Escher
Babby
I was thinking just now - there are quite a few OLL cases that involve taking out a pair and 'messing' with it. Just using wiki algs, OLLs 1, 7, 8, 9, 17, 29, 33, 34, 35, 36, 38, 41, 42, 52, and 57 could all be done from the standard Winter Variation starting point.
This is only 15 cases (there are probably more) but just being aware of these would be useful.
What do you guys think?
Oh, and I just realised that (setup:F' L' U L F) is a less obvious one - and that also makes me think that what you would be doing here is just 'getting lucky' with VH algorithms. I'm not sure.
Also, I had a good and ridiculous idea - its called X-OLL or HyperOLL, I havent decided yet
you learn 6 or less cases for every OLL algorithm (yes, I know), so that while solving OLL you also solve corners. This means that just like normal COLL, you get a PLL skip 1/12 of the time, and an EPLL every time.
Apart from the normal COLL H, there are 5 more cases with H corners for which you would only have to learn 3 more algorithms (if you knew full OLL).
There are 7 more cases than the normal COLL Pi for which recognition would be barely more difficult than normal COLL.
Learning these alone would be quite useful (Pi = 5x8 algorithms, H = 3x5 total = only 55 algorithms), as recognition of these cases is really easy (though I admit, the algorithms might not be as friendly as the normal OLL cases).
Anyway, I think that the total number of algorithms you would have to learn, assuming you knew full OLL and not COLL, would be
H: 5x3 = 15
Pi: 5x8 = 40
Sune: 5x8 = 40
Anti Sune: 5x8 = 40
(I cant be bothered to count up any more, I'm assuming that the rest have equal likelihood)
Triple Sune: 40
U: 40
T: 40
Im not sure how many cases the corners solved OLLs would have - the same amount as the other OLLs? If so, you would learn 15 more algorithms.
Including the corners solved OLL, that brings the grand total of algorithms to learn up to 270! Which is less than ZBF2L alone, I think
I'm not sure that this number is correct: it probably isn't, I wasn't ever that great at maths (or cube theory, for that matter).
I know that this idea is ridiculous, and has probably been thought of before, but its a fun idea to play with.
Anyway, I intend at some point to learn the 'extended winter variation' (provided that there is more to it than just getting a OLL skip with a VH alg - though i was considering learn VH anyway), and I will probably eventually learn the X-OLL for Pi and H cases.
Any thoughts? I expect plenty of criticisms, but remember I'm not actually proposing that either is a good system, its just a couple of ideas.
Oops, I also just counted up the number of cases I accounted for in X-OLL and I think im missing some (probably because of my laziness). Anyway, the number is probably around 270-300.
This is only 15 cases (there are probably more) but just being aware of these would be useful.
What do you guys think?
Oh, and I just realised that (setup:F' L' U L F) is a less obvious one - and that also makes me think that what you would be doing here is just 'getting lucky' with VH algorithms. I'm not sure.
Also, I had a good and ridiculous idea - its called X-OLL or HyperOLL, I havent decided yet
you learn 6 or less cases for every OLL algorithm (yes, I know), so that while solving OLL you also solve corners. This means that just like normal COLL, you get a PLL skip 1/12 of the time, and an EPLL every time.
Apart from the normal COLL H, there are 5 more cases with H corners for which you would only have to learn 3 more algorithms (if you knew full OLL).
There are 7 more cases than the normal COLL Pi for which recognition would be barely more difficult than normal COLL.
Learning these alone would be quite useful (Pi = 5x8 algorithms, H = 3x5 total = only 55 algorithms), as recognition of these cases is really easy (though I admit, the algorithms might not be as friendly as the normal OLL cases).
Anyway, I think that the total number of algorithms you would have to learn, assuming you knew full OLL and not COLL, would be
H: 5x3 = 15
Pi: 5x8 = 40
Sune: 5x8 = 40
Anti Sune: 5x8 = 40
(I cant be bothered to count up any more, I'm assuming that the rest have equal likelihood)
Triple Sune: 40
U: 40
T: 40
Im not sure how many cases the corners solved OLLs would have - the same amount as the other OLLs? If so, you would learn 15 more algorithms.
Including the corners solved OLL, that brings the grand total of algorithms to learn up to 270! Which is less than ZBF2L alone, I think
I'm not sure that this number is correct: it probably isn't, I wasn't ever that great at maths (or cube theory, for that matter).
I know that this idea is ridiculous, and has probably been thought of before, but its a fun idea to play with.
Anyway, I intend at some point to learn the 'extended winter variation' (provided that there is more to it than just getting a OLL skip with a VH alg - though i was considering learn VH anyway), and I will probably eventually learn the X-OLL for Pi and H cases.
Any thoughts? I expect plenty of criticisms, but remember I'm not actually proposing that either is a good system, its just a couple of ideas.
Oops, I also just counted up the number of cases I accounted for in X-OLL and I think im missing some (probably because of my laziness). Anyway, the number is probably around 270-300.
