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I am trying to learn 2GR and am having trouble understanding how slice moves (and wide moves, by extension) affect EO. For ZZ I learned the definition that an edge is good if it is in the <R, U, L, D, F2, B2> group. Unfortunately, following an M, E, or S move, there are two ways in which a solved edge can be defined. The first definition is that an edge is solved iff it is adjacent to both centers of its component colors and one of its colors is adjacent to the same colored center, and the second definition is that an edge is solved iff it would satisfy the previous definition assuming that all center pieces are in WCA orientation.

Here are the two metrics on a solved cube in WCA orientation:

Relative to Centers Definition:
M': All edges in the M slice are still solved, and therefore can be solved in <R, U, L, D, F2, B2>, and the remaining eight edges can be trivially solved in 2 FTM in <R, L>, and therefore can be solved in <R, U, L, D, F2, B2>. Thus M' does not affect EO. Therefore (by induction) <M> does not affect EO.

E': All edges in the E slice are still solved, and therefore can be solved in <R, U, L, D, F2, B2>, and the remaining eight edges can be trivially solved in 2 FTM in <U, D>, and therefore can be solved in <R, U, L, D, F2, B2>. Thus E' does not affect EO. Therefore (by induction) <E> does not affect EO.

S: All edges in the S slice are still solved, and therefore can be solved in <R, U, L, D, F2, B2>, but the remaining eight edges cannot be solved in <R, U, L, D, F2, B2>. Thus S flips the orientation of the F and B edges.

Relative to Cube Orientation Definition:
M': All edges in the M slice are now bad because performing F2 B2 leaves the four M edges permuted correctly, but oriented incorrectly. The remaining eight edges are already in their solved position and therefore can be solved in <R, U, L, D, F2, B2>. Thus M' flips the orientation of the M edges.

E': All edges in the E slice are now bad because performing F2 B2 leaves the four E edges permuted correctly, but oriented incorrectly. The remaining eight edges are already in their solved position and therefore can be solved in <R, U, L, D, F2, B2>. Thus E' flips the orientation of the E edges.

S: All edges in the S slice are now bad because performing L2, U2, R2, xor D2 leaves the Yellow-Orange, Orange-White, Red-White, and Red-Yellow E edges respectively permuted correctly, but oriented incorrectly. The remaining eight edges are already in their solved position and therefore can be solved in <R, U, L, D, F2, B2>. Thus S flips the orientation of the S edges.

So an r move, would either leave EO the same, or change the orientation of the M-slice edges.

A u move would either leave EO the same, or change the orientation of the E-slice edges.

An f move would either maintain the orientation of the F-face edges (F or F' changes F-face EO, but S according to the Relative Centers definition would also change F-face EO, so it cancels out) while flipping the orientation of B-face edges and maintaining the orientation of the S-slice edges, or change the orientation of the F-face and S-slice edges.

So which is it, and how do you keep track of it in a solve? I've been going through some of the example solves in the 2GR doc, and the EO-Pair solutions seem incredibly arcane to me; even more-so than EO-Line, which I can at least follow even though I can't plan it in inspection of find an optimal solution pretty much ever on my own. Why do the solutions rely so much on <Rr, Uu, Ff> to solve EO-pair? Do any ZZ solves do EO-Line like that? How does one begin to understand how teoidus (the creator of 2GR) up with his EO-Pair solutions? Even if he used a computer solver, I still think it's reasonable to assume he wouldn't have used those solutions to EO-pair in his example solutions if he didn't think that they were reasonable examples of how to solve for a human. So while a human solution might take a few more moves with only 15 seconds to plan, it should still use some wide moves most of the time based on the examples provided.

In ZZ, the first step is to orient edges so that the cube is the <R, U, L, D, F2, B2> group. Is this the same as the <R, U, L, D> group?

In other words, is it possible to solve a cube with the scramble F2 using only <R, U, L, D> ?

Edit: After typing this I tried it with brute force and confirmed that it is possible, just by doing a regular ZZ solve. Follow-up question: What is the optimal solution to solving F2 with <R, U, L, D>?

In ZZ, the first step is to orient edges so that the cube is the <R, U, L, D, F2, B2> group. Is this the same as the <R, U, L, D> group?

In other words, is it possible to solve a cube with the scramble F2 using only <R, U, L, D> ?

Edit: After typing this I tried it with brute force and confirmed that it is possible, just by doing a regular ZZ solve. Follow-up question: What is the optimal solution to solving F2 with <R, U, L, D>?

U L' U R U R' U2 L2 D' L D L U' L (14f*)
U' R U' L' U' L U2 R2 D R' D' R' U R' (14f*)
R U' R D R D' R2 U2 L' U L U R' U (14f*)
R' D R' U' R' U R2 D2 L D' L' D' R D' (14f*)
D R' D L D L' D2 R2 U' R U R D' R (14f*)
D' L D' R' D' R D2 L2 U L' U' L' D L' (14f*)
L D' L U L U' L2 D2 R' D R D L' D (14f*)
L' U L' D' L' D L2 U2 R U' R' U' L U' (14f*)

In ZZ, the first step is to orient edges so that the cube is the <R, U, L, D, F2, B2> group. Is this the same as the <R, U, L, D> group?

In other words, is it possible to solve a cube with the scramble F2 using only <R, U, L, D> ?

Edit: After typing this I tried it with brute force and confirmed that it is possible, just by doing a regular ZZ solve. Follow-up question: What is the optimal solution to solving F2 with <R, U, L, D>?

I know this has already been answered, and this isn't the solution asked for, but this reminded me of Chris Hardwick's signature: R L F2 B2 L' R' U R L B2 F2 L' R'