1LLL

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1LLL
1LLL.jpg
Information
Proposer(s): Bernard Helmstetter
Proposed: 2000
Alt Names:
Variants: ZBLL
No. Steps: 1
No. Algs: 1211
Avg Moves: 12.58 HTM
Purpose(s):
  • Not useful in practice.

1 Look Last Layer means completing the last layer using only one look. On the 3x3x3, currently ZBLL is the only known practical system for achieving a 1-look last layer, but other experimental approaches such as 1-look 2-alg may have potential.


Number of 1LLL Cases

Assuming the last layer is an outer layer, the number of cases is calculated as:

corner orientations * edge orientations * corner permutations * edge permutations / 2

With the 2 at the end because of odd-even parity.

Numerically this is:

3^3 * 2^3 * 4! * 4! / 2 = 62208
  • The 3^3 at the beginning is because there are 4 corners on the last layer while the final one depends on the remaining three.
  • The 2^3 is because there are 4 edges on the last layer while the final one depends on the remaining three.

Treating cases which are the same, but rotated by 90, 180 or 270 degrees as the same case, the number of cases becomes:

62208/4 = 15552

Although these cases may be regarded as unique, some of them can be solved by applying the same algorithm from a different angle. For cases with no rotational symmetry, there are 3 equivalent cases which may be solved with the same alg. For cases with 180 degree rotational symmetry, there is 1 equivalent case which may be solved with the same alg. Cases with 90, 180 and 270 degree rotational symmetry are unique.

As well as rotationally symmetrical cases, there are also reflectively symmetrical cases. These may be solved by applying a reflection of the algorithm. Finally, some cases are inversions of others. These can be solved by reversing (inverting) the algorithm solving the original case. Exactly which cases are mirrors of each other requires a case-by-case analysis. Work by Bernard Helmstetter established that the number unique 1LLL cases (excluding mirrors and inverses) is:

1212

The number of algorithms a solver would require to solve the last layer in one look depends on the solver's ability to work out mirrors and/or inverses. If a solver can work out the mirror/inverse of any alg 'on the fly', then the number of algorithms they would need to learn would be:

1212 - 1 = 1211 algorithms

See also

External links

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