# Belt Method



## Hodari (Jan 18, 2011)

Basically, the idea is to solve the middle layer first, then sort the remaining pieces into the top and bottom layers and solve each layer separately. However there are a few special "parity" cases where one piece in each layer is incorrectly placed or oriented. It may be possible to use full OLL and PLL to do this, however due to these parity cases I imagine recognition would be extremely difficult, so 2 look versions are probably much easier.

(Optional) Place the 4 bottom layer edge pieces first, oriented correctly but not necessarily permuted. This gets a bit away from being a pure belt method but avoids most of the "parity" problems and greatly simplifies the remaining steps.

The middle layer can be solved very easily so shouldn't need to go into detail there. Since there's no F2L step to worry about, color neutrality is also much easier here which is one advantage to this method.

Next the top/bottom edges need to be sorted into their correct sides. If there are an even number of edges on the wrong side, this can be done easily by placing 2 incorrect edges on the top right and bottom right, do R2, rotate top and bottom layers to align another pair of incorrect edges, and finally another R2 to place the edges and fix the middle layer. If there are an odd number of edges that need to be switched, for the final pair, simply place then both on the front and do:
M' U2 M

Next orient all of the edges. If there are an even number incorrectly flipped on each side, this can be done normally. If there are an odd number, you can get 3 oriented correctly on each side, place the remaining 2 edges on the front, and use the following algorithm:
U M U M U2 F2 M' U M' U' F2

Next the corners need to be sorted onto their correct sides. Place one incorrect corner from each layer in the front right and do:
(R U R' U') x3

You can also switch 2 pairs of corners at once by placing them on UL and DR and doing:
R2 F (R U R' U') x3 F' R2

Then orient the corners. This can be done using normal OLL algorithms, however again there is the possibility of having one corner incorrectly flipped on each side. To fix this, use normal OLL to orient everything on one side except for one corner, place this corner on the bottom front right, and use one of the following algorithms:

If the bottom color is on the right side:
R U2 R U R' U R U2 R2

If the bottom color is on the front:
R2 U2 R' U' R U' R' U2 R' 

This will correct the corner parity on both sides and then normal OLL can be used to finish the other side.

Next is corner permutation(no parity issues here  ) and then finally edges, which do have a parity that can be fixed with the following algorithm:
M2 U2 M2.


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## da25centz (Jan 18, 2011)

interesting. i played around with it for a few minutes, then then timed a solve, around 55 seconds, and i avg 28. its not particularly difficult, and it could be fun just to play around with. i believe there was a thread about it a long time ago, but i havent really searched for it. the way i would do it is

solve the "belt"
separate edges using M' U M
separate corners using (R U R' U')x3
2look oll + R U2 R U R' U R U2 R2 for corner parity, U M U M U2 F2 M' U M' U' F2 for edge parity
PLL + M2 U2 M2 for parity


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## Julian (Jan 18, 2011)

Pretty cool. I use 2 sunes for corner orientation parity. Got a 45.31 with only layer sorting parities. OLL and PLL went smoothly.


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## Hodari (Jan 19, 2011)

da25centz said:


> separate edges using M' U M


 
I believe that should be M' U*2* M with the edges in front? Looks like that is a much better way of doing that case


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## da25centz (Jan 19, 2011)

Hodari said:


> I believe that should be M' U*2* M with the edges in front? Looks like that is a much better way of doing that case


 
i do one at a time. the edge on top should be to the right, with the edge on bottom at DF


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## Cool Frog (Jan 19, 2011)

Nope, M' U M works fine
However M' U2 M' is great for partial edge control. I find that F2l like insertions to effect corner orientation eliminates parity and gives you no oll for D face.
31.86 Need to refresh my OLL Have not seen that case in over 6 months.
You can force some skips and easy D face Pll's really easy.


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## da25centz (Jan 19, 2011)

Cool Frog said:


> Nope, M' U M works fine


 
its just a preference as to where you want to put the edges to be swapped, and whether youd rather do U or U2 turns


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## AvGalen (Jan 19, 2011)

If you like solving this way, you should look into domino, square-1 puzzles and ortega/guimond methods. Also, create a cube with only 3 colored stickers, 1 for the bottom layer (not just face, so 21 stickers), another one for the middle layer (12 stickers) and another one for the top layer (21 stickers again) and then try to solve it middle layer first, then last two layers at the same time. Sub 10 moves is the goal, but I don't know if that is always possible. I normally don't need more than 11 though

This method will not be fast though, too many moves and steps and things to do on the bottom. "seperation" works well if there are a limited amount of possibilities like the 10*10 = 100 on square-1 edges, but for 3x3x3 21*21 is too much (ignoring parity) and having to do middle layer, seperation, 2 * OLL and 2 * PLL is just too many big steps. for easy seperation half turns would also be very often used and they are slower than quarter turns


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## Ranzha (Jan 19, 2011)

I'd solve middle layer and then orient all pieces. Separate and permute corners. FL edges and EPLL.
MOSIE.
Middle Layer
Orient
Separate and permute corners (3x3 PBL)
Initial edges (FL edges)
EPLL.


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## cuBerBruce (Jan 19, 2011)

AvGalen said:


> Also, create a cube with only 3 colored stickers, 1 for the bottom layer (not just face, so 21 stickers), another one for the middle layer (12 stickers) and another one for the top layer (21 stickers again) and then try to solve it middle layer first, then last two layers at the same time. Sub 10 moves is the goal, but I don't know if that is always possible. I normally don't need more than 11 though



My analysis for AvGalen's tri-color cube is given in the spoiler. I calculate step 1 taking up to 5 moves, step 2 taking up to 8 moves. But the whole puzzle can always be solved in 8 moves or less.



Spoiler





```
Step 1 optimal:
moves   positions
  0          1
  1          4
  2         50
  3        286
  4        152
  5          2
           ---
           495

Step 2 using <U,D,L2,R2,F2,B2>
moves   positions
  0          1
  1          4
  2         22
  3         82
  4        292
  5        986
  6       2001
  7       1312
  8        200
          ----
          4900

Step 2 optimal:
moves   positions
  0          1
  1          4
  2         22
  3         82
  4        292
  5       1074
  6       2521
  7        892
  8         12
          ----
          4900

Whole puzzle optimal:
moves   positions
  0          1
  1         12
  2        150
  3       1882
  4      22824
  5     241494
  6    1359451
  7     799130
  8        556
       -------
       2425500
```


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## AvGalen (Jan 19, 2011)

cuBerBruce said:


> My analysis for AvGalen's tri-color cube is given in the spoiler. I calculate step 1 taking up to 5 moves, step 2 taking up to 8 moves. But the whole puzzle can always be solved in 8 moves or less.


I thought "middle first, then the rest was an efficient technique for this puzzle. But with roughly 2/3 of the solution solvable in 6 moves or less I will try to really FMC it in the fuutre. Also, good to hear that some of the stage 2 cases do require so many moves. I have had a couple of situations where I needed 4 or 5 moves for step 1, but couldn't finish step 2 in 6 moves or less, not even after sometimes cancelling a move with step 1.

If step 1 is the E-slice, are there any positions that are solved with a quarterturn of FRBL? (aka, non domino)


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## AvGalen (Jan 19, 2011)

cuBerBruce said:


> My analysis for AvGalen's tri-color cube is given in the spoiler. I calculate step 1 taking up to 5 moves, step 2 taking up to 8 moves. But the whole puzzle can always be solved in 8 moves or less.


I thought "middle first, then the rest was an efficient technique for this puzzle. But with roughly 2/3 of the solution solvable in 6 moves or less I will try to really FMC it in the fuutre. Also, good to hear that some of the stage 2 cases do require so many moves. I have had a couple of situations where I needed 4 or 5 moves for step 1, but couldn't finish step 2 in 6 moves or less, not even after sometimes cancelling a move with step 1.

If step 1 is the E-slice, are there any step 2 solutions that are solved with a quarterturn of FRBL? (aka, non domino)


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## pi.cubed (Jan 19, 2011)

Great description. I just put a link to here on the wiki page.

EDIT: This would be good as THE wiki page. Mind if I copy it and paste?


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## abctoshiro (Jan 19, 2011)

I do:

EObelt
remove your so-called "parity" by putting the top and bottom layers into possible OLL cases (<R, U> subset, but I mostly use <R2, U>)
Orient everything.
Separate.
PBL.


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## Hodari (Jan 19, 2011)

pi.cubed said:


> Great description. I just put a link to here on the wiki page.
> 
> EDIT: This would be good as THE wiki page. Mind if I copy it and paste?


 
I'd be honored to have it on there actually once all the best algorithms have been worked out for it. M' U2 M definitely looks better than my original one for switching the last 2 edges and R U2 R U R' U R U2 R2 might be better for corner parity, but doesn't seem to work in all cases, so I'm guessing either a second algorithm is needed to handle those cases or maybe just a little more detail on exactly how the corners should be positioned to use it. Also that algorithm messes up the orientation of other pieces as well, so it needs to be done before OLL. One which preserves the orientation of the other pieces might be better if a reasonably short algorithm can be found which does this. Further discussion of any ways to avoid the parities or force skips on any other steps might be good as well(sorry if these are obvious to most of you but again, I'm fairly new at this)



> This method will not be fast though, too many moves and steps and things to do on the bottom. "seperation" works well if there are a limited amount of possibilities like the 10*10 = 100 on square-1 edges, but for 3x3x3 21*21 is too much (ignoring parity) and having to do middle layer, seperation, 2 * OLL and 2 * PLL is just too many big steps. for easy seperation half turns would also be very often used and they are slower than quarter turns



Yeah, I think I acknowledged in my OP that this wasn't going to be the fastest method in the world and this was mostly just for fun and trying something different. Odds are it will be tough to find anything much faster than Fridrich anyway but on the other hand, the only way anything ever would be found is by experimenting with different stuff like this. Anyway, if someone got close to 30 seconds on it within an hour or so of me posting this, it can't be TOO bad and I imagine with practice and some further improvements, 15-20 seconds at least should be possible for some people.

I think this method does have 2 things going for it though. First, as I said before, color neutrality is much easier since there's no F2L to worry about so all you really need to know is which colors are opposite each other. Second, F2L is probably the hardest part of Fridrich method to master, so taking that out and replacing it with another OLL and PLL may well be useful for at least some people.

Finally, consider this to be the equivalent of a beginner's method. I imagine if people really wanted to take it seriously and get faster with it, several things could be done like developing new algorithms that handle the parity cases and solve at least one side's OLL in a single step or using some of the techniques referred to by others in this topic to avoid those parities in the first place.


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## da25centz (Jan 19, 2011)

Hodari said:


> R U2 R U R' U R U2 R2 might be better for corner parity, but doesn't seem to work in all cases, so I'm guessing either a second algorithm is needed to handle those cases or maybe just a little more detail on exactly how the corners should be positioned to use it. Also that algorithm messes up the orientation of other pieces as well, so it needs to be done before OLL.


 
the way i use that alg is i do OLL on one side, then use that alg to fix parity on that side, then do OLL to the other side. That alg fixes parity on DFR, do i would do oll on the u face, then do a z2 to put the prity corner in that spot, do the parity alg, and then do OLL on what is now the U face, then PLL*2


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## Hodari (Jan 19, 2011)

da25centz said:


> the way i use that alg is i do OLL on one side, then use that alg to fix parity on that side, then do OLL to the other side. That alg fixes parity on DFR, do i would do oll on the u face, then do a z2 to put the prity corner in that spot, do the parity alg, and then do OLL on what is now the U face, then PLL*2


 
Thanks, though still looks like that only works if the bottom color is on the right side with the corner in that position and would have to do it twice when the bottom color is in front instead or use another alg for that case. Would R2 U2 R' U' R U' R' U2 R' be best there?


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## cuBerBruce (Jan 19, 2011)

cuBerBruce said:


> My analysis for AvGalen's tri-color cube is given in the spoiler. I calculate step 1 taking up to 5 moves, step 2 taking up to 8 moves. But the whole puzzle can always be solved in 8 moves or less.


 


AvGalen said:


> If step 1 is the E-slice, are there any step 2 solutions that are solved with a quarterturn of FRBL? (aka, non domino)


 
My analysis found that solving optimally allows you to solve 1074-986 = 88 positions in 5 moves that require at least 6 moves to be solved domino style. For example, consider scramble F B L B' F.

Similarly, there are step 2 positions (meaning E layer solved) at depth 6 (600), and 7 (164) that require more moves to solve domino style than solving optimally. Any step 2 position requiring 4 moves or less can be solved optimally domino style.

(I note that my way of counting positions treats positions differing only by a y rotation as distinct, so I get 8!/(4!*4!) = 70 corner positions and (12!/(8!*4!))*(8!/(4!*4!)) = 34650 edge positions, or 70*34650 = 2425500 total positions.)


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## da25centz (Jan 19, 2011)

Hodari said:


> Thanks, though still looks like that only works if the bottom color is on the right side with the corner in that position and would have to do it twice when the bottom color is in front instead or use another alg for that case. Would R2 U2 R' U' R U' R' U2 R' be best there?


 
yeah that would be the CW parity fix


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## Hodari (Jan 20, 2011)

Thanks for the help..updated the original post with a few of these algorithms.


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## Ordos_Koala (Jan 22, 2011)

i looked at it too  it's nice idea for something different than LBL, once solved the cube using this method, but never did it again... but nice work, I'm totally going to learn it for sub-1 min  btw try Petrus, it's way different from LBL method, so when you do it just sometimes when you get bored, it's really fun (just intuitive block-building, last layer is just last layer :/)


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## Hodari (Jun 24, 2011)

With the recent discussion on the LCBM method, I thought I'd take another look at this and came up with a couple more ideas for it. First of all, I added an algorithm for swapping 2 adjacent corners on each layer to the correct sides. Still need to work out algorithms for the remaining cases(diagonally opposite corners on each side, opposite on one side and adjacent on the other, 3 corners on each side, and all 4 corners on each side). 

Secondly, instead of starting with the belt, what about first placing all the bottom color edges in the D layer, oriented but not necessarily permuted correctly(permutation will get messed up in later steps anyway). Then continue with the belt. Placing the D layer edges like this should probably only take about 4 or 5 moves(don't know the exact number but obviously <= the number needed for a true cross and being color neutral should reduce this a bit as well) and this eliminates the first couple parity cases described in my original method and simplifies many of the other remaining steps as well.


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