# 5x5x5 fewest moves for concentric rings



## unsolved (Dec 19, 2015)

I managed to create this in 24 moves in the single slice turn metric. I was wondering if anyone could solve it in fewer turns?


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## henrysavich (Dec 20, 2015)

I'm assuming you also just spammed +cancelled commutators? Also not entirely sure what "Single slice turn metric is" but I assume it is like stm but wide slice moves count as multiple slice moves.

anyway, best I could get was 31:

m r' F l F' r m' F2 r' F' l' F2 l m F' r U2 r' F' l' m' F r m F r' F' m' F r F'


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## unsolved (Dec 20, 2015)

Well first I move the 4 centers at the corners simultaneously with

2R U2 2R 2U2 2D2 2R' 2L 2D2 2U2 2L' U2 2R'

https://alg.cubing.net/?setup=2R_U2_2R_2U2_2D2_2R-_2L_2D2_2U2_2L-_U2_2R-&puzzle=5x5x5&view=playback

Then I moved the 4 middle centers simultaneously

3R2 D' 3F2 B' 2R 3F2 3U2 2R' 3U2 B D 3R2

https://alg.cubing.net/?setup=3R2_D-_3F2_B-_2R_3F2_3U2_2R-_3U2_B_D_3R2&puzzle=5x5x5&view=playback


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## Christopher Mowla (Dec 20, 2015)

unsolved said:


> I managed to create this in 24 moves in the single slice turn metric. I was wondering if anyone could solve it in fewer turns?


Yes. I just found a 22 by hand:

 2L U' 3L D 2L2 3F2 2L2 3F2 D' 3L' D F2 D' U 2L' 2R' U' D F2 U D' 2R


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## henrysavich (Dec 20, 2015)

Christopher Mowla said:


> Yes. I just found a 22 by hand:
> 
> 2L U' 3L D 2L2 3F2 2L2 3F2 D' 3L' D F2 D' U 2L' 2R' U' D F2 U D' 2R



Derivation please?


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## Christopher Mowla (Dec 20, 2015)

henrysavich said:


> Derivation please?


Oh, I didn't think unsolved was interested, so I didn't give one, but here it is since you asked.



Spoiler: Derivation



[1]
Start with qqwref's old (but cool) adj. center swap alg applied to the 5x5x5 (converted to the face turn metric for fewest single slice turn moves):
2L
U' D F2 D' U
2L'
2R'
U' D F2 U D'
2R

[2]
We see that this is just a swap of two + centers away from being the desired result. Using the commutator:
2L2 3F2 2L2 3F2

[3]
So, start executing the alg above.
2L
U'

We can see that if we insert a move 3L, then the two front + center pieces are on the opposite side of the cube as the top two + center pieces. (These are the four we need to cycle.)

2L
U'
3L

[4]
Executing the next move of qqwref's alg, they are now not only on opposite sides, but directly lined up as needed to be for 2L2 3F2 2L2 3F2.

2L
U'
3L
D

[5]
So execute that 4 move commutator now:
2L
U'
3L
D
2L2 3F2 2L2 3F2

[6]
Now, when we inserted the move 3L, we have created a problem which we must "undo the effects of" before the move F2.

Since 2L2 3F2 2L2 3F2 does nothing except affect the centers, we "undo" the effects of what we did with D' 3L' D (since 3L D + D' 3L' D = D = the move before the commutator) and then execute the rest of the sequence.
2L
U'
3L
D
2L2 3F2 2L2 3F2
*D' 3L' D*
F2 D' U
2L'
2R'
U' D F2 U D'
2R

[7]
Writing on one line, we're done.

2L U' 3L D 2L2 3F2 2L2 3F2 D' 3L' D F2 D' U 2L' 2R' U' D F2 U D' 2R


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## unsolved (Dec 20, 2015)

The functional solving range of my 5x5x5 brute force solver is roughly 12 moves.






You can see from above it processed 147,990,866,406,820,695 (147 quadrillion) nodes in under half an hour. Even those speeds can't compete with the huge branching factor inherent in the 5x5x5 cube.

Despite that, occasionally it finds a cool algorithm that I can't locate with intense googling. Like solving the 4 corner centers simultaneously as shown in my previous post. There isn't a collection of 5x5x5 algs anywhere with links to that awesome viewing tool made by Mr. Garron.

So I thought the fastest solve might entail linking two different optimal solves together, moving 4 centers simultaneously each time. Now I know there is a faster solution. Thanks.


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## Christopher Mowla (Dec 20, 2015)

unsolved said:


> Now I know there is a faster solution. Thanks.


You're welcome.

Also note that my alg is 22 fewer qtm and 10 fewer ftm (OBTM) than your combo as well.


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## unsolved (Dec 21, 2015)

Christopher Mowla said:


> You're welcome.
> 
> Also note that my alg is 22 fewer qtm and 10 fewer ftm (OBTM) than your combo as well.



I never really had all of the different move metrics sink in. I originally wrote the 4x4x4 program because I forgot how to solve it, then it ballooned into the huge project it is today with parallel processing capabilities and the ability to do 5x5x5 solves as well.

Every now and then, I execute a solve incorrectly when I am curbing, and I'm left with some strange edge-cycling position, like the one shown above. Is that a "6-cycle," if there is such a thing?

I have thought about having the 5x5x5 program work backwards from the solved state and export any "algs of interest," such as the 8-center swap at the top of the post.


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## Christopher Mowla (Dec 21, 2015)

It's actually a 4-cycle. There is someone who found many 2 2-cycle "seed algorithms" as well as 3-cycle "seed algorithms". He posted his results here. (Go to Cube Design-> CubeAlgorithms.)

One of the 2 2-cycle seeds he found was one which we can build my nxnxn btm optimal unoriented opp PLL parity algorithm, F2 2L e F2 e' 2L' 2R' e F2 e' 2R F2//Safe, from. (He found this seed among others the year after I found and posted this alg.)

However, he didn't provide 4-cycle algorithm sets. He also has many algorithms for center pieces, but I don't think he has any algorithm sets like the two you used to get your 24.

Now, about 4-cycles of wing edges, I found many brief algorithms for several of the 110 last layer 4-cycle cases, but I never published the document online because it's incomplete and there wasn't a demand for it.

I do note, however, that finding 4-cycle move sequences is "trivial" if you use a center setup algorithm such as:

3r u2 l r u2 r2 U2 r U' l U2 for the 5x5x5 (this sets up the centers in slice 2L) 
2F2 2U' 2R2 u2 s' for the 4x4x4 (this sets up the centers in slice 2R). 

That is, you simply use outer layer turns to insert edges in the 2L or 2R slices, respectively, (while preserving the center formation in that slice) to achieve any desired 4-cycle by conjugating. For example, on the 4x4x4 parity algorithms wiki page, I have (2F2 2U' 2R2 u2 s') 2R (s u2 2R2 2U 2F2).

Of course, these "trivial" algs are not brief (for the most part), but I thought I would mention this for those who might be interested regardless.


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## unsolved (Dec 21, 2015)

So are there any 5x5x5 algs that come up that cubers wish had shorter solutions? Or are they all essentially "safe" versions of 4x4x4 algs, like the one you discovered?


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## Christopher Mowla (Dec 21, 2015)

unsolved said:


> So are there any 5x5x5 algs that come up that cubers wish had shorter solutions? Or are they all essentially "safe" versions of 4x4x4 algs, like the one you discovered?


At this point, unless they are new and are unaware of all of the available algorithm sets, I highly doubt it. However, if they do, they will post their request in the "Request an Alg Thread" (where you should have probably posted about this topic to begin with, unless you were going to come up with a conjecture of some kind if no one could beat 24 moves. I don't know why no one else had a crack it your request or attempted to beat my 22).

If you really want to "get your hands dirty", you can find optimal solutions (in the single slice turn metric, of course, because that's what your solver does) to all 110 last layer 4-cycle cases (on pages 13, 15, 17, and 19 of that PDF).

Of course, we have move optimal algorithms to a few of those already, but there's many cases for which optimal solutions have never been found to my knowledge.


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## unsolved (Dec 22, 2015)

Can you provide me with a "longest known" 4-cycle alg so I can see if the current version of my program can improve on it?

With very little coding, I can have my program just spit out every move sequence that produces an "n-cycle" position. It would be easy enough to reverse them to get the "solution" once presented with such a position. I should be able to solve every such case requiring 16 moves or fewer in one calendar year. I imagine that would generate quite a few algs we've never seen before!


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## Christopher Mowla (Dec 22, 2015)

unsolved said:


> Can you provide me with a "longest known" 4-cycle alg so I can see if the current version of my program can improve on it?


Here's a 23 I came up with: 2R U2 2R R U' 2L U F2 U' 2R U F2 U' 2L' 2R' U R' U 2R U' 2R' U2 2R'

Note that I didn't spend too much time on any particular case, but this is one of a few of my longest.


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## unsolved (Dec 22, 2015)

And 4-cycles are the most numerous due to all of the possible permutations? Cubers tend to forgo learning any larger cycle count because there's too much to memorize and the likelihood of encountering a specific case is small I would imagine.

Once my program finishes looking at a swap + flip of two wing edges I'll load up your 23-mover.


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## clement (Jan 4, 2016)

Christopher Mowla said:


> Here's a 23 I came up with: 2R U2 2R R U' 2L U F2 U' 2R U F2 U' 2L' 2R' U R' U 2R U' 2R' U2 2R'



My (4x4x4) solver gave a 19 moves solution: 2L2 D2 F 2L2 F' 2R F 2L2 F 2R' F2 2L D2 B2 U' 2R U B2 2L
EDIT: Well, I got a 17 moves solution: 2F2 R2 D2 U2 L2 U 2B L2 D2 2F' U2 R2 2F2 U2 2F' U 2F2
EDIT2: 15 moves solution: 2F2 U' R2 2B U2 2F L2 2F L2 2B' U2 R2 2F' U 2F2


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## Christopher Mowla (Jan 4, 2016)

clement said:


> My (4x4x4) solver gave a 19 moves solution: 2L2 D2 F 2L2 F' 2R F 2L2 F 2R' F2 2L D2 B2 U' 2R U B2 2L
> EDIT: Well, I got a 17 moves solution: 2F2 R2 D2 U2 L2 U 2B L2 D2 2F' U2 R2 2F2 U2 2F' U 2F2
> EDIT2: 15 moves solution: 2F2 U' R2 2B U2 2F L2 2F L2 2B' U2 R2 2F' U 2F2


Excellent. I think most of the 4-cycles I found are 15 moves, but this is impressive.


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