New Idea: The Partial Cross Method

[Jason Baum]

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

 

[Erik]

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

 

[Kenneth]

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.

 

[watermelon]

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.

 

[Kenneth]

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.

 

[Jason Baum]

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.

 

[watermelon]

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

 

[masterofthebass]

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

 

[Dene]

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?

 

[Lt-UnReaL]

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.

 

[pete]

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)

 

[Lt-UnReaL]

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

 

[Jason Baum]

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.

 

[Lt-UnReaL]

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

 

[dChan]

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.

 

[qqwref]

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).

 

[watermelon]

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.

 

[Speedy McFastfast]

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...

 

[Lt-UnReaL]

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?

 

[watermelon]

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) .