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dudefaceguy

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Feb 17, 2019
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This is similar to @dudefaceguy's intuitive 4x4 method, with some variations. His method uses commutators for your 5th step, to stay intuitive, but maybe you can find a faster alg based approach.
Yup, that's my method! I solve the inner slice after completing both blocks, but you can really do it either way. Hm, maybe I will experiment with switching some of the steps around. Very cool that you remembered :)

Seems to me that it has potential as a speed method, but I don't really know since I'm not a speed solver. The most obvious problem is that it uses completely different skills compared to 3x3, so it's not as easy to leverage your existing skills. I designed it this way on purpose, because I wanted my 4x4 solves to be different than my 3x3 solves.

Recognition is also difficult when pairing opposite wing edges in the inner slice - you need to identify which blue/white edge goes with which blue/yellow edge, even though they have the same colors.

But I am getting good times with this method, i.e. 4x slower than my 3x3 times. This is about what 4x4 times should be for a casual solver. So, a dedicated speed solver who is not an old man could probably get competitive times. Over time, I've come to do some of the steps exactly the same way, effectively making them algorithmic even though I'm technically using commutators. There are certainly some gains to be had by further refining algorthmic steps.

Edit: many of the steps are already used in other speed methods, for example Lewis and Sandwich. The thing that distinguishes it from these two methods is solving 3/4 of one inner slice, and using the other single slice to solve wing edges.
 
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dudefaceguy

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5. Solve the rest of the cube using U and the right inner slice (still working on this part)
By the way, I have tried to do this, and you CANNOT solve both centers and edges using only U and r (unless I have really missed something). EDIT: Actually you should be able to, since you can scramble the same pieces with U r. It just seems like the movecount would be very high.

It's either centers first and then edges using commutators (Lewis) or edges first and then centers using commutators (QTPI and Sandwich). You can do some center control while solving edges to get a few extra center pieces solved, but I'm not sure that this is worth it. There are 10 center pieces left if you solve edges first, or 8 if you also solve the two centers in the l slice while solving edges. 1/4 of these will usually be solved by accident, so there are usually 7 or 8 center pieces left, or 6 if you solve the extra 2 center pieces while solving edges. This is the difference between 2 and 3 commutators (or 1 4-move commutator cycling 6 pieces). Center commutators/algs can be really fast, but you have to look at the bottom and back faces to recognize the case.

Anyhow, I am obviously very excited to talk about this method but I will stop now and go to sleep.
 
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Alex Shih

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By the way, I have tried to do this, and you CANNOT solve both centers and edges using only U and r (unless I have really missed something). It's either centers first and then edges using commutators (Lewis) or edges first and then centers using commutators (QTPI and Sandwich). You can do some center control while solving edges to get a few extra center pieces solved, but I'm not sure that this is worth it. There are 10 center pieces left if you solve edges first, or 8 if you also solve the two centers in the l slice while solving edges. 1/4 of these will usually be solved by accident, so there are usually 7 or 8 center pieces left, or 6 if you solve the extra 2 center pieces while solving edges. This is the difference between 2 and 3 commutators (or 1 4-move commutator cycling 6 pieces). Center commutators/algs can be really fast, but you have to look at the bottom and back faces to recognize the case.

Anyhow, I am obviously very excited to talk about this method but I will stop now and go to sleep.
Do you know any specific cases where this isn't possible, or the specific reason this isn't possible? There might be a workaround (although I have a feeling that the workaround would probably be algorithmic). Also, if you want to continue this discussion, we should probably move to a different thread.
 
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Skewbed

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Here's an idea for finishing Hexagonal Francisco solves using pseudoslotting during the last edge.

Pseudoslotting ZBLL Finish:

1. Hexagon on D
2. Solve 3 E-slice edges like normal (RUru-gen)
3. Insert the last one with the corner by using pseudoslotting
4. Insert DF edge while doing EO (MU-gen)
5. ZBLL or such

Pseudoslotting OLL PLL Finish:

1. Same
2. Same
3. Same
4. Insert DF edge (MU-gen)
5. OLL
6. PLL

Example solve using OLL PLL Finish:

Scramble: U' L' D2 U2 B' D2 B2 L2 B R2 B L2 F2 U L' D R U' R2 B'

y2 // inspection
L D' L' // 3/4 cross, probably inefficient way to build hexagon
U' L' U' L // corner
R2 U' L U L' // corner
(D' U') L' U L // corner
u R U R' r U r' // edge
u r U r' F' U' F // pseudoslot
M' U' M // setup to LL
D' U R U R' U R d' R U' R' F' // OLL
U' L' U R' z R2 U R' U' R2 U D // PLL
R' // AUF (or ARF I guess)
 
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1.) What do you mean by X? an XCross?
2.) You forgot the rest of F2L
3.) There's a thread especially for proposing new methods, which can be found here.
I meant that you would make a literal X. Not practical, but somehow helps with F2L.
Btw for the other people talking about where's f2l, I kind of forgot to say that you did f2l with it, as I've used it so much that it's been pretty much forgotten as a step and more a part of the cross
 

dudefaceguy

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Do you know any specific cases where this isn't possible, or the specific reason this isn't possible? There might be a workaround (although I have a feeling that the workaround would probably be algorithmic). Also, if you want to continue this discussion, we should probably move to a different thread.
Yes, let's move to the thread for this method: https://www.speedsolving.com/threads/intuitive-4x4-method-with-parity-avoidance.73049/

I will post a reply there.
 

ProStar

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I meant that you would make a literal X. Not practical, but somehow helps with F2L.
Btw for the other people talking about where's f2l, I kind of forgot to say that you did f2l with it, as I've used it so much that it's been pretty much forgotten as a step and more a part of the cross

By X do you mean inserting all F2L pairs? If so then that's a really bad version of PCMS @CodingCuber
 
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Dunno if this exists but here it goes.
This is a wacky method that uses lots of algorithms so if you don't like learning algs then don't use this method.
I've posted this in a thread called "Pillman something something" and discussed this with other peoples, now I want to share it.

Step 1: it has two substeps to make inspection more feasible, step 1a is to make an FB + EO and plan it in inspection, I know that sounds crazy hard and it is, but with practice though, people can do it, I can guess some brave people can do it :p. Step 2b is to blockbuild an edge with 2 corners, Thus making one layer + EO.
tl;Dr: solve a layer + EO in two steps.

Step 2: solve the U layer edges into its place, and also, if all of your U layer edges are on the E slice, you can use one alg to solve them, kinda like L4EP but outside of last layer. tl;Dr: solve the U layer edges into its place. This step can obviously be improved to have more freedom, I'm open to all suggestions!

Step 3: Now, this alg set solves everything else, the LL corners and the oriented E slice edges in one algorithm while preserving the D layer and the U layer edges' orientation and permutation, this is similar to ZBLL, but, here's the catch, I think, this algorithm set has better recog and a bit lower algcount, now why better recognition? well ZBLL solves 8 pieces on the last layer, and the pieces are all combined together making it hard to recognize, but in this alg set, the 8 pieces, the corners and edges are separated, the E slice edges are not in the same layer as the LL corners, the part where it has a bit fewer algs? well, I just estimate that so take that with a grain of salt.
tl;Dr: solve the LL corners + the already oriented E slice edges.

I appreciate it if anyone calculates the average movecount for this method.

Step 4: Your cube is solved!

I'm open to any criticisms or corrections.

also, I'm trying this Chris Tran vibe kind thing, let me know haha
 
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So the FB is actually on D, not on L. That makes a lot more sense, and is also bad.
If you don't like FB + EO on D, you could do this, EOCross, insert the D layer corners in a way that prevents the U layer edges from being on the E slice, now on the insertion of the last D layer corner, you permute all of the U layer edges, kinda like LPELL, but it's for one corner, not a pair.

also, just to clarify, the alg set which does CLL + E slice edges preserves the U layer edges, orientation and permutation but permutes the E slice edges.
 
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PapaSmurf

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Yeah, it's definitely worse than doing ZZ. You're doing layer+EO which is already kinda not good, then you're gonna do a step with not super recog then you're gonna do a step with not super recog or algs. As a general rule of thumb, any diag corner swap+edges that aren't in U are going to be majority awful algs.
 
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