# UWR - Largest Computer Cube Solve Ever!

Discussion in 'General Speedcubing Discussion' started by qqwref, May 20, 2013.

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1. ### qqwrefMember

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I solved the 111x111x111!

It took a total of just under 30 hours of work, done over 10 days or so on the newest version of IsoCubeSim. Some more detailed statistics:
- Total solve time: 29:51:02.641, or 107,462.641 seconds.
- Total moves: 201,858*
- Number of pieces: 72,596
- Seconds per piece: 1.480 (!!!)
- Moves per piece: 2.781 (!!!)
- Moves per second: 1.878
- Real time spent: about 10 days 6 hours 48 minutes (from 2:59 PM on the 9th, until 9:47 PM on the 19th).
- Percent of time spent working: 12.1%

*The game says 199430 moves, but it actually forgot to count the moves from my first save due to a programming bug which I fixed as soon as I noticed it.

If you're interested in some more details, check out the gallery of progress screenshots or download all the replay files.

I also want to describe the method I used. I really put a lot of thought into making a method for every part that would scale well - remain easy even with extremely large cubes. It's going to be a bit long, so I'll put it in a spoiler...
First, some basics. I solved most faces in two steps: "cleaning" and finishing up. The idea of cleaning a face is to solve as many pieces as possible as quickly as possible. After that step, which typically solves 50% or more of the face, I can go through the longer step of putting each of the remaining pieces of that color into the center, one at a time or in blocks. As for notation, I'll be using something like SiGN (xR means turn the xth layer on the R face, or the first layer if x is absent), except that I'll also be using lowercase p and q as variables to better describe classes of commutators.

1) Solve the edges. I do edges first for large computer cubes because reduction (centers first) scales badly, and because having solved edges actually lets me use them to keep track what part of the puzzle I'm working on. A lot of the time I do a pR' move or something to start working on a row of one center, and the incorrect edge reminds me which row it is. With so many layers it's impossible to do it by counting or visual inspection alone. Anyway, for the first 9 edges I did basically the same thing as with other cubes. For the last three edges I used moves like pR U2 pR' (for the 10th edge only), pR2 U2 pR U2 pR2, and pR' F U' R F' U pR. Instead of solving one orbit at a time like on my keyboard solves, I went for an edge at a time and solved pieces in the same position in groups. For instance on the last 3 edges there are only 6 possible positions for each piece, so with 109 pieces in each edge there are a lot of pieces in the same position which can be solved in the same way, all at once, by replacing a pR move with a whole bunch of slice moves along the same layer. This took about an hour in total.

2) Clean the first center. I used white. Hold the white face on U, and then for each row (perpendicular to the F face), we do something like this: move white centers out of the way on the corresponding row on F using pU moves, then qR', then solve as many white centers into that layer as possible using only pU-axis moves, then qR. By "solve as many white centers ... as possible" I mean I basically look at that row on R, B, and L, and for every white center I see, move that layer the right amount. Note that these rows are vertical, which is something I did because of the way Iso is set up (when zoomed in, it's easy to scroll through a vertical row on F or R using just the up/down arrow keys). You could do any of this from a different angle if you wanted. A typical layer might be something like (2U 4U 7U ...) 3R' y (3U 6U 8U ...) y (U2 4U2 8U2 ...) y (2U' 3U' 4U' ...) y 3R. Each center on white corresponds to 5 stickers, and a white center on any of those 5 stickers means the center gets solved. I think roughly 60% of the center ends up solved.

3) Finish the first center. Now I can hold the center on F and look at R. If I see a white center anywhere, I can solve it using pF qU' pF' qU. Four-move commutators are fast! However, I can't just do that randomly, but instead, I have to do it so that the center moves into a place on the white center that didn't have a solved center already. To check if there is room I do qU' qU, and if there isn't, I rotate the white center and repeat as necessary. You'll see this type of checking often as I continue. Note that this requires centers to be adjacent, and since we're solving from all five, we have to swap two adjacent centers. On the 5x5x5 the algorithm for this is 2R U2 2R' 2L' U2 2L U 3R U' 3R', which you then have to properly undo later.

4) Clean the second center. I used yellow. This is the same idea as the first center - hold yellow on U (so, white on D) and proceed row by row. However, instead of using pR' moves to move a row onto the F face, I used pR U2 pR' moves. This does solve the layers in a different order, but that's okay. For the middle layer I used U pR U' pR'. Each center on yellow again corresponds to 5 stickers (any yellow piece in one of those 5 places means that piece gets solved), and I think roughly 67% of the center ends up solved after this.

5) Finish the second center. Again I swap yellow with another center, so I can hold yellow on F, white on U, and the solved center on R. The commutator this time is pF2 qU' pF2 qU (or pF2 qU pF2 qU' for the mirror). Since I can't just do x rotations and keep going, I have to do some extra center swaps to bring each of the four non-white/yellow centers into the right place. There was also plenty of the qU' qU moves to check if there's room to put a piece.

6) Clean the third center. I used orange (for visibility). Place orange on U and white/yellow on L/R. Now we can do pU2 moves but not pU moves. So I do something like the white center cleaning, where I do pU2 moves to move orange centers out of the way, then a qR' move to bring the row down, then do more pU2 moves to solve as many orange centers as possible, then a qR move. A typical row clean might look something like (2U2 4U2 7U2 ...) 3R' y2 (2U2 3U2 4U2 ...) y2 3R. Each center on orange corresponds to only 3 stickers, solving about 57% of the center.

7) Solve the third center. This is just like what I do on yellow, with the pF2 qU' pF2 qU moves, since they keep U and D solved, and again I use plenty of qU' qU moves. In fact, there are fewer center swaps than last time, since the D center can stay where it is.

8) Clean the fourth center. As a general rule you want this to be adjacent to the third center, so I chose green (since it was brighter). Unfortunately the cleaning setup I use only really lets us clean a center opposite of the third one, so we place orange on D and green on F, then swap F and U, and begin. The cleaning itself is just like the third center except, again, we use pR U2 pR' rather than pR'. Each center on green corresponds to 3 stickers, solving about 70% of the center.

9) Solve the fourth center. Things got pretty messy here and it took a lot longer than I expected; most of that was probably because I can't use 4-move commutators any more and thus had to switch to 8-move ones. I held green on F and orange on D, and then did commutators like [pU2, qR U qR']. In fact I did all of the pU2 moves ahead of time, so that for each row I could immediately do a qR move for each piece I wanted to solve (using pU2 pu2 to check), and then solve many piece in that row at once. The problem, apart from the large number of qR moves, is that I couldn't stray too far from U, because I had to do U turns. So I ended up having to stay on the top half of F, and doing a lot of F and B moves to be able to solve things. Oh yeah, and once I got rid of all the green pieces on B, I had to swap U and B and do it again.

10) Solve the last two centers. No cleaning here - I couldn't see any way to quickly solve most of the center. I held the two centers on F and R, and solved one vertical row of R at a time. The commutator looks something like [pU', R' qU' R]. Basically, I would do a pU' for every center in that row that I could solve, then do the R' qU' R, undo the pU moves, and finish the commutator. Then rotate F and go again - in at most 4 separate block commutators I would solve all pieces in that row. (With one caveat - if the x-center closer to D is unsolved, this commutator won't work on it. When that happened I had to do it individually.) After each finished row I did a qF2 move to get it out of the way and make the next row easy to see; since I had blue on R, I ended up with a whole bunch of green rows on that face. Since I had to be in view of the F-R edge to do R tuns, I only did the first half of the blue face before undoing all those green rows, turning that center by 180 degrees, and doing it again.

And that's it! It might sound kind of complicated, but each step is pretty straightforward once you know what's going on, and I spent the vast majority of my time doing the same kind of thing over and over. The low move/piece count was mostly due to the cleaning stages, where I could solve a ton of pieces in one or two moves each. Those stages would have an absurdly low movecount in axial turn metric / snyder metric, with something like 400 moves solving 5000+ pieces. And in fact, you can do a cleaning stage on the same center more than once - just turn the center to a different orientation and continue. With four of these cleaning stages you should be able to solve every piece on the affected centers. I didn't do this because the returns do diminish, and so I don't think it's worth it to clean more than once for any individual center. Maybe for the fourth center it's worth it though!

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3. ### That70sShowDudeMember

Sep 8, 2008
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o____________________________________________________________O

Jul 10, 2008
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xAsianRejectx
¿¿¿wat???

5. ### Blake4512Member

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oh. my. god.

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Sep 17, 2009
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4EverTrying
No one messes with Mr. Gottlieb! Epic. I nominate qqwref the craziest (in a good way) member of the year.

7. ### uniactoMember

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will you be reconstructing this solve any time soon?

8. ### qqwrefMember

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Thanks everyone!

@uniacto: Nope

9. ### OwenMember

Nov 2, 2009
Incredible.
This must be discouraging to that guy in the middle of a 105x105 solve..

10. ### RubiksfreakMember

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Feb 26, 2012
St.Cloud MN
Based on the world record for 3x3, if you were moving at high speeds to complete this cube it would take around 1220 hrs and that's at world record speeds. Are you sure you completed it in only 29 hrs?

Sent from my LG-L38C using Tapatalk 2

11. ### qqwrefMember

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A 3x3x3 has 20 pieces, so the WR spent ~0.28 seconds per piece. This one had 1.48. Admittedly, by my standards on other large cubes, that IS pretty darn fast. I think it had to do with lots of block commutators (with low moves per piece, but some overhead per row) as opposed to the large number of 4- and 8-move commutators I normally do.

EDIT: Also I'd like to point out that if this took 1220 hours I couldn't exactly have started on May 9th Even including non-solving time, I only spent ~247 hours working on this.

Last edited: May 20, 2013

Dedication, brains and years of computersolving experience put together. Interesting read about the cleanup phase.
But I think you avoided doing a 112x112x112 because the secret parity on that is too much for you

13. ### IggyMember

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Woahh... that's awesome. Great job. :tu

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wow gratz

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ucrazy >.<

16. ### ben1996123Banned

May 17, 2009
Ponyville
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yae you finished it sub30

also, soup has done 20 hours of his 105x105 solve... hes half way through the first centre

17. ### BaconCuberMember

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Jan 29, 2013
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Um...um...um...*stares at screen with mouth open in shock*

18. ### soupMember

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Mar 12, 2012
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I hereby quit my 105x105 solve. 120x120 now.

19. ### uniactoMember

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