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 Post subject: How to solve a 2x3x4?
PostPosted: Thu Aug 09, 2012 6:51 am 
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I'm still waiting for my mf8 2x3x4, but I've got a fully functional 2x3x4 from Garrett and another (semi full functional) from Traiphum.
(Traiphum's is restricted after a turn of the 2x4 faces, still it does shapeshift and can reach all possible configurations.)
Many will consider the 2x3x4 an easy puzzle compared with the well discussed 3x4x5.
Still, a decently scrambled 2x3x4 provides a similar challenge as the 3x4x5: How do you do a 3-cycle of edge wings?
As I do not have a good video equipment, I'll try it with photo sequences. (You can click onto the pictures to enlarge them.)

I'll use WCA notation where small letters refer to inner slice layers below the face with the same capital letter:
e. g. r is the slice below R and it turns like R. If I turn both layers at once I'll write (Rr). (Rr is a syonym to Rw)
Whenever I want to show all pieces I'll use a similar representation of frontview / backview as in Gelatinbrain.

Image

Because the same sequence cannot be performed on a Traiphum 2x3x4 I use this even shorter sequence:

Image

The result of the same sequence on Garrett's
Image

You will recognize the heart of these sequence (U Rw2 U' Lw2)x2 as the well known "corner sieres" (e.g. used in the Ultimate solution) in a slightly adapted form.
You can use the logic for a pure corner 3-cycle as well.

Did you find a pure 3-cycle of edge wings (or corners) in less than 18 moves?

EDIT: I want to show the difference between Traiphum's and Garrett's regarding "fully functional"

Image
The left part of the picture shows just a U turn of both.
The right part shows that an Rw2 turn is possible on Garret's (the smaller one) only.

In this case both puzzles can perform an Rw2 turn:
Image

BTW, Traiphum's has some kind of locking too: You would expect whenever half of the puzzle is in cuboid form, that you can make an inner slice turn. This is not always true.
On Garrett's I recognize no locking at all, sometimes pieces catch a bit, but a 2x3 face is never locked as described for the mf8 puzzle.

EDIT Sept 11th: The term "edge wings" refers to the eight edges on the 2x4 faces. The term "edges" would include the four edges on the 2x3 faces as well. I'll edit my following posts changing the word `edge` to `edge wings` wherever it is appropriate.

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Last edited by Konrad on Tue Sep 11, 2012 2:52 am, edited 3 times in total.

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 Post subject: Re: How to solve a 2x3x4?
PostPosted: Mon Aug 13, 2012 9:17 am 
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BUMP
Because nobody has replied to my post, I want to elaborate a bit more.
I think everybody is very occupied by the locking/bandaging issue discussed here. So, not many had really the opportunity to try solving this puzzle. Those who have got a custom made 2x3x4 , will have solved it ages ago. This post adresses novices regarding a fully functional 2x3x4, but being familiar with a 2x2x3.

I can nothing contribute to the "bandaged" 2x3x4 issue, because I'm still waiting for my mf8 version.
I'll try the method of unblocking it with "soft skills" first, but I'm really hoping for a hardware solution (new parts.)

When I googled "How to solve the 2x3x4" a few days ago, I did not find much that helps a beginner.
If you have never solved a 3x4x5 before, you may find the 2x3x4 to be not so easy.

A beginner can try to start with non-shape-shifting turns first (180 degree turns only).
Maybe, early versions of a "fully functional 2x3x4" allowed for 180 degree turns only, anyway.
I found an entry in the museum saying something like this
twistypuzzle museum wrote:
2x3x4 (bandaged 4x4x4)
It turns out there is additional unanticipated layer of complexity in the solution. A true cuboid 2x3x4 would allow only 180°-twists for every layer and dimension. In contrast this bandaged 4x4x4 sometimes allows 90°-twists.
If you restrict yourself to 180 degree turns when scrambling the puzzle, you will most probably find an easy way to solve it.
You can start with the corners, followed by the centres and then you need some pretty trivial permutations for the edges and edge wings. Still you could perceive some situations as special (I would not talk about "parity" and just say "special").
My pictures describe six such special situations, each with frontview/backview as in Gelatinbrain:
Image
Things become a bit harder when you do shape-shifting turns (90 degree turns of the 2x4 faces).
I see three ways to solve it:
1. Reduce it to 2x2x3, joining each outer piece with an fitting inner piece (as described by Burgo in the other thread)
2. do the inner 2x2x3 not caring about outer pieces (corners, edges, edge wings)
3. start with the outer 2x2x3 including the centres
Image
The greyish pieces are to be placed later.

In any case, you will need some permutations of the edges and edge wings.
Personally, I find it visually easier to solve the corners first (case 3 above)
When you are back to a cuboid shape with the corners solved, you may or may not be able to solve the rest with 180° turns.
In case you see the need to do a 3-cycle of edge wings on 2x4 faces, you'll find a pretty short pure 3-cycle in my first post above.

EDIT Sept 11th: The term "edge wings" refers to the eight edges on the 2x4 faces. The term "edges" would include the four edges on the 2x3 faces as well. I'll edit my following posts changing the word `edge` to `edge wings` wherever it is appropriate.

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Last edited by Konrad on Tue Sep 11, 2012 2:56 am, edited 1 time in total.

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 Post subject: Re: How to solve a 2x3x4?
PostPosted: Thu Aug 23, 2012 3:34 pm 
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I am pretty new to all this twisty stuff and need a little help!

My 2x3x4 arrived an I duly scrambled it. My original plan was to see if I could use the same approach as I do for the 3x4x5. This technique I learned from SuperAntonioVivaldi and it is pretty simple.

I quickly found both ends blocked and by fairly random movements I unblocked them and then managed to return it to cuboid shape with all corners correctly placed. Subsequent moves has returned everything to position except for three edge pieces. I was careful to make sure I only used 180deg turns to avoid taking it out of cuboid form and blocking up again. I cannot seem to get these last pieces into place. To swap these last 3 pieces seems to be impossible using Jon's techniques! My question is: are there any easy techniques to swap these last 3 cubbies? None are opposite each other they are all a 180deg face turn fom each other.

Thanks guys - I guess if I have to take it out of cuboid and it blocks up then I'll need to wait for the fix from Calvin.

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 Post subject: Re: How to solve a 2x3x4?
PostPosted: Thu Aug 23, 2012 4:12 pm 
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Hi Kevin,

EDIT: With non-shape-shifting turns (180° only) you can NOT perform a 3-cycle of edge wings.
You need to perform a 3-cycle similar to those in my posts described.
You have to take care that the edge wing glued to the core does not interfere.


If you find out which 2x4 edge wing is glued to the core, you can use one of my two move sequences above.
Just make sure that the r2 (in the first case) or b2 (in the second case) will not involve the glued edge wing.
You can use the method described by drswirly to identify the "rooted piece" glued to the core.

Let's look at U R2 U' L2 U R2 U' b2 U R2 U' L2 U R2 U' L2 b2 L2 (the second sequence):
If the rooted piece is at DLb, the 18 move sequence will not lock up the puzzle.
If you start with an unlocked, cuboid state the result is unblocked too and the edge wings perform a pure 3-cycle as shown in my photos (the photos does not show the mf8 version, but trust me it works on my mf8 puzzle)
ULb -> DLf -> DRb -> ULb

On my mf8 puzzle, the orange/yellow edge wing is the glued one. This will be very likely different on yours. (There are eight possibilities.)

You need to inverse my move sequence if you need to cycle the edge wings in the other direction.
You could assume that a L2 setup followed by the 18 moves would work, but this locks the cuboid afterwards.

EDIT Sept 11th: The term "edge wings" refers to the eight edges on the 2x4 faces. The term "edges" would include the four edges on the 2x3 faces as well. I'll edit my posts changing the word `edge` to `edge wings` wherever it is appropriate.

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Last edited by Konrad on Tue Sep 11, 2012 3:00 am, edited 3 times in total.

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 Post subject: Re: How to solve a 2x3x4?
PostPosted: Fri Aug 24, 2012 7:10 am 
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I hope my explanation above helped.
Anyway, I made a photo serie of the move sequence using the mf8 2x3x4:
Image
The rooted piece, the orange yellow edge wing adjacent to the blue/orange/yellow corner is glued to the core in my case.
I have marked it in the "Start" situation.
After the first 7 turns it is located at ULf (I marked in again after move 7.)
(You might need to click on the picture to enlarge it.)

Turn b2 in 8 does nothing to the core.
The following 7 moves are the inverse of the first 7.
So, all inner edges are correct again.

I hope, my photo sequences helped.
This a strange thread (one reply so far).
Probably the discussion suffers because the unwanted, unexpected locking interferes with the solution attempts?

What about my challenge?
How do others solve a situation that requires a 3-cycle?

EDIT Sept 11th: The term "edge wings" refers to the eight edges on the 2x4 faces. The term "edges" would include the four edges on the 2x3 faces as well. I'll edit my posts changing the word `edge` to `edge wings` wherever it is appropriate.

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Last edited by Konrad on Tue Sep 11, 2012 3:04 am, edited 1 time in total.

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 Post subject: Re: How to solve a 2x3x4?
PostPosted: Sun Aug 26, 2012 12:09 am 
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Hi Konrad! Today I repaired my MF8 2x3x4 so that it is not bandaged. I was solving away until I ended up with a case that I have not seen before on this puzzle, where 3 edges are out of place. You covered it on August 13th. Your algorithm and photographs worked perfectly and now my 2x3x4 is again solved. Thank you for posting it.

I've been solving larger cuboids, such as the 3x4x5 but I'm a little lost on the 2x3x4, probably because I haven't solved a 2x2x3. Really, the 2x3x4 is not difficult to restore to cuboid, though I got stuck for quite some time with two corners swapped.

Is there a way to solve the 2x3x4 that eliminates or reduces the cases of 3 edges out of place?

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 Post subject: Re: How to solve a 2x3x4?
PostPosted: Sun Aug 26, 2012 6:14 am 
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Pete the Geek wrote:
...Is there a way to solve the 2x3x4 that eliminates or reduces the cases of 3 edges out of place?
Hi Pete, as I wrote on Aug 13th:
Konrad wrote:
....I see three ways to solve it:
1. Reduce it to 2x2x3, joining each outer piece with an fitting inner piece (as described by Burgo in the other thread)
2. do the inner 2x2x3 not caring about outer pieces (corners, edges, edge wings)
3. start with the outer 2x2x3 including the centres
....
If you reduce it first as in 1. (Burgo's method), you will have a simple 2x2x3 afterwards. While you are reducing it, in many cases you have to come up with something similar to my move sequences above.
You can look at the discussion about "How to solve a 3x4x5?" It is essentially the same.

Method 2. might be easier for the locking version: It is simpler to take care of the "rooted piece"

I hope nobody has problems with solving a 2x2x3? It is not much different from solving the corners of a Domino.
I have not checked it, but I suppose that you will find a tutorial for a 2x2x3, easily.

EDIT Sept 11th: The term "edge wings" refers to the eight edges on the 2x4 faces. The term "edges" would include the four edges on the 2x3 faces as well. I'll edit my posts changing the word `edge` to `edge wings` wherever it is appropriate.

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Last edited by Konrad on Tue Sep 11, 2012 3:06 am, edited 1 time in total.

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 Post subject: Re: How to solve a 2x3x4?
PostPosted: Mon Aug 27, 2012 7:37 am 
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I repeat here what I posted on Aug 21 on the other thread, because it has to do with solving the real mf8 2x3x4. Many people have expressed that the bandaging is most unexpected and unwanted, but you have either to deal with it or wait for a replacement:
Konrad wrote:
This photo shows the eight hidden edges in a solved state where all 3x2 faces can be turned.
They are below the circles.
I show the puzzle in frontview /backview.

I'll use the same terminology as drswirly: "good" edges (indicated by a +) mean split edges that allow the 3x2 face to turn, bad edges (-) will not.

Image
Good edges are at UL, UR and DL, DR.
Bad edges are at UF, UD, DF, DB.
Burgo had already shown a usual Domino sequence to exchange UF and UR.
You just have to take care of the "rooted piece" (again drswirly terminology; in my picture the marked orange/yellow edge adjacent to the blue /orange/yellow corner; you need to find it on your puzzle using drswirly's hints) and have it not involved in this sequence:
[(Rw2, U)x2 (Rw2, U2)x2 Rw2, U, Rw2, U', Rw2]
The puzzle looks now:
Image
Besides the swap of outer pieces two hidden edges at UF and UR are swapped!
UR has become a bad end and UF a good end.
The swapped visible pieces can be swapped back by cuboid turns (18o° moves) after you have corrected all bad edges.

Without peering into the interior of the puzzle you can now use a trial and error plan to swap recognized bad edges with supposed good edges.
Whenever you recognize one or more bad ends, there must be one or more good edges at places where a bad edge is OK.(UF, UD, DF, DB in my picture)

The inner edge adjacent to the rooted piece will travel always with the rooted piece.
So, there can be a maximum of three bad edges to be replaced by good edges.

Everybody may decide by herself / himself if he /she views this behaviour as fun or a "nice challenge".
I can be mistaken, but my guess is that close to 100% would prefer a "normal" fully functional 2x3x4.
Should we make a voting? :lol:


EDIT Sept 11th: In this post the term `edge` refers to internal, hidden edges which are part of the mechanism not to visible cubies.

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Last edited by Konrad on Tue Sep 11, 2012 3:14 am, edited 1 time in total.

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 Post subject: Re: How to solve a 2x3x4?
PostPosted: Mon Sep 10, 2012 3:14 am 
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I haven't been following this thread so would have chimed in earlier.

...from the other thread...
Konrad wrote:
Would you be ready to share your 3,1 commutator and "some healthy thought processes along the way" over in the solution thread?

Puzzlemad wrote:
Subsequent moves has returned everything to position except for three edge pieces. I was careful to make sure I only used 180deg turns to avoid taking it out of cuboid form and blocking up again. I cannot seem to get these last pieces into place. To swap these last 3 pieces seems to be impossible using Jon's techniques! My question is: are there any easy techniques to swap these last 3 cubbies? None are opposite each other they are all a 180deg face turn fom each other.

Konrad wrote:
What about my challenge?
How do others solve a situation that requires a 3-cycle?

Pete the Geek wrote:
Is there a way to solve the 2x3x4 that eliminates or reduces the cases of 3 edges out of place?

Well all this seem to be focussing on the same thing, which is moving around 3 pieces. I don't know if I'll be able to explain it that well, and whether it will seem obvious. (In fact, re-reading my post, it is difficult to explain.)

With this puzzle, it's clear that returning to cuboid is not too hard, and solving a reduced 2x2x3 is easy. So the only real problem is pairing corners with edges. I wanted to try and do this without resorting to the roughly 22 move commutator that I use for the 3x4x5, which Konrad posted on the 3x4x5 thread. I found that it is possible on the 2x3x4 because you don't have to worry about keeping the edges in place on the 2x3x4, because there are none. I mean the edges on the 3x4x5 on the 3x5 faces.

I found that once it's returned to cuboid, there are sort of 3 main cases, which can be achieved with simple turns (that is, without shapeshifting).

Case 1. Two reduced corners in different orbits (meaning they can be placed at, say, DFR and DBR), and the remaining 6 unreduced corners can be reduced with something like L2 U2 L2 U2. Obviously, this is the holy grail of cases.

Case 2. Four reduced corners which can be placed in positions like UFR, UBR, DFR and DBR. When this happens, L2 will reduce the remaining corners every time. (I should point out that the L2 only works if the inner edge colour on front and back face is the same, otherwise it breaks those edges.)

Case 3. Four reduced corners which can be placed in positions like UFR, UBR, DFR and DFL. This last case is the annoying case. (I'm fairly sure this is equivalent to the max. 5 reduced corners people are talking about above, since a turn like L2 will make 5.) This means one of the pieces is in the wrong orbit. When this happens I could find no way to get that 4th piece into the DBR position using simple moves. So I shapeshifted it there. (So Kevin will have to wait until he gets a working puzzle :wink: )

So let's say I have 3 reduced corners in UFR, DFR and DBR, and the 4th in DBL. I want the 4th in UBR. So I do D, then (U' Ll2 U Rr2) x2 then D', which (with the D setup) is the corner 3-cycle from Ult. sol. which I use for things like 3x3x2s. Now those reduced corners are in the right position for me, but of course it's shapeshifted. So I shapeshift it back using 3 unreduced corners. So I do y' U2 so that my 3 corners involved are unreduced. And then do the same cycle: (U' Ll2 U Rr2) x2. From there the puzzle's in cuboid form and I can use simple 180* turns to get the pieces to case 2 above.

Of course, once all corners are reduced, after going through one of the 3 cases above, it's 2 or 3 turns to place the middle edges of the 2x2x3, and then all that's left is to place the (reduced) corners, using things like (U' Ll2 U Rr2) x2. Yes, this will shapeshift it, but it will unshapeshift by the end. If I get two 2x2x3 corners needing to swap, then I turn the upper (2x4) face one turn and just solve corners from there.

Solving it in the way I describe above means the biggest alg I have to remember is 8 moves. It works every time and is an enjoyable puzzle (particularly now that it doesn't culturally block).

There are some videos if anyone's interested on my blog or my youtube. I'm also happy to make a video showing what I try to describe in case 3 above.

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Last edited by rline on Mon Sep 10, 2012 5:55 pm, edited 1 time in total.

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 Post subject: Re: How to solve a 2x3x4?
PostPosted: Mon Sep 10, 2012 2:49 pm 
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Hi rline, Kevin and all others interested,
I find the terminology `reduced edges` in your post a bit misleading. I think I understand what you mean, though. I'll try to clarify the terminology.
rline wrote:
...Well all this seem to be focussing on the same thing, which is moving around 3 pieces. I'm confused because I can't tell whether the issue is single corners, or reduced edges.
Whenever I talk about `pieces` or `cubies` without adding the word "reduced", I mean just plain un-reduced pieces. Let us look at a 2x3x4 in the following setup: U, D are 2x4 faces; F,B are 3x4 faces; L,R are 2x3 faces as in my picture below.

We have 8 corners, 8 edges on the 2x4 faces, 4 edges on the 2x3 faces and 4 centres.
In my post August 13th I have shown 3 different outlines to solve this puzzle.
In the following text I will concentrate on
Quote:
1. Reduce it to 2x2x3, joining each outer piece with an fitting inner piece (as described by Burgo in the other thread)
I hope we can agree on the following definitions:
a) reduced 2x2x3 corners = `reduced corners`refer to a pair of pieces, a corner and its adjacent fitting 2x4 edge
b) reduced 2x2x3 edge = `reduced edge` refers to a pair of pieces, a 2x3 edge and its adjacent fitting centre piece.
Because we have two different types of `edges`, I have introduced in my first post the term `edge wing` for the eight 2x4 edges of an un-reduced 2x3x4.
Whenever we have completed the reduction phase, we have a 2x2x3 puzzle with 8 reduced corners and 4 reduced edges. Usually we will finish the reduction in a cuboid shape.
rline wrote:
...Case 1. Two reduced edges in different orbits (meaning they can be placed at, say, DFR and DBR), and the remaining 6 unreduced edges can be reduced with something like L2 U2 L2 U2. Obviously, this is the holy grail of cases.
I have a problem with the term "reduced edges" here. You mean obviously the edge part of a "reduced corner"?
I understand the reduction method as "reducing a given scrambled 2x3x4 puzzle to a 2x2x3 puzzle".
An ordinary 2x2x3 consists of two piece types only: corners and edges.
So, we pair two 2x3x4 pieces to "reduced edges" and "reduced corners".
rline wrote:
...Case 3. Four reduced edges which can be placed in positions like UFR, UBR, DFR and DFL. This last case is the annoying case. (I'm fairly sure this is equivalent to the max. 5 reduced edges people are talking about above, since a turn like L2 will make 5.) This means one of the pieces is in the wrong orbit. When this happens I could find no way to get that 4th piece into the DBR position using simple moves. So I shapeshifted it there. (So Kevin will have to wait until he gets a working puzzle :wink: )....
There are certain situations in all three different approaches (as discussed in my post from August 13th) that can never be solved by `simple` moves (180° turns; cuboid turns).
Your case 3 is describing the problem: Whenever you reach a situation with 5 reduced corners, obviously 3 corners are not yet paired. I have described this in my first post as a problem to 3-cycle pieces.
Actually, I use the same [3,1] commutator that you call the corner sequence of the Ultimate Solution (CPS)
(U R2 U' L2)x2. (OK, you wrote the mirrored version of this.) Just my setup is a bit different.
In my post from August 24th I describe the following sequence in a step by step picture sequence (unfortunately this was a bit overloaded by the locking problem stuff):
(U R2 U' L2)x2 L2 b2 (U R2 U' L2)x2 b2 L2 (Because the 9th turn reverses the 8th turn, I have shortened this to 18 moves alltogether.)
You can view this as a [[3,1],3] commutator or even [[[1:1],1],[1:1]]: [[[U:R2],L2],[L2:b2]]. Because 5 moves are reduced to one in #8 = b2, we count 18 moves.

In a reduction method I use this to solve your "case 3" problem.

You can replace the b2 above by B2: (U R2 U' L2)x2 L2 B2 (U R2 U' L2)x2 B2 L2 and you will get a pure 3-cycle of 2x3x4 corners.

I hope the folowing picture clarifys the terminology and we can agree upon the terms "reduced corner" for pair "A" in diagram 3 and "reduced edge" for pair "B".
Reduced pieces are broken up if you turn l and r slices NOT simultaneosly with their L and R layers.
My second mf8 2x3x4 (with 7 unsplit internal edges) is an unbreakable 2x2x3 :lol:

Image

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Last edited by Konrad on Tue Sep 18, 2012 8:13 am, edited 2 times in total.

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 Post subject: Re: How to solve a 2x3x4?
PostPosted: Mon Sep 10, 2012 6:05 pm 
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Hi Konrad

Quote:
I find the terminology `reduced edges` in your post a bit misleading. I think I understand what you mean, though. I'll try to clarify the terminology.

Yes, guilty as charged. I've edited my post above to refer to "corners" now. In my defence, I think the reason I was confused as well was due to comments like
Puzzlemad wrote:
Subsequent moves has returned everything to position except for three edge pieces.
and
Pete the Geek wrote:
Is there a way to solve the 2x3x4 that eliminates or reduces the cases of 3 edges out of place
which I assume are all talking about the same problem. (Perhaps they're talking about something completely different.)

Konrad wrote:
I hope we can agree on the following definitions:
a) reduced 2x2x4 corners = `reduced corners`refer to a pair of pieces, a corner and its adjacent fitting 2x4 edge
b) reduced 2x2x4 edge = `reduced edge` refers to a pair of pieces, a 2x3 edge and its adjacent fitting centre piece.

For sure. But you mean reduced 2x2x3, don't you?

Quote:
I understand the reduction method as "reducing a given scrambled 2x3x4 puzzle to a 2x2x3 puzzle".

Yes, me too. The terminology is the only issue.

Quote:
An ordinary 2x2x4 consists of two piece types only: corners and edges.

But then I get confused by this because you've done it twice now (refer to 2x2x4) and I'm not sure where a 2x2x4 fits into things. :?

Quote:
I hope the folowing picture clarifys the terminology and we can agree upon the terms "reduced corner" for pair "A" in diagram 3 and "reduced edge" for pair "B".

As almost always, the pictures tell the story and that's certainly what I was referring to. :)

(But even in the picture, your number 4 has the title "Reduced 2x2x4". :? What am I missing? I'm really hoping all these references to 2x2x4 are just meant to be 2x2x3, otherwise there's definitely something I'm not understanding.

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 Post subject: Re: How to solve a 2x3x4?
PostPosted: Tue Sep 11, 2012 3:19 am 
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rline wrote:
....(But even in the picture, your number 4 has the title "Reduced 2x2x4". :? What am I missing? I'm really hoping all these references to 2x2x4 are just meant to be 2x2x3, otherwise there's definitely something I'm not understanding.
Sorry for the typo :oops:
I made it so consistently that it became really confusing :oops:

Regarding the names:
I hope that I have used the term `edges` and `corners` without the prefix `reduced` consistently for single cubies on the 2x3x4. (I went back and changed `edges` to `edge wings`where appropriate. I hope this has not confused anybody.)

Following the WCA notation including slice layers we find (U=2x4; F= 3x4):
8 corners: UFL, UFR, UBR, UBL, DFL, DFR, DBR, DBL
8 edge wings (=2x4 edges): UFl, UFr, UBr, UBl, DFl, DFr, DBr, DBl
4 edges (2x3 edges): FL, FR, BR, BL (all in the E layer)
4 centres: Fl, Fr, Br, Bl

We could use a similar naming scheme for `reduced edges` and `reduced corners`(a 2x3x4 viewed as a 2x2x3):
8 reduced corners: UFLl, UFRr, UBRr, UBLl, DFLl, DFRr, DBRr, DBLl
4 reduced edges: FLl, FRr, BRr, BLl
We could allow names without the lower case letters whenever the context makes it clear that we are speaking about `reduced` elements.

I think we have reached now a clear common understanding.

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 Post subject: Re: How to solve a 2x3x4?
PostPosted: Fri Sep 14, 2012 5:17 pm 
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rline wrote:
So let's say I have 3 reduced corners in UFR, DFR and DBR, and the 4th in DBL. I want the 4th in UBR. So I do D, then (U' Ll2 U Rr2) x2 then D', which (with the D setup) is the corner 3-cycle from Ult. sol. which I use for things like 3x3x2s. Now those reduced corners are in the right position for me, but of course it's shapeshifted. So I shapeshift it back using 3 unreduced corners. So I do y' U2 so that my 3 corners involved are unreduced. And then do the same cycle: (U' Ll2 U Rr2) x2. From there the puzzle's in cuboid form and I can use simple 180* turns to get the pieces to case 2 above.


What do you mean with: "y' U2"?


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 Post subject: Re: How to solve a 2x3x4?
PostPosted: Fri Sep 14, 2012 5:25 pm 
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robin wrote:
rline wrote:
So let's say I have 3 reduced corners in UFR, DFR and DBR, and the 4th in DBL. I want the 4th in UBR. So I do D, then (U' Ll2 U Rr2) x2 then D', which (with the D setup) is the corner 3-cycle from Ult. sol. which I use for things like 3x3x2s. Now those reduced corners are in the right position for me, but of course it's shapeshifted. So I shapeshift it back using 3 unreduced corners. So I do y' U2 so that my 3 corners involved are unreduced. And then do the same cycle: (U' Ll2 U Rr2) x2. From there the puzzle's in cuboid form and I can use simple 180* turns to get the pieces to case 2 above.


What do you mean with: "y' U2"?

Turn the cube anticlockwise about its vertical axis, then turn the upper face 180*. You can see pretty pictures of it all here.

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 Post subject: Re: How to solve a 2x3x4?
PostPosted: Tue Sep 18, 2012 6:48 am 
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In this post I want to show the similarities between the 3-cycle approach and rline's method:
rline wrote:
.....Case 3. Four reduced corners which can be placed in positions like UFR, UBR, DFR and DFL. This last case is the annoying case. (I'm fairly sure this is equivalent to the max. 5 reduced corners people are talking about above, since a turn like L2 will make 5.) This means one of the pieces is in the wrong orbit. When this happens I could find no way to get that 4th piece into the DBR position using simple moves. So I shapeshifted it there. (So Kevin will have to wait until he gets a working puzzle :wink: )

So let's say I have 3 reduced corners in UFR, DFR and DBR, and the 4th in DBL. I want the 4th in UBR. So I do D, then (U' Ll2 U Rr2) x2 then D', which (with the D setup) is the corner 3-cycle from Ult. sol. which I use for things like 3x3x2s. Now those reduced corners are in the right position for me, but of course it's shapeshifted. So I shapeshift it back using 3 unreduced corners. So I do y' U2 so that my 3 corners involved are unreduced. And then do the same cycle: (U' Ll2 U Rr2) x2. From there the puzzle's in cuboid form and I can use simple 180* turns to get the pieces to case 2 above.....
I show an example as described by rline: UFR, DFR and DBR, and the 4th in DBL (see situation 1). Each situation is described by two photos showing frontview and backview of the puzzle in Gelatinbrain fashion. You can click onto the image to enlarge it.
Image
In my approach I do either L2 or l2 and get five reduced corners. The remaining three corners must be paired with their corresponding wedges (i.e. 2x4 edges) by one 3-cycle.
Step 2 shows how the three corners have to be cycled.
In step 3 I have turned around the whole puzzle by 90° counterclockwise.
In step 4 I have the right setup for the 3-cycle of corners by my 18 move sequence. The required 3-cycle is shown in the backview. It is originally a [[[1:1],1],[1:1]] commutator (which would be 22 moves) shortened to 18 moves, because 5 consecutive moves are folded into one.

In step 5 the reduction is completed.
(The reverse of the R2 setup is the first step to solve the reduced 2x2x3 in this example.)

Here is rline's method.
Image
We start with four reduced corners and group them all in the right half of the puzzle (steps 2 to 6).
Then we pair all remaining corners with their wedges in step 7.

The total move count is quite similar.
This means that a pure 3-cycle ist not very costly.
You can try to solve a scrambled 2x3x4 by
1. back to cuboid
2. try to solve it with cuboid turns (180° turns)
3. If necessary do a 3-cycle at the end

I switch between reduction method and the outline above.
It is probably a matter of taste, what you will like better.

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 Post subject: Re: How to solve a 2x3x4?
PostPosted: Sun Nov 11, 2012 1:26 pm 
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I finally got on board and purchased an mf8 2x3x4 and used the extra pieces to make it fully functional and unbandaged. After some experimentation and documentation I was able to solve it, but did not have a fully developed strategy in place yet. I might mention that I do not have either a 2x2x3 or a 3x4x5, but I do have a 4x4x6 and doubtless scrambled and solved it as a 2x2x3 at some point. I've also made a sticker mod of a 4x4x4 to turn it into a 2x2x4. Solving the inner layers first makes it like a 2x2x3 also. Anyway, before fully developing a strategy based upon my initial experimentation, I wanted to see how others were approaching the 2x3x4. I have now incorporated some of rline's ideas and one of Konrad's ideas into a new strategy that I am happy with.

Basically I solve the 2x2x3 first, then the 8 corners, then the middle layer edges. I never have to think about returning it to cuboid shape with the method I use. I never have to think about 3-cycling edge wings that make up the 2x2x3.

My question is: what is the typical move count for the different methods used by our community? Can you all take a break from your current puzzle projects and revisit this one? My goal for the day will be to solve it as many times as I can, keeping track of how many moves it takes each time.....

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 Post subject: Re: How to solve a 2x3x4?
PostPosted: Sun Nov 11, 2012 1:48 pm 
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.

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 Post subject: Re: How to solve a 2x3x4?
PostPosted: Sun Nov 11, 2012 6:21 pm 
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Quote:
Basically I solve the 2x2x3 first, then the 8 corners, then the middle layer edges. I never have to think about returning it to cuboid shape with the method I use. I never have to think about 3-cycling edge wings that make up the 2x2x3.

My question is: what is the typical move count for the different methods used by our community? Can you all take a break from your current puzzle projects and revisit this one? My goal for the day will be to solve it as many times as I can, keeping track of how many moves it takes each time.....
I count M2 as 1 twist. Just like R2 is 1, and U is 1. I do not count whole puzzle turns as twists, so if I turn the whole puzzle in my hands to use a different side as F, for example, it is not a twist. With my method I never twist only r or l or f or b. The only slice move I ever have to do is the layer between the 2x4 layers. So far I have counted twists for 5 solves. the average is 30 twists per solve, with an average of 16 twists per solve to get the inner 2x2x3, and 14 twists per solve to get the 8 corners and middle layer edges. I have not had an opportunity yet to 3-cycle corners, which would drive the average up. I am curious how this compares with the twist counts of others. Perhaps someone would be interested in comparing twist counts for some given scrambles?

(Edit)
I finally had a scramble that required 2 3-cycles to get the corners, not to mention more twists to get the inner 2x2x3 to begin with. 22 twists to get the 2x2x3, and 38 twists to get the corners, then 4 to get the middle layer edges, for a total of 64 twists. So now my average is 36 twists per solve over 6 solves.

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 Post subject: Re: How to solve a 2x3x4?
PostPosted: Mon Nov 12, 2012 4:00 am 
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Hi Robert,

I hardly ever consider move counts, but I think you're doing pretty well in that area. I just did a few solves: I reduce the 234 to a 223.

A solve will generally go something like:
[5-8 moves] [to reduce the shape] most times I will use: (U R2 U R2 U) as the backbone to what I do. I have shape patterns to reduce the shape of the 345 as well.
[3-5 moves] [to begin reducing corners] Rr Ll twists with F2 or U2 moves dispersed sometimes an E twist to start getting some centres reduced.
[6 moves] [setup for reducing centres- at the same time as corners] (U2 F2)x3: Sometimes this might be necessary before the last R2 corner reduction twist, sometimes just Ll2 E twists will suffice.
[13 move sequence (+ 2-3 setups)] D2 (R2 U R2 U' R2) D' (R2 U R2 U' R2) D' sometimes a 3cycle is needed, usually a few twists associated with a setup.
[10-20 moves] [to solve the reduced 223]. The average can be more towards the 10-16 because I have a semi-efficient 222 method, and you can combine strategic E twists. The most common `backbone` sequence I use is R2 U R2 U2 y' R2 U R2 combined with a few strategic E twists if needed, sometimes (R2 U2)x3 needs to be added at the end to fix the E layer.

I'd say it's around a 50 move average? If you don't need the corner 3cycle, it can knock a bunch off.

Cheers,
Burgo.

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 Post subject: Re: How to solve a 2x3x4?
PostPosted: Mon Nov 12, 2012 4:29 am 
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Burgo wrote:
Hi Robert,

I hardly ever consider move counts, but I think you're doing pretty well in that area. I just did a few solves: I reduce the 234 to a 223.
I hardly ever consider move counts myself, but when I made a blogspot page explaining how I am solving it, it seemed so long that it made me curious.

Burgo wrote:
[10-20 moves] [to solve the reduced 223]. The average can be more towards the 10-16 because I have a semi-efficient 222 method, and you can combine strategic E twists. The most common `backbone` sequence I use is R2 U R2 U2 y' R2 U R2 combined with a few strategic E twists if needed, sometimes (R2 U2)x3 needs to be added at the end to fix the E layer.
R2 U R2 U2 y' R2 U R2 !!! That is what I use to solve the 2x2x2 portion of the 2x2x3 that is the puzzle minus the corners and middle layer edges. That is why I do not have to first get it to cuboid shape. After using this algorithm on the middle section of the puzzle, it not only solves 8 pieces, it puts it in cuboid shape too.

Burgo wrote:
I'd say it's around a 50 move average? If you don't need the corner 3cycle, it can knock a bunch off.
Thank you Burgo for this. So my method of solving it from the inside out without reducing does not seem to be unreasonably longer than the reduction method you use. I wonder how it compares to Konrad's method of working from the outside in, and, say, rline's method.

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 Post subject: Re: How to solve a 2x3x4?
PostPosted: Mon Nov 12, 2012 4:58 am 
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If anyone is up to it, try these scrambles with a 2x4 face on top, and a 3x4 face on front:

1. L2 R2 U2 D2 F2 Rr2 D2 F2 U Rr2 U' Ll2 D2 F2 U Rr2

2. R2 U2 R2 F2 R2 U2 R2 F2 U2 R2 Ll2 U2 R2 F2 U Rr2 U Rr2 D Rr2 F2 D' Rr2 U F2

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 Post subject: Re: How to solve a 2x3x4?
PostPosted: Mon Nov 12, 2012 5:33 am 
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I was trying out your 1st sequence when I had a major internal pop. Turns out the little locking piece from the unsplit edge came out and was loose internally, I'll be gluing for a bit :roll:


Attachments:
234 internal pop.jpg
234 internal pop.jpg [ 369.96 KiB | Viewed 5738 times ]

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 Post subject: Re: How to solve a 2x3x4?
PostPosted: Mon Nov 12, 2012 6:08 am 
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Burgo wrote:
I was trying out your 1st sequence when I had a major internal pop. Turns out the little locking piece from the unsplit edge came out and was loose internally, I'll be gluing for a bit :roll:


eek. nightmare. it was bad enough putting it together the first time :(

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 Post subject: Re: How to solve a 2x3x4?
PostPosted: Mon Nov 12, 2012 8:08 am 
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Hi Robert.. what an ordeal! Because I had it well and truely scrambled I needed to take it back to 0 to rebuild it, this has got to be my favourite puzzle to assemble :x .

1st sequence:-----------------------------2nd sequence:
4 shape------------------------------------3 shape
2 first 4 corners---------------------------1 + 6move + 2 (U2 x L2 R2)
2 F2 R2 next 2 corners-------------------(reduced to 223)
4 + 6move + 1 last corners + edges-----2 U&D colour matched
(reduced to 223)--------------------------7+2 our shared algo
2 U&D colour matched--------------------4 to finish E slice
7+2 our shared algo----------------------(27 moves)
6move to finish E slice
(36 moves)

Maybe I got your second sequence wrong? You were mixing up R2s & Rr2s I think. What I did was R2 unless R2 `had to be` Rr2, in which case if both layers were locked I was forced to do that move. Neither scramble required a 3cycle of corners to reduce.
PS I had a few twists on the 345 after and it felt like a giant!
Cheers,
Burgo.

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 Post subject: Re: How to solve a 2x3x4?
PostPosted: Mon Nov 12, 2012 4:03 pm 
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Burgo wrote:
Maybe I got your second sequence wrong? You were mixing up R2s & Rr2s I think. What I did was R2 unless R2 `had to be` Rr2, in which case if both layers were locked I was forced to do that move. Neither scramble required a 3cycle of corners to reduce.
Until the first 90˚ U, every R was a single layer, and every Rr was a double layer. After the first 90˚ U, every R was double layer. In fact, after the first 90˚ U every turn I made when scrambling was a double layer, whether it was R or F. Should I have continued to call them (Rr) and (Ff)?

Thanks for all the bother you put into this Burgo. Very helpful. :D

On the second sequence you say it took 3 twists to do "shape". Does shape mean restore it to cuboid shape? If I am not mistaken, the original scrambled shape is what I call pinwheels. And I don't see how to get to cuboid in 3. :?

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 Post subject: Re: How to solve a 2x3x4?
PostPosted: Tue Nov 13, 2012 4:39 am 
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Sorry Robert, I obviously got the second scramble wrong.. I think I was a bit confused by the R / Rr mixup and I tried for a bit this afternoon to retrace my steps but there's just too many variables.. so I resolved it and recorded the entire solve:

Scramble 1: L2 R2 U2 D2 F2 Rr2 D2 F2 U Rr2 U' Ll2 D2 F2 U Rr2
R2 U2 F2 D -----------------------shape
R2 L2 -----------------------------first 4 corners
F2 R2 -----------------------------break 2 corners and make 2 (switch)
U2 Rr2 F2 (U2 B2)x3 R2 ---------reduced to 223
D' F2 ------------------------------U&D colour matched
y2 (R2 U R2 U2 y' R2 U R2) U' ---solved
(28 moves)

Scramble 2: R2 U2 R2 F2 R2 U2 R2 F2 U2 R2 Ll2 U2 R2 F2 U Rr2 U Rr2 D Rr2 F2 D' Rr2 U F2
Rr2 U2 F2 U ------------------shape
R2 L2 first 4 ------------------corners
F2 U2 R2 ---------------------break 2 corners and make 2 (switch)
U2 Rr2 F2 [(U2 F2)x3] R2 --reduced to 223
F2 U2 Rr2 --------------------U&D colour matched
F2 Rr2 F2 D2 ---------------- U&D solved
(Ll2 U2)x3 223 ---------------solved
(31 moves)

I hope that's better, sorry for the confusion.
Cheers,
Burgo.

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 Post subject: Re: How to solve a 2x3x4?
PostPosted: Tue Nov 13, 2012 5:54 pm 
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Thanks for the clarification on notation, Burgo. One thing though. Should the later F's also be Ff's?

After seeing how many moves it was taking me to do a double swap of corners using 2 corner 3-cycles (36 moves for just the 4 pieces!) I decided to rethink things. After reading what rline put on his blog about it, and experimenting a little, the 36 moves is down to 12 to accomplish the same thing! :lol:

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 Post subject: Re: How to solve a 2x3x4?
PostPosted: Wed Nov 14, 2012 1:54 am 
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Hi Robert,
robertpauljr wrote:
After seeing how many moves it was taking me to do a double swap of corners using 2 corner 3-cycles
Did you notice what I was doing with this: U2 Rr2 F2 (add the redundant Rr2 to see it visually).

Another thing that I thought about for quite a short 3cycle of corners (works great with my reduction method and works on the 345 too):
R2 (U Rr2 D' Rr2 U') R2 (7 moves) With that I think I could achieve a (30-35 move) average?

That's pretty efficient I think. If you want to bring it back to the shape and make it pure add (U Rr2 D Rr2 U') but I wouldn't need that. How I use it is to use the first R2 to make the first one (and break 2 made ones). Then the sequence exchanges the one I made with the new one, then the second R2 (returns the 2 broken ones) and completes the last 2. What did you think of my version of reducing it? I must admit I can't imagine yours.. Could you write down your solves for the same scrambles? Use your most improved method.
robertpauljr wrote:
Thanks for the clarification on notation, Burgo. One thing though. Should the later F's also be Ff's?
I just thought if anyone's dropping in and reading along it might be easier for them.. I see Konrad is using a mixture of shorthand and longhand notation in his posts too, it's kind of hard to think in the longhand way. Maybe we should Use Ff and Rr whenever we do an inner twist and there's cubies in the outer layer, I suppose it would be unnessesary if there `were no cubies present in the outer layer` :lol: ?

Cheers,
Burgo.

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 Post subject: Re: How to solve a 2x3x4?
PostPosted: Wed Nov 14, 2012 6:14 am 
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Burgo wrote:
.... I see Konrad is using a mixture of shorthand and longhand notation in his posts too, it's kind of hard to think in the longhand way. Maybe we should Use Ff and Rr whenever we do an inner twist and there's cubies in the outer layer, I suppose it would be unnessesary if there `were no cubies present in the outer layer` :lol: ?

Cheers,
Burgo.
If I hold on a solved cuboid the 3x4 as F and 2x4 as U, I can never make an f turn or F turn independently after scrambling. Therefore, I think a Ff notation is unnecessary. We could use the face name f instead of F and always write Ff2 instead of F2. Does this make it clearer?
If you want to simulate a 2x3x4 on a 4x4x4 this becomes a necessity: We are looking at the f surface where the outer layer F is removed and we must use Ff2 turns always on the 4x4x4 simulating a 2x3x4.
BTW Rr2 is a short hand, too, for R2, r2. (On a physical 4x4x4 Rr2 becomes ambiguous and can be interpreted as R, r2. Therefore I use brackets (Rr)2 or Rw2. The latter is WCA notation, too, and stands for "R wide")

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 Post subject: Re: How to solve a 2x3x4?
PostPosted: Wed Nov 14, 2012 6:53 am 
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Hi Konrad,

Thanks for commenting, you're much better at notation than me.

Until trying the scrambles I had no problems with the notation, but I'm not very good at following scrambles.

You can have an occasion where you need the notation for F and f to twist independently, because a series of moves may include U E D for example, or y. I often use E and y twists inside other algorithms, sometimes to solve the E layer along with the outer layers (although I can't imagine explaining doing sequences like that). I really like your idea to use f or r etc for occasions when there are no cubies in the outer layer, because it differentiates between the outer turn, which otherwise you are `looking for`.

I see what you mean by the confusion with R r2 and Rr2, if I use Rr2 I always leave a space in between, I like spaces, I think it makes it look clearer :) .

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 Post subject: Re: How to solve a 2x3x4?
PostPosted: Wed Nov 14, 2012 5:47 pm 
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Quote:
Did you notice what I was doing with this: U2 Rr2 F2 (add the redundant Rr2 to see it visually).

No, I hadn't. Let's see…
U2 (Rr)2 F2 does a double swap. Swaps 2 2x1 blocks and 2 2x2 blocks.

Quote:
Another thing that I thought about for quite a short 3cycle of corners (works great with my reduction method and works on the 345 too): R2 (U Rr2 D' Rr2 U') R2 (7 moves) With that I think I could achieve a (30-35 move) average?
I tried this from solved. Saw the 3 cycle. Noticed it also scrambled things up a bit. Tried to go back starting with R2 then the algorithm. Didn't work. So I tried to go back just stepping through it backwards and inversing. Ended up with a pure 3-cycle of corners. FRD > BRU > BLD. I must have zigged when I was supposed to zag somewhere along the line. :lol:

Quote:
That's pretty efficient I think. If you want to bring it back to the shape and make it pure add (U Rr2 D Rr2 U') but I wouldn't need that.
Maybe this is what I did? :D

Quote:
How I use it is to use the first R2 to make the first one (and break 2 made ones). Then the sequence exchanges the one I made with the new one, then the second R2 (returns the 2 broken ones) and completes the last 2. What did you think of my version of reducing it?
Very nice.

Quote:
I must admit I can't imagine yours.. Could you write down your solves for the same scrambles? Use your most improved method.

Scramble 1: L2 R2 U2 D2 F2 Rr2 D2 F2 U Rr2 U' Ll2 D2 F2 U Rr2
Get the centers. Done.
Separate whites and yellows (my U and D colors). y' (Rr)2 D (Rr)2 U (Rr)2 [I think I got the y' thing right. I don't usually think about it or count it as a move. What I did was turn the front to the right.]
Solve the 2x2x3. x2 D' [There is one pair. Put them on the lower left.] (Rr)2 U (Rr)2 U' (Rr)2 U (Rr)2 U' (Rr)2. E (or is it E' ?).
At this point I have been solving the middle layer edges last, but since the corners are trivial in this current solve, I will swap the middle layer edges now.
Middle layer edges. (Ll)2 F (Ll)2 U2 (Ll)2 U2 (Rr)2 U2 (Rr)2 U2 (Ll)2 F (Ll)2 D2
Corners. L2 R2

Before I looked at some of rline and Konrad's ideas I was developing a solution in which I reduced the corners then solved the reduced 2x2x3. Perhaps I should go back to my earlier experimentations and see where it leads.

Maybe I'll just shelve the 2x2x3 for awhile and open up the CT Bandage Kit and play with it. :D

The scramble I solved above didn't include any corner solving at the end really. Here is another.

Scramble: U2 R2 F2 U2 R2 F2 L2 (Rr)2 U2 (Rr)2 F2 U (Rr)2 U (Rr)2 D (Rr)2 D (Rr)2 D2 (Rr)2 F2

Solve:
B2 (Rr)2 [centers]
U' (Ll)2 U' (Ll)2 [separate U/D colors]
D' (Rr)2 U (Rr)2 U2 F2 U F2 D' [Inner 2x2x3 solved]
R2 U2 L2 U2 L2 [corners all solved]
turn the puzzle so F goes to L
L2 F2 L2 U2 L2 U2 R2 U2 R2 U2 L2 B2 L2 D2 [swap 2 middle layer edges]

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 Post subject: Re: How to solve a 2x3x4?
PostPosted: Fri Nov 16, 2012 6:38 am 
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Thanks for sharing Robert,

Really interesting, so different to how I was thinking about it.
robertpauljr wrote:
Bugro wrote:
Did you notice what I was doing with this: U2 Rr2 F2 (add the redundant Rr2 to see it visually).
No, I hadn't. Let's see… U2 (Rr)2 F2 does a double swap. Swaps 2 2x1 blocks and 2 2x2 blocks.
Because you're working with 2 seperate corner orbits, it's a handy little exchange out: for keeping the corners from one orbit (on R) and bringing the corners from the other orbit (from L) `in`. It's really handy for a break, exchange, rebuild like in my solves ^^. On a 222, it's just a checker pattern.

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