The New Method / Substep / Concept Idea Thread

[Teoidus]

That's about what I thought. One thing about the median that I personally don't like is if there's an equal amount of 2 movecounts (i.e. same amount of 7's as 9's) and you get a median movecount that doesn't exist. What would you want to do to fix the problems of median? Look at the pool of highest probability cases and see the median from there so it's more exact?

The most accurate way to go about it is to compute the expected value of the movecount. So basically, you want, for an algset of size N, expected movecount = sum(i = 1 to N) of p_i M_i, where M_i is the movecount of the ith case and p_i is the probability that it'll show up

 

[Neuro]

The most accurate way to go about it is to compute the expected value of the movecount. So basically, you want, for an algset of size N, expected movecount = sum(i = 1 to N) of p_i M_i, where M_i is the movecount of the ith case and p_i is the probability that it'll show up

Good to know, might start implementing this into my algsheets!

 

[efattah]

A good example of algorithm move count average vs. weighted average is edge orientation of the M-slice when the L and R slices are solved. There are four algorithms, they take 0, 9, 9, 13 moves to solve. The average movecount is (0+9+9+13)/4 = 7.75. But if you weigh the averages by their probabilities:
1/8 * 0 + 4/8 * 9 + 2/8 *9 + 1/8 * 13 = 8.375.

 

[AlphaSheep]

Oh and one quick question: Why don't most alg makers include median movecount into their statistics? It seems like it'd be at least as if not more valuble than the average movecount of the algs because it shows us what to expect for that step (at least in the realm of probability). Therefore, we get a more accurate representation as to what to expect from the algset.

I don't think the median is a useful statistic at all.

What you really want is what statisticians call the expected value, which, if the probability of each case is equal, then that is exactly equal to the mean. For most larger alg sets, the majority of cases have the same probability, and there are usually only a handful of cases that have a reduced probability (almost always due to some form of symmetry). Also, apart from the skip case (which usually has the maximum symmetry), the symmetric cases tend to have a similar movecount to the rest of the set. Therefore assuming equal probability for all cases is not an unreasonable approximation, and the mean is usually quite close to the expected value. Its probably worth noting that any alg at that solves cases that start with a only single piece of a certain type outside the layer in which it belongs cannot have any symmetry, and all such cases are guaranteed to have equal probability.

For smaller alg sets, there's really no excuse not to work out the case probabilities and work out the actual expected movecount. Even for larger sets, it's not too difficult if you know how to use Excel, and if you know some programming, it's really really easy to do.

I did it for TSLE (105 algs), and it only took about 10 minutes.

 

[Neuro]

Hey I thought of a (potentially) new way to end CFOP with TTLL: Pair LE with any corner and perform VLS as a substititute to TSLE. Might be good for really advanced CFOP users.

 

[Thermex]

@Neuro for a while I've been thinking of ways to 2-look the last slot+last layer in a CFOP solve, and I did think of the way you just said. But instead I thought maybe...

1. Insert your last corner and orient all the pieces
2. Insert your last edge and solve the rest of the cube

Not sure how many algs there would be here, but it seems like it would probably be A LOT and also would be really helpful in OH/big cubes.
Anybody have any idea how many algs there would be and how well this would work?

 

[Neuro]

@Neuro for a while I've been thinking of ways to 2-look the last slot+last layer in a CFOP solve, and I did think of the way you just said. But instead I thought maybe...

1. Insert your last corner and orient all the pieces
2. Insert your last edge and solve the rest of the cube

Not sure how many algs there would be here, but it seems like it would probably be A LOT and also would be really helpful in OH/big cubes.
Anybody have any idea how many algs there would be and how well this would work?

You'd end up with more orientation algs than VLS because of the lack of a pre-oriented edge (432+), and the LE+LL edge is ZZ-HW. Chris claims it has 72 cases, but I'm not really sure b/c HKPLL (also inserts 1 edge) has over 100 algs. So yeah there'd be well over 500 cases.

 

[2180161]

Is there a petrus style 4x4 method?

 

[Neuro]

Not that I know of anyway, go for it!

 

[2180161]

All right, so its kind of strange, but I just beat my PB AO5 with it (granted, I average about 1:40 on 4x4)

anyway....

1. 2 Opposite Centers

2. Build 3x3x1 squares in BDL and BDR

3. Finish Centers while solving BD edge

4. Finish Edge Pairing without disturbing the 3x3x4 block that is built

5. 3x3 finish

Now, in the example solve, please note that I am extremely inefficient, as I don't know how to notate single slice turns other than R r'.

Example Solve: here

Spoiler

Scramble: Fw2 B2 R B' F U' D' R' Fw L2 R2 Fw R Uw' R' L' B' D2 Uw' B' Uw' R Uw Rw F' L2 B2 U2 Uw2 R B' Rw2 Uw B2 F2 D U L' B' Uw'

z' u' B u' z' R2 u r U2 r' y' u2 U' l' U2 l // opposite centers

z L2 U' l' U L2 U' F' r' R'U R U' R' U2 B U R U r' U' R2 r' U' r U' R' U R U' R' U R // 2 Squares

m U m2 U r2 R2 F m U R2 r2 U2 R2 r2 m' R' r U' R r' m2 U2 m2 L l' U L' l // centers+ BD

F2 U' F R' F' R U2 u L F' L' F u' U L' U2 L U' u F' L F L' u' F U F' u' R U R' F R' F' R u // finish edge pairing

y' r2 B2 U2 l U2 r' U2 r U2 F2 r F2 l' B2 r2 R U R' U' y R U' R' y' U2 R U R' U' R U2 R' U R U' R' U' R' U R U R' U2 R U2 F R U' R' U R U2 R' U' R U R' U' F' 2R2 U2 2R2 u2 2R2 2U2 R2 U R U R' U' R' U' R' U R'// 3x3 stage

Edit: Found out how at the end, was just too lazy to change the rest.

 

[Thermex]

You'd end up with more orientation algs than VLS because of the lack of a pre-oriented edge (432+), and the LE+LL edge is ZZ-HW. Chris claims it has 72 cases, but I'm not really sure b/c HKPLL (also inserts 1 edge) has over 100 algs. So yeah there'd be well over 500 cases.

Actually I just figured this out, this is super cool. If you pretend the edge attached to the corner is a good edge (50% of the time it actually is) then you can just insert the corner using a VLS with slightly altered recognition. At this point you should have the corner in its slot and every piece on the top layer yellow (or whatever the u-layer color is) except one. Then you look at the RF edge. If the edge attatched to the corner was a good edge when you did your VLS, then RF should be facing towards you and all edges are oriented. Now just finish off the solve with a ZZ-HW (I never knew about ZZ-HW until you told me). If the edge attatched to the corner was bad when you did your VLS, then at this stage you should have all pieces oriented except two edges. This is easy to recognize because RF will be facing to the right. Now you can just do a y rotation and mirror your ZZ-HW case. Seems like that would work. The only algs that would have to be generated would be the irregular cases when your corner is either twisted in its slot (Twisty OLL?) or when the white sticker is facing up (assuming you start on white) and also when you get to ZZ-HW and you already have the edge in its slot but its flipped. Are there any documents with the ZZ-HW algs? I would need that to learn them and to be able to mirror the algs. In total this method would be slightly less algs then full ZB if you don't count the mirrors as new algorithms. It would also be slightly more efficient and easier than ZB since you wouldn't need to pair your last pair up.
On a side note, one alg set that I think definitely needs to exist for either of our methods to work is a set that solves VLS when the corner is on the left (this might already exist, ignore this statement if it does).

@2180161 that seems like a great idea for a method, especially if it was paired with my or someone elses 2-look last slot algs. But could you elaborate more on steps 2 and 3 more? I don't really get it.

 

[2180161]

Actually I just figured this out, this is super cool. If you pretend the edge attached to the corner is a good edge (50% of the time it actually is) then you can just insert the corner using a VLS with slightly altered recognition. At this point you should have the corner in its slot and every piece on the top layer yellow (or whatever the u-layer color is) except one. Then you look at the RF edge. If the edge attatched to the corner was a good edge when you did your VLS, then RF should be facing towards you and all edges are oriented. Now just finish off the solve with a ZZ-HW (I never knew about ZZ-HW until you told me). If the edge attatched to the corner was bad when you did your VLS, then at this stage you should have all pieces oriented except two edges. This is easy to recognize because RF will be facing to the right. Now you can just do a y rotation and mirror your ZZ-HW case. Seems like that would work. The only algs that would have to be generated would be the irregular cases when your corner is either twisted in its slot (Twisty OLL?) or when the white sticker is facing up (assuming you start on white) and also when you get to ZZ-HW and you already have the edge in its slot but its flipped. Are there any documents with the ZZ-HW algs? I would need that to learn them and to be able to mirror the algs. In total this method would be slightly less algs then full ZB if you don't count the mirrors as new algorithms. It would also be slightly more efficient and easier than ZB since you wouldn't need to pair your last pair up.
On a side note, one alg set that I think definitely needs to exist for either of our methods to work is a set that solves VLS when the corner is on the left (this might already exist, ignore this statement if it does).

@2180161 that seems like a great idea for a method, especially if it was paired with my or someone elses 2-look last slot algs. But could you elaborate more on steps 2 and 3 more? I don't really get it.

Basically, the second step is solving four of your edges, and two corners. forming almost Roux F2B, however, you are missing the FR and FL pairs. The third step is where because you have your L and R centers solved, and your blocks, you are solving your centers while also pairing the BD edge, as that completes your 3x3x4 once you place that edge. Then, you can finish pairing edges however you like, as you have 7. Once those seven are finished, you can finish as per CFOP, Petrus, or any method you choose (however due to the first few steps, ZZ would just be Petrus, unless you used the free F-face to solve EO,which may actually be good, and Roux would be inefficient if you were to disregard the M-slice.) If you look at the example solve, it should clear it up more.

 

[Thermex]

Okay, here are all my method ideas in my head right now for each puzzle:

2×2 method- VL5C (V, last 5 corners)
1. Solve a "V" (full layer minus one corner) on the d-layer
2. Put the last corner in and solve CLL in one algorithm (L5C)
~11-12 moves HTM

3×3 method- Waterman VH (VH are my initials, this method is a variant of waterman)
1. Solve a "hexagon" (full layer minus one edge and one corner) on the d-layer
2. Insert the last corner and solve CMLL at the same time (L5C)
3. Solve three edges on the u-layer in one algorithm (TEUL)
4. Rotate the cube so that the "m-ring", UL and UR edges are unsolved and solve the last six edges (L6E). Finally, permute the midges
~39-40 STM

3×3 OH method (the reason I developed a seperate method for OH is because 1. I had another idea for a method, and 2. This method uses mostly R and U moves as opposed to my waterman varient that uses lots of M moves, which can be difficult in OH)
1. Do a regular CFOP solve until 3 F2L pairs are solved and your last F2L slot is open. Stop here
2. Insert your last corner into the last slot and orient all the pieces
3. Solve the rest of the cube by using 1 of 72 ZZ-HW algorithm (for a more detailed explanation of the full method, read my post about two posts above this one)
~46-48 HTM

Pyraminx method- L5E (last 5 edges)
1. Solve the centers and one edge on the d-layer
2. Solve the rest of the pyraminx in one algorithm (does anybody know how many algs there would be for this step?)
~11-12 moves HTM

Skewb method- TCLL (twisty complete last layer)
1. Solve 3 corners around their center
2. If the last corner is twisted in its place, solve the rest of the skewb using one algorithm. If not, insert the corner normally and proceed normally with Sarah's advanced method.
This is not a standalone method, but a method just like 2×2 TCLL that saves 4-5 moves per solve when the corner is twisted in place.

Square-1 method- OBL (orient both layers)
I know this already sort of exists, but I don't think anybody's actually generated algs for every case yet. Anyway:
1. Cubeshape
2. Orient both layers using one algorithm, opposed to the normal 2-look CO&EO strategy (can anobody calculate how many OBL cases there would be?)
3. Finish off the solve normally

Megaminx method- LSELL (last slot edges last layer)
1. Solve everything normally up until you have one last open slot without a solved F2L pair
2. Insert your last corner into themslot and orient all the edges (LCEO)
3. Insert your last edge into the slot and permute the remaining edges (L6E)
4. You should be left with all edges completely solved now. Solve the corners now (COLL) (How many COLL algs would there be on a megaminx? Are they already generated?)

An alternitive to this if there are WAY too many (>200) cases for COLL:

Megaminx alternate method- LSCLL (last slot corners last layer)
1. Insert the last corner and orient all the last layer corners (basically just WV on megaminx without the edge attached to your last corner)
2. Insert the last edge and solve corner permutation (LECP)
3. Finish off the solve with an ELL (only 40 algs)
Both these methods cut off ~15 moves from your last slot+last layer

Okay, so those are all of my ideas. Keep in mind these are all just ideas and I haven't generated algs or anything for any of these methods. How many of these seem like good methods that might work? What do you think is best idea here? Which could use the most revision/which one seems the least likely to work? Feedback would be greatly appreciated

 

[Teoidus]

I'll comment on what I'm familiar with:

2x2 : This is a superset of TCLL. There will be a LOT of algorithms (42 * 3 if last corner in D layer + 27 * 6 * 3 if not = 612 total).

3x3 : On top of the algs as mentioned above, TEUL will also be a lot of algorithms ((8 nCr 3) * 8 = 2688. If you allow AUFs and AEFs then it will be less; i'm too lazy to do the math, but my hunch is the number will still be pretty big)

3x3 OH : Last corner + orient everything is a lot of algs (at least 3 times more than VLS, which I believe is already 432 algs). On top of that ZZ-HW is also a fair amount of algs (if you allow AUFs, 2 CP * 5!/2 = 120)

Square-1 OBL : This is already a thing. I can't remember the name off the top of my h ead, but it's mentioned in this video:

 

[shadowslice e]

Okay, here are all my method ideas in my head right now for each puzzle:

2×2 method- VL5C (V, last 5 corners)
1. Solve a "V" (full layer minus one corner) on the d-layer
2. Put the last corner in and solve CLL in one algorithm (L5C)
~11-12 moves HTM

This is VOP more or less with the last two step I one alg which will be a pretty big number and I think EG is still better.

3×3 method- Waterman VH (VH are my initials, this method is a variant of waterman)
1. Solve a "hexagon" (full layer minus one edge and one corner) on the d-layer
2. Insert the last corner and solve CMLL at the same time (L5C)
3. Solve three edges on the u-layer in one algorithm (TEUL)
4. Rotate the cube so that the "m-ring", UL and UR edges are unsolved and solve the last six edges (L6E). Finally, permute the midges
~39-40 STM

Again, L5C has a lot of cases. Also, why not just solve the block on R and then not have to rotate?

I'm also quite skeptical of that movecount as well.

3×3 OH method (the reason I developed a seperate method for OH is because 1. I had another idea for a method, and 2. This method uses mostly R and U moves as opposed to my waterman varient that uses lots of M moves, which can be difficult in OH)
1. Do a regular CFOP solve until 3 F2L pairs are solved and your last F2L slot is open. Stop here
2. Insert your last corner into the last slot and orient all the pieces
3. Solve the rest of the cube by using 1 of 72 ZZ-HW algorithm (for a more detailed explanation of the full method, read my post about two posts above this one)
~46-48 HTM

This method would be reasonable where it not for the stupidly high alg count for the last step. Pretty sure it has more algs that ZBLL and gives a similar iF not slightly higher movecount with worse recognition and ergonomics.

Pyraminx method- L5E (last 5 edges)
1. Solve the centers and one edge on the d-layer
2. Solve the rest of the pyraminx in one algorithm (does anybody know how many algs there would be for this step?)
~11-12 moves HTM

I think there are already people doing a variation of this one.

Skewb method- TCLL (twisty complete last layer)
1. Solve 3 corners around their center
2. If the last corner is twisted in its place, solve the rest of the skewb using one algorithm. If not, insert the corner normally and proceed normally with Sarah's advanced method.
This is not a standalone method, but a method just like 2×2 TCLL that saves 4-5 moves per solve when the corner is twisted in place.

I don't do skewb at all but this sounds a lot like what people are doing already (at least when they get really advanced).

Square-1 method- OBL (orient both layers)
I know this already sort of exists, but I don't think anybody's actually generated algs for every case yet. Anyway:
1. Cubeshape
2. Orient both layers using one algorithm, opposed to the normal 2-look CO&EO strategy (can anobody calculate how many OBL cases there would be?)
3. Finish off the solve normally[/quote]
There would be roughly 400 cases.

Megaminx method- LSELL (last slot edges last layer)
1. Solve everything normally up until you have one last open slot without a solved F2L pair
2. Insert your last corner into themslot and orient all the edges (LCEO)
3. Insert your last edge into the slot and permute the remaining edges (L6E)
4. You should be left with all edges completely solved now. Solve the corners now (COLL) (How many COLL algs would there be on a megaminx? Are they already generated?)

The recognition for the first step would suck and the second case has 5!*3^4= 1944 cases

An alternitive to this if there are WAY too many (>200) cases for COLL:

Megaminx alternate method- LSCLL (last slot corners last layer)
1. Insert the last corner and orient all the last layer corners (basically just WV on megaminx without the edge attached to your last corner)
2. Insert the last edge and solve corner permutation (LECP)
3. Finish off the solve with an ELL (only 40 algs)
Both these methods cut off ~15 moves from your last slot+last layer

Again my main grip is recognition and alg count for the method (and personally corner comms are nicer than edge comms)

Okay, so those are all of my ideas. Keep in mind these are all just ideas and I haven't generated algs or anything for any of these methods. How many of these seem like good methods that might work? What do you think is best idea here? Which could use the most revision/which one seems the least likely to work? Feedback would be greatly appreciated

Most of them aren't horrible ideas but the alg count for most of them is insanely high which is why they're not used. The most useful ones would probably be the skewb and Squan ideas though I think they are already a thing. The 3x3 method seems like it would be good if you find ways to reduce the alg count.

 

[efattah]

I'd imagine L5C would have pretty bad recognition, which means it would be useful only as a 1-look solution to solve the corners only right off the inspection (which can't happen if you have to build a whole face first). Regardless, my personal opinion is that you would be better off with CLL, EG1, EG2, TCLL-, TCLL+, and TEG1. TEG1 has never been generated and has lots of algorithms (320), but the combination of all these (126+43+43+320=532) is still less than L5C (612). From my own experience it is easier to create a TEG1 face than three correct corners required for L5C. Considering that 2x2 experts can one look the 2x2 with just CLL/EG1/EG2, the added value of hundreds of extra algorithms seems questionable, although it is easier to one look if you know TCLL & TEG1.

For Waterman VH, the movecount would be higher than 39-40 for sure. I am open minded and there might be some potential to the method but some really new method would be needed to solve multiple U-layer edges. In LMCF I solve U-layer edges as I explain in my tutorial video, and indeed it is almost always possible to rapidly solve at least 2 U-layer edges; but there are some cases in which it is very difficult to do so quickly.

 

[Thermex]

Okay, my responses to each one of you:

2×2 method (L5C)-
I completely screwed up my math here, I thought there would only be 162 cases because I heard Effatah say in his LMCF pdf "theres a 1 in 162 chance that you get a CMLL skip", which led me to think that there were only 162 new algs to add on to TCLL, CLL, and EG. I just realized then there would be the reflections for when the corner's on the left and the other cases when the corner is facing up in the u-layer. That's over 400 algs for a method that might not work anyway. As for TEG, I thought about that, but one of the main reasons I picked L5C over this was because it wouldn't work for my 3×3 method. But now that I think about it, I should probably generate algs for TEG before someone else does. Now one question I have about TEG is: wouldn't you have to create four different subsets of TEG, since the corner can be twisted one of two ways (+ or -) and you can also get adjaceant swaps in either the back or on the left?

3×3 method (Waterman VH)
Okay, so I'm really going to have to revise this method if L5C won't work, but could TEG work with this method? Would it be less moves than solving a hexagon? I really hope it is, because then this method definitely could be under 40 moves. Now I understand you guys are skeptical about that figure, but here's the original thought I had in my head:
1. Hexagon ~9 moves
2. L5C ~10 moves
So at this point we have two steps that can easily be done in under 20 moves, as long as the last half of the solve (solving the last 9 edges) can be done in under 20 moves, this would be sub 40. Okay, for TEUL, there would probably be lots of cases and I would need help with lots of it, but i seriously doubt it would anywhere around the ~2500 figure Teoidus provided. I thought TEUL could be divided into 4 different subsets:
1. All three edges on u-layer (only 60 algs)
2. Two edges on u-layer (I'm pretty sure there would be ~60 algorithms to solve the two edges alone, then if the E-layer edge can be in two different orientation when you move it to RF, and it can go in two different spots, I get the number 240 for this set, but I doubt that's exact)
3. One edge on u-layer (no idea how to calculate the number of cases here, but it would probably be inbetween the number for sets 1 and 2.)
4. No edges on u-layer (probably only like 100 cases or so)
Which means total I estimate there's about 550 cases, although this figure may be way off. This would definitely be the most alg heavy step of the method, as the last step is only about 90 cases.

Pyraminx method (L5E)
Who's working on this? I've never heard anything about it

Skewb method (TCLL)-
Again, who's working on this? I've never heard anyone mention it before.

Squan method (OBL)-
Yeah, I'd seen that seminar video before. I just wasn't sure if he'd actually generated all the cases yet.

Megaminx methods-
Are you sure about those numbers? 1944 cases seems like WAY too many cases just to permute 6 pre-oriented edges.

Hopefully my responses clear some things up, and thanks for the feedback.

 

[efattah]

TEG1 is a useful set, useful both for 2x2 and for 3x3 LMCF. Its advantage is not so much in lower movecount but easier 1-look of the solution in 15 seconds. There are 8 sets, two twists of each of the 4 bottom corners. Before generating it I would calculate the average number of moves to create a TEG1 face. I believe for full EG the average is 3.8? If full EG plus TEG1 results in sub-3 average moves to make a face then it is worth it.

 

[Karl Ferber]

So I was thinking about this new method. I ce across the idea when I messed up a Roux solve lol.

The general idea is like this:

Step one: Build a 1x2x3 block in the DL or DR area.

Step two: Build a 1x2x3 block in the UR or UL area (opposite to the first block but rotated 180° on the R/L layer.)

Step three: Orient all edges in the M slice between the two blocks + permute the FM and BM edges in one algorithm.

Step four: Solve the rest of the cube in one alg (UL edge+corners, DL edge+corners)

Please let me know what you think about this, I am still figuring out algorithms and optimal ways to create blocks but I would really appreciate your opinions!
Peace out!

 

[shadowslice e]

Okay, my responses to each one of you:
3×3 method (Waterman VH)
Okay, so I'm really going to have to revise this method if L5C won't work, but could TEG work with this method? Would it be less moves than solving a hexagon? I really hope it is, because then this method definitely could be under 40 moves. Now I understand you guys are skeptical about that figure, but here's the original thought I had in my head:
1. Hexagon ~9 moves
2. L5C ~10 moves
So at this point we have two steps that can easily be done in under 20 moves, as long as the last half of the solve (solving the last 9 edges) can be done in under 20 moves, this would be sub 40.

The problen here is the L5C the movecounts for speed optimised algs will be far higher than the pure comms so the movecount will be closer to 15 than 10 I'm pretty sure

.Okay, for TEUL, there would probably be lots of cases and I would need help with lots of it, but i seriously doubt it would anywhere around the ~2500 figure Teoidus provided. I thought TEUL could be divided into 4 different subsets:
1. All three edges on u-layer (only 60 algs)
2. Two edges on u-layer (I'm pretty sure there would be ~60 algorithms to solve the two edges alone, then if the E-layer edge can be in two different orientation when you move it to RF, and it can go in two different spots, I get the number 240 for this set, but I doubt that's exact)
3. One edge on u-layer (no idea how to calculate the number of cases here, but it would probably be inbetween the number for sets 1 and 2.)
4. No edges on u-layer (probably only like 100 cases or so)
Which means total I estimate is there's about 550 cases, altought this figure may be way off.

There would be
4!*2^3/4=48 cases for the first set,
(4!/2)*5*2^3= 480 cases for the second set,
4*5!/3!*2^3=640 cases for the third set and
5!/2*2^3=960 cases for the fourth set giving a naive total of 2128 though this would probably be reduced to something like 1500 cases so less but still quite lot.

Just so you know L5C has about 5!*3^4/4/2=~1000 cases.

This would definitely be the most alg heavy step of the method, as the last step is only about 90 cases.

Unless you do L6E LSE style, it will have 6!*2^5/4/2/2=1440 cases hence why roux solvers do not learn as most would if it was "only" 90 algs.

Pyraminx method (L5E)
Who's working on this? I've never heard anything about it

It was talked about for a bit a while back but was abandoned because other method were found to be better for one looking and stuff even if they were slightly less efficient. By all mean pursue it if you want though

Skewb method (TCLL)-
Again, who's working on this? I've never heard anyone mention it before.

Basically all 2x2 methods have had people try to apply them so skewb with varying levels of success like skewb EG and stuff.

Btw, TCLL stands for Twisty Corner Last Layer not Twisty Complete Last Layer

Megaminx methods-
Are you sure about those numbers? 1944 cases seems like WAY too many cases just to permute 6 pre-oriented edges.

I calculated the alg count for the COLL and that number of algs is why the best people learn EOLL/OCLL/PLL rather than EOLL/COLL/EPLL.

Incidentally, there would be 6!/5=144 cases for EPLS.