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Learn 5-style: You can do it!

mark49152

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There are many challenges in the world that seem impossible (send a man to the moon, internet over the entire planet, curing cancer) but with a few years of invested time and collective effort, it becomes mainstream and widely adapted.
Such things involve judgment, planning and innovation as well as investment of effort. There are reasons we haven't built a city on the moon yet.

As I'm sure I've said before, I'm excited to see you pushing the limits and would love to see you come up with something useful. The best contribution I can make today is to point out the folly of trying to brute-force 5-style, and encourage you to approach it differently.

Personally I think the most promising way to explore 5-style would be to focus on the 4-movers plus a systematic method of setting up other cases to those, with the objective of maximising the number of cases covered. That way, you stand a better chance of achieving a reasonable move count benefit while keeping an intuitive approach to aid with learning and recall. I'm just not sure a significant enough proportion of cases could be covered with manageable effort, but it's worth a try. You could also try cancelling 4-movers into adjacent 3-cycles, e.g. solve ABCD as (ABCE)(ED).

Edit with example:
[M', U] solves IADM speffz. IADF could be solved with (IADM)(MF) where MF is [U: [M', U L' U'].

(M' U M U')(U M' U L' U' M U L U2) = M' U2 L' U' M U L U2 = [M', U2 L' U']
 
Last edited:

Seniorcuber

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Dec 9, 2019
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I find the idea good, but not the approach.

Also, the number of cases should not be the main argument. Rather, is it possible to find a well structured method that enables us to tackle all cases with just a few algorithms and less or equal than three or max four setup moves.

I don't think someone learned an algorithm for each of the 440 edge 3-cycle perms. I am using a variation of the TuRBo method. 3 algorithms with their mirror and inverse (2+2+4=8), 3 edges (=6 stickers) affected, of which each can coincide with the buffer (8x6=48) cover 48 out of 440 possible permutations, and the 392 remaining can be solved with obvious setup moves (1 or 2, at max 3).
I learned 3 algorithms to solve the 440 permutations.

Nobody will learn the sheer number of 126 720 permutations. But it should be possible to reduce that number to a set of algorithms of a reasonable size, together with simple rules to determine the algorithm, its orientation and setup moves.

The algorithm MBM'B' has 8 variations (a mirror, an inverse and a double MBM'B'MBM'B'). It affects 5 edges with 10 stickers and can thus solve 80 permutations. With an AUF-move (B, B' and B2), it can solve 80x4 = 360 permutations.

There's another 'simplification' I'm thinking of : instead of considering sticker-permutations, we could consider piece-permutations, and keep track of orientation separately. The number of permutations solved with one algorithm increases : there are 2^5 = 32 sticker-perms for one piece-perm. The 8 variations cover 4 piece-perms, they affect 5 pieces and with an AUF-move, this covers 4×5×4×32 = 2560 sticker-perms.

Here is an outline of the method :
  1. Solve piece-permutation, minimizing edge flips.
    • Looking at the 5 pieces involved (buffer + letter quartets), determine the setup moves needed to have four pieces on the same plane and one on the opposite side.
    • Determine the AUF-move and the variation of the algorithm required. There are two solutions for each piece-perm. Choose the one that results in less remaining flipped edges.
    • Keep track of the edges that have the wrong orientation (11 bits of information for the whole cube. Should be doable).
  2. If there's a remaining 3-cycle, do it.
  3. Orient flipped edges with algs of the 3EO method.
  4. Handle parity, if necessary.

There are still missing pieces :
  • - How to handle breaking into a new cycle
  • - There are 6 permutations of 4 pieces on the same plane, given the starting point. This alg only covers 4 of them.

Nevertheless, I think this is a good starting point for tackling the problem of 5-style.
What is your opinion?
 
Last edited:
Joined
Jun 29, 2019
Messages
460
I find the idea good, but not the approach.

Also, the number of cases should not be the main argument. Rather, is it possible to find a well structured method that enables us to tackle all cases with just a few algorithms and less or equal than three or max four setup moves.

I don't think someone learned an algorithm for each of the 440 edge 3-cycle perms. I am using a variation of the TuRBo method. 3 algorithms with their mirror and inverse (2+2+4=8), 3 edges (=6 stickers) affected, of which each can coincide with the buffer (8x6=48) cover 48 out of 440 possible permutations, and the 392 remaining can be solved with obvious setup moves (1 or 2,at max 3).
I learned 3 algorithms to solve the 440 permutations.

Nobody will learn the sheer number of 126 720 permutations. But it should be possible to reduce that number to a set of algorithms of a reasonable size, together with simple rules to determine the algorithm, its orientation and setup moves.

The algorithm MBM'B' has 8 variations (a mirror, an inverse and a double MBM'B'MBM'B'). It affect 5 edges with 10 stickers and can thus solve 80 permutations. With an AUF-move (B, B' and B2), it can solve 80x4 = 360 permutations.

There's another 'simplification' I'm thinking of : instead of considering sticker-permutations, we could consider piece-permutations, and keep track of orientation separately. The number of permutations solved with one algorithm increases : there are 2^5 = 32 sticker-perms for one piece-perm. The 8 variations cover 4 piece-perms, they affect 5 pieces and with an AUF-move, this covers 4×5×4×32 = 2560 sticker-perms.

Here is an outline of the method :
1. Solve piece-permutation, minimizing edge flips.
1a. Looking at the 5 pieces involved (buffer + letter quartets), determine the setup moves needed to have four pieces on the same plane and one on the opposite side.
1b. Determine the AUF-move and the variation of the algorithm required. There are two solutions for each piece-perm. Choose the one that results in less remaining flipped edges.
1c. Keep track of the edges that have the wrong orientation (11 bits of information for the whole cube. Should be doable).
2. If there's a remaining 3-cycle, do it.
3. Orient flipped edges with algs of the 3EO method.
4. Handle parity, if necessary.

There are still missing pieces :
- How to handle breaking into a new cycle
- There are 6 permutations of 4 pieces on the same plane, given the starting point. This alg only covers 4 of them.

Nevertheless, I think this is a good starting point for tackling the problem of 5-style.
What is your opinion?
I think this is a very good idea (although it got kinda deep to where I didnt understand to well but we’ll ignore that.) That would make it a lot less algs and not be as fast but still faster than 3-style if there was enough practice put into it.
 

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