The New Method / Substep / Concept Idea Thread

[Suratha]

In CFOP,what if we do ELS in the last slot to orient all edges and insert the corresponding corner while permuting the last layer corners so that we can do 2-gll to finish the last layer? will the move count be lower than MGLS and PLL? or atleast we can avoid some bad pll cases such as v-perm and e-perm?

 

[TDM]

In CFOP,what if we do ELS in the last slot to orient all edges and insert the corresponding corner while permuting the last layer corners so that we can do 2-gll to finish the last layer? will the move count be lower than MGLS and PLL? or atleast we can avoid some bad pll cases such as v-perm and e-perm?

ELS+CPLS. Also mention in the second paragraph.

 

[Dapianokid]

Or you could just use Petrus. Even if it's entirely different.

To put Roux into the shortest defintion I can think of... "Sexy M slices."
Petrus, in short, would be: "Not really but still CFOP"

 

[Athefre]

Apples: Not really but still oranges

 

[TheNextFeliks]

If I recall correctly, it takes only 16 algorithms to orient the top 8 edges on a 4x4. The permutation of the edges, however, may take many algorithms because of the permutation possibly requiring and odd number of 2-cycles to solve. This bring me to the idea making the permutation of the edges even while orienting them. There are 2 possibilities between even and odd for both edge cycles making this require 4x16 or 64 algorithms. A large amount of these would be mirrors. 64 is definitely a learnable amount. After the permutation has been made even, the edges would have (4!/2)x(4!/2) or 144 permutations requiring 143 algorithms. This, I think is learnable, but I'm just questioning weather it is efficient to have to recall 1 out of 143 algorithms along side the fact that you cant master all 143. This is the only reasonable system for a 2 look last 8 edges though.

As a side note - I've been wandering why there hasn't been some version of Roux for the 4x4? I've been messing with it some and it seems pretty efficient.

KCBM or Meyer/Stadler

 

[Michael Womack]

How possible is it to reduce a scrambled 4x4 into a 2x2?

 

[Username]

How possible is it to reduce a scrambled 4x4 into a 2x2?

Probably not good for speed

 

[Hypocrism]

How possible is it to reduce a scrambled 4x4 into a 2x2?

http://www.youtube.com/watch?v=CGgwWiRWhto

It looks clumsy to me, it would require blockbuilding or what seems like a slow LBL method for the first two layers, and then the bandaged 4x4 solve which looks slow.

 

[Michael Womack]

Probably not good for speed

I was thinking of doing it for fun.

 

[elrog]

http://www.youtube.com/watch?v=CGgwWiRWhto

It looks clumsy to me, it would require blockbuilding or what seems like a slow LBL method for the first two layers, and then the bandaged 4x4 solve which looks slow.

If you take parity to mean mathematical parity, then the video wrongly claims that parity is eliminated in a 4x4x4 solve if you solve it by reducing it down to a 2x2x2 puzzle. The only difference between reducing the 4x4x4 to a 3x3x3 and a 2x2x2 is that you solve the parity before you finish reducing the 4x4x4 to a 2x2x2 unlike reducing it to a 3x3x3.

Ex: Do the single edge flip OLL parity to a 4x4x4 and then scramble it like a 2x2x2. To solve this, you would get those 2 corners beside each other, solve the parity, then solve like a 2x2x2. You may also have to watch out for messing up the centers while solving the parity.

Whether you view parity as reduction parity or mathematical parity, you still have to solve the mathematical parity.

 

[Hypocrism]

If you take parity to mean mathematical parity, then the video wrongly claims that parity is eliminated in a 4x4x4 solve if you solve it by reducing it down to a 2x2x2 puzzle. The only difference between reducing the 4x4x4 to a 3x3x3 and a 2x2x2 is that you solve the parity before you finish reducing the 4x4x4 to a 2x2x2 unlike reducing it to a 3x3x3.

Ex: Do the single edge flip OLL parity to a 4x4x4 and then scramble it like a 2x2x2. To solve this, you would get those 2 corners beside each other, solve the parity, then solve like a 2x2x2. You may also have to watch out for messing up the centers while solving the parity.

Whether you view parity as reduction parity or mathematical parity, you still have to solve the mathematical parity.

As I understand it parity only applies to after the reduction has been done. If, once reduction is done, the cube is always solvable as the reduced puzzle, that's solved without parity. In the case of 3x3, half the time it isn't solvable after reduction so that there is parity.

Parity doesn't exist in the 2x2 reduction because it's an even puzzle!

 

[Dapianokid]

Hypocrism, and in th case of a twisted 2x2x2 corner...?

 

[TheNextFeliks]

Hypocrism, and in th case of a twisted 2x2x2 corner...?

Impossible. That would mean one corner is twisted in place which is impossible.

 

[elrog]

As I understand it parity only applies to after the reduction has been done. If, once reduction is done, the cube is always solvable as the reduced puzzle, that's solved without parity. In the case of 3x3, half the time it isn't solvable after reduction so that there is parity.

Parity doesn't exist in the 2x2 reduction because it's an even puzzle!

I specified that what I said only applies if you consider parity in a mathematical sense rather than a reduction sense. It seems you are viewing parity as reduction parity and trying to tell me my statement is false.

In a mathematical sense, parity is determined by the amount of 2-swaps required to solve a group of pieces (such as edges or corners or any other kind of piece). If the group requires an odd number of 2 cycles, it is in an odd permutation and is considered parity.

Do a single U turn on the 2x2x2 and solve it using only 3-cycles (which are made up of 2 2-swaps). It is impossible.
If you don't understand that, try mixing the 2x2x2 up using an odd number of moves, and solve it using an even number of moves, but remember that you must count moves in QTM. It is impossible.

 

[rybaby]

COLL on the R-face.
Proposed by Ranzha V. Emodrach (Brandon Harnish) and Ryan Przybocki (me).
We have been generating algorithms especially for recognizing COLLs on R. This is very useful for the Waterman method (my plan), and can be implemented by cross-on-left solvers. For Waterman, this gives a clearer look ahead for the next step) inserting redges. Perhaps Ranzha could share his spreadsheet of algorithms.

 

[elrog]

Cool. I've always wandered why Waterman didn't just solve the corners on the right side in the first place.

 

[Ranzha]

https://docs.google.com/spreadsheet...6bWVPSTJHRW1hRFUyLXlIcE9sSUE&usp=docslist_api

Here ya go.
It's not done, and a fair bit of the algs suck. Open to suggestions, just PM or FB msg.

This oughta be interesting not only for Waterman solvers, but for COL CFOP solvers.

 

[JackJ]

https://docs.google.com/spreadsheet...6bWVPSTJHRW1hRFUyLXlIcE9sSUE&usp=docslist_api

Here ya go.
It's not done, and a fair bit of the algs suck. Open to suggestions, just PM or FB msg

This oughta be interesting not only for Waterman solvers, but for COL CFOP solvers.

So is this new, or just regular coll from a different angle?

 

[TheNextFeliks]

So is this new, or just regular coll from a different angle?

Basically just coll from a different angle. But not same algs if that's what you meant.

 

[Ranzha]

So is this new, or just regular coll from a different angle?

COLL with last layer on the R face, so yeah, it's a new angle. New algs for a bunch of the cases. Simply translating COLLs that already exist usually ends up in crappy algs.