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I've been thinking it over, and thanks to the link xyzzy gave me I could see some good arguments on it (btw xyzzy's post wasn't very convincing and would have veered me away even further if not for the other thread). An EG version would be pretty nice as the "V" would be very easy to make. Of course this is only if you want to go beyond Full-EG which already has a lot of algorithms.

I've been thinking it over, and thanks to the link xyzzy gave me I could see some good arguments on it (btw xyzzy's post wasn't very convincing and would have veered me away even further if not for the other thread). An EG version would be pretty nice as the "V" would be very easy to make. Of course this is only if you want to go beyond Full-EG which already has a lot of algorithms.

lol ok, i only know almost140 of the algs but the 1LLS algs (60-ish) are very rare and the Waterman (80ish algs) are full 2-look L5E and other algs for solving edges before L5E.

This may be unrelated but I am developing a method called triangular waterman and when i was doing example solves, I got a 37 mover lol.

D2 B' D2 L2 F2 R2 B' R2 F' R2 F2 R2 L' U2 B2 R2 B F2 R2 D'

Inspection: x2 y' (0/0)
122 Block Skip lol (0/0)
Finish Triangle: R' U F2 R (4/4)
1LLS: U' R2 F' R' F R' U2 R' F' (9/13)
2 Edges: y U M' U M (4/17)
2 Edges: z M2 U' M' U (4/21)
Waterman Set 2 Alg: r2 R' U' M2 R U M' U' R' M' U (11/33)
Permute Midges: x' U2 M' U2 R' (4/37)

Well its Waterman but more algs lol. Steps are as follows.

1. Triangle: Just like the 1st step in Triangular Francisco. The 1st substep is to solve a 1x2x2 Block in DBR and have the flat side at bottom. The 2nd substep is to solve the DFL and BDR corners. Average movecount is like 8 or something like that.

2. 1LLS: 579 algs and I know about 140 of them. This is what lowers the movecount more than solving the last corner than CLL. There are 6-8 move 1LLS algs that most people dont know. While I was doing example solve, most of the algs I saw were 8 moves long so not that bad alg length.

3. Solve 3-5 More Edges: This is the difference between a 40 average movecount with all algs and ZZ average movecount with only L5E algs. Extensions can lower the movecount alot more. Use this time very wisely and lookahead (believe it or not) isnt bad here. You can solve 1-3 Edges before doing a z/z' rotation.

4. L5E, L6E, L7E: There are 2 substepss. The first one is to solve the last 1-3 L/R Edges WHILE orienting the midges (close to 400 algs full not counting 1 look L4E and L5E). This can be hard sine the L and R layers might be mis-aligned here. The 2nd substep is to permute the midges which is like 4 algs and are all 3 moves long.

I have no idea what the average movecount is but its lower than Roux is you learn all the algs (1k-ish). The beginner version which is 32 algs (DFL set for direct L5E (8 algs), OLFC (16 algs) and PLFC (6 algs) which is 2 look last slot) is a movecount kinda between ZZ and CFOP moveconut. THe more algs you know, the lower the movecount. The problem is there are ALOT of algs.

For the 1LLS method on 2x2, the method is 579+ algs and the average movecount is like 11-12.

Or do MRUG variant E which is this, but CP. I have also considered being extension neutral so solving the V so that LS is in dbr or dfr, or solving V on right so that LS is in ufl or ubl
It only needs like 110~ algs I think

Hey, I want to share my 2x2x2 method (MRUG) which I've developed 11 years ago. The method consists of 2 phases: PHASE 1: orient the cube so that DBL is solved (depending on what color you want to start with) solve DFL while reducing the cube to the {RU} group Phase 1 has average of 3.52 moves...

Hey, I want to share my 2x2x2 method (MRUG) which I've developed 11 years ago. The method consists of 2 phases: PHASE 1: orient the cube so that DBL is solved (depending on what color you want to start with) solve DFL while reducing the cube to the {RU} group Phase 1 has average of 3.52 moves...

ok so I finally finished the U cases I want to finish T CLL by today. btw I think I set the permission to public I found out so thats progression on 1LLS doc.

I honestly like the idea of this method. 80% of the time you will have three of color or it's opposite on a side of the cube. 33 percent of the time of that 80% you will have the same three colors on a time. Thus, you will only have one algorithm left to solve the cube 26.66% of the time, which is over a fourth of the time. Also makes one looking more feasible

I honestly like the idea of this method. 80% of the time you will have three of color or it's opposite on a side of the cube. 33 percent of the time of that 80% you will have the same three colors on a time. Thus, you will only have one algorithm left to solve the cube 26.66% of the time, which is over a fourth of the time. Also makes one looking more feasible