<?xml version="1.0"?>
<feed xmlns="http://www.w3.org/2005/Atom" xml:lang="en">
	<id>https://www.speedsolving.com/wiki/api.php?action=feedcontributions&amp;feedformat=atom&amp;user=Kc+kennylau</id>
	<title>Speedsolving.com Wiki - User contributions [en]</title>
	<link rel="self" type="application/atom+xml" href="https://www.speedsolving.com/wiki/api.php?action=feedcontributions&amp;feedformat=atom&amp;user=Kc+kennylau"/>
	<link rel="alternate" type="text/html" href="https://www.speedsolving.com/wiki/index.php/Special:Contributions/Kc_kennylau"/>
	<updated>2026-11-02T19:16:22Z</updated>
	<subtitle>User contributions</subtitle>
	<generator>MediaWiki 1.43.8</generator>
	<entry>
		<id>https://www.speedsolving.com/wiki/index.php?title=Template:H-PLL&amp;diff=23971</id>
		<title>Template:H-PLL</title>
		<link rel="alternate" type="text/html" href="https://www.speedsolving.com/wiki/index.php?title=Template:H-PLL&amp;diff=23971"/>
		<updated>2014-06-14T11:57:54Z</updated>

		<summary type="html">&lt;p&gt;Kc kennylau: /* H Permutation */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;===H Permutation===&lt;br /&gt;
{{case&lt;br /&gt;
|image=H.gif&lt;br /&gt;
|name=H-PLL, X-PLL&lt;br /&gt;
|methods=[[EPLL]], [[CPLL]], [[PLL]], [[ELL]], [[ZBLL]], [[ZZLL]]&lt;br /&gt;
|optimal=10 [[HTM]], 7 [[STM]]&lt;br /&gt;
|text=X-PLL is H-PLL + U2.&lt;br /&gt;
}}&lt;br /&gt;
{{Alg|M2&#039; U M2&#039; U2 M2&#039; U M2&#039;|PLL|vidurl=http://www.youtube.com/watch?v=RkznplJdcAI}}&lt;br /&gt;
{{Alg|M2&#039; U&#039; M2&#039; U2&#039; M2&#039; U&#039; M2&#039;|PLL|vidurl=http://www.youtube.com/watch?v=4YqOYb6pHSs}}&lt;br /&gt;
{{Alg|M2&#039; U2 M2&#039; U&#039; M2&#039; U2 M2&#039;|PLL}}&lt;br /&gt;
{{Alg|(y&#039;) M2&#039; U2 M2&#039; U M2&#039; U2 M2&#039;|PLL}}&lt;br /&gt;
{{Alg|R2 U2 R U2 R2 U2 R2 U2 R U2 R2|PLL}}&lt;br /&gt;
{{Alg|R2&#039; U2 R&#039; U2 R2&#039; U2&#039; R2&#039; U2 R&#039; U2 R2&#039;|PLL|vidur2=http://www.youtube.com/watch?v=HI9ff1ACklU}}&lt;br /&gt;
{{Alg|R2 U2 R2 U2 R2 U R2 U2 R2 U2 R2|PLL}}&lt;br /&gt;
{{Alg|R U2&#039; R&#039; U&#039; R&#039; U&#039; R2 U&#039; R2&#039; U2&#039; R2 U2&#039; R&#039; (U)|PLL}}&lt;br /&gt;
{{Alg|L2 U2 L&#039; U2 L2 U2 L2 U2 L&#039; U2 L2|PLL}}&lt;br /&gt;
{{Alg|L R U2 L&#039; R&#039; F&#039; B&#039; U2 F B|PLL}}&lt;br /&gt;
{{Alg|L R U2 L&#039; R&#039; (y) L&#039; R&#039; U2 L R|PLL}}&lt;br /&gt;
{{Alg|(M2&#039; U)6|PLL}}&lt;br /&gt;
{{Alg|(M2&#039; U&#039;)6|PLL}}&lt;br /&gt;
{{Alg|(R&#039; S R u) 6|PLL}}&lt;br /&gt;
{{Alg|S R U2 R2 U2 R2 U2 R S&#039;|PLL}}&lt;br /&gt;
{{Alg|F2 M2&#039; F2 U&#039; F2 M2&#039; F2|PLL}}&lt;br /&gt;
{{Alg|(x) U2 M2&#039; U2 B&#039; U2 M2&#039; U2 (x&#039;)|PLL}}&lt;br /&gt;
{{Alg|L R U2 L&#039; l&#039; U&#039; u&#039; L2 U (z) L|PLL}}&lt;/div&gt;</summary>
		<author><name>Kc kennylau</name></author>
	</entry>
	<entry>
		<id>https://www.speedsolving.com/wiki/index.php?title=1LLL&amp;diff=23970</id>
		<title>1LLL</title>
		<link rel="alternate" type="text/html" href="https://www.speedsolving.com/wiki/index.php?title=1LLL&amp;diff=23970"/>
		<updated>2014-06-14T10:54:33Z</updated>

		<summary type="html">&lt;p&gt;Kc kennylau: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;{{Method Infobox&lt;br /&gt;
|name=1LLL&lt;br /&gt;
|image=1LLL.jpg&lt;br /&gt;
|proposers=Bernard Helmstetter&lt;br /&gt;
|year=2000&lt;br /&gt;
|anames=&lt;br /&gt;
|variants=[[ZBLL]]&lt;br /&gt;
|steps=1&lt;br /&gt;
|algs=1211&lt;br /&gt;
|moves=12.58 [[HTM]]&lt;br /&gt;
|purpose=&amp;lt;sup&amp;gt;&amp;lt;/sup&amp;gt;&lt;br /&gt;
* Not useful in practice.&lt;br /&gt;
}}&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;1&#039;&#039;&#039; &#039;&#039;&#039;L&#039;&#039;&#039;ook &#039;&#039;&#039;L&#039;&#039;&#039;ast &#039;&#039;&#039;L&#039;&#039;&#039;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 may have potential.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
== Number of 1LLL Cases ==&lt;br /&gt;
&lt;br /&gt;
Assuming the last layer is an outer layer, the number of cases is calculated as:&lt;br /&gt;
: corner orientations * edge orientations * corner permutations * edge permutations / 2&lt;br /&gt;
With the 2 at the end because of odd-even parity.&lt;br /&gt;
&lt;br /&gt;
Numerically this is:&lt;br /&gt;
: &amp;lt;span style=&amp;quot;color:red&amp;quot;&amp;gt;3^3&amp;lt;/span&amp;gt; * &amp;lt;span style=&amp;quot;color:orange&amp;quot;&amp;gt;2^3&amp;lt;/span&amp;gt; * &amp;lt;span style=&amp;quot;color:green&amp;quot;&amp;gt;4!&amp;lt;/span&amp;gt; * &amp;lt;span style=&amp;quot;color:purple&amp;quot;&amp;gt;4!&amp;lt;/span&amp;gt; / 2 = &#039;&#039;&#039;62208&#039;&#039;&#039;&lt;br /&gt;
* The &amp;lt;span style=&amp;quot;color:red&amp;quot;&amp;gt;3^3&amp;lt;/span&amp;gt; at the beginning is because there are 4 corners on the last layer while the final one depends on the remaining three.&lt;br /&gt;
* The &amp;lt;span style=&amp;quot;color:orange&amp;quot;&amp;gt;2^3&amp;lt;/span&amp;gt; is because there are 4 edges on the last layer while the final one depends on the remaining three.&lt;br /&gt;
&lt;br /&gt;
Treating cases which are the same, but rotated by 90, 180 or 270 degrees as the same case, the number of cases becomes:&lt;br /&gt;
: 62208/4 = &#039;&#039;&#039;15552&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
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.&lt;br /&gt;
&lt;br /&gt;
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:&lt;br /&gt;
: &#039;&#039;&#039;1212&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
The number of algorithms a solver would require to solve the last layer in one look depends on the solver&#039;s ability to work out mirrors and/or inverses. If a solver can work out the mirror/inverse of any alg &#039;on the fly&#039;, then the number of algorithms they would need to learn would be:&lt;br /&gt;
: 1212 - 1 = &#039;&#039;&#039;1211&#039;&#039;&#039; algorithms&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
* [[Last Layer]]&lt;br /&gt;
* [[:Category:3x3x3 last layer methods]]&lt;br /&gt;
* [[2LLL]], [[3LLL]], [[4LLL]]&lt;br /&gt;
* [[ZBLL]]&lt;br /&gt;
* [[Orientation]]&lt;br /&gt;
* [[Permutation]]&lt;br /&gt;
&lt;br /&gt;
== External links ==&lt;br /&gt;
* [http://www.ai.univ-paris8.fr/~bh/cube/ Bernard Helmstetter&#039;s last layer algorithms]&lt;br /&gt;
* [http://www.speedcubing.com/docs/uppertable.zip All 1211 last layer algorithms(1MB zipped)]&lt;br /&gt;
* Speedsolving.com: [http://www.speedsolving.com/forum/showthread.php?t=12815 3x3 last layer: number of positions]&lt;br /&gt;
* Speedsolving.com: [http://www.speedsolving.com/forum/showpost.php?p=440250&amp;amp;postcount=31 Derivation of unique PLL cases]&lt;br /&gt;
&lt;br /&gt;
[[Category:3x3x3 methods]]&lt;/div&gt;</summary>
		<author><name>Kc kennylau</name></author>
	</entry>
	<entry>
		<id>https://www.speedsolving.com/wiki/index.php?title=User:Kc_kennylau&amp;diff=23969</id>
		<title>User:Kc kennylau</title>
		<link rel="alternate" type="text/html" href="https://www.speedsolving.com/wiki/index.php?title=User:Kc_kennylau&amp;diff=23969"/>
		<updated>2014-06-14T10:21:07Z</updated>

		<summary type="html">&lt;p&gt;Kc kennylau: Created page with &amp;quot;I&amp;#039;m not so good at the cube, generally &amp;lt;40 secs.&amp;quot;&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;I&#039;m not so good at the cube, generally &amp;lt;40 secs.&lt;/div&gt;</summary>
		<author><name>Kc kennylau</name></author>
	</entry>
</feed>