Hi from a LEGO Speedcuber robot creator

BTW you didn't answer my question about whether your improved 19 bhtm, 29 bqtm solution was shorter than your previously know solution and whether it is "speedcuber friendly". I'm just curious... :)

Well, it was certainly shorter than the one posted earlier. I don't personally know an optimized way to do that position since it can't come up in my solves. It does seem speedcuber-friendly although adding some rotations could help out (although I don't expect an algorithm-generating software to care about that).

Thanks.

Do you happen to have a particular position for which you'd like to explore different algorithms? If so would you mind describing it and give me an idea of the turn metrics for your best algorithm for it? I'd like to have a collection of real "test cases" if I'm really going to consider trying to create a good algorithm explorer.
 
I don't personally have a ton of difficult positions I'm interested in, but different "permutation parity" cases would be nice. For instance maybe you could find a better solution for the case that swaps the entire UF and UR edges (both oriented properly), or the cases that swap two oriented adjacent or opposite corners on the top layer only (without affecting anything else). The common algorithms for those cases are essentially
F' U' F (swap UF and UB) F' U F
(swap UF and UB) y' (R U R' U' R' F R2 U' R' U' R U R' F')
L U L' (swap UF and UB) y' (R U R' U' R' F R2 U' R' U' R U R' U' F')
There's a specific 'speed-optimal' UF and UB swap, Uw2 Rw2 U2 r2 U2 Rw2 Uw2, that is used. It's 16 block qtm, as is the one your program used earlier.

You could also see if you can optimize some of Randelshofer's patterns. Their solutions aren't in the metric you're used to, but a different one where any turn of some group of consecutive layers in the same direction counts as one move. Still, your program could probably optimize several of them. Start with the ones that have the highest movecounts, of course, as those are hardest for humans :)
 
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Apologies...

Hi All,

Sorry I haven't been very active on speedsolving.com for a while... I am still investigating 4x4x4 algorithm generation in my spare time and hope to have some interim results to share soon...

But in the mean time I thought some of you might be interested in my latest LEGO MultiCuber 777 robot that can be seen on the ARMflix channel on YouTube solving a V-CUBE 7 :)

http://www.youtube.com/watch?v=b5b9BIBuOd4
 
I wondered what you were upto these days. I see non-cubing related videos on your channel from time to time...

Nice use of an Android :D

Looking forward to updates ^_^
 
Don't know exactly what BHTM is.

My current solution is 18 moves where I count any of r R r2 R2 Rw Rw2 as a single move.

"Block Half-turn Metric" I believe the "block" means any number of layers together including one face turned together and "half-turn" means both 90 and 180 degree turns count as one move so I think R R2 Rw Rw2 all count as one move but (I may be wrong here) in BHTM r and r2 would count as two because they would need to be done as Rw R' and Rw2 R2 respectively.

Can anyone here confirm my understanding?

Anyway, thanks. The 18 moves gives me some idea of what to aim for.

My investigations of the ULF<->DBR exchange, I believe my software has proved that no solution shorter than 16 BHTM exists and has not yet been able to find a 16 BHTM solution. My guess is that an 18 BHTM solution could well be the shortest although I'm still working on this.

I suspect that the minimum number of moves for a UFr <-> URf swap is probably the same as a ULF<->DBR exchange so it may well be that an 18 move solution is the shortest. However, there are likely to be a number of solutions of this length...

Is your current solution "nowhere near fast enough" because it is an awkward sequence to manipulate or just that you feel it is too many moves?
 
I currently use;

R2DB2l'U2l'U2xU2l'U2rx'U2r'U2l2B2D'R2

18 moves in BSTM, 24 moves in BHTM.

It's an awkward sequence to execute.

@Kirjava,

Did you notice my solution I posted in my parity thread? Is that faster than the alg you use?
Here is a K4 Alg for swapping two adjacent wing edges (21q/15h)
On Even Cubes: (notice that for the 4X4X4, you can merge the last two slice turn moves as M' l')
2L'2 U l U2 2R' U2 r' U2 x U' 2R' U x' U2 x U' 2R 2L'2
On Odd Cubes:
2L'2 U l U2 2R' U2 3r' U2 x U' 2R' U x' U2 x U' 2R 2L'2

Edit:

@IAssemble
According to qqwref, block quarter turn metric is:
There are for mayor metrics

QTM, single face quarter turns
HTM, single face half turns
SQTM, slice quarter turns
STM, slice turn metric

If you do not specify I (and most people) assumes you are talking about face turns. This because the "normal" metric is HTM.

If you use a M, S or a E in FMC it is always counted as two turns because there turns are counted in the normal metric...

For cubes greater than 3x3, this post is wrong and completely nonsensical.

Apart from the distinction between half and quarter turn metrics (since every metric can be considered with either quarter or half turns), there are three useful metrics on larger cubes:
- block turn metric, where any group of contiguous slices moving the same way is counted as one move
- layer turn metric, where the only allowed moves are moves of a single slice layer
- multislice turn metric, where the only allowed moves are moves that move slices 1 through n on a given face in the same way
I believe what you are using, cmowla, is the block quarter turn metric. I prefer the block (half) turn metric for pattern-making because it is the least restrictive without adding moves that don't feel like moves. It reduces to STM/SQTM on 3x3, as does the layer turn metric; the multislice metric reduces to HTM/QTM on 3x3.
So BHTM is the same thing as BQTM, but 180 is counted as one turn instead of two in BQTM.

So, what I think you are referring to is not BHTM, but half turn plane metric: ftm (full turn moves I guess). BHTM and ftm are equivalent as long as there are no inner-slice turns involved (as you have implied), but BHTM counts r2, r, etc., as one move each.

And in ftm, when you refer to UFr <-> URf and ULF<->DBR swaps, are you referring to non-center preserving algorithms or are you referring to pure algorithms (no centers discolored)?


Edit:
@IAssemble,
How do you figure that the minimum for UFr <-> URf and ULF<->DBR swaps are equal (both 18ftm)?

The least I could find for a pure UFr<-->URf is 20 ftm (the algorithm in this post).
The least I could find for a pure ULF<->DBR is 18 ftm (I found several of those through conventional parity algorithms), which is what I assume your solver has found too.

I agree with your software's conclusion that below 16ftm is impossible for ULF<->DBR, but, I don't think anything below 18ftm for ULF<->DBR is possible either (I agree with you...and I think your software will eventually agree with us). Based on all the ways I know to form pure 2-cycle algorithms, indeed this very strict move count cannot manifest in less than 18 ftm for ULF<->DBR and definitely not for UFr<-->URf.

Do you have an 18ftm for UFr<-->URf? Because I cannot imagine that being lower than 20 ftm (the alg I found).

Going to another category,from my findings, for non-center preservation and non 3X3X3 preservation, the (absolute) minimum amount of ftm to swap ULF<-->DBR (while preserving all other dedges) is 9 ftm, for UFr<-->DBR 11ftm, for the single edge flip (both double parity and solely OLL parity) is 11ftm, UFr<-->URb is 9ftm, etc.
 
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Hi All,

Sorry I haven't been very active on speedsolving.com for a while... I am still investigating 4x4x4 algorithm generation in my spare time and hope to have some interim results to share soon...

But in the mean time I thought some of you might be interested in my latest LEGO MultiCuber 777 robot that can be seen on the ARMflix channel on YouTube solving a V-CUBE 7 :)

http://www.youtube.com/watch?v=b5b9BIBuOd4

just saw it on a science website. awesome:)
 
I currently use;

R2DB2l'U2l'U2xU2l'U2rx'U2r'U2l2B2D'R2

18 moves in BSTM, 24 moves in BHTM.

It's an awkward sequence to execute.

@Kirjava,

Did you notice my solution I posted in my parity thread? Is that faster than the alg you use?
Here is a K4 Alg for swapping two adjacent wing edges (21q/15h)
On Even Cubes: (notice that for the 4X4X4, you can merge the last two slice turn moves as M' l')
2L'2 U l U2 2R' U2 r' U2 x U' 2R' U x' U2 x U' 2R 2L'2
On Odd Cubes:
2L'2 U l U2 2R' U2 3r' U2 x U' 2R' U x' U2 x U' 2R 2L'2

Edit:

@IAssemble
According to qqwref, block quarter turn metric is:
There are for mayor metrics

QTM, single face quarter turns
HTM, single face half turns
SQTM, slice quarter turns
STM, slice turn metric

If you do not specify I (and most people) assumes you are talking about face turns. This because the "normal" metric is HTM.

If you use a M, S or a E in FMC it is always counted as two turns because there turns are counted in the normal metric...

For cubes greater than 3x3, this post is wrong and completely nonsensical.

Apart from the distinction between half and quarter turn metrics (since every metric can be considered with either quarter or half turns), there are three useful metrics on larger cubes:
- block turn metric, where any group of contiguous slices moving the same way is counted as one move
- layer turn metric, where the only allowed moves are moves of a single slice layer
- multislice turn metric, where the only allowed moves are moves that move slices 1 through n on a given face in the same way
I believe what you are using, cmowla, is the block quarter turn metric. I prefer the block (half) turn metric for pattern-making because it is the least restrictive without adding moves that don't feel like moves. It reduces to STM/SQTM on 3x3, as does the layer turn metric; the multislice metric reduces to HTM/QTM on 3x3.
So BHTM is the same thing as BQTM, but 180 is counted as one turn instead of two in BQTM.

So, what I think you are referring to is not BHTM, but half turn plane metric: ftm (full turn moves I guess). BHTM and ftm are equivalent as long as there are no inner-slice turns involved (as you have implied), but BHTM counts r2, r, etc., as one move each.

And in ftm, when you refer to UFr <-> URf and ULF<->DBR swaps, are you referring to non-center preserving algorithms or are you referring to pure algorithms (no centers discolored)?

Ah - so are all my metrics "MHTM" (multislice, not BHTM) as my software only considers turns of contiguous layers always including a face layer? Or is this what you refer to as just FTM? BTW I think about it this way as this naturally fits with the way my LEGO robots manipulate cubes larger than 3x3x3 :)

Edit:
@IAssemble,
How do you figure that the minimum for UFr <-> URf and ULF<->DBR swaps are equal (both 18ftm)?

I wasn't thinking when I wrote this (I missed the point of the lower-case letters refering to edge pieces rather than corner pieces).

My software has analysed sequences for ULF<->DBR swaps up to and including 16 FTM moves and I have convinced myself that no solutions exist as short as this but I don't have any exhuastive analysis for more that 16 moves.

The least I could find for a pure UFr<-->URf is 20 ftm (the algorithm in this post).
The least I could find for a pure ULF<->DBR is 18 ftm (I found several of those through conventional parity algorithms), which is what I assume your solver has found too.

I agree with your software's conclusion that below 16ftm is impossible for ULF<->DBR, but, I don't think anything below 18ftm for ULF<->DBR is possible either (I agree with you...and I think your software will eventually agree with us). Based on all the ways I know to form pure 2-cycle algorithms, indeed this very strict move count cannot manifest in less than 18 ftm for ULF<->DBR and definitely not for UFr<-->URf.

Do you have an 18ftm for UFr<-->URf? Because I cannot imagine that being lower than 20 ftm (the alg I found).

I have started to analyse the UFr <-> URf case now and don't have any interesting results yet.

No, I don't have a solution as short as 18ftm. Sorry if I gave the impression that I did!

Going to another category,from my findings, for non-center preservation and non 3X3X3 preservation, the (absolute) minimum amount of ftm to swap ULF<-->DBR (while preserving all other dedges) is 9 ftm, for UFr<-->DBR 11ftm, for the single edge flip (both double parity and solely OLL parity) is 11ftm, UFr<-->URb is 9ftm, etc.

Interesting.

I really must devote more time to my software and stop playing with LEGO! :D
 
Ah - so are all my metrics "MHTM" (multislice, not BHTM) as my software only considers turns of contiguous layers always including a face layer? Or is this what you refer to as just FTM? BTW I think about it this way as this naturally fits with the way my LEGO robots manipulate cubes larger than 3x3x3 :)
That would make sense, because in order for it to do, say, r, it would need to do Rw R'.

Interesting.

I really must devote more time to my software and stop playing with LEGO! :D
Come on, don't spoil the fun!:)
 
Ah - so are all my metrics "MHTM" (multislice, not BHTM) as my software only considers turns of contiguous layers always including a face layer? Or is this what you refer to as just FTM? BTW I think about it this way as this naturally fits with the way my LEGO robots manipulate cubes larger than 3x3x3 :)
That would make sense, because in order for it to do, say, r, it would need to do Rw R'.
Exactly. Thinks.... maybe I could design a robot capable of slice turns?!?
Interesting.

I really must devote more time to my software and stop playing with LEGO! :D
Come on, don't spoil the fun!:)
Yes - fun is what it's all about :)
 
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