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I know but that isnt exactly what i meant, like if you have sune on top and pi on bottom thats say 1 alg, but you counted pi on top and sune on bottom as another when you could do a z2

D'oh!
Yes, you're right.
There are 4 symmetric cases, 4*3 =12, 63-12 = 51 cases.
 
whats the most common plls other then the 2 look ones... cause those are the ones i know
G perms


Ok, how many algs would it take to Orient all corners on a 2x2, as in all U/D colors are on the U/D faces, and then how many algs to solve it in one look from there?
Starting with the standard OLL cases, there's 8*8=64 total cases, one of them being both faces already oriented.
Then there's a whole lot more for the 'parity' cases where an individual face isn't orientable by itself. I don't feel like actually going through the cases. It's not worth it. With 63 algorithms already, your method is going to suffer from terrible recognition.
After orienting all the corners, Ortega's last step will suffice. That's 6 algorithms.

A few things, he never said anything about separation, and it would probably only be like half that # since there would be a few reflections/ mirrors.
Read the edit.
I did the numbers assuming no "AUF," which was probably not realistic, you're right. Mirrors are generally considered distinct algorithms, though. For example, G permutations.
Well i was basing this off being a 2 step method where the first step could be very easily planned in inspection. So you orient all the corners in 1 alg, but at this point all U/D colors are interchangable, then you do PBL and separation in 1 step. Does that help at all?
 
Well i was basing this off being a 2 step method where the first step could be very easily planned in inspection. So you orient all the corners in 1 alg, but at this point all U/D colors are interchangable, then you do PBL and separation in 1 step. Does that help at all?

Sounds like a faster version of Guimond. Maybe I'm not thinking this through right, but you combined the second and third step of Guimond together? That's a lot of cases I think :eek:
 
whats the most common plls other then the 2 look ones... cause those are the ones i know
G perms


Ok, how many algs would it take to Orient all corners on a 2x2, as in all U/D colors are on the U/D faces, and then how many algs to solve it in one look from there?
Starting with the standard OLL cases, there's 8*8=64 total cases, one of them being both faces already oriented.
Then there's a whole lot more for the 'parity' cases where an individual face isn't orientable by itself. I don't feel like actually going through the cases. It's not worth it. With 63 algorithms already, your method is going to suffer from terrible recognition.
After orienting all the corners, Ortega's last step will suffice. That's 6 algorithms.

A few things, he never said anything about separation, and it would probably only be like half that # since there would be a few reflections/ mirrors.
Read the edit.
I did the numbers assuming no "AUF," which was probably not realistic, you're right. Mirrors are generally considered distinct algorithms, though. For example, G permutations.
Well i was basing this off being a 2 step method where the first step could be very easily planned in inspection. So you orient all the corners in 1 alg, but at this point all U/D colors are interchangable, then you do PBL and separation in 1 step. Does that help at all?
Guimond method is very similar to what you're talking about...You find a face where there are 3 colors completed (counting opposite colors as the same), and do some steps to get the final U/D colors oriented...Hope this helps.
 
Well i was basing this off being a 2 step method where the first step could be very easily planned in inspection. So you orient all the corners in 1 alg, but at this point all U/D colors are interchangable, then you do PBL and separation in 1 step. Does that help at all?

Sounds like a faster version of Guimond. Maybe I'm not thinking this through right, but you combined the second and third step of Guimond together? That's a lot of cases I think :eek:

Yeah that's pretty much it, my recognition for the first step would be to see the OLL case on top and then the OLL case on bottom, i didn't say it would be a good method, i'm just trying to thing of new ideas

Guimond method is very similar to what you're talking about...You find a face where there are 3 colors completed (counting opposite colors as the same), and do some steps to get the final U/D colors oriented...Hope this helps.

lol, i know what guimond is...
 
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Well i was basing this off being a 2 step method where the first step could be very easily planned in inspection. So you orient all the corners in 1 alg, but at this point all U/D colors are interchangable, then you do PBL and separation in 1 step. Does that help at all?

Sounds like a faster version of Guimond. Maybe I'm not thinking this through right, but you combined the second and third step of Guimond together? That's a lot of cases I think :eek:

Yeah that's pretty much it, my recognition for the first step would be to see the OLL case on top and then the OLL case on bottom, i didn't say it would be a good method, i'm just trying to thing of new ideas

Guimond method is very similar to what you're talking about...You find a face where there are 3 colors completed (counting opposite colors as the same), and do some steps to get the final U/D colors oriented...Hope this helps.

lol, i know what guimond is...
Well, it basically sound just like guimond, except no intuition for the first step with algs, and what I predict to be 1260 algs for the final step (8! positions of cubies, divided by 8 orientations, making it 5040, divided by 4 AUFs, making it 1260). I could be wrong with the final number, but I know the answer is huge.
 
Well i was basing this off being a 2 step method where the first step could be very easily planned in inspection. So you orient all the corners in 1 alg, but at this point all U/D colors are interchangable, then you do PBL and separation in 1 step. Does that help at all?

Sounds like a faster version of Guimond. Maybe I'm not thinking this through right, but you combined the second and third step of Guimond together? That's a lot of cases I think :eek:

Yeah that's pretty much it, my recognition for the first step would be to see the OLL case on top and then the OLL case on bottom, i didn't say it would be a good method, i'm just trying to thing of new ideas

Guimond method is very similar to what you're talking about...You find a face where there are 3 colors completed (counting opposite colors as the same), and do some steps to get the final U/D colors oriented...Hope this helps.

lol, i know what guimond is...
Well, it basically sound just like guimond, except no intuition for the first step with algs, and what I predict to be 1260 algs for the final step (8! positions of cubies, divided by 8 orientations, making it 5040, divided by 4 AUFs, making it 1260). I could be wrong with the final number, but I know the answer is huge.
Yay, i thought of a fail method, oh well, practice makes perfect.
 
Well, it basically sound just like guimond, except no intuition the first step with algs, and what I predict to be 1260 algs for the final step (8! positions of cubies, divided by 8 orientations, making it 5040, divided by 4 AUFs, making it 1260). I could be wrong with the final number, but I know the answer is huge.

You're actually totally wrong with the final number. Even assuming you don't want to have to flip your cube and thus cases which are similar up to z2 are different cases, you get:
- with separation done (just PBL): 8 cases (not counting skip)
- with one wrong corner in each layer: that corner can be chosen arbitrarily, and you get 6 permutations of the other 3, hence 6*6=36 cases
- with two wrong corners in each layer, forming bars on both: choosing one of them arbitrarily, you have 3 possibilities for the other one and 2 permutations of the last two. That's 6 cases for each layer, and given that some cases may be identical up to a y2 you actually have only 21 different cases.
- with bar on top and diagonal on bottom: choose arbitrarily one cube of each layer to put on the top layer. You have 3 possible choices to fill each remaining slots, and with symmetries up to a U2 it yields to 6 different cases. For the bottom layer, you have only 2 possible choices (which differ by a Y-perm), so 12 cases.
- with diagonal on top and bar on bottom: similar as above
- with diagonal on both: again only 12 cases by the same reasoning.
That's 101 algs to learn, the number is quite large but not unmanageable. Recognition would be quite messy though.
 
I've been hearing people mention that they've forced PLL/OLL skip. While I can force PLL skip with a couple OLL algs (T and fish patterns), I can't generalise it. Any links?
 
I've been hearing people mention that they've forced PLL/OLL skip. While I can force PLL skip with a couple OLL algs (T and fish patterns), I can't generalise it. Any links?

You can orient all the LL edges by using EOLine
 
I've been hearing people mention that they've forced PLL/OLL skip. While I can force PLL skip with a couple OLL algs (T and fish patterns), I can't generalise it. Any links?

You can orient all the LL edges by using EOLine

He's talking about forcing PLL.

What most people do in a speed solve is, they know two different algorithms for a case, one gives them a skip, one a PLL. This is how Erik got a WR.
 
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