# New Idea: The Partial Cross Method



## Jason Baum (May 18, 2008)

Hey everyone,

I've recently been playing around with new idea I had. I was talking to some people at the Washington DC Open about it, and they all seemed to like it, so let me know what you think. The idea is to eliminate all tricky crosses. Sometimes, it's much easier to make your cross with one piece misoriented. So, if a case like this comes up, make your cross this way and finish your F2L normally. When you get to the LL, hold the misoriented edge at DF. The LL is done similarly to the Roux method: orient edges, solve corners, solve edges. For orienting edges, you will always have either two or four edges to orient. You can orient DF, UL, UF, and UR by simply doing M' U' M. So you want to make sure to always get this case, because it's much easier to do this than it is to orient two edges. So, use some parial edge control/ZBF2L if necessary to make sure to get this case. Next, solve corners using COLL. Now you solve the remaining edges in one step. You will either have a 3 cycle, a 5 cycle, or two 2 cycles. There are 39 cases in this step (40 including the solved case). The algs are very short (longest is 11 moves STM), and they are all MU two gen algs. Here are the 39 cases:

3 Cycles:
UF-UR-DF: M U2 M U M’ U2 M’ U’ (8)
UF-DF-UR: U M U2 M U’ M’ U2 M’ (8)
UF-UB-DF: U2 M’ U2 M (4)
UF-DF-UB: M’ U2 M U2 (4)
UF-UL-DF: M U2 M U’ M’ U2 M’ U (8)
UF-DF-UL: U’ M U2 M U M’ U2 M’ (8)

5 Cycles:
UF-UR-UB-UL-DF: M2 U M2 U M U2 M’ (7)
UF-UR-UB-DF-UL: U2 M U2 M' U M2 U M2 U2 (9)
UF-UR-UL-UB-DF: U2 M’ U2 M’ U’ M’ U2 M U’ M2 (10)
UF-UR-UL-DF-UB: M' U2 M U M' U2 M U' (8)
UF-UR-DF-UB-UL: U M U2 M' U M2 U M2 U' (9)
UF-UR-DF-UL-UB: U2 M U2 M' U M2 U M U2 M (10)

UF-UB-UR-UL-DF: U2 M’ U2 M U’ M’ U2 M U’ (9)
UF-UB-UR-DF-UL: M U2 M' U' M2 U' M U2 M U2 (10)
UF-UB-UL-UR-DF: U2 M' U2 M U M' U2 M U (9)
UF-UB-UL-DF-UR: M’ U2 M’ U’ M2 U’ M U2 M’ U2 (10)
UF-UB-DF-UR-UL: U' M' U2 M U M' U2 M (8)
UF-UB-DF-UL-UR: U M’ U2 M U’ M’ U2 M (8)

UF-UL-UR-UB-DF: U2 M' U2 M' U M' U2 M U M2 (10)
UF-UL-UR-DF-UB: M’ U2 M U’ M’ U2 M U (8)
UF-UL-UB-UR-DF: M2 U' M2 U' M U2 M' (7)
UF-UL-UB-DF-UR: U M2 U’ M2 U’ M U2 M’ U’ (9)
UF-UL-DF-UR-UB: U2 M’ U2 M’ U M2 U M U2 M’ (10)
UF-UL-DF-UB-UR: U2 M2 U’ M2 U’ M U2 M’ U2 (9)

UF-DF-UR-UB-UL: M U2 M’ U M2 U M2 (7)
UF-DF-UR-UL-UB: U’ M’ U2 M U’ M’ U2 M U2 (9)
UF-DF-UB-UR-UL: M2 U’ M’ U2 M U’ M U2 M U2 (10)
UF-DF-UB-UL-UR: M2 U M’ U2 M U M U2 M U2 (10)
UF-DF-UL-UR-UB: U M’ U2 M U M’ U2 M U2 (9)
UF-DF-UL-UB-UR: M U2 M’ U’ M2 U’ M2 (7)

2 Cycles: 
UF-UR, UB-DF: U2 M U2 M' U' M' U2 M' U M2 U2 (11)
UF-UR, UL-DF: U' M U2 M' U M' U2 M' U' M2 U (11)
UF-UL, UB-DF: U2 M U2 M' U M' U2 M' U' M2 U2 (11)
UF-UL, UR-DF: U M U2 M' U' M' U2 M' U M2 U' (11)
UF-UB, UR-DF: U M U M2 U2 M2 U M' U' (9)
UF-UB, UL-DF: U' M U M2 U2 M2 U M' U (9)
UF-DF, UR-UB: M U2 M' U M' U2 M' U' M2 (9)
UF-DF, UB-UL: M U2 M' U' M' U2 M' U M2 (9)
UF-DF, UR-UL: M U M2 U2 M2 U M' (7)

Example solve (cross on bottom):
D' B F' D L' D' L' R' F U' L2 R2 U2 D L2 U' D' R U' B R2 B F' D' L2

Partial Cross: y2 U' R2 F D L2
First Pair: U2 L U' L2 U L
Second Pair: U' R' U R U2 y R U R'
Third Pair: y U2 L U' L
Fourth Pair (with reversed edge control): y' U2 R' U' R U' R B' R' B
Orient Edges: y2 M' U' M
CLL: U2 F U' L' U R2 U' L U R2 F'
Solve Edges: M' U2 M U M' U2 M


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## Erik (May 18, 2008)

Wouldn't it be easier if your LL was just: CLL/ELL without orienting the edges at all?
It's only a 'couple' times learning ELL


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## Kenneth (May 18, 2008)

I use this already but I do F2L, CLL, EO + place FD, end in EPLL, 8:12 is crazy fast U-PLL, 1:12 skips.

But I have gone one step futher, I even leave the missing edge in LL or F2L, it moves around and sometimes you can easily solve it on the way (r U2 r'). This gives easier cases for EO + place FD.

Erik, I had that idéa but the problem is recognition and some algs are tricky, EO + place FD is really fast (I'm learning algs for the cases to do it in one look at the moment, try M' U M =) and EPLL is as you know the same.


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## watermelon (May 18, 2008)

I wonder what the average number of moves to solve a partial cross would be (hopefully 1-2 moves lower). Do those few moves saved in the beginning outweigh the partial edge control + alternate LL?

Perhaps someone would be willing to compare a few normal solves to partial cross solves with the same scrambles so that we could see the difference between move counts in HTM/STM.


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## Kenneth (May 18, 2008)

Tricky crosses are slow and I think you can save a little more than 1-2 turns, it is the last edge that is the hardest to solve when doing all four.


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## Jason Baum (May 18, 2008)

I forgot to add that you can also just leave a yellow edge piece in as your final cross piece, just make sure that it is misoriented.

Example (cross on bottom): F' R L B' F2 U' D' F R2 F2 R D2 F2 B2 U2 B R D' F' B' R' L' B2 F' U'
Cross: L2 U' R F R

The rest of the solve works the same. So, if nothing else, this method gives you a lot more freedom when deciding what your cross will be.


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## watermelon (May 18, 2008)

As a quick example, I got 55 moves on the posted scramble, whereas the partial cross solve was 53/59 moves (STM/HTM).

Scramble (from original post): D' B F' D L' D' L' R' F U' L2 R2 U2 D L2 U' D' R U' B R2 B F' D' L2

Solution:
Cross: D2 U R U2 R B' F' D'
First Pair: R2 U2 R2 U2 R2
Second Pair: U2 R U R' y' U R' U' R
Third Pair: U' R U' R' U' L' U' L
Fourth Pair: R U2 R'
OLL: R' F' R L' U' L U R' F R
PLL: U2 y R2 U' R' U' R U R U R U' R


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## masterofthebass (May 18, 2008)

Mitchell, it didn't hurt that you had an edge cycle, one of the shortest PLLs


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## Dene (May 18, 2008)

So you'd rather go through all this hassle rather than work on colour neutrality which requires a total of 0 extra algorithms, and less effort on recognition?


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## Lt-UnReaL (May 18, 2008)

This is a really cool idea. It's a lot more helpful to have a yellow edge in the cross than a white edge, because then when you orient edges, half the time you can have your D layer solved, and recognition will be a lot easier for the LL.

Seems like the orienting edges step cancels out with solving the 4th edge of the cross. So compared to regular CFOP, the LL is a lot less moves, easier finger tricks, also a 1/12 chance of skipping PLL.

I might learn this, it should be looked into a lot more for optimization. There are some cases where I don't know how to have 4 edges misoriented for the last F2L pair.


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## pete (May 19, 2008)

watermelon said:


> As a quick example



wouldn't it be more appropriate (in this thread) to show the partial cross solve example instead ?

(everybody knows how to solve CFOP)


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## Lt-UnReaL (May 19, 2008)

Hmm, wouldn't this method be just as fast as CFOP with COLL + EPLL, though?


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## Jason Baum (May 19, 2008)

It probably is just as fast. I think this should be used as a supplement to Fridrich, not as a replacement of it. The nice thing about it is you make 7-8 move crosses into 4-5 move crosses. This also makes it easier to spot your first pair during preinspection. The moves that you save during the cross should (theoretically) cancel out with the M' U M edge orientation part. I'm not sure what the average COLL move count is, but it is probably somewhere around 11. The average number of moves for the final step is 8.69 (STM). So, if my COLL estimate is right, that is a sub 20 move count for the LL on average, excluding the edge orientation.


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## Lt-UnReaL (May 19, 2008)

So basically it could be faster than COLL/EPLL, because you get a better look ahead to F2L because of easier crosses. Sounds good.


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## dChan (May 19, 2008)

I have learned 4 COLL cases(24 algorithms out of 40) so far since starting last Wednesday. Looking at at another group of 40 algorithms is just painful right now, lol.

Bot, on a more serious note, this is pretty interesting. An average Fridrich LL solve is about 24 moves(10 moves for OLL and 14 for PLL) so a 20 move LL is not bad. I think, though, that this is something for the experts to learn. You guys are already so fast that any decrease in move count allows you to get times that are even more insane than what you get now. But for everyone else, I would say that you should stick to mastering your regular method before learning this. It isn't like COLL which really will give you a bit of decrease in solving time if you know how to control LL edges efficiently with your last pair. This is more like something that will only be good when you get tricky cases for your cross. 

So, for experts, this is good, but for everyone else, I think we should all wait until we become experts.


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## qqwref (May 19, 2008)

This reminds me of Doug Li's method a lot, although I don't know if anyone has ever successfully used it. As I recall the basic idea of that method was to make a 2x2x3 block with the middle piece missing (and replaced by either a white or a yellow edge in any orientation, typically), then to expand it to F2L with one edge missing, and then to finish with CLL and then EO/EP using M turns. Of course I could be misremembering, but that's what I thought of when I heard of this idea.

As for the speed of the method, it might be fast for some people, but I don't think I would ever be one of them. You would have to be very fast at executing <M,U> algs, and if you're good enough at <M,U> that you would be happy to do it every solve, you'd probably do even faster times on a method like this: solve F2L minus DB and DF, CMLL, and last 6 edges 

So yeah, I don't think this would be much faster than normal Fridrich. You say you can save 3-4 moves on the cross, but then have to use a completely different last layer, so even if you get an easy solve that way it seems like it's more algs to practice and learn for little benefit. I'd agree with Dene that it would be no more work to just learn to be fully color neutral, or to always start with an xcross (like Hardwick).


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## watermelon (May 19, 2008)

qqwref said:


> This reminds me of Doug Li's method a lot, although I don't know if anyone has ever successfully used it. As I recall the basic idea of that method was to make a 2x2x3 block with the middle piece missing (and replaced by either a white or a yellow edge in any orientation, typically), then to expand it to F2L with one edge missing, and then to finish with CLL and then EO/EP using M turns.


According to Ryan Heise's description, the method is as you described up until LL. Once the F2L has been solved with one missing edge, the remaining edge is inserted and the LL edges are oriented in one step, then COLL, then EPLL.


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## Speedy McFastfast (May 20, 2008)

The idea sounds good, but is it really worth it to try to learn all those tricks for the LL? It would take longer for me to learn stuff for the LL, so my times would really drop. It seems easier to just take a bad cross then try to get a whole new LL...


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## Lt-UnReaL (May 20, 2008)

watermelon said:


> Once the F2L has been solved with one missing edge, the remaining edge is inserted and the LL edges are oriented in one step


How many algs would that be?


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## watermelon (May 20, 2008)

I would guess 16 cases.

Assume the empty F2L edge is at DF, and the edge that belongs there is at UF. There are two orientations that edge can be in (correct/flipped). Looking at the other U edges, there are 2 possible orientations for each, making 2^3 = 8 cases. At this point we can ignore the orientation of the edge at DF since it is already determined by looking at the four edges on U. Since there are 8 cases for the other U edges, and the edge belonging at DF can have 2 orientations, that's a total of 8*2 = 16 cases.

By the way, I think you misquoted the above statement (I said it, not qqwref) .


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## Johannes91 (May 20, 2008)

Lt-UnReaL said:


> qqwref said:
> 
> 
> > Once the F2L has been solved with one missing edge, the remaining edge is inserted and the LL edges are oriented in one step
> ...





watermelon said:


> I would guess 16 cases.


Yep, it's 2^4 = 16 when the missing edge is in the LL.

But it can also be in its correct place. If it's solved, there are 4 cases. If it's flipped, there are 2. So the total count is 22 (including solved).

I remember seeing a list of algs (displayed using Java applets) somewhere. But because there are so few cases it would be easy to generate them. [Edit: This is the list I was thinking of: http://web.archive.org/web/20070811005317/http://www-personal.umich.edu/~dlli/NewAlgSet.html.]


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## fanwuq (May 21, 2008)

Looks like corner first and layer methods are merging. There's going to be so many Roux variations if I can could this one. 
Mine idea is to orient all edges except 2 LL edges (ignore them, doesn't matter weather they are oriented or not). Then use ELL. Only about 10 cases I think excluding reflections and inverses. I know all from TuRBo except ones with flip that act like Z perm. These algs are typically as short as the EPLLs.


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## Kenneth (May 21, 2008)

Basicly you need this:

P = M' U2 M

O = M' U M
O' = M' U' M
O2 = P

From that plus some U turns you can create algs to solve any case, example O U P = (M' U M) U (M' U2 M)

The algs for some cases can be shorter if you also add this to the system:

F() = F2 ( moves )  F2

The U turn can be any U, U' U2 or no U turn depending on how U moves inside the parentesis of the "function". F(U P) makes the usual 7 turn U-PLL = F2 (U M' U2 M) U F2. The alg F(U2 O) U P solves the worst ELL (there is also a mirror case)... it was nice, before I used O' U O U O U2 O' to solve it =)

Using this system it gets wery intuitive to create algs that are easy to remember, try it


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## Kenneth (May 24, 2008)

Jason Baum said:


> 3 Cycles:
> UF-UR-DF: M U2 M U M’ U2 M’ U’ (8)
> UF-DF-UR: U M U2 M U’ M’ U2 M’ (8)
> UF-UB-DF: U2 M’ U2 M (4)
> ...



Jason, you do not need that many algs. If you do AUF then it is only three 3-cycles (of which one is a mirror). three 2-cycles (one mirror) and six 5-cycles (3 are mirrors).

I found some nice algs for the cases too, I will list all here later today but here is an example:

(y') R U2 R2 U2 R2 U2 R .... solves the easy two cycle

Now it's later and I'm back from my day outside. The rest of the algs:

R2 d M U2 M' d' M' U2 (r' R') ... the tricky 2-cycle, do the last two moves as one. The alg is basicly a U-PLL with a insertion.
L2 d' M U2 M' d M' U2 (l L) ... mirror

M' U2 M ... the "P" case = the easy 3-cycle, to bad it's only a 1:30 chance.

R2 d' M' U2 M' d M' U2 (r' R') ... tricky 3-cycle, basicly the same alg as the 2-cycle but direction is diffrent for d and the first MU2M.
L2 d M' U2 M' d' M' U2 (l L) ... mirror

(M' U2 M) U (M' U2 M) ... H5 PLE (P Last Edges) Two "P"'s and a U = P U P, really fast one and recognition is tow opposite edges in the U-layer, looks just like H from that direction, start from the one not opposite to the FD pice as the UB pice, do P and then move the FD pice to UF = the U turn (or U' in the mirror alg), then do another P.
M' U2 M U' M' U2 M ... mirror

M' U2 M' U M' U2 M U M2 ... U5 PLE, one opposite edge in LL like U-PLL. FD at UF at start. The U or U' (mirror) turn moves the opposite edge initially found in LL to UB. The alg is a P followed by a U-PLL = M2 U M' U2 M U M2
M' U2 M' U' M' U2 M U' M2 ... mirror

M2 U M2 U M U2 M' ... Z5 PLE, as Z-PLL no edges are opposite in LL, move U so it has the side at F that is the same as the pice in the FD position and do a "side swap" = M2 U/U' M2. This moves the case from the F side to the b side and also transformes it to a P cycle
M2 U' M2 U' M U2 M' ... mirror

Preceed all algs with a AUF (if needed) and also end in another AUF (if needed =)

That's the algs I found, some are the same as Jasons.

I think the P case is a lucky case if I do not do anything else to solve the edges. I actully think all times when I solve the last five edges in less than six turns is a lucky case, it may be P, O or O' = [AUF] M' U/U' M [AUF] = the easiest "OLE".

This is a part of my current work. I won't hi-jack this thread anymore when writing about it, I'm planning to start one about my new "floating edge" approach a little later.


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## Kenneth (May 29, 2008)

Kenneth said:


> R2 d' M' U2 M' d M' U2 (r' R') ... tricky 3-cycle, basicly the same alg as the 2-cycle but direction is diffrent for d and the first MU2M.
> L2 d M' U2 M' d' M' U2 (l L) ... mirror



Here is a much faster one I found yesterday:

(y') r2 U M' U2 M U r2

(y') r2 U' M U2 M' U' r2 ... mirror


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## Kenneth (May 30, 2008)

Solving like this gives new possibilitys, I'm currently looking for 2x2x2 style algs to solve CLL, algs that do not preseve the FD edge (almost the same as CMLL). I tried the 5-turn R2 U2 R U2 R2 but it destroys middle layer edges... But you can do it like this:

(x) U M U F2 U' F2 U M' U (x) ... well, it's shorter than 2xSune but not faster =)

But the two algs used for T and L in Ortega 2x2x2 are nice:

r U' r' U' F' U F ... for the L case, also mirror and it makes two CCL's
F' U' F U r U r' ... for the T case, also mirror and it makes two CLL's

The usual algs for these CLL's are also short so you can alter to do edge control during CLL, I use these and many others to do that, usaly the normal alg with a M slice setup, works fine for many CLL/OLL algs. Go for all oriented if the FD pice is oriented initially or the 1 + 3 case (M' U M) if its not.

Method:

1, F2B a la Roux ... also do M-slice + BD edge, it's intuitive and uses at the maximum 5 turns.

2, CLL ... with edge control if possible.

3, orient the edges that are left.

4, permute last edges ... use the algs above.

Or this:

1, F2B
2, Orient corners + solve M-slice and place BD edge .... 7 cases, the rest intuitive or algs for all, it's not many cases if you allow AMS (adjust M slice).
3, Permute corners and orient last edges ... ca 25 cases
4, Permute last edges ... 16 cases

The last method is as you can see easy to learn, it uses less than 50 algs.


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