Petrus Variation- does it have any potential?

[Nukoca]

Okay, friends, I believe I've come up with a pretty good Petrus variation.

The beginning of the solve is pretty much all that is in common with Petrus. Solve the 2x2x2, and then the 2x2x1. Whereas Petrus would fix the bad edges, solve the second 2x2x1, finish the F2L, and then OLL/PLL, this method will not deal with the last layer at all.

When solving with Petrus, when you finish the 2x2x2 block, you have 3 directions that you can expand to a 2x2x3. You would choose the most optimal, and continue solving.

However, with this method, you would obtain the 2x2x2 and expand your 2x2x1 in ALL THREE directions. You would solve them one at a time, like so:

That last 2x2x1 will be tricky, but it’s not hard once you get the hang of it.

*rotates cube 180* You would then solve the remaining “Y” shaped unsolved section in one algorithm.

The alg for this case is: U F’ U2 F L’ B2 R’ B R’ B’ R B2 L U’
The only problem I see with such a method is the number of algorithms for that last bit. Here’s what I came up with:

(Corner Permutation x Corner Orientation)(Edge Permutation x Edge Orientation)/Parity – Mirrors
(4x3x2x1x9)(6x4)/Parity = 5184/Parity – Mirrors

What is the math concerning parity? As 5184 is a substantial number of algorithms to deal with, I hope that parity and mirrors will cancel out a fair number of the algorithms necessary for the last step. If it turns out to be too many, we can always go with a two-step solution, e.g. corner permutation first and then CO and edges. (I will edit in the extra math after I am informed)

Last but not least… what shall I call it (that is, if no one has come up with this before)? How about… the Coke method (after Nukoca)?

 

[Johannes91]

Tripod. It has been discussed several times.

 

[qqwref]

This method very strongly resembles the "Tripod" method that I came up with (although many people probably also came up with it on their own) - the name represents what the last 7 pieces look like. With Tripod I'd build the 2x2x2 block and then add on three times, as you would, although I also allowed the last two additions to be made so that you had (for instance) the FR, UR, and UB edges (with associated corners) free. This gave a lot more freedom in the last few blockbuilding steps. After that, my way to solve the remaining pieces was to put in one corner-edge pair (there are two or three possible pairs to try depending on which three edges are free), mostly using R'FRF' type moves, and then solve the last layer (5 pieces) in one look, with about 60 algs. In real life, when I try this method, I solve it in two looks because I don't know all those algs.

The number of positions for the last 7 pieces is (4! * 3^3)(3!/2 * 2^2) = 7776, but if you allow mirrors, inverses, and rotations around URF, there are approximately 7776/(2*2*3) = 648 separate algs. It's way too much to do in one step.

 

[Nukoca]

The number of positions for the last 7 pieces is (4! * 3^3)(3!/2 * 2^2) = 7776, but if you allow mirrors, inverses, and rotations around URF, there are approximately 7776/(2*2*3) = 648 separate algs. It's way too much to do in one step.

So how would you explain the mathematics there? I thought I understood the concept of cube math, but apparently not.

 

[byu]

The number of positions for the last 7 pieces is (4! * 3^3)(3!/2 * 2^2) = 7776, but if you allow mirrors, inverses, and rotations around URF, there are approximately 7776/(2*2*3) = 648 separate algs. It's way too much to do in one step.

So how would you explain the mathematics there? I thought I understood the concept of cube math, but apparently not.

4! stands for corner permutation (4 positions)
3^3 stands for corner orientation (3 corners that can be oriented, 4th given)
3!/2 stands for edge permutation (3 positions, 2 for parity)
2^2 for edge orientation (2 edges that can be misoriented, 2nd given)

 

[Johannes91]

From http://games.groups.yahoo.com/group/petrusmethod/message/780:

[In] the final step (which literally leaves a "Tripod" of cubies
to be solved) our program gave 2610 cases modulo symmetries, and only
794 if redundant inverses are removed as well.

There's also an explanation on how to calculate it by hand in that thread.

I guess that many of the optimal algs are (R U' R' U) followed by another alg or similar. Learning 1-look tripod doesn't sound impossible to me, but it'd take a long time and a lot of work.

 

[Nukoca]

Arg... it gets so annoying when you spend a long time making a very useful post, complete with pictures, only to find that someone's already invented the method.

 

[fanwuq]

Arg... it gets so annoying when you spend a long time making a very useful post, complete with pictures, only to find that someone's already invented the method.

Why is it annoying? I would actually be happy that there are more resources for the method than I expected.
Something similar happened when I first joined the forum. I was trying to reinvent TuRBo.

 

[blah]

How's the recognition for 1-look tripod?

 

[EmersonHerrmann]

When solving with Petrus, when you finish the 2x2x2 block, you have 3 directions that you can expand to a 2x2x3. You would choose the most optimal, and continue solving.

I would?

 

[Am1n-]

Mabe you should go 2-look, or corners first or somthing like that. The corners can easily be done with comutators

 

[qqwref]

How's the recognition for 1-look tripod?

Awful. Even though you can see where every piece is by looking at U, R, and F, you have to recognize so many things at once (corner and edge permutation and orientation, on so many different pieces, with no AUF to help you out). Also, you essentially have to be color neutral. In my opinion it's way harder than ZBLL recognition.