ZBLL algs?

[4Chan]

Ah, ZB_FTW, I don't think I've heard of that variation. 0:

Also, Eastamazonanitdote, for around 10 or so cases, I can identify them with blocks of color, because they form distinctive blocks. These cases are in the minority, but you can, "in theory" recognize with blocks. It would be rather difficult and would take up a lot more memory than the intuitive way.

 

[Tim Major]

Ah, ZB_FTW, I don't think I've heard of that variation. 0:

Also, Eastamazonanitdote, for around 10 or so cases, I can identify them with blocks of color, because they form distinctive blocks. These cases are in the minority, but you can, "in theory" recognize with blocks. It would be rather difficult and would take up a lot more memory than the intuitive way.

Should I make a new thread, because I think it would require fewer algorithms, though they are quite fast cases? Plus, Winter Variation would be similar to VHf2l (correct me if I'm wrong) with only 27 algorithms, plus mirrors. The OLLs used with all corners oriented, are very fast, and easy to learn. Plus you can make inverses for all cases, taking the amount of algs way down. One problem is, I don't know how to make the cube pictures, as I would make a site with all the algos. I could even use some ZBf2ls for this, as they affect corners. Recog would be rather easy.

Is this a new method?

 

[4Chan]

Kinda reminds me of Summer Variation, now that I think about it.

http://www.speedsolving.com/forum/showthread.php?t=8820

 

[guitardude7241]

oh, mr. zb dude..

uhh, how do you get all the algs for one set of zbll? isn't it the subset, then the permutation of the corners, and all permutations of the edges?

would there be 12 different edge permutations?

would there be 9 different corner permutations?

 

[4Chan]

Haha, MisterZB is my name on cubemania.

Ah, i always believed there were 6 different corner permutations.
(except in H set)

I used to know why, but I forgot.

Oh and yes, 12 edge permutations per subset.

 

[miniGOINGS]

There are 6 different corner permutations.
-All permuted
-Front two switched
-Right two switched
-Back two switched
-Left two switched
-Two diagonally switched

 

[guitardude7241]

There are 6 different corner permutations.
-All permuted
-Front two switched
-Right two switched
-Back two switched
-Left two switched
-Two diagonally switched

lets call BL corner 1, BR 2, FL 3, and FR 4

1-2 means 1 and 2 are swapped

correct
1-2
1-3
1-4
2-3
2-4
3-4
1-2; 3-4
1-3; 2-4

there's 9, i believe.

 

[Tim Major]

Kinda reminds me of Summer Variation, now that I think about it.

http://www.speedsolving.com/forum/showthread.php?t=8820

Considering that's not really the same, should I make a new thread? I probably won't learn it, though once I've finished OLL, I'll learn the rest of WV (about 23 algs) plus the R U set of algs (about 26) and maybe, maybe then I'll learn the rest. Also, I have a question for you, how do you work out how many algs for these cases? I used to think there would be 22 algs for each set of the ll algs, but according to you, that doesn't work.

I need to work out

OLL solved :22 (1 being solved)
No edges flipped correctly: 1 alg already learned.
2 edges flipped in H pattern: 2 algs already learned.
And 2 edges flipped with other pattern: 1 alg already learned.

I know full PLL, so the first set is out of the way, how many after that please? Or would I have to manually count?

 

[4Chan]

Some of those cases you listed are analogous.
They're the same.

EDIT: That was in reply to guitardude.

 

[miniGOINGS]

lets call BL corner 1, BR 2, FL 3, and FR 4

1-2 means 1 and 2 are swapped

correct
1-2
1-3
1-4
2-3
2-4
3-4
1-2; 3-4
1-3; 2-4

there's 9, i believe.

1-3; 2-4 = correct
1-3 = 2-4 = 1-2; 3-4

kthnxbai

 

[guitardude7241]

lets call BL corner 1, BR 2, FL 3, and FR 4

1-2 means 1 and 2 are swapped

correct
1-2
1-3
1-4
2-3
2-4
3-4
1-2; 3-4
1-3; 2-4

there's 9, i believe.

1-3; 2-4 = correct
1-3 = 2-4 = 1-2; 3-4

kthnxbai

i don't get how it's correct, both of them are swapped.
i understand that the 1-3; 2-4 is diagonal switched corners.

 

[4Chan]

It just is.
Thats like a U2 move does the same effect.

EDIT: Sorry ZB_FTW!!!
I'm not sure about that method. D:
Ummm, yeah, i think if you make a thread, a puzzle theorist might come along and offer better advice than I can give.

 

[miniGOINGS]

lets call BL corner 1, BR 2, FL 3, and FR 4

1-2 means 1 and 2 are swapped

correct
1-2
1-3
1-4
2-3
2-4
3-4
1-2; 3-4
1-3; 2-4

there's 9, i believe.

1-3; 2-4 = correct
1-3 = 2-4 = 1-2; 3-4

kthnxbai

i don't get how it's correct, both of them are swapped.
i understand that the 1-3; 2-4 is diagonal switched corners.

It means that you just have to do U2 at the end of the ZBLL alg.

 

[guitardude7241]

It just is.
Thats like a U2 move does the same effect.

EDIT: Sorry ZB_FTW!!!
I'm not sure about that method. D:
Ummm, yeah, i think if you make a thread, a puzzle theorist might come along and offer better advice than I can give.

i don't get it, but i'll keep that in mind. so there's 84 different cases for each subset?

 

[4Chan]

Seventy two.

Unless its the H set.

 

[guitardude7241]

Seventy two.

Unless its the H set.

please explain how there's only 72, not 84.

and about the h set.

 

[miniGOINGS]

Seventy two.

Unless its the H set.

please explain how there's only 72, not 84.

and about the h set.

Or you could just use the COLL + Edge Cycle recognition system.

 

[4Chan]

Ah, okay.

Here is the list of 40 COLLs.
http://www.cubezone.be/coll.html

Each COLL case has 12 edge permutations.
Depending of the corners, being an even or odd cycle, the edges will also have a corresponding even or odd edge cycle.

I forgot why H set has so few.
I used to know, but I forgot.

EDIT: I forgot to add, there are 7 sets of COLL.
Each set has 6 subsets. (Except H)

So, 6x12=72.

 

[miniGOINGS]

Ah, okay.

Here is the list of 40 COLLs.
http://www.cubezone.be/coll.html

Each COLL case has 12 edge permutations.
Depending of the corners, being an even or odd cycle, the edges will also have a corresponding even or odd edge cycle.

I forgot why H set has so few.
I used to know, but I forgot.

Because two of the H COLL's are mirrorable (by doing U2) so the number of those cases are cut in half.

 

[4Chan]

Haha, thats right!
Thats why.

+1 to you.