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God's Number for CFOP and Roux

NigelTheCuber

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no offence but i don't think that question makes sense. you can just use extremely long pairs or extremely long lse for roux, so they both have infinite.


if u meant which method has less moves, if you count slice turns as 1 move, roux, but if you use half turn metric, cfop
 

Silky

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Yeah as said before this is a bit of a poorly worded question. Technically if Roux and CFOP where completely optimized they would solve the cube in 20 moves or less as this is the most optimal way to solve the cube. Therefor they'd be equal. This isn't really a satisfying answer given that its really just the definition of God's number. To answer your question a bit more directly, if you were to optimize CFOP and Roux they'd just end up being other methods. A very optimized CFOP would use edge control and finish with ZBLL, which is just ZB ( another option would be to solve with a 1LLL which isn't really CFOP ). Similarly, Roux would use edge and corner control to reduce the cube to a DR state, making it very similar to something like HTA. Optimal-ness ends up blurring the line between methods since you don't really define this outside of fewest moves.
Best answer is use this program. Comes up with very efficient solutions while staying within the defined parameters for each method ( with options to change those parameters e.g. OLL/PLL vs ZBLL vs 1LLL ).
 

xyzzy

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My guess is Roux it depends. (Yes, even if we restrict to face turn metric. I don't care much for STM either. Everything else in this post will use FTM. I'll explain why I think Roux is generally better later.)

"CFOP" and "Roux" are not methods so rigorously defined that you could ask two different people to code implementations of these methods, without further instructions, and end up with solvers that always produce the same solutions. I'm much more familiar with CFOP than with Roux, so I'll use that as an example first.

CFOP is just cross, F2L, OLL, then PLL, right? Right? Wrong.

Very often, you'll have multiple choices of cross colour that take the same number of moves. Which do you choose? Do you pick one arbitrarily, or do you pick one that has the best continuation?

That's a trick question; you don't pick cross colours for their continuations, you pick cross solutions. Do you generate all cross solutions, then pick the one that has the best first pair? What if e.g. white cross has a 5-move solution that leads into 3-move first pair, while yellow cross has a 4-move solution but 7-move first pair?

Moving on to F2L, do you always pick the best pair to solve? What about multislotting? Are algs that exploit free slots allowed? Are adjacent pairs prioritised over diagonal pairs, and if so, to what extent?

OLL and PLL are the mechanical parts involving no degree of freedom, and those aren't really up for dispute.

---

You know where this is going already, but just for the sake of formality…

Roux is just FB, SB, CMLL, then LSE, right? Right? Wrong.

FB runs into the same CN-related issue as above.

SB and CMLL have no degree of freedom (assuming we're not using alt solutions for better continuations).

LSE as done by beginners is usually broken into three steps: EO ("4a"), UL UR ("4b"), L4E ("4c"). As done by experts, EO and UL UR are (partially) combined into a single step, EOLR. (EOLR usually involves setting up EO to the arrow case first, but there's also full EOLR which solves 4a and 4b directly.)

If we're to compare methods of similar alg counts, it would only be fair to include EOLR as part of Roux.

(edit As GodCubing pointed out, I misunderstood what EOLR is. This does not materially affect the numbers presented here; EOLR proper uses more algs, but also has slightly lower move count than the variant I described above.)

Also, since we're using FTM as the metric of choice, of course we'll also be using solutions for the individual steps that are optimal in FTM, not STM.

---

Let's say we solve the steps individually, without regard for continuations.

CFOP is, in reality, a seven-step method:
1. Cross
2. First pair
3. Second pair
4. Third pair
5. Last pair
6. OLL
7. PLL (and AUF)

Roux is only a six-step method:
1. First block
2. Second block
3. CMLL
4. EOLR setup (either set up to arrow or solve 4a)
5. EOLR
6. 4c

Assuming we allow using free slots, the worst case for each F2L pair is at least 8 moves (corner twisted in place and edge flipped in place), except for the very last pair having a worst case of 9 moves (corner solved, edge flipped in place). That makes F2L take 8+8+8+8+9 = 41 moves. OLL's worst case is 12 moves (OLL 20), while PLL's worst case is 14 moves (V perm at any AUF). This gives a total of 41+12+14 = 67 moves.

I don't know the exact numbers for Roux, but my guesstimates are:
FB: 9
SB: 12
CMLL: 11 (known to be the worst case)
EOLR setup: 8
EOLR: 13
L4E: 8
Total: 61

There's enough slack that I could be slightly wrong in a few of these and Roux's worst case would still be better than CFOP's worst case.

I imagine that if you restrict to purely MU algs for LSE, or if LSE has to be done as 4a-4b-4c, Roux would lose by a significant margin, but for reasons I already stated above, these would be silly restrictions to impose.
 
Last edited:

GodCubing

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Roux is only a six-step method:
1. First block
2. Second block
3. CMLL
4. EOLR setup (either set up to arrow or solve 4a)
5. EOLR
6. 4c
This is not how roux works. LSE is done in 3 steps, EOLRa, insert ULUR edges, 4c. Your high end estates are way off.

FB 9
SB 17
CMLL 13
EOLRa 9
Insert ULUR 3
L4E: 5
This is worst case scenario for roux
 

xyzzy

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This is not how roux works. LSE is done in 3 steps, EOLRa, insert ULUR edges, 4c. Your high end estates are way off.

FB 9
SB 17
CMLL 13
EOLRa 9
Insert ULUR 3
L4E: 5
This is worst case scenario for roux
On one hand, you're right, I misremembered what EOLR is like. Consider my post to be with an EOLR variant rather than standard EOLR, then.

On the other hand, this does not materially affect my argument. Your numbers are also wildly inaccurate. I know for a fact that CMLL is 11 optimal because I checked it as I wrote my above post; L4E's worst case (dots) also needs 8 moves. Inserting ULUR needs 4 moves (AUF, M2, AUF) and at most one move at the end can cancel with L4E. And how would you do EOLRa for this case in only 9 moves?

(Also, my numbers all never accounted for cancellations to begin with, because cancellations make this substantially more complicated. Where do you draw the line between using a longer alg that cancels more moves versus just combining the steps?)
 
Last edited:

PiKeeper

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On one hand, you're right, I misremembered what EOLR is like. Consider my post to be with an EOLR variant rather than standard EOLR, then.

On the other hand, this does not materially affect my argument. Your numbers are also wildly inaccurate. I know for a fact that CMLL is 11 optimal because I checked it as I wrote my above post; L4E's worst case (dots) also needs 8 moves. Inserting ULUR needs 4 moves (AUF, M2, AUF) and at most one move at the end can cancel with L4E. And how would you do EOLRa for this case in only 9 moves?

(Also, my numbers all never accounted for cancellations to begin with, because cancellations make this substantially more complicated. Where do you draw the line between using a longer alg that cancels more moves versus just combining the steps?)
The difference is that you are trying to do it is HTM while godcubing is doing his numbers in STM. Also, the case you gave should have either the white or yellow center on top, since otherwise you have to do an unneccesary extra move to orient centers.
 

GodCubing

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Your numbers are also wildly inaccurate. I know for a fact that CMLL is 11 optimal because I checked it as I wrote my above post; L4E's worst case (dots) also needs 8 moves. Inserting ULUR needs 4 moves (AUF, M2, AUF) and at most one move at the end can cancel with L4E. And how would you do EOLRa for this case in only 9 moves?
1. CMLL is on average like 11.5 but you wanted worst case and my longest CMLL is 13 moves it's also symmetrical so no pre auf is necessary.
2. Dots is done E2 M' E2 M which is 4. HTM is a lie
3. Auf, M2 auf seems you disproved yourself there. That's 3
4. As @PiKeeper mentioned centers are always oriented before LSE, and the solution to that assuming centers are oriented with an M' is: U M U' M' U' M U' M' U2 M' that is 10 so I was off by one, but I inserted ULUR edges too even though putting them in the D layer it would still be 10.

I don't mean any hate, but when I feel obligated to spread truth. If you are doing this in HTM I disrespectfully disagree with that metric for movecount evaluation.
 

xyzzy

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The difference is that you are trying to do it is HTM while godcubing is doing his numbers in STM. Also, the case you gave should have either the white or yellow center on top, since otherwise you have to do an unneccesary extra move to orient centers.
I don't mean any hate, but when I feel obligated to spread truth. If you are doing this in HTM I disrespectfully disagree with that metric for movecount evaluation.
I said that I'd be using FTM right at the very start of my post, and so did the OP of this thread (albeit not in the top post). It's not my fault if GodCubing can't read and decides they want to answer a question different from the one asked by the OP. ;)

Is it not enough for you that Roux still seems to be more efficient than CFOP even when measured in FTM?

1. CMLL is on average like 11.5 but you wanted worst case and my longest CMLL is 13 moves it's also symmetrical so no pre auf is necessary.
Your longest CMLL is not using a move-optimal alg for it then. We're not discussing move count of speed-optimal algs here (that would be silly; you should just measure execution time instead); we're discussing move count of move-optimal algs.

CLL preserving F2L (i.e. not just F2B) has been known since before you were born to need at most 11 moves.
 

GenTheSnail

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This depends on how you define CFOP and Roux.

For example, for CFOP, this moves might go down significantly if you use
xxxcross, pseudoslotting, ZBLL, EO etc.
Same with roux - you could solve a super efficient F2B with Psedo or NM and then do an ACMLL that forces an LSE skip...

But that's not what was asked here - I would suggest you read the previous posts in this thread as an answer has already been found.
 

Manxkiwi

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I don't know as much as most on this site, but I would have thought 'God's number' is, by definition method absent. It's simply the lowest number of moves to a solution. A computer generated GN solution would/could probably be full of all kinds of whacky stuff that wouldn't fit into 'method' type algs. I may be wrong??
 
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