The New Method / Substep / Concept Idea Thread

[Hypocrism]

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.

I realise you know more about what parity means than I do however I thought that, excluding 2x2, it's always impossible to do an odd number of 2-swaps on any cuboid puzzle? By your definition, this would make all 4x4 states parity-less by the definition you gave, because there's no such thing as a state requiring an odd number of 2-swaps. That's what I don't understand about your post.

And since on even puzzles, a "3x3 2-swap" of edges is actually an even cycle (edges on even puzzles are even, edges on odd puzzles are odd, obviously) then edge swaps on the pseudo-3x3 count as even edge swaps on the bigger puzzle, which is why it's not soluble with the basic pseudo-3x3 sides.

However if we go back to the reduction to 2x2, since both puzzles involved are even there isn't parity, because every pseudo-2x2 move on the 4x4 is an even swap, making all the cases soluble.

I'm sure you're right, I do like getting to know puzzle theory though

 

[bobthegiraffemonkey]

Looked through this thread and for some reason started working on inserting pair+making a 2x2 block on U. I started filling a big table, this is the first subset that ended up finished (2 oriented edges, 1 oriented corner):

Spoiler

UL UB UBL RU'R'
UL UB UBR U'L'U2RU2R'U2L
UL UB UFR U'R'FR2UR'U'RUR'U'F'
UL UB DFR U'R'FRURU'R'F'
UL UR UBL F'L'U'LF
UL UR UBR RUR'U'RU'R'
UL UR UFR y'rDr'U'rD'r'
UL UR DFR F'RU2R'U'RU'R'FRU'R'
UL FR UBL L'U2RUR'U2L
UL FR UBR RU'R'U'R'U'RU'R'U2R
UL FR UFR RUR'U'RU'R'
UL FR DFR URU'R'URU2R'
UB UR UBL URU2R'
UB UR UBR U'R'FRF'
UB UR UFR URU'R'URU2R'
UB UR DFR RUR'U'RU'R'
UB FR UBL URU'R'U'R'FRF'
UB FR UBR U'L'URU'LR'
UB FR UFR R2DR'URD'R2'U'RU'R'
UB FR DFR RUR2U'R2U'R'UR'UR
UR FR UBL U'R'F2R2U'R'U'RUR'F2
UR FR UBR RU'R'
UR FR UFR RU'R'UR'U'RU'R'U2R
UR FR DFR RUR'U2M'URU'r'

I doubt anyone is interested, I doubt I'll ever look at this again, but sometimes I just feel like doing stuff. Also, some of the algs might not be the fastest but w/e.

 

[elrog]

1: I realise you know more about what parity means than I do however I thought that, excluding 2x2, it's always impossible to do an odd number of 2-swaps on any cuboid puzzle?

2: By your definition, this would make all 4x4 states parity-less by the definition you gave, because there's no such thing as a state requiring an odd number of 2-swaps. That's what I don't understand about your post.

3: I'm sure you're right, I do like getting to know puzzle theory though

Sorry for picking your post apart, but I wanted to reply to certain parts of it.

1:Well, it is possible to swap just 2 pieces on the many cubes. On a 2x2, doing a single U turn does a 4-cycle which is composed of 3 2-swaps. In other words, you'd have to switch 2 pieces 3 times to move all 4 U layer corners to their correct positions. I did say this already, but I'll say it again. A 3-cycle (rotating 3 pieces like in an A perm) is made up of 2 2-swaps. So, by doing a single U turn (odd number of moves), then doing an A perm (even number of moves) you end up in a position that has 2 pieces needing swapped and an odd number of moves to solve the puzzle because an even number plus an odd number gives you an odd number. So, every position that is an odd number of moves away from solved needs an odd number of 2-swaps. Where parity comes in is being able to swap 2 pieces independently of other pieces. On the 3x3, you cant move edges without moving corners (unless you have a void cube), but on the 4x4, a slice move can move 4 edges in a 4-cycle without affecting the corners. This also doesn't affect the center pieces because you rotate 8 centers which can be solved in an even number of moves. The centers also have multiples of the same color letting you solve them in an odd or even number of moves, meaning that you can have the edges needing a 2-swap independently of the rest of the pieces.

2:I believe number 1 answers this, but just for an example, the single edge flip parity on the 4x4 swaps just those 2 adjacent edge pieces. This is possible because the edges can be moved independently of the corners while centers don't matter (just like on a void cube).

3: If you are really interested in cube theory, we have a section of the forums dedicated to it and we also have people willing to answer your questions about it. cmowla is very, very knowledgeable about cube theory. He also has come up with dozens of parity algs an the sort for the 4x4. He'd be a great person to consult with your questions.

 

[TheNextFeliks]

Looked through this thread and for some reason started working on inserting pair+making a 2x2 block on U. I started filling a big table, this is the first subset that ended up finished (2 oriented edges, 1 oriented corner):

Spoiler

UL UB UBL RU'R'
UL UB UBR U'L'U2RU2R'U2L
UL UB UFR U'R'FR2UR'U'RUR'U'F'
UL UB DFR U'R'FRURU'R'F'
UL UR UBL F'L'U'LF
UL UR UBR RUR'U'RU'R'
UL UR UFR y'rDr'U'rD'r'
UL UR DFR F'RU2R'U'RU'R'FRU'R'
UL FR UBL L'U2RUR'U2L
UL FR UBR RU'R'U'R'U'RU'R'U2R
UL FR UFR RUR'U'RU'R'
UL FR DFR URU'R'URU2R'
UB UR UBL URU2R'
UB UR UBR U'R'FRF'
UB UR UFR URU'R'URU2R'
UB UR DFR RUR'U'RU'R'
UB FR UBL URU'R'U'R'FRF'
UB FR UBR U'L'URU'LR'
UB FR UFR R2DR'URD'R2'U'RU'R'
UB FR DFR RUR2U'R2U'R'UR'UR
UR FR UBL U'R'F2R2U'R'U'RUR'F2
UR FR UBR RU'R'
UR FR UFR RU'R'UR'U'RU'R'U2R
UR FR DFR RUR'U2M'URU'r'

I doubt anyone is interested, I doubt I'll ever look at this again, but sometimes I just feel like doing stuff. Also, some of the algs might not be the fastest but w/e.

Can you explain this? So like what do the pieces before the algs mean? Looks interesting.

 

[elrog]

Can you explain this? So like what do the pieces before the algs mean? Looks interesting.

To me, it appears that he is first stating the position of the edge that goes in the position counterclockwise of the corner followed by the edge that goes on the other side of the corner, then the corner that will be in the 2x2 block, and finally the algorithm for the case.

When there are 2 faces (FL, or UR, or DB) included in a pieces position, it is an edge piece. If there are 3 (UFR, or ULF, or BLD) faces included in a pieces position, it is a corner piece. The faces stated (Front=F, B=back, R=Right, and so on) are the faces that that piece intersects. So FR is the edge that is both in the U layer and the R layer.

He did not state anything about the orientation of the pieces, so I'm just assuming they are supposed to be oriented already.

Some case like: R U' R' are shown twice because they can form a 2x2 block in the U layer when the pieces to make it are in any multiple positions.

 

[bobthegiraffemonkey]

Can you explain this? So like what do the pieces before the algs mean? Looks interesting.

(2 oriented edges, 1 oriented corner)

If you have the FR/DFR pair at UF/UFL and you have 2 oriented LL edges and 1 oriented LL corner, you can look up those pieces in the list and perform the given alg to make those pieces form a 2x2 block on LL.

 

[TheNextFeliks]

If you have the FR/DFR pair at UF/UFL and you have 2 oriented LL edges and 1 oriented LL corner, you can look up those pieces in the list and perform the given alg to make those pieces form a 2x2 block on LL.

Ohhh. Ok.

 

[IQubic]

NEW ZZ LAST LAYER APPROACH????? ZZ-D Profane Koala + Phasing

I kinda come up with a recognition method for Asmallkitten's Profane Koala. See that thread. But the amount of 2GLLs is unbelievable; really 80+ algs!!! So I thought Why not permute corners, with Profane Koala, then use Phasing.

I believe this shall reduce the number of 2GLLs you need to learn from 80+ to 28. I think that is the right number, because phasing removes the 8 U-Perms from the possible edge permutations.

Correct me if I'm at all wrong. I feel i have a new concept, but maybe the wrong logic.
I also want to get some Algs generated for this approach. If i get a guide, I would totally put in the time and effort to generate the Algs myself.
If this approach needs a name, I propose ZZ Profane Phasing.

-IQubic

 

[TDM]

I kinda come up with a recognition method for Asmallkitten's Profane Koala. See that thread. But the amount of 2GLLs is unbelievable; really 80+ algs!!! So I thought Why not permute corners, with Profane Koala, then use Phasing.

I believe this shall reduce the number of 2GLLs you need to learn from 80+ to 28. I think that is the right number, because phasing removes the 8 U-Perms from the possible edge permutations.

Correct me if I'm at all wrong. I feel i have a new concept, but maybe the wrong logic.
I also want to get some Algs generated for this approach. If i get a guide, I would totally put in the time and effort to generate the Algs myself.
If this approach needs a name, I propose ZZ Profane Phasing.

-IQubic

Yeah... not new.

EDIT: Also, why not Prophased Koala?

 

[StachuK1992]

IQubic:

Don't take this the wrong way, but you probably
a) shouldn't bother with any ZZ-d variant until your average is at /least/ sub20.
b) expect that most of your ideas are taken and/or bad. Have fun playing around, but don't suspect that any good will come of it

 

[aznanimedude]

i too kinda had this thought to phase->reduced 2GLL set the moment porky talked about porky v1/ ZZ-rainbow, but i figured it was a good stepping stone, but really, the alg count for 2GLL is not that big i felt honestly that it was not really worth it to add in the recognition time to phase, you might as well take any recognition time used for phasing to just recognize the edge permutation

 

[TheNextFeliks]

Idea:
1. Solve cross+2 opposite pairs
2. Solve 3rd pair and place it in the wrong slot
3. Solve 4th pair in wrong slot
4. Swap the two pairs and solve ocll
5. PLL
Somewhere eo needs to be completed. Or use zz. CFOP is better but fun idea.

 

[Michael Womack]

Would someone Develope better ALGs for this 4x4 method and a name for it? http://www.youtube.com/watch?v=Kr37rWsEZMY

 

[aznanimedude]

Idea:
1. Solve cross+2 opposite pairs
2. Solve 3rd pair and place it in the wrong slot
3. Solve 4th pair in wrong slot
4. Swap the two pairs and solve ocll
5. PLL
Somewhere eo needs to be completed. Or use zz. CFOP is better but fun idea.

i'm trying to visualize how/why this would work. your swap two pairs and solve OCLL, after reinserting the 3rd pair in the correct spot, don't you essentially just have a VHLS case?

 

[Christopher Mowla]

Would someone Develope better ALGs for this 4x4 method and a name for it? http://www.youtube.com/watch?v=Kr37rWsEZMY

Hi Michael,

If you want algorithms to pair up the dedges in the U and D faces, there are 16! possible cases. Ollie made a complete set of 3-cycles of wing edges, so you can use that. However, this method is probably not new, as you have just left a different portion of the cube unsolved than K4, for example. Anyone who is fluent in commutators can solve any portion of the 4x4x4 they wish.

Regardless, here is a list of all possible 2 2-cycles for the last layer:
2 2-cycles.pdf , and here are 2 2-cycles in more than one face Algs_Sorted.pdf.

Perhaps my K4 LL method could give you further insight:
cmowla'sK4LLMethod.pdf

For the 3-cycle and 2-cycle oriented cases, I have made faster (and shorter) algorithms, but I never updated my document: here.
I have also introduced a conjugate method which creates 4-cycles. So the algorithms for the 4-cycle cases can be replaced with "prettier" but longer algorithms (in btm) like: [B2 U2 Rw' Uw2 U Rw' U2 Lw U Rw' U2: r].

I have also made algorithms which involve more than 4 pieces, such as: [F2 r2 F2 U' M2 U: M2] and l r (U' y) L R u L' R' (U y') M2 (U' y) L' R' u L R (U y') l r
Lastly, I made documents full of 2 2-cycles and 2 3-cycles for solving both pieces of a F3L dedge simultaneously. Perhaps this could give you further insight: F3L.

 

[Michael Womack]

Hi Michael,

If you want algorithms to pair up the dedges in the U and D faces, there are 16! possible cases. Ollie made a complete set of 3-cycles of wing edges, so you can use that. However, this method is probably not new, as you have just left a different portion of the cube unsolved than K4, for example. Anyone who is fluent in commutators can solve any portion of the 4x4x4 they wish.

Regardless, here is a list of all possible 2 2-cycles for the last layer:
2 2-cycles.pdf , and here are 2 2-cycles in more than one face Algs_Sorted.pdf.

Perhaps my K4 LL method could give you further insight:
cmowla'sK4LLMethod.pdf

For the 3-cycle and 2-cycle oriented cases, I have made faster (and shorter) algorithms, but I never updated my document: here.
I have also introduced a conjugate method which creates 4-cycles. So the algorithms for the 4-cycle cases can be replaced with "prettier" but longer algorithms (in btm) like: [B2 U2 Rw' Uw2 U Rw' U2 Lw U Rw' U2: r].

I have also made algorithms which involve more than 4 pieces, such as: [F2 r2 F2 U' M2 U: M2] and l r (U' y) L R u L' R' (U y') M2 (U' y) L' R' u L R (U y') l r
Lastly, I made documents full of 2 2-cycles and 2 3-cycles for solving both pieces of a F3L dedge simultaneously. Perhaps this could give you further insight: F3L.

Oh wow thanks.

 

[Mohammadahmadi]

My FF2L Method

i designed one method with my algorithm maker software again! HEHE! and i named it FF2L. it's mean First F2L
(First First 2 layer?! omg )

in F2L you solve FR and FRD without any effect on BR and BRD, but sometimes there is a faster way to solve FR but it can move BR too. so when BR isn't true, you can use FF2L to solve FR. so you can use it 1-3 times in each solve

you can see my method here
please give your comments and ideas about my method

 

[Mohammadahmadi]

I already designed an method for 2x2 first layer.

we have 90 cases for the first face (not first layer). we just do the first face and then we use EG method to solve the cube
Notice that sometimes you should make this cases with one move
the average of moves is 4.05 so we can solve the first layer in average with 4 moves

you can see my method here
Please give your comments and ideas about my method

 

[elrog]

As many knowledgeable cubers know, corner permutation is preserved (to a point) if you only do 2-gen R and U moves. Whenever you place the bottom 2 corners, the top corners are always permuted correctly in relation to each other and will be no more than a U move away from fully permuted. Solving a cube 2-gen is desirable because the algorithms are easier to preform usually.

The drawback of finishing a solve 2-gen is that it is difficult to get the corners into the correct permutation so that you can actually solve them. The recognition is horrid.

This all applies to cube-shaped cubes of course. On the Mega minx, you can do corner 3-cycles with only R and U moves. We do we not use this?

 

[porkynator]

4x4 LL & Beginner's Roux

Two random ideas:

4x4 LL + both parities:
- CLL + OP (not COLL)
- ELL + PP

This should give a reasonable amount of algs to learn

A Roux variation (possibly for beginners):
- First Layer minus one edge (or 3x2x1 + 2 corners)
- CLL on Right (Left)
- Second Roux Block using <R, M, U>
- LSE

The 4x4 LL method can be good for speed (assuming good algs exist), the Roux variation is just for fun.