rachmaninovian
Veteran
let me propose perhaps a better form of your revised 4SCS:
for step 1, allow adjacents too. according to my brute force speed approach, adjacent on average has better times than opposites, which are harder to execute and recognize.
actually to be honest, I would really like to direct solve everything, because that way you have a clearer view of what to solve next.
In fact, after my exams (like 2 months later) I would try to find combos...like the "double niklas" (case 3,adjacent, or the opposite counterpart).
regardless of whether to preserved solved pieces or not, you would only need to find cases for step 2 and 3, because step 1 and 4 would use the same algs. based on my experience, allowing step 1 to include adjacent cases would be much faster in general. some centres cases would be better to be done adjacent than opposite.
to be efficient and to perhaps to decrease the number of algs needed, I suggest only finding cases for centres with already 2 centre pieces done.
right now I guess I might just as well come up with a similar approach, but with more freedom: 2 look 1st centre, 1 look 2nd centre, finish last 2 centres (be it adjacent/opposite). argh never mind.
for step 1, allow adjacents too. according to my brute force speed approach, adjacent on average has better times than opposites, which are harder to execute and recognize.
actually to be honest, I would really like to direct solve everything, because that way you have a clearer view of what to solve next.
In fact, after my exams (like 2 months later) I would try to find combos...like the "double niklas" (case 3,adjacent, or the opposite counterpart).
regardless of whether to preserved solved pieces or not, you would only need to find cases for step 2 and 3, because step 1 and 4 would use the same algs. based on my experience, allowing step 1 to include adjacent cases would be much faster in general. some centres cases would be better to be done adjacent than opposite.
to be efficient and to perhaps to decrease the number of algs needed, I suggest only finding cases for centres with already 2 centre pieces done.
right now I guess I might just as well come up with a similar approach, but with more freedom: 2 look 1st centre, 1 look 2nd centre, finish last 2 centres (be it adjacent/opposite). argh never mind.
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