Difference between revisions of "1LLL"

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|name=1LLL
 
|name=1LLL
 
|image=1LLL.jpg
 
|image=1LLL.jpg
|proposers=Bernard Helmstetter
+
|proposers=[[Bernard Helmstetter]]
 
|year=2000
 
|year=2000
|anames=
+
|anames= OLLCP-EP
 
|variants=[[ZBLL]]
 
|variants=[[ZBLL]]
 +
[[M-CELL]]
 
|steps=1
 
|steps=1
|algs=1211
+
|algs='''3915'''
 
|moves=12.58 [[HTM]]
 
|moves=12.58 [[HTM]]
 
|purpose=<sup></sup>
 
|purpose=<sup></sup>
* Not useful in practice.
+
* [[Speedsolving]], however the multitude of algorithms makes it much less useful in whole
 +
* [[FMC]]
 
}}
 
}}
  
'''1''' '''L'''ook '''L'''ast '''L'''ayer means completing the [[last layer]] using only one look. 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.
+
'''1''' '''L'''ook '''L'''ast '''L'''ayer 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 (such as [[M-CELL]]) may have potential.
  
 +
A non-optimised database for all 1LLL algorithms & cases can be found at http://birdflu.lar5.com
  
 
== Number of 1LLL Cases ==
 
== Number of 1LLL Cases ==
  
 
Assuming the last layer is an outer layer, the number of cases is calculated as:
 
Assuming the last layer is an outer layer, the number of cases is calculated as:
 
 
: corner orientations * edge orientations * corner permutations * edge permutations / 2
 
: corner orientations * edge orientations * corner permutations * edge permutations / 2
 
+
With the 2 at the end because of odd-even parity.
  
 
Numerically this is:
 
Numerically this is:
 
+
: <span style="color:red">3^3</span> * <span style="color:orange">2^3</span> * <span style="color:green">4!</span> * <span style="color:purple">4!</span> / 2 = '''62208'''
: 3^3 * 2^3 * 4! * 4! / 2 = '''62208'''
+
* The <span style="color:red">3^3</span> at the beginning is because there are 4 corners on the last layer while the final one depends on the remaining three.
 
+
* The <span style="color:orange">2^3</span> 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:
 
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'''
 
: 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. By solving multiple cases with the same algorithm from different angles, the number of cases becomes:
 +
: '''3916''' cases (including the solved state / [[AUF]] states and subsets such as [[PLL]], [[ELL]], [[2GLL]], [[L4C]] and [[ZBLL]]) which can be solved with '''3915''' algorithms
  
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''' cases (including the solved / [[AUF]] states) which can be solved with '''1211''' algorithms
  
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:
+
Since most speed solvers regard reflective symmetry and inversions as different cases (e.g. [[PLL]] is thought of as 21 algorithms rather than 13 algorithms plus mirrors and inverses), 1LLL should be thought of as '''3915''' algorithms.
  
: '''1212'''
+
[[PLL]], [[ELL]], and [[ZBLL]] are all subsets of 1LLL, albeit a small proportion of the algorithms. Knowing some of these subsets means less algorithms need to be learnt for a 1LLL:
  
 +
*If the solver already knows all 21 [[PLL]] cases, which is a subset of [[ZBLL]], there would be '''3894''' algorithms to learn.
 +
*If the solver already knows all 493 [[ZBLL]] cases, there would be '''3422''' algorithms to learn.
 +
*If the solver already knows all 29 [[ELL]] cases, there would be '''3886''' algorithms to learn.
 +
*If the solver already knows all 518 [[ELL]] and [[ZBLL]] cases (4 [[EPLL]] cases appear in both sets), there would be '''3397''' algorithms to learn.
  
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:
+
It should be noted that additional subsets of 1LLL also exist such as [[L3C]] (subset of L4C, which is a subset of ZBLL) , Pure OLL (orientation without affecting permutation), [[ZZLL]] (subset of ZBLL), [[ZZ-Blah]] (subset of ZBLL), [[Tripod]] LL and more.
  
: 1212 - 1 = '''1211''' algorithms
+
== List of 1LLL subsets ==
 +
* [[ZBLL]] (includes [[PLL]])
 +
* [[ELL]]
 +
* [[Tripod LL]]
 +
* [[Line]]
 +
* [[Flipped Line]]
 +
* [[Pure OLL]]
 +
{{work}}
  
 
== See also ==
 
== See also ==
 
* [[Last Layer]]
 
* [[Last Layer]]
* [[Last Layer Methods]]
+
* [[:Category:3x3x3 last layer methods]]
 
* [[2LLL]], [[3LLL]], [[4LLL]]
 
* [[2LLL]], [[3LLL]], [[4LLL]]
 
* [[ZBLL]]
 
* [[ZBLL]]
 
* [[Orientation]]
 
* [[Orientation]]
 
* [[Permutation]]
 
* [[Permutation]]
* [[Puzzle Theory]]
 
  
 
== External links ==
 
== External links ==
 
* [http://www.ai.univ-paris8.fr/~bh/cube/ Bernard Helmstetter's last layer algorithms]
 
* [http://www.ai.univ-paris8.fr/~bh/cube/ Bernard Helmstetter's last layer algorithms]
* [http://www.speedcubing.com/docs/uppertable.zip All 1211 last layer algorithms(1MB zipped)]
+
* [http://www.speedcubing.com/docs/uppertable.zip 1211 last layer algorithms(1MB zipped)]
 +
* Speedsolving.com: [http://www.speedsolving.com/forum/showthread.php?t=51999 1LLL/OLLCP-EP algorithms]
 +
* Speedsolving.com: [http://www.speedsolving.com/forum/showthread.php?t=55518 BindeDSA's Algorithms]
 
* Speedsolving.com: [http://www.speedsolving.com/forum/showthread.php?t=12815 3x3 last layer: number of positions]
 
* Speedsolving.com: [http://www.speedsolving.com/forum/showthread.php?t=12815 3x3 last layer: number of positions]
 
* Speedsolving.com: [http://www.speedsolving.com/forum/showpost.php?p=440250&postcount=31 Derivation of unique PLL cases]
 
* Speedsolving.com: [http://www.speedsolving.com/forum/showpost.php?p=440250&postcount=31 Derivation of unique PLL cases]
 +
* Speedsolving.com: [http://www.speedsolving.com/forum/showthread.php?t=53675 Hierarchy of Last Layer Sub-Steps, Subsets of OLLCP and ZBLL]
  
 
+
[[Category:3x3x3 last layer substeps]]
 
 
[[Category:3x3x3 methods]]
 
[[Category:Cubing terminology]]
 

Revision as of 09:16, 3 June 2020

1LLL method
1LLL.jpg
Information about the method
Proposer(s): Bernard Helmstetter
Proposed: 2000
Alt Names: OLLCP-EP
Variants: ZBLL

M-CELL

No. Steps: 1
No. Algs: 3915
Avg Moves: 12.58 HTM
Purpose(s):
  • Speedsolving, however the multitude of algorithms makes it much less useful in whole
  • FMC


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 (such as M-CELL) may have potential.

A non-optimised database for all 1LLL algorithms & cases can be found at http://birdflu.lar5.com

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. By solving multiple cases with the same algorithm from different angles, the number of cases becomes:

3916 cases (including the solved state / AUF states and subsets such as PLL, ELL, 2GLL, L4C and ZBLL) which can be solved with 3915 algorithms

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 cases (including the solved / AUF states) which can be solved with 1211 algorithms

Since most speed solvers regard reflective symmetry and inversions as different cases (e.g. PLL is thought of as 21 algorithms rather than 13 algorithms plus mirrors and inverses), 1LLL should be thought of as 3915 algorithms.

PLL, ELL, and ZBLL are all subsets of 1LLL, albeit a small proportion of the algorithms. Knowing some of these subsets means less algorithms need to be learnt for a 1LLL:

  • If the solver already knows all 21 PLL cases, which is a subset of ZBLL, there would be 3894 algorithms to learn.
  • If the solver already knows all 493 ZBLL cases, there would be 3422 algorithms to learn.
  • If the solver already knows all 29 ELL cases, there would be 3886 algorithms to learn.
  • If the solver already knows all 518 ELL and ZBLL cases (4 EPLL cases appear in both sets), there would be 3397 algorithms to learn.

It should be noted that additional subsets of 1LLL also exist such as L3C (subset of L4C, which is a subset of ZBLL) , Pure OLL (orientation without affecting permutation), ZZLL (subset of ZBLL), ZZ-Blah (subset of ZBLL), Tripod LL and more.

List of 1LLL subsets

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See also

External links