oh, damn. They're closing the library. I'll check out the links after baseball practice and homework, but I really appreciate the help.

Edit: Checked the links and they are awesome!!! Thanks you guys. Just what I was looking for.