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I have this new redux/blockbuild/psuedoblock method for big (5x5+) cubes that I call "Blocs Nouveau." It is based on the principle that optimal edges are prie-fixed in pairs, for instance on 5x5 an optimal edge is 2/3 of the tredge, or on even bigger cubes, the n-edge with the most solved edges. This can't be applied to 3x3, since 3x3 is the final step of blocs nouveau. It can be applied to 4x4, but it wouldn't be "optimal" because optimality is either solved or unsolved; there isn't a partial n-edge on a 4x4; there are only two possibilities for an n-edge state.

Now, I will talk about the BN optimality rule: utilise this; only try to solve for 1/3 solved tredges if there are no 2/3 or 3/3 tredges. However, when I try the method in my free time, there is usually always a 2/3 tredge, and sometimes 3/3 tredges. You will save a lot of time and cuts on movecount. Nothing needs to be solved or matched into a "final state" until 3x3 step (such as fully solving a block), so don't worry about pre-permutations of 3x3 solved at any time during the solve. Also, corners mean nothing until the final step.

Steps (the sample here is if you were applying BN to a 5x5, which is the cube I derived the method from:):

1: BNF1B
1a: inspect for a partially solved or fully solved tredge that is, if it is either 2/3 or 3/3 solved. If partial, match it up to solve for one tredge. If already 3/3, skip to 1b
1b: Solve any center next to your solved tredge (doesn't need to match any colors on your tredge), and form another tredge. Follow optimality rule. This solves a BN 1x2x2. Keep the BN 1x2x2 on LF.
1c: Solve any center; keep it on the D layer.
1d: Solve another center adjacent to the BN 1x2x2 on the F face. Simultaneously create a tredge adjacent to the new center and D face center. Follow optimality rule. This will create a BN 2x2x2.
1e: Solve another center on R(or B; this example follows on R), solve two new tredges following the optimality rule, slotting them into RD and RF. This creates a BN 2x2x3.

2:BNFNL
2a: rotate your cube so that your unsolved centers are on U and F. Solve two new tredges following optimality rule, place them on FR and FL.
2b: Solve your two remaining centers and a tredge on DF.

3:edge cycles, parity
3a: your remaining unsolved tredges should be on the U layer. Just do edge cycles to solve them. As you would on any big cube method.
3b; if you get parity, solve it Now, it will help you with the last step.if no parity, just skip to the final step.

4: 3x3. By now, your cube should be in a scrambled 3x3 state. Just solve it as you would a normal 3x3, and you are complete.

Pros:
Use of optimal edges can lower movecount.
Easy center pairing
Doesn't regard any final permutations until the final step

Cons
BN blocks look like scrambled 3x3 blocks, which may appear confusing.
Recognition would probably need to be higher because of BN optimality rule.
Potentially high amount of rotations, but that may just be me.


I am curious to see if someone more seasoned In 5x5 can give this method a go.
 
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I don’t totally understand the BN rule, becausee surely all tredges are 1/3 solved instantly? Tell me if I’m misunderstanding this.

Secondly, for steps 2 and 3, wouldn’t it be better to solve L2C then the edges? It just seems more intuitive and better.
 
2b: Solve your two remaining centers and a tredge on DF.
whoa hold up on those slice moves

I've tried doing a solve based on your method outline, and while it's not absurdly inefficient, I don't see it being any better than freeslice in any sense. In terms of move count, it's worse. In terms of ergonomics, it needs considerably more slice moves, and you need to solve centres all over the place, so it's a lot worse. (It's not strictly necessary to do rotations, but good luck keeping track of centres on U, R and F simultaneously without rotating at all.)

Unlike Yau/Yau5/Hoya5 which trade off a bit of move count efficiency for ergonomics, lookahead and an easier 3×3×3 stage, your method trades off move count efficiency for (seemingly) nothing useful.

Pros:
Use of optimal edges can lower movecount.
Easy center pairing
Doesn't regard any final permutations until the final step

Cons
BN blocks look like scrambled 3x3 blocks, which may appear confusing.
Recognition would probably need to be higher because of BN optimality rule.
Potentially high amount of rotations, but that may just be me.
I have no idea what these pros/cons are supposed to be relative to. The pros are all not pros when compared to freeslice, and only the first and sort of kind of the third are relevant to Yau5.
 
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I don’t totally understand the BN rule, becausee surely all tredges are 1/3 solved instantly? Tell me if I’m misunderstanding this.

Secondly, for steps 2 and 3, wouldn’t it be better to solve L2C then the edges? It just seems more intuitive and better.

When solving for a new tredge, since tredges are made of three smaller edges, find a unit in which two of the three edges to form a tredge are solved.
 
Has Anyone thought of influencing 3-2-3 edge pairing during L4C in Yau? By doing a few setup moves it might be possible to "skip" one or two edges, resulting in a quicker edge pairing. I think that this would be beneficial as the movecount during L4C would only have to increase by 10 or so to place 2 wings and match them up whilst doing L4C.
 
Has Anyone thought of influencing 3-2-3 edge pairing during L4C in Yau? By doing a few setup moves it might be possible to "skip" one or two edges, resulting in a quicker edge pairing. I think that this would be beneficial as the movecount during L4C would only have to increase by 10 or so to place 2 wings and match them up whilst doing L4C.
Pair 3 solved edges and put them on DR, FR and BR if your UD centres are on LR and your 3cross edges are on LF, LD and LB. But then you will have to use slices for the centres. It looks great on paper but the reduced ergonomics mean it is not worth it
 
Could CFOP-CT be a thing? You could do EO after F2L-1. Then TSLE then TTLL. That makes it the same steps as full CFOP though. EO+TSLE+TTLL vs Insert pair + OLL + PLL. How many algs would TSLE be if edges weren't orientated?
 
Has anyone ever thought to make VLS but you don’t have to have the pair made?

There are about 500 ols algs (what you’re saying, but with eo). To add eo into there, you’re saying multiply it by 8, so ~4000 algs. Probably not worth it.
This is called OLS. There is a wiki page on it. According to that it has 17k algs though I'm not sure where that number comes from.
 
Could CFOP-CT be a thing? You could do EO after F2L-1. Then TSLE then TTLL. That makes it the same steps as full CFOP though. EO+TSLE+TTLL vs Insert pair + OLL + PLL. How many algs would TSLE be if edges weren't orientated?

I think 840 for TSLE + EO. It could be a thing but ZZ-CT isn't as great of a method as you might think.It's one more Step as CFOP but as many steps as VH.
You basically have a 3LLS


According to that it has 17k algs though I'm not sure where that number comes from.

Oriented OLS has 502 Cases * 8 by adding EO = 4016. This is only for the FR slot This means * 4 = 16064

If we add OLL and the complete OLS step I'm exactly 1590 cases away from their number
 
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Could CFOP-CT be a thing? You could do EO after F2L-1. Then TSLE then TTLL. That makes it the same steps as full CFOP though. EO+TSLE+TTLL vs Insert pair + OLL + PLL. How many algs would TSLE be if edges weren't orientated?
I do VHLS to do EO and COLL+EPLL
ZBLL would be slightly faster
 
Here's an idea: NMF2L (non-matching F2L)
Here's an example of what I mean:

setup: B2 L' R' B' D2 R B2 D B L' R F' R' D' F R' D z2

You can see that there is a 4-move white cross. But, look at the Right side: There's a 2x2x1 block. Let's go about the solve and pretend that the yellow/orange edge is solved, and the corner-edge pair with it is also solved. Now we have a 1-move X-cross! Here's how you would go about doing this (make sure you did the z2 in the setup!)

L // X-cross
y' R' U' R // F2L-2
R U' R' U R U R' // F2L-3
d R U R' U' R U R' // F2L-4
U S' L' U' L U L F' L' f // OLL (recognition is the same if you pretend white and yellow are the same)
Nb perm // PLL
Now just AUF and fix the F2L with an L2!
 
Here's an idea: NMF2L (non-matching F2L)
Here's an example of what I mean:

setup: B2 L' R' B' D2 R B2 D B L' R F' R' D' F R' D z2

You can see that there is a 4-move white cross. But, look at the Right side: There's a 2x2x1 block. Let's go about the solve and pretend that the yellow/orange edge is solved, and the corner-edge pair with it is also solved. Now we have a 1-move X-cross! Here's how you would go about doing this (make sure you did the z2 in the setup!)

L // X-cross
y' R' U' R // F2L-2
R U' R' U R U R' // F2L-3
d R U R' U' R U R' // F2L-4
U S' L' U' L U L F' L' f // OLL (recognition is the same if you pretend white and yellow are the same)
Nb perm // PLL
Now just AUF and fix the F2L with an L2!
Non-matching f2l (usually called Non-matching last layer/NMLL) has already been thought before but maybe developing non-matching last layer recognition system(s) could be useful.
 
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