Can someone please give me a 2-gen alg. for this set of algorithms.

Lotsofsloths

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I'm trying to get at a set of algorithms(2Gen Please) that both orients corners(without disturbing permutation) and solves the permutation of the edges..

Also, how many algorithms it would take to accomplish that would be greatly appreciated, if its over 200 forget about posting all the different solutions and tell me its 200+..
 
ok according to Petrus' site there are 35 plus mirrors and inverses...
Anyway there are 12 edge permutations for each orientation case so 12x7= 84 algs. But the H case most likely has loads of symmetry so it may have less cases, so the number is <84. I have the Pi and U cases done with 1-5 algs for each cases to pick from.
I found them all using Ron's CubeSolver, Jason showed me how to use ACube but it may be easier to still use this since its 2-gen and will be very easy to change between the different cases.
 
H case is nice, 8:12 is a double Sune, mirror, inverse or mirror inverse.

Set up corners to correct position, if one edge is solved then it is one of the double Sunes. Orient the cube so the solved edge is at FU and then do the alg. If corners are at R/L sides then it is a Sune or mirror, if corners are at F side then it is inverse or mirror inverse. Mirror or not comes from the direction of the U PLL that is either RU-LU-BU-RU or RU-BU-LU-RU.

Mirror does not have to be done on the left side = LU, RU is also useful. If it starts like L' U' L... then do (U2 or y2) R' U' R... instead.

The H case where all edges are correct is the worst LL case of them all, I'm not sure but I can guess a RU 2-gen is more than 20 turns for that case :p

I'm using a QTM strait of 18 turns for the case (16 HTM is optimal):

F 2x(U R d' R U' R' d R') F'

(wery nice triggers =)

BTW: The pi case with solved edges is also 16 HTM optimal but I can do that one in 15 STM :p
 
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http://www.ai.univ-paris8.fr/~bh/cube/
There are 35 unique cases, and 50 without inversing.

Btw, has anyone done/thought of doing this, but using QTM instead of FTM? I'm thinking of giving it a try (but for ZBLL only, not 1-look Fridrich LL).

F 2x(U R d' R U' R' d R') F'
I use R U' R U R U R' U' R' U' R2 U' R U R' U' R U' R'.

And for the pi/Bruno: R U2 R2 U' R B' D' R D B F' L' U' L F.
I execute it more or less like this: R U2 R2 U' R f' R' U R S L' U' r U.
 
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