System By System and big cubes

[shadowslice e]

Do when I say a system, I am talking about where each piece can go on a cube. For example 2x2 has only one system (corners) as any piece on the cube can go anywhere else. A 3x3x3, on the other hand, has 2 systems (corners and edges)*. in this way, it becomes quite easy to identify different systems when a cube is taken apart- each different type of piece is part of a different system.

Thus, when solving system by system (SBS), each different type of piece is solved individually. On a 2x2x2, an SBS is essentially any method of solving as all will solve the single system- corners. On a 3x3x3 an SBS would be a corners or edges first method. A 4x4x4 would be much the same although with 3 systems- the methods could be centres first, corners first or edges first.•

We can thusly project that if we solve in an SBS style, the last system must always be solved using commutators so as not to destroy any of the work that has already been done. What is the feasibility of creating such method for the big cubes so the major steps are essentially solving each system (although they could be broken into smaller substeps)? And, in addition, how would the move count stack up against more orthodox techniques such as reduction or Yau?

*the 3x3x3 could be considered to have three systems (the addition of centres). However I have decided not to include them as they are fixed on relation to each other so are essentially always a solved system not worthy of further thought.
•the number of systems will increase with every size bigger a cube gets- a 5x5x5 with have more systems than a 4x4x4 (corners, outer edges, inner edges, outer corner centres, outer edge centres and corners (and the inner centres although as I have said this will always be a solved centre like in a 3x3x3

 

[qqwref]

I think these are usually called "orbits". The kind of solving methods you mention are very possible. For relatively easy puzzles like 3x3x3 or 4x4x4, these methods are generally slow because when an entire orbit is solved you don't have a lot of freedom to do moves without messing up what you already have, which means you have to solve the rest of the pieces using relatively long algorithms that don't affect much of the rest of the puzzle (such as 8-move commutators).

However, in blindfolded solving, pretty much every method solves the pieces one orbit at a time, because it is very important to affect as few pieces as possible at each part of the solve. And in harder puzzles like Master Pentultimate, Starminx, Curvy Copter III, etc. it is common to do orbit-by-orbit solves because it is just too difficult to solve pieces from multiple orbits at the same time without disturbing the pieces you have already solved.

 

[Chree]

I thought an "orbit" was a group of states not reachable from one another. Like the group starting from a solved state is one orbit, but if you twist 1 corner that would be another orbit... and if you flip a single edge that's another. Obviously I could be wrong here, as it's been a long time since I brushed up on any such terminology.

 

[qqwref]

I think there are just two ways of applying the concept of an orbit. From the perspective of a piece, the orbit is everywhere that piece can go on the cube (thus we have a corner orbit, an edge orbit, a center orbit, etc.); from the perspective of the entire cube's position, the orbit is every position you can get to from the current position (thus we have a normal orbit, a permutation parity orbit, an edge twist orbit, etc.).

 

[Stefan]

We can thusly project that if we solve in an SBS style, the last system must always be solved using commutators so as not to destroy any of the work that has already been done.

Why must it be solved using commutators?

 

[Chree]

I think there are just two ways of applying the concept of an orbit. From the perspective of a piece, the orbit is everywhere that piece can go on the cube (thus we have a corner orbit, an edge orbit, a center orbit, etc.); from the perspective of the entire cube's position, the orbit is every position you can get to from the current position (thus we have a normal orbit, a permutation parity orbit, an edge twist orbit, etc.).

Cool. That makes sense. Thanks man.

 

[shadowslice e]

Why must it be solved using commutators?

lwell, it must be solved using commutators or at least algs derived from commutators so as not to "unsolve" (if that is a word) the other systems.

 

[Stefan]

lwell, it must be solved using commutators or at least algs derived from commutators so as not to "unsolve" (if that is a word) the other systems.

What do you mean with "derived from commutators"? And can you show any algorithm that can't be "derived from commutators"?

 

[shadowslice e]

What do you mean with "derived from commutators"? And can you show any algorithm that can't be "derived from commutators"?

I mean derived from commutators that only affect one system. For example, in an edge first method, the commutators can only affect corners.

 

[Christopher Mowla]

We can thusly project that if we solve in an SBS style, the last system must always be solved using commutators so as not to destroy any of the work that has already been done.

If you are solving an nxnxn system/piece orbit/cluster by system/piece orbit/cluster, certainly it's not just the last system/piece orbit/cluster which would "require" commutators (if you don't know how to solve the last system/piece orbit/cluster without commutators, that is), right?

lwell, it must be solved using commutators or at least algs derived from commutators so as not to "unsolve" (if that is a word) the other systems.

What do you mean by "derived from commutators"?

Do you mean algorithms which are conjugated commutators such as F2 [2R, U R U'] F2, or do you mean an algorithm which can be written as a nested product of commutators with, perhaps, an extra quarter turn (if it's an odd parity algorithm)? For example, [3l E 3d: [3d, 3b' 3R' 3b] [2-3r 3d: 3R'] ].

Or do you mean something else?

 

[shadowslice e]

If you are solving an nxnxn system/piece orbit/cluster by system/piece orbit/cluster, certainly it's not just the last system/piece orbit/cluster which would "require" commutators (if you don't know how to solve the last system/piece orbit/cluster without commutators, that is), right?

yes, it is true that each of the systems can be solved by commutators. However, it is not always a necessary step- hence the corners first methods whicch do not use commutators as the would not always need to be preserved.

What do you mean by "derived from commutators"?

Do you mean algorithms which are conjugated commutators such as F2 [2R, U R U'] F2, or do you mean an algorithm which can be written as a nested product of commutators with, perhaps, an extra quarter turn (if it's an odd parity algorithm)? For example, [3l E 3d: [3d, 3b' 3R' 3b] [2-3r 3d: 3R'] ].

Or do you mean something else?

i think I may have slightly changed the definition of commutators. When I say commutators, I mean algs or moves that only affect pieces in a specific orbit/system/cluster/etc

 

[qqwref]

i think I may have slightly changed the definition of commutators. When I say commutators, I mean algs or moves that only affect pieces in a specific orbit/system/cluster/etc

That's not a commutator, that's a pure algorithm.

A commutator is something of the form P Q P' Q' (for some move sequences P and Q).

 

[Christopher Mowla]

yes, it is true that each of the systems can be solved by commutators. However, it is not always a necessary step- hence the corners first methods which do not use commutators as the would not always need to be preserved.

I understand that completely. I was just wondering why you mentioned that we need to use algorithms which only affect one system only when it's the last system. What about the second-to-last system? The third-to-last? We are talking about the nxnxn cube, which has this number of systems, not just two, right?

Spoiler: Example

Suppose you are solving the 5x5x5. You can use a 3x3x3 solution to solve the corners and middle (central) edges. Now you have three systems remaining: two systems of centers (X centers and + centers) and one wing edge system.

Suppose you choose to solve the wing edge system without algorithms which only affect that system (somehow). Now you have to solve the two center systems. How can you solve either one with algorithms which affect all systems?

Or are you arguing that we can technically solve either of the last two systems using algorithms only affect those two systems.

Spoiler: Example

If we choose to solve the middle edge system and both center systems on a 5x5x5 so that we are only left with the corner and wing edge system, we can use an algorithm like this [r' F2 :[U 2R' U' R' B R, F2] [ U 2R' U' 2L' U2: 2R] ] to solve either the corners or wing edges. We then finally use an algorithm which affects only the corner or wing edge system to solve the last system.

If so, this is quite redundant because algorithms which affect only a subset of systems in a given slice are analogous to algorithms which only affect a single system in regards to approximate length and feasibility.

If this is not what you meant, can you explain?

 

[shadowslice e]

That's not a commutator, that's a pure algorithm.

A commutator is something of the form P Q P' Q' (for some move sequences P and Q).

True, i know that but comms, when used in an intuitive format, will only usually affect one system

 

[Christopher Mowla]

will only usually

Make up your mind and face the consequences!

Seriously though, I don't see why the commutator [U 2R' U' R' B R, F2] isn't in an "intuitive" form, where I define "intuitive form" to be from piece isolation commutators.

BTW, no comments on my last post!?

 

[shadowslice e]

Make up your mind and face the consequences!

Seriously though, I don't see why the commutator [U 2R' U' R' B R, F2] isn't in an "intuitive" form, where I define "intuitive form" to be from piece isolation commutators.

BTW, no comments on my last post!?

yea, I know I used the wrong word becuase at the time I posted it was very late and I was asleep so in my head commutators commutators essentially became single piece three cycles (ie. Not cycling pairs or triplets etc).
So I acknowledge I used the wrong word in the given context.

Besides, i was only really interested in the applications of SBS in big cubes or other harder puzzles in which it would be difficult to solve pieces from multiple systems at once- hence what the earliest comments on this thread were about.