# Method Substeps and Subgroups



## whauk (Nov 19, 2013)

a random thought that came across my mind today:
(nearly) all methods with steps work by reducing the cube group to smaller and smaller subgroups until only identity (=solved cube) is left. (between the steps the subgroup may be left again. however that is not of our interest in this topic.)

i found one exception though: when solving the inverse scramble with a "normal" method (and therefore finding a solution for the actual scramble by inverting the solution) you actually do some fancy stuff with cosets. but i am too lazy to think about it in detail right now. however it clearly doesn't work by reduction to subgroups on the normal scramble and still has steps.
are there other exceptions that i missed? or can somebody construct a method that doesn't work by reducing to subgroups *and* can be understood "directly". (my example doesn't fulfill this since, when watching a solution of the inverse scramble being executed on the normal scramble it most likely looks like random moves to you; apart from the last ~6 turns that just happen to solve everything)

maybe we didn't think about many other solving methods yet because we were too fixated on the approach using subgroups. however i do not really believe in this. 

inb4 pseudoblocks is essentially the same concept as with the inverse-scramble-stuff.


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## Lucas Garron (Nov 19, 2013)

It would probably be useful to have a definition of a subgroup stage: A stage is defined by a subgroup if the possible states of the cube at the end of the stage form a group. (In particular, they must be closed under composition, e.g. two OLL states in a row give an OLL.)

Whenever I go down this line of reasoning, the annoying things are:

1) While many steps are defined using subgroups, we *constantly* leave them for speedsolving. (Sortega has a notable counterexample.)
2) Since, say, F2L can be solved in any order, subgroup descriptions are not especially useful. Cube rotations mess with this a lot.

It's easy to come up with a stage that is not described using a subgroup. The first step of ZZ-TOP comes to mind (orient all edges belonging in the E or D layers).

Also, NISS usually still uses subgroup steps; I'd say it just makes it easier for humans to explore the space by trading branches (e.g. try lots 2x2x3 solutions) for a more flexible solving.


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## Stefan (Nov 19, 2013)

whauk said:


> when solving the inverse scramble with a "normal" method (and therefore finding a solution for the actual scramble by inverting the solution) you actually do some fancy stuff with cosets. but i am too lazy to think about it in detail right now. however it clearly doesn't work by reduction to subgroups on the normal scramble



I don't understand what you mean, can you rephrase or give an example of such a solve?


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## whauk (Nov 19, 2013)

OK let's say we have the scramble U L' B D F D2 R2 D' L' F R2 L2 F2 D' B2 D F2 D' F2 L2 U. we can invert it and get: U' L2 F2 D F2 D' B2 D F2 L2 R2 F' L D R2 D2 F' D' B' L U'. now we can solve this inverted scramble with normal techniques:
step 1, 2x2x3: F' U' B' L F' U L2 (reducing to the subgroup where this 2x2x3 is solved)
step 2, F2L: R' B2 R B D2 (reducing to the subgroup where this F2L is solved)
step 3, LL: R' B U2 F' U B U' F U2 B' R (reducing to the subgroup containing only identity). in case you are wondering this is a recent FMC solution by me.

so we now have a solution for the inverted scramble. if this 3-step-solution is inverted we get another 3-step-solution for the uninverted (=*normal*) scramble, that is:
step 1, "LL": R' B U2 F' U B' U' F U2 B' R 
step 2, "F2L": D2 B' R' B2 R 
step 3, "2x2x3": L2 U' F L' B U F

so let's look at what step1 does:
the moves in the new step1 make absolutely no sense to someone watching the solution and are also no reduction to a subgroup (at least i don't see one). the only thing we know about it is, that when the rest of the solution (let's call it H) is performed after it, a solved cube will be the result. shortly:

scramble * step1 * H = identity. which is equivalent to: scramble * step1 = H'

so step1 actually makes sense when we know that the whole cube group was "shifted" by H' first. (this means that some element "x" from the group looks like "H' * x" now, explicitly the identity is now H' * id = H'.) because when id is H' now step1 is actually just a solution to the whole cube (and "makes sense" as in "making progress in the solving process" as in "solves the cube"). so mathematically speaking this shifting is actually a left coset of the whole cube group with the fixed element H'.

but this H' changes after every step so things get very weird (at least for me) and i described this with the words "fancy stuff with cosets". probably cosets might not the best/most natural way to describe this phenomenon but it was the first thing that came across my mind. 

do things get clearer now?


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## Stefan (Nov 20, 2013)

Yes, thanks, now I understand what you mean (*). But quite complicated stuff and I don't quite know what to think of it 

(*) I had thought the "fancy cosets stuff" refers to the "normal" solve, i.e., _"when solving the inverse scramble with a "normal" method you actually do some fancy stuff with cosets"_ (parentheses should _"contain material that could be omitted without destroying or altering the meaning of a sentence") and that didn't make sense to me. Still not sure whether you meant that or not, but whatever, the example shows what you mean._


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