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I really don't this is going to be a good idea. In the past cfop-roux hybrids have been proposed countless times (my first proposal was also a roux-cfop hybrid actually) and they are almost always worse than vanilla roux or vanilla cfop. I don't think an argument can be made for "transition methods" as the best way to transition to a new method is often just to start using the new method, breaking down hard to understand steps into simpler components as you go.
 
Since phasing isn't really good with LLOB, what I came up with as a progression to full ZBLL was to orient 2 corners adjacent to each other during the insertion of the last F2L pair, with no regards to edge orientation, resulting in a T or U ZBLL or PLL. Once you are comfortable with those sets, you add in L, then S and AS, and finally H and Pi. Here's what I ended up with. For recognition, the pair is oriented in the UFL position, and the first letter refers to what face the U sticker of UBR is on, and the second letter refers to the U sticker's face on UFR.
U, U
L’ U2 R U R’ U2 L

U, R
U’ R’ F R F’

U, F
R U’ R’

B, U
U’ R' F R2 U R' U' F'

B, R
l U F' U' l'

B, F
R U' R2 U' R U' R' U2 R

R, U
U R U' R' U R U2 R'

R, R
U R U2 R’

R, F
R2 D R' U R D' R' U2 R'
As for the credit for the method, I think it goes to CriticalCubing and Teoidus because they are the ones who suggested using EO and ZBLL, which is what really made it viable.
 
Possible ZZ variant method:

Setup: F2 D2 F2 R2 D F2 U R2 U L2 B2 F' D2 U' L' B' D L2 R D' F'

z2 F2 R2 D' R' B D F' // EOLine (7/7)
U R L U' D2 R U L U2 D2 R // Layer (11/18)
U2 R2 U' R U2 D2 L' U' D2 R2 // Separate edges into correct layers (10/28)
U' R2 D R' U2 R D' R' U2 R' // OLL (9/37)
U' L2 U2 L2 U2 L2 // PLL setup (6/43)
y' L' U' L F L' U' L U L F' L2 U L // PLL (13/56)

(Usually I would skip the PLL setup unless I had PLL parity, then just solve the E layer afterwards)
 
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This has already sort-of-been proposed
This is the thread:
https://www.speedsolving.com/forum/threads/zz-blah.8849/

And this is the post where that comes up
https://www.speedsolving.com/forum/threads/zz-blah.8849/#post-125479
I s'pose, but mine is exclusively for LLOB, as it ignores EO, and is meant as a method of progression.
Possible ZZ variant method:

Setup: F2 D2 F2 R2 D F2 U R2 U L2 B2 F' D2 U' L' B' D L2 R D' F'

z2 F2 R2 D' R' B D F' // EOLine (7/7)
U R L U' D2 R U L U2 D2 R // Layer (11/18)
U2 R2 U' R U2 D2 L' U' D2 R2 // Separate edges into correct layers (10/28)
U' R2 D R' U2 R D' R' U2 R' // OLL (9/37)
U' L2 U2 L2 U2 L2 // PLL setup (6/43)
y' L' U' L F L' U' L U L F' L2 U L // PLL (13/56)

(Usually I would skip the PLL setup unless I had PLL parity, then just solve the E layer afterwards)
The thing with this is that by building your D layer right after EOLine, you're restricting your moveset to <U>, essentially making anything non-algorithmic impossible. The effects of this are shown in the high alg count and high-ish movecount.
 
Could you point out where Hexagonal/ZZ-CT were first proposed? I've heard of Roux-CFOP hybrid and some other things being reinvented hundreds of times (LLOB included) but thought those two were relatively original.
The basic concepts are laid out in the M-CELL post where the ideas of leaving a corner or an edge of the first layer (usually both) and then solving it with the final layer. Although the documentation was definitely nowhere near the level which ZZ-CT is so I don't mind so much that I am not credited for it (though I believe that it is still very close to an intermediate variation of the ZZ variant of M-CELL). However, Hexagonal Francisco is much closer to M-CELL (and an almost identical approach is proposed in one of the M-CELL variants) I actually participated in the fb group for a while and it was actually acknowledged that M-CELL and HF where more or less one and the same. "Hawaiian Kociemba" is also very similiar.

With regards to LLOB, M-CELL is near identical (especially in one of the variants) and the "hyper phasing" concept is the key concept I based the method off (I wanted to create a "true" 1LLSLL). I never really documented the method with all necessary algs much as it never gained much attention (though L5E was an already existing alg set and ZZ-CT now has most of the other algs needed).

(incidentally of all the methods I proposed Briggs and B2 (Briggs2) are the only ones which had nothing to so with the method when I first proposed it though I decided that "Briggs" was just going to refer to any CO methods I came up with hence "Briggs2").
Possible ZZ variant method:

Setup: F2 D2 F2 R2 D F2 U R2 U L2 B2 F' D2 U' L' B' D L2 R D' F'

z2 F2 R2 D' R' B D F' // EOLine (7/7)
U R L U' D2 R U L U2 D2 R // Layer (11/18)
U2 R2 U' R U2 D2 L' U' D2 R2 // Separate edges into correct layers (10/28)
U' R2 D R' U2 R D' R' U2 R' // OLL (9/37)
U' L2 U2 L2 U2 L2 // PLL setup (6/43)
y' L' U' L F L' U' L U L F' L2 U L // PLL (13/56)

(Usually I would skip the PLL setup unless I had PLL parity, then just solve the E layer afterwards)
This is something that I considered while trying to improve the "second phase" of SSC though I concluded I couldn't get it to work though I wouldn't say that it would be impossible to do so I would really encourage you to have a look at how to improve it. You may also find it interesting to look at L2L or L2L2 or L2L3 (however many there are of those. I think that there's maybe 4 main ones that some other variants like L2L4K).
e: Here's the wiki link anyway
 
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trigger style reduction method

a better inspection attempt 4

1a find out how to make a 1x1x3 block

1b imagine making the block and keep track of the corners

1c imagine placing corner A and corner F in the A and F slots (could be swapped) and find the inverse of the moves to get there

1d find the swap pair and use the inverse moves while keeping track of swap pair transformation
the current swap pair is the needed swap pair
 
One again I'll recommend the 2GR recognition & tracking system (some links in my sig also explain the theory behind why they work). The method is more efficient in both time and working memory
 
trigger style reduction method

2gr recognition and why it won't work

i never really understood it, although looking at it, studying it, for more than an hour, when i first found it

tsr does not permutate corners the same way 2gr does

if we do create a way for it to work, it would not have as an efficient movecount because rotating the right side (not including diagonal swap), permutation can be solved in an average of 1 move, yes 1 move average.(1.5 including diagonal swap)

i don't see a way for it to work using this recognition (i barely understand it, so of course i don't see a way)
if you do see a way (because you understand more than i do) please explain how this is better for movecount AND recognition speed.
 
Essentially, you're not reducing CP to <R,U>. You're reducing CP to <B,U>.

Actually, you're not really even doing that, you're reducing CP to a certain subset of key swaps that include the solved state wrt <B,U>; this subset is closed under triggers.

Therefore what you can do is find all key swaps that are in this subset, recognize CP wrt <B,U> in inspection, track key swap wrt BU throughout the 3x1x1, and then use a single R move to send this swap to the subset you want.

When I speak of time/space efficiency I'm actually talking about inspection. The movecount will be the same (1 move average), and with the above you might be able to see why: while most CP methods try to reduce to 1 set of states (i.e. "solved CP"), your method makes it possible to reduce to a whole family of sets of states (i.e. those reachable by triggers).
 
trigger style reduction method

Therefore what you can do is find all key swaps that are in this subset, recognize CP wrt <B,U> in inspection, track key swap wrt BU throughout the 3x1x1, and then use a single R move to send this swap to the subset you want.

can you explain this? i still don't get it.

what you have incorrect about my method
you fix cp after building the 2x2x3 block not the 1x1x3 block
and no, "reducing to a whole family of sets of states", these states still occupy only 1/6 of combinations, same as normal corner permutation.
 
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Using the 2GR recognition system, you can determine what CP fix is necessary in inspection by treating B and U as R and U and recognizing CP that way.

Once you have a solution to the 1x1x3, you can determine what will happen to the corners by tracking CP through the 1x1x3 as in 2GR, with respect to the B and U faces. This will determine what CP fix is necessary after the 1x1x3 is solved.

Since you're performing reduction after 2x2x3, you will have to track this swap (or "transform" it, as you say) through the 2x2x3 solution. You can do this with the 2GR tracking system as well.

And yes sorry, you're right, triggers reach 1/6 of states. What I meant was more that, in the space of all possible key swaps, more of them are permissible under trigger reduction (compared to the single fully reduced state in normal CP reduction), and this is why movecount is lower.
 
trigger style reduction method

recognition v2 attempt 1

ok, i understand 2gr corner recognition now (i think, i have something good either way). i actually came up with something exactly like it (without auf)

the recognition i use orders the corners on top ccw in ascending order (1-4)
the FDR piece and spot is called "A", the BDR piece and spot is called "F"
imagine solving the bar. find piece A and F, find witch case it matches, then find at most a single swap that will cause the specific pieces to be in mod 4 increasing order.

A in spot A-F in spot F
any top 3 in ccw form

A in spot F-F in spot A
any top 3 in cw form

A above spot A-F above spot F
top 2 ccw+spot A

A above spot F-F above spot A
A+top 2 cw

A on top diagonal F on top
spot A+ccw piece A+spot F

A in spot A-F on top
ccw piece F+cw piece F+spot F

A in spot F-F on top
cw piece F+ccw piece F+spot A

F in spot F-A on top
spot A+ccw piece A+cw piece A

F in spot A-A on top
spot F+cw piece A+ccw piece A


i used the term transform due to R move effecting the swap pair in strange ways
 
Lol, yeah, that's identical to 2GR recognition. For tracking, you may find that the next section on the site helps to describe the "strange ways" in which R moves transform things.
 
An interesting way to recognize the relative permutation of 3 LL corners: methods for this usually revolve around using the identities of the corners or using color patterns where you imagine where bars and opposites would be if the corners were to be oriented properly.

I'm not sure how people are going about the latter, but I just realized that it can be adapted into the following fairly efficient process:
For corners A1 .. An:
Find the sticker counterclockwise of the yellow/white sticker on A1, and compare it with the sticker clockwise of the yellow/white sticker on A2.
For i = 2 .. n-1, compare the unexamined sticker on Ai to the sticker clockwise of the yellow/white sticker on A(i+1).

This way the skills needed to be practiced are "locating stickers clockwise of yellow/white" and "locating unexamined stickers," but of which could be made pretty mechanical.
 
trigger style reduction

(recognition v2 attempt 1) responding

this is not exactly 2gr recognition. the 3 pieces you need to look at are not the same because we are not permutating them the same way (not because of the fact i went ccw)

i already understand how the R moves affect the swap pair, due to my own studies

i won't be using this recognition style, because i prefer another recognition style i made (attempt 3)
i even explained how R moves affected the swap pair in this post.
 
@ShyGuy1265 You can solve the Ribbon Slot at any point in the F2L. The later you do it, the less efficient it becomes, but you could technically just wait until the last slot and throw the edge in. I usually solve it either first or second if I see a full pair.

I've got my average within a second of my CFOP average now with only knowing about 40 new cases, with a PB of 9.63. Learning full TTLL first, then diving into the orientation algorithms. Also, Ribbon OLL is referred to as TOLS now, for Twisted Orientation and Last Slot.
 
trigger style reduction method

skip chances

one target 1x1x3 block formed from the scramble
1/12096

1x1x3 block formed from scramble for colour neutral solvers
i dont know how to do this

at most 1 move from 1x1x3 block for colour neutral solvers
i dont know how to find this

one target F2L pair formed anywhere on the cube after step 1
1/18

any F2L pair formed anywhere on the cube after step 1
i dont know how to do this

both F2L pairs formed anywhere on the cube after step 1
i dont know how to do this

one F2L pair in place + another F2L pair anywhere else on the cube after step 1
1/2160

one F2L pair anywhere on the cube after placing first F2L pair
11/180

DF+eo skip after step 2
1/80

2gll skip after step 3
1/324

i will update this when i find another skip chance
the reason i put this in this section is because this is information about my method
 
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