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As with most methods the problem is the number of years needed to master it. I'd love to see the true potential of this method in a few months-- which brings up the question, how many (if any?) people know full ZBLL (and know it well) and also can do F2B at sub-10 Roux speed? I hope there is at least one person, but I suspect zero...

EDIT: Perhaps a temporary measure of the method could be to execute ZBRoux Ao100 by someone who only knows partial ZBLL. If you end up with a ZBLL case you don't know, delete that solve from the average. Not great, but still gives an idea.

Me, in two or three months. I'm sub 10 with roux, and know about 190 ZBLL algorithms, and am learning it at a rate of 12 algorithms per two/three days.

To be honest, I would prepare to be disappointed. It will be a long time before I get the recognition up to speed, and also, sub 10 level blocks probably won't get me to world class level. Even if I get sub 9 or 8 averages, it's not that impressive since those kind of times can be achieved using CFOP + 4LLL at this stage.

Well theoretically we really only need to time the last 10 pieces. In other words, start with a cube that has the first two blocks already built. Now finish the solve with CMLL+LSE, time 100 solves, then do another hundred finishing with DFDBEO+ZBLL. That already gives a good idea; the real question is if DFDBEO+ZBLL is faster than CMLL+advanced LSE, then this method would beat regular Roux, which Kian already proved is at least equal to the best of CFOP. Do keep in mind that for a fair comparison we need to compare vs. advanced Roux, i.e. multiple CMLL's to improve edge orientation then full EOLR. Still, I believe that DFDBEO+ZBLL would be sufficiently faster to be very obvious.

Assuming the best Roux solvers can identify the CMLL case (including edge orientation) in 0.5 s (this is the slowest part of Roux), then execute the algorithm in 0.9 s average, and finish advanced LSE in 1.5 seconds average, then the total time is 2.9 seconds. So the average time to beat for DFDBEO+ZBLL is less than 2.9 seconds. Assuming 0.9 seconds for DFDBEO and 1.25 seconds for 15 move ZBLL at 12 tps, theoretically DFDBEO+ZBLL should take 2.15 seconds, assuming zero recognition time for ZBLL hoping that recognition occurs during M/U spam. Even adding another 0.25 for recognition still puts it at 2.4.

Well theoretically we really only need to time the last 10 pieces. In other words, start with a cube that has the first two blocks already built. Now finish the solve with CMLL+LSE, time 100 solves, then do another hundred finishing with DFDBEO+ZBLL. That already gives a good idea; the real question is if DFDBEO+ZBLL is faster than CMLL+advanced LSE, then this method would beat regular Roux, which Kian already proved is at least equal to the best of CFOP. Do keep in mind that for a fair comparison we need to compare vs. advanced Roux, i.e. multiple CMLL's to improve edge orientation then full EOLR. Still, I believe that DFDBEO+ZBLL would be sufficiently faster to be very obvious.

Assuming the best Roux solvers can identify the CMLL case (including edge orientation) in 0.5 s (this is the slowest part of Roux), then execute the algorithm in 0.9 s average, and finish advanced LSE in 1.5 seconds average, then the total time is 2.9 seconds. So the average time to beat for DFDBEO+ZBLL is less than 2.9 seconds. Assuming 0.9 seconds for DFDBEO and 1.25 seconds for 15 move ZBLL at 12 tps, theoretically DFDBEO+ZBLL should take 2.15 seconds, assuming zero recognition time for ZBLL hoping that recognition occurs during M/U spam. Even adding another 0.25 for recognition still puts it at 2.4.

Some issues: ZBLL would take at least 1.5 times as long as CMLL simply due to movecount and given that most CMLLs are more fingertricky than most ZBLL the difference would be even greater.

In addition, you would undoubtedly have a pause when doing DFDBEO because you would have issues recognising what cases you have during second block- sone thing which is usually done in CMLL for Roux as it is just an alg which has the same effect so recognition becomes a lot easier.

Lastly, it is much easier to switch into MU solving than it is out of it.

Some issues: ZBLL would take at least 1.5 times as long as CMLL simply due to movecount and given that most CMLLs are more fingertricky than most ZBLL the difference would be even greater.

In addition, you would undoubtedly have a pause when doing DFDBEO because you would have issues recognising what cases you have during second block- sone thing which is usually done in CMLL for Roux as it is just an alg which has the same effect so recognition becomes a lot easier.

Lastly, it is much easier to switch into MU solving than it is out of it.

Good points. So increasing ZBLL to 1.35 seconds and adding 0.3 second DFDBEO recognition gives 0.3+0.9+1.35 = 2.55. I suppose there is also the issue of AUF time. I'm not sure if top ZBLL solvers predict AUF or just look ahead.

EDIT: How many DFDBEO cases are there? I couldn't find it in the LLOB document. I assume around 60? So ZBRoux would need 60+494 = 554, vs. Roux would need 84 CMLL's plus 60 EOLR or 144.

I will be learning ZBLL's in the future and seeing how this method fares(currently busy with work, so cannot accommodate time for it). Currently with what I have done, FB+SB+EODFDB can be done pauselessly. The undoubt-able pause during EODFDB isn't that big of a issue if you know how your SB inserts are affecting the other pieces. Its probably easy for me because I was a CFOP user, but I don't have any problems with the first 3 steps of this method.
Now comes the inevitable pause: ZBLL Recognition. However, using the concept of LSE and solving EO+DFDB like you solve EO+ULUR, you can easily predict where your edges are going to end up, thus predicting ZBLL case isn't going to be difficult as you only need prediction of 2 edges which isn't a problem. This style of EO+DFDB solving may be a bit inefficient, but you barter some recognition time which may/may not make it worth it the extra moves. Switching out of MU is again not a problem as it just needs some getting use to. Doing LSE scramble, then solving LSE and using RU Ua perm+Ub perm to get the cube back solved, coming out of MU is not difficult to do.
Only thing remaining is learning ZBLL's and testing this out. (which I shall do as soon as I start learning ZBLL's in due time)

Kian says it takes him 2.9 to 3.5 seconds to do CMLL+LSE. So DFDBEO+ZBLL needs to 'beat' 2.9 - 3.5 seconds to be an improvement over advanced Roux. I had estimated DFDFEO+ZBLL at 2.55 seconds, so even if that is aggressive, there is great potential.

Although it is't quite enough to be considered accurate, I have done 20 example solves using the OLL/PLL and ZBLL variants of this method as well as completed there Roux counterparts. The solves are from FB, so no solve should have an advantage over another (and all movecounts are NOT including FB) All algs used are speed-optimized, but all of the non-algorithmic steps were done using HARCS so no human error. As expected, ZBLL had the lowest average and OLL/PLL had the highest.

ZBLL=31.6 STM from FB
Roux=34 STM from FB
OLL/PLL=39.05 from FB

I have had yet to test the Ribbon Alpha and CLS/PLL variants of the method. ZBLL from what I can tell has a low variance, and while OLL/PLL's is higher, both Roux and OLL/PLL have fairly large variances. It should be noted that on the Roux solves LSE was done by hand using vanilla Roux and only minor influence on efficiency, so the average given would be lower if EOLR and other tricks were used. My current hypothesis is that ZBLL will have the lowest movecount, and Roux/Ribbon Alpha will be second. Here's a link to the sheet I am using to track solves/calculate stats. If anyone would like to help me complete the test, please PM me your email and I'll add you

Although these numbers are extremely interesting, since the real benefit of ZBRoux is the elimination of the CMLL recognition time, I don't think move count is the main factor in determining which variant is the best one. Sadly I don't think there is any way to really determine it without actually mastering the method...

Although these numbers are extremely interesting, since the real benefit of ZBRoux is the elimination of the CMLL recognition time, I don't think move count is the main factor in determining which variant is the best one. Sadly I don't think there is any way to really determine it without actually mastering the method...

I agree with this, but I still feel like getting a true avg movecount for each of the variants is useful and will help determine the "best" variant.

Once I have completed 50 solves, I am going to add a stats page and provide a conclusion based not only on the stats, but also looking at things like ergonomics, recog time, and alg count to determine the pros/ cons of each and the overall best variant. May take a while to get them all completed, but I'll come back with my findings soon enough.

I agree with this, but I still feel like getting a true avg movecount for each of the variants is useful and will help determine the "best" variant.

Once I have completed 50 solves, I am going to add a stats page and provide a conclusion based not only on the stats, but also looking at things like ergonomics, recog time, and alg count to determine the pros/ cons of each and the overall best variant. May take a while to get them all completed, but I'll come back with my findings soon enough.

38 move solve using ZBLL from the example solve thread. May not seem super impressive, but sub-40 is actually pretty common with LLOB using ZBLL (may I suggest ZR Method (Zbignew-Roux) or vice-versa?) On one scramble, I got a 28 but unfortunately I no longer have the solution. It ended in lefty Niklas and F2B was done simultaneously with a 3 move EODFDB.

F2 R2 U L2 D F2 R2 B2 U' B2 R B' D2 R' B D' L F2 U L B'

y2 x U M U' R L U' L B2 x2//FB (8/8)
R U R U2 F R F' R2 U R' U' R//SB (12/20)
U M2 U M U M U2 M' U' M//EODFDB (10/30)
F R' F' r U R U' r'//ZBLL (8/38)

So we have the possibility of reaching intermediate FMC movecounts in speedsolves fairly consistently

Well if you consider that the 3x3 WR average had ZBLL in it, then one could argue that ZBRoux could be a method-neutral alternate solve to CFOP-ZBLL. If you see an easy XCross, go CFOP/ZBLL, otherwise use ZBRoux if you see an easy first block.