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 Post subject: How to solve a 3x4x5?
PostPosted: Sat Jan 28, 2012 9:02 am 
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Can somme one help me how to solve the 3x4x5? I can put it back to it cubic form, and solve the 3x3x4 in it. But it at the 2 extra layers the (3x4 layers), that I get lost, cause I have no idear what to do/or more what approch to comme with.

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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Sat Jan 28, 2012 9:56 am 
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There is a tutorial on youtube.


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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Sat Jan 28, 2012 12:15 pm 
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Here is the tutorial http://www.youtube.com/watch?v=6Yjjya1BhFQ

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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Sat Jan 28, 2012 6:19 pm 
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Hi Drake,
If you don't want a complete spoil by watching a tutorial, you might try to [ reduce it to a 3x3x4 before you solve the 3x3x4 ].
Cheers,
Burgo.

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PB 3x3 55sec Jan 2011 (When I was a kid 1:30 was speedcubing so I'm stoked).
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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Sun Jan 29, 2012 4:31 pm 
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Burgo wrote:
Hi Drake,
If you don't want a complete spoil by watching a tutorial, you might try to [ reduce it to a 3x3x4 before you solve the 3x3x4 ].
Cheers,
Burgo.
I understand Burgo's intention not to be too much of a spoiler. Still, I'm wondering if this "spoiler" was a little bit cryptic for somebody seeking for help.
May I ask who got the `Ahaa` effect (is this understandable in English? :lol: ) reading this little hint?

EDIT: Nobody is posting here anymore?
Is it really so easy?
I had a brief look at the mentioned tutorial (extracts only). It has three parts and did not sound so easy to me.
Maybe Luke could post something here about his irritatingly easy method? :lol:
Luke wrote:
...As for the challenge, I found it to be irritatingly easy. I would compare the difficulty to around the 4x4x5. Nevertheless, it is a brilliant collection piece, and currently my favourite cuboid to play with.
I have revisited the 3x4x5 after quite a long time and still think it is not so easy. The shapeshifting adds some confusing factor. For me it is harder than the 4x4x5.
I had not documented my method very well and had to reinvent the wheel. This time I'll make it better. :)
And I'll try to improve my method.

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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Wed Feb 01, 2012 2:14 am 
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Hi Konrad,
I'd like to give this a little nudge because I wouldn't mind seeing a bit of conversation on it.

Think of what I said [reduce it to a 3x3x4 before you solve the 3x3x4] in terms of [reduce the 4x4x4 to a 3x3x3 and then solve the 3x3x3]. It wasn't intended to be hugely cryptic, but a very succinct answer for Drake's specific situation. I am surprised to see that Drake has not made a follow up post? Although I do think it's time to elaborate a bit:

Reduce the cube to a 3x3x4 and then solve the 3x3x4:
1. Make the cuboid shape.
2. Pair the inner slice 3x4 pieces with their matching outer layer 3x4 pieces.
3. Solve the 3x3x4.

It has surprised me to find out that this is perhaps not the common method?
Cheers,
Burgo.


Attachments:
3x4x5 reduction.jpg
3x4x5 reduction.jpg [ 106.23 KiB | Viewed 8074 times ]

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PB 3x3 55sec Jan 2011 (When I was a kid 1:30 was speedcubing so I'm stoked).
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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Fri Feb 03, 2012 9:20 am 
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Burgo wrote:
Hi Konrad,
I'd like to give this a little nudge because I wouldn't mind seeing a bit of conversation on it.

....
Hi Burgo, another nudge. Are we riding a dead horse? :lol:
It is surprising that nobody is interested in this thread besides the two uf us, who can solve it already.

I find this even more surprising because I had another brief look at Superantoniovivaldi's tutorial at Youtube.
No offense meant, especially if he is one of the TP forum, but the tutorial does not seem to be easy to follow. He refers a lot to his known 3x3x4 tutorial but I could not find this on Youtube.
He is confusing himself sometimes ("Here is a parity, oh no it's not") and talks about "sandwich formation" without explaining it further. Maybe, it is clearer if I could view his 3x3x4 tutorial (I do not think he means the Crazy Witeden 3x3x4?)
I suspect that most people use a different method solving a 3x3x4? At least I do. (I solve it as two Dominos).
Anyway, his outline is clear:
1. Solve the outer 3x3x4 (do not care about the inner slices yet)
2. Solve the inner 3x3x4 slices (done in part 3, but I have not yet watched this)

I see now three different outlines:
A 1. inner 3x3x4
2. outer pieces
B 1. outer 3x3x4
2. inner slices
C 1. make an ordinary cuboid
2. reduce it to a 3x3x4 by pairing outer pieces with their correct neighbours
3. solve the 3x3x4

Maybe, Luke's "irritatingly easy" method is number four?
The common problem for all three methods is the fact that in most cases you cannot turn the 3x4 faces independent on the slice layer. If you look at a 4x5 face as F and have the 3x5 as U, you can turn (Rr)2 always, but not always r2 or R2 independently. So in methods A and B you have to come up with move sequences for moving outer pieces independent on slice pieces and vice versa. E.g. a pure 3-cycle for corners or inner edges (e.g. UrF I call an inner edge) is not so simple as on the 4x4x5.

I'm not sure if your method C is better here?
I have tried it, but I could not find simple ways to pair pieces. On a 4x4x4 or 5x5x5 reduction is simple, but still some parity handling is needed.
I think you cannot avoid parity handling by your method?

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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Fri Feb 03, 2012 5:55 pm 
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Konrad wrote:
The common problem for all three methods is the fact that in most cases you cannot turn the 3x4 faces independent on the slice layer. If you look at a 4x5 face as F and have the 3x5 as U, you can turn (Rr)2 always, but not always r2 or R2 independently. So in methods A and B you have to come up with move sequences for moving outer pieces independent on slice pieces and vice versa. E.g. a pure 3-cycle for corners or inner edges (e.g. UrF I call an inner edge) is not so simple as on the 4x4x5.
Hi Konrad,

I see what you are thinking here. I don't think you have the full idea of what I was implying, which is why we need the conversation, I think. I honestly thought at the start that this was obvious and that this was what everybody else would be doing (sorry if I appeared cryptic). I will elaborate a little on the steps of my method to give you the idea:

Reduce the cube to a 3x3x4 and then solve the 3x3x4:
For most situations I am picturing the puzzle with [4x5 F, 3x5 U]. Sometimes temporarily for a 3x3x4 sequence use [3x4 F, 3x5 U] but then return to the standard position.

1. Make the cuboid shape.
The reason I have reduced it to a cuboid shape is because then it is possible to turn both 3X4 outer layers and those turns are no longer blocked by the shape shifted puzzle.

2. Pair the inner slice 3x4 pieces with their matching outer layer 3x4 pieces.
I did this in a very similar way to how you might reduce a crazy 3x3 Jupiter. Because I'm right handed I kept the [3x4 Rr face] in the correct shape and used it as the `exchange face`, and I used the Ll face as the `store face`. First I did edges and then I did corners.

1. Edges
Edges can mostly be moved around intuitively with u2 slice turns or with the domino algos (Rr2 U2)X3 or (Rr2 U)X2 (Rr2 U2)X2 Rr2 U Rr2 U' Rr2. But they are exchanged to pair up with R2 turns. For the last few you may need to insert 2 opposing corret edges of the same colour and make the R2 turn.

2. Corners
Set up the puzzle so that it has opposing same coloured edges in Rr. Eg From a solved puzzle: Ll2 F2 U2, now you can see that a R2 turn will not disturb the solved states of the edges. Domino corner algos [(Rr2 U Rr2 U' Rr2) U' D (Rr2 U' Rr2 U Rr2) D' -Switch 2 corner groups FUR<>BUR] and repair algos for the edges (with respect to the reduction state) will allow corner solving with R2 turns. They will be exchanged in the opposing X pattern. It is not important to conserve the correctly shaped puzzle on the left.

For the last few it may be necessary if 3 corners remain: to place 2 solved corners in opposing positions (on this part / of the X) and to solve 1 and temporarily break the bond in the other 2 already paired corners. This allows for a setup to solve the last 2 by exchanging out the `newly solved one` with the `stored unsolved one` and doing an R2 turn (perhaps Konrad can explain this better, I would need to `show` it- sorry).

3. Solve the 3x3x4.
Solve the 3x3x4 by not making any R or L turns.

I hope this explanation is OK.

Cheers,
Burgo.

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1st 3x3 solve Oct 2010 (Even though I lived through the 80s).
PB 3x3 55sec Jan 2011 (When I was a kid 1:30 was speedcubing so I'm stoked).
1st 3x3 Earth (nemesis) solve Jan 2011 My You Tube (Now has ALLCrazy 3X3 Planets with Reduction)


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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Sat Feb 04, 2012 5:48 am 
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Burgo wrote:
...I see what you are thinking here. I don't think you have the full idea of what I was implying, which is why we need the conversation...
Hi Burgo, I think I understood exactly the general idea of your method. I had not looked into the details seriously, though.
I just wanted to express my doubts if the reduction method has big advantages compared with the other two methods (A and B as above).
Let us take a 3-cycle of corners as an example. (You will agree that this is sometimes necessary.)
I'm doing this by (F is 4x5; R is 3x4):

R2 F2 [U (Rr)2 U‘ (Ll)2]*2 F2 R2 F2 [(Ll)2 U (Rr)2 U‘]*2 F2 (22 moves)
It is a pure 3-cycle of corners URF -> ULB -> DRB -> URF
You sent via PM something similar (you talked about breaking 3 corners). (I'm sure you don't mind if I mention here this PM and the fact that it contained a certain move sequence.)

You described a sequence of 40 moves starting with a solved cube.
I had written in my last post:
Konrad wrote:
...So in methods A and B you have to come up with move sequences for moving outer pieces independent on slice pieces and vice versa. E.g. a pure 3-cycle for corners or inner edges (e.g. UrF I call an inner edge) is not so simple as on the 4x4x5.

I'm not sure if your method C is better here?
I have tried it, but I could not find simple ways to pair pieces. On a 4x4x4 or 5x5x5 reduction is simple, but still some parity handling is needed.
I think you cannot avoid parity handling by your method?
I should NOT have used the word parity :wink: , instead I should have spoken about 3-cycles needed in special situations.
This diagram shows 6 special situations that could come up in similar ways in the different methods A,B and C
Image
You can click on it to enlarge it.
EDIT: Please keep in mind that my diagrams resemble my physical puzzle which has the old colour scheme white opposite blue!
I'm not in the position to judge which method is shorter or better. Different people came up with different outlines. May each solver decide which one suits him /her best.
And it might be the case that something that is obvious and easy to you or me is not so obvious to others.
Luke has said the 3x4x5 is irritatingly easy. I'm still wondering if he does it by method number 4 or just thinks the 3x4x5 is so easy because it is so close to a 3x3x4.
At least for me, the three part tutorial of Superantoniovivaldi is NOT irritatingly easy to follow :lol:

For me it was not so easy to come up with something like the 22 move sequence quoted above:
R2 F2 [U (Rr)2 U‘ (Ll)2]*2 F2 R2 F2 [(Ll)2 U (Rr)2 U‘]*2 F2
Maybe I should analyze it briefly:
The part F2 [U (Rr)2 U‘ (Ll)2]*2 F2 (10 moves) is a 3-cycle of paired corners on both, the inner and the outer 3x3x4 URF -> ULF -> ULB -> URF.
F2 [(Ll)2 U (Rr)2 U‘]*2 F2 is the inverse sequence of this.
The first R2 puts the outer corner URB into the URF position.
The following 10 moves are the mentioned 3-cycle of paired corners.
R2 in move 12 puts the former ULB corner (just placed by the 3-cycle) to the URB position.
The last 10 moves just reverse the paired corner 3-cycle and we end up with URF -> ULB -> DRB -> URF
The diagram pair in the upper left corner of my picture shows this mirrored.

EDIT:
If you exchange R by r and U by u, you get a pure 3-cycle of slice pieces:
r2 F2 [u (Rr)2 u‘ (Ll)2]*2 F2 r2 F2 [(Ll)2 u (Rr)2 u‘]*2 F2
If you change U by u only, you'll get a 3-cycle of outer edge pieces
R2 F2 [u (Rr)2 u‘ (Ll)2]*2 F2 R2 F2 [(Ll)2 u (Rr)2 u‘]*2 F2

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Last edited by Konrad on Sat Feb 04, 2012 10:29 am, edited 1 time in total.

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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Sat Feb 04, 2012 7:23 am 
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I haven't watched any tutorials, but my method goes by this:

1) Solve 3x3x4 with outer pieces representing pieces for the 3x3x4.
2) Pair edges with two short algorithms.
3) Solve any unsolved centres using variants of the same algorithms.

One of the algorithms that I use is a God send. It can be applied to pretty much any cuboid, and the main variation I use is: Rw2 R2 B2 U2 Rw2 R2 U2 B2 Rw2 R2. If you apply this and variants of this algorithm to most cuboids (ones with at least one axis of even numbers are preferable), then you can solve many different kinds of pieces using the algorithm. Perhaps next weekend I'll produce a walk through video.

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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Sat Feb 04, 2012 8:10 am 
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Hi Luke & Konrad,

Welcome to the conversation Luke :)
Sorry for my ignorance, what is Rw2 and the name of the notation?

Thanks for sharing your algorithm Konrad.

I will include the sequence I gave Konrad in my PM. I didn't want to confuse people too much with it because it's not a sequence that I would `use`, it's just an example of `how I might link things` and be generally treating the puzzle during the corners reduction, rather than to be giving a short sequence.

Burgo wrote:
From a solved position and 4X5 as F, 3X5 as U perform:

Ll2 U2 F2 -Place edges in exchange face (the consistent setup position for corners)
R2 -exchange 4 corners (I will conserve URB & DRF)
U2 -setup
(Rr2 U Rr2 U' Rr2) U' D (Rr2 U' Rr2 U Rr2) D' -Switch 2 corners FUR<>BUR
U -setup
(Rr2 U Rr2 U' Rr2) U' D (Rr2 U' Rr2 U Rr2) D' -Switch 2 corners FUR<>BUR
(Rr2 U2)X3 –edge fix
U –setup
R2 –return first 2 corners and break 1 more = 3 corners broken


(Rr2 U Rr2 U' Rr2) U' D (Rr2 U' Rr2 U Rr2) D' U is a corner 3cycle with a U turn (left undone):so this can be used to make a more efficient 3cycle of corner groups, if you feel the need to make a complete algorithm. But the beauty of the reduction method used like this is that you don't need to learn any other sequences other than the 3x3x4 sequences you already use. I just find it intuitive to trade out some corner groups, repair some edges and then do an R2, for this step.

Cheers,
Burgo.

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1st 3x3 solve Oct 2010 (Even though I lived through the 80s).
PB 3x3 55sec Jan 2011 (When I was a kid 1:30 was speedcubing so I'm stoked).
1st 3x3 Earth (nemesis) solve Jan 2011 My You Tube (Now has ALLCrazy 3X3 Planets with Reduction)


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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Sat Feb 04, 2012 8:22 am 
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Rw2 means both layers of the side. So to do a checkerboard pattern on a 5x5x5, one could use: Rw2 R2 Lw2 L2 Uw2 U2 Dw2 D2 Fw2 F2 Bw2 B2. The term "w" is part of the WCA standard notation.

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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Sat Feb 04, 2012 9:05 am 
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Rw = (Rr)
When Burgo writes Rr2 he means R2 r2 or Rw2 or as I write it (Rr)2.
Thanks Luke for sharing your move sequence.
I'll look at it later.

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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Sat Feb 04, 2012 9:28 am 
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Luke wrote:
Rw2 R2 B2 U2 Rw2 R2 U2 B2 Rw2 R2
Hi Luke,
It's a variant of (R2 U2)X3 It's effectively that with a U2 turn at the end so that it `appears` to affect corners in a 2+2 swap (different E layer edges switch out too).
I look forward to seeing your video. Unfortunately I don't have time to do videos at the moment.
Cheers,
Burgo.

_________________
1st 3x3 solve Oct 2010 (Even though I lived through the 80s).
PB 3x3 55sec Jan 2011 (When I was a kid 1:30 was speedcubing so I'm stoked).
1st 3x3 Earth (nemesis) solve Jan 2011 My You Tube (Now has ALLCrazy 3X3 Planets with Reduction)


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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Sat Feb 04, 2012 10:00 am 
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I'm not sure what you mean. Try it on a 4x4x4 and see what happens.

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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Sat Feb 04, 2012 11:14 am 
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Luke wrote:
I haven't watched any tutorials, but my method goes by this:

1) Solve 3x3x4 with outer pieces representing pieces for the 3x3x4.
2) Pair edges with two short algorithms.
3) Solve any unsolved centres using variants of the same algorithms.

One of the algorithms that I use is a God send. It can be applied to pretty much any cuboid, and the main variation I use is: Rw2 R2 B2 U2 Rw2 R2 U2 B2 Rw2 R2. If you apply this and variants of this algorithm to most cuboids (ones with at least one axis of even numbers are preferable), then you can solve many different kinds of pieces using the algorithm. Perhaps next weekend I'll produce a walk through video.
This is a simple sequence indeed. If I'm not interpreting something incorrectly it just does a double swap of four `slice` pieces.
Image
I achieve the same result by Lw2 B2 U2 l2 U2 B2 Lw2.
I would write your sequence as r2 B2 U2 r2 U2 B2 r2 and I count 7 moves as in my sequence.
Your notation resembles more what we are actually doing.

This one sequence does not save the day of a solver. :wink: How do you do 3-cycles of slice pieces? (I need some in almost any solve.)

BTW has anybody thought about naming the pieces?
Using the paradigm of an inner 3x3x4 and an outer 3x3x4 we could name the slice pieces:
- inner corner (a corner of the inner 3x3x4
- inner edges (an edge of the inner 3x3x4) equal to a corner of the innermost 3x3x2 Domino.
- inner centres (only four of them are visible)
Respectively we would talk about outer corners, edges, centres.
Still, we would have sometimes the need to differentiate between different edge types.

EDIT:
Burgo wrote:
Luke wrote:
Rw2 R2 B2 U2 Rw2 R2 U2 B2 Rw2 R2
Hi Luke,
It's a variant of (R2 U2)X3 It's effectively that with a U2 turn at the end so that it `appears` to affect corners in a 2+2 swap (different E layer edges switch out too).
I look forward to seeing your video. Unfortunately I don't have time to do videos at the moment.
Cheers,
Burgo.
I do not understand this: (R2 U2)x3 swaps 10 pieces (besides the uneven rotation of centre pieces), Luke's sequence affects only four. Why do you say that it is comparable?

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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Sat Feb 04, 2012 11:34 am 
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Do you know an algorithm that can give me a 3 cycle of edges? I rarely seem to get situations like that (out of the twenty or so solves I've done, I've only had a 3 cycle once, and I didn't keep track of what I did). If there is a way I can easily get into a 3 cycle I'll show you what I do then (it may be part of the other algorithm I use, which is all the more similar to the (R2 U2)x3 case).

Also, I never really like to use the term "r2". I always use to interpret that as Rw2, and so I don't like using small letters on even layered cubes.

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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Sat Feb 04, 2012 12:21 pm 
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Luke wrote:
Do you know an algorithm that can give me a 3 cycle of edges? I rarely seem to get situations like that (out of the twenty or so solves I've done, I've only had a 3 cycle once, and I didn't keep track of what I did). If there is a way I can easily get into a 3 cycle I'll show you what I do then (it may be part of the other algorithm I use, which is all the more similar to the (R2 U2)x3 case).

Also, I never really like to use the term "r2". I always use to interpret that as Rw2, and so I don't like using small letters on even layered cubes.

A 3-cycle of 3x5 wedges (or "corners" of the inner 3x3x4):
r2 F2 [U (Rr)2 U' (Ll)2]*2 F2 r2 F2 [(Ll)2 U (Rr)2 U']*2 F2
or as in WCA too:
r2 F2 [U Rw2 U' Lw2]*2 F2 Rw2 R2 F2 [Lw2 U Rw2 U']*2 F2
The result on a solved 3x4x5:
Image
I'm confused by
Quote:
I never really like to use the term "r2". I always use to interpret that as Rw2
r is the name of the inner slice below R. This is WCA too. Rw2 ("R wide"2) is not equal to r2 but R2 r2.
I have interpreted your Rw as in WCA
Quote:
Double Outer Slice Moves (outer slice plus adjacent inner slice):
•12a5) Clockwise, 90 degrees: Fw, Bw, Rw, Lw, Uw, Dw. (see 12a1).
•12a6) Counter clockwise, 90 degrees: Fw', Bw', Rw', Lw', Uw', Dw' (see 12a5).
•12a7) Clockwise, 180 degrees: Fw2, Bw2, Rw2, Lw2, Uw2, Dw2 (see 12a5).
•12a8) Counter clockwise, 180 degrees: Fw2', Bw2', Rw2', Lw2', Uw2', Dw2' (see 12a5).

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Last edited by Konrad on Sat Feb 04, 2012 1:03 pm, edited 1 time in total.

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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Sat Feb 04, 2012 12:37 pm 
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Konrad, what I meant was when I was a beginner the tutorials I watched used the term "r" as "Rw". For some reason that has still stuck with me, hense why I don't like to use it, because chances are that I'm using it incorrectly. :lol:

As for the 3 cycle... Interesting. I have only ever had that case once (I think), and I didn't keep track on how I solved it. I shall ponder on it more for now.

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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Sat Feb 04, 2012 5:27 pm 
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Anyone know an algorithm to swap...

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these pieces?


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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Sat Feb 04, 2012 5:36 pm 
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Luke wrote:
3) Solve any unsolved centres using variants of the same algorithms.

One of the algorithms that I use is a God send. It can be applied to pretty much any cuboid, and the main variation I use is: Rw2 R2 B2 U2 Rw2 R2 U2 B2 Rw2 R2. If you apply this and variants of this algorithm to most cuboids (ones with at least one axis of even numbers are preferable), then you can solve many different kinds of pieces using the algorithm.

Burgo wrote:
It's a variant of (R2 U2)X3 It's effectively that with a U2 turn at the end so that it `appears` to affect corners in a 2+2 swap (different E layer edges switch out too).

Konrad wrote:
I do not understand this: (R2 U2)x3 swaps 10 pieces (besides the uneven rotation of centre pieces), Luke's sequence affects only four. Why do you say that it is comparable?
Hi Konrad and Luke,

Luke's original post included a disclaimer that said he used it in variants (various positions, different slices and different cuboids) as an all rounder algorithm.

In my solve it is comparable to (R2 U2)X3 and how I would use that. I use [3x4 U, 4x5 F] or [3x4 U, 3x5 F]: (R2 Uu2)X3 to move edges around easily after I have made a corner exchange decision, for example.

You would have to add (or take away) a U2 turn [(R2 U2)X3 U2] to make it resemble it (or add a U2 to the end of his). Moved to the same inner slice:
r2 B2 U2 r2 U2 B2 r2 U2
r2 U2 r2 U2 r2 (U2)
You can see that (R2 U2)X3 moves centres (I thought this is possibly what he means in part 3 of his solve where he uses variants to move centres). You will see if he moves his algo to the outer layer variant R2 B2 U2 R2 U2 B2 R2 U2 the pieces that are being hidden by the slice turns on this cuboid are being revealed, and the similarities to (R2 U2)X3 are more obvious.

Solving-wise, I think they function similarly and are variants of the same type of thing.

Cheers,
Burgo.

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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Sat Feb 04, 2012 5:52 pm 
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Hi Vykeria,

Variants of (R2 U2)X3
Like: [3x4 U, 4x5 F] or [3x4 U, 3x5 F]: (R2 Uu2)X3
Or: [3x4 U, 4x5 F] or [3x4 U, 3x5 F]: (R2 U2)X3
depending on the depth.
They will move a few other pieces, but looking at what I can see of the stage of your solve it may not matter?

Cheers,
Burgo.

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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Sat Feb 04, 2012 6:01 pm 
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Okay new problem. It's all solved except 3 edge pieces on the 3x3 side that need to be cycled.

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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Sat Feb 04, 2012 6:07 pm 
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Vykeria wrote:
Anyone know an algorithm to swap...

Image

these pieces?
(F is 3x4; U is 3x5) (F2 R2)x3 L2 D2 L2 (F2 U2)x3 L2 D2 L2

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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Sat Feb 04, 2012 6:16 pm 
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I would be stretching here, but it's possible to try [3x4 U, 3x5 F]: (R2 U2)X3 and throw in an F2 turn depending on your needs and come up witha 3 cycle. You are going to throw out your (5) edges, but it is possible to keep exchanging them back with thoughtful setups.
Luke's algo: R2 B2 U2 R2 U2 B2 R2 U2 is a variant that moves different edges, it's possibly better for this. You may end up in a 3x3x4 parity in which you solve it the same way as the 3x3x4. But that's how I would approach it from there. Basically with these types of algos, if it doesn't work, you can just repeat the algo and try something else.

I have paired those already in my reduction method.


EDIT: yeah, I just did it with Luke's algo: [3x4 U, 3x5 F] (R2 B2 U2 R2 U2 B2 R2 U2) F2 (R2 B2 U2 R2 U2 B2 R2 U2) F2
Cheers,
Burgo.

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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Sat Feb 04, 2012 6:31 pm 
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I tried Luke's algorithm Burgo, and it didn't seem to work. I played around more and got it to the same situation but on the 4x4 edges instead of the 3x3 edges.


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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Sat Feb 04, 2012 6:34 pm 
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Konrad wrote:
Vykeria wrote:
Anyone know an algorithm to swap...
these pieces?
(F is 3x4; U is 3x5) (F2 R2)x3 L2 D2 L2 (F2 U2)x3 L2 D2 L2
Nice work Konrad, very smooth.
It also works with [F 3x5, U 3x4] (R2 U2)X3 y' (R2 B2 U2 R2 U2 B2 R2 U2) y

Quote:
I tried Luke's algorithm Burgo, and it didn't seem to work. I played around more and got it to the same situation but on the 4x4 edges instead of the 3x3 edges.
That would be because I made a typo and gave the wrong face to use as F, I quickly changed it but it was too late, you must have read it.. (you can do it on any of the faces for the same 3cycle, hold the solved edge on BD). Just repeat it 2 more times and the edges will go back :) .

Cheers,
Burgo.

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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Sun Feb 05, 2012 4:34 am 
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Thanks for creating this topic! It isn't a dead horse to me, and I found the 3x4x5 much more confusing than the 4x4x5, mainly because I am so used to finding sequences that can use 90 degree moves without changing the shape of the puzzle and hence restricting the range of later possible moves. The first few times I solved the 3x4x5 I didn't have a method and I didn't really know what I was doing. Here is the method I use now:

1. Restore the cuboid shape. This can be done by holding the puzzle with [U&D = 3x5] and permuting edges and corners of the inner 3x3x4 like a double-decker (inner and outer) Domino/2x3x3.

2. Solve the corners and the 3x5 edges adjacent to the corners relative to the center pieces of the 3x5 faces. I start intuitively and finish with cycles. If a 3-cycle is needed, I use a variation of my edge-cycling algorithm for higher cubes, with three setup moves to avoid losing the cuboid shape:

[F = 4x5, U = 3x4] (B² R B²) [(Uu)² L (Uu)² L' (Uu)², R'] (B² R' B²)

Or in shorter notation, [B² : R] : [(Uu)² L : (Uu)², R']

This algorithm cycles 3 corner-edge pairs. Commutating with U2 gives a 3-cycle of corners only. Sometimes just two corners need to be swapped; diagonal corners of a 3x5 face can be swapped the same way as with the last layer of a Domino/2x3x3.

3. Solve the rest of the puzzle using 180 degree moves only. Again I start intuitively and finish with cycles or double swaps of groups of pieces or single pieces (groups of "oblique" pieces, or single "T-junction" pieces, if we think of the different types of center cubies in the faces of higher cubes). I use two types of sequence:

[A C B C]*2 and [A B]*3

A and B are an outer layer or an inner layer or both together, and C is an outer layer.

--------------------------------------------------------

Reducing to a 3x3x4 is an interesting idea and I will try it next time.

Edit -- forgot about the case of two swapped corners after restoring the cuboid shape, so added this to my outline. Also added details on the initial cuboid-restoring stage and removed an unnecessary stage following that.


Last edited by Julian on Sun Feb 05, 2012 6:26 pm, edited 1 time in total.

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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Sun Feb 05, 2012 6:29 am 
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Julian wrote:
Thanks for creating this topic! It isn't a dead horse to me, and I found the 3x4x5 much more confusing than the 4x4x5....
Thanks for your contribution, Julian. I'm glad to hear from somebody like you that you found it confusing as well :) (I will look at your post tomorrow, because I will be away from my computer for the rest of the day).

Here are a bunch of move sequences handling edges (they should be sufficient if beginners scramble the 3x4x5 by 180 degree turns only):
Image
You can click on the picture to enlarge it a bit.
I hope I have not mad any typos. Please correct me if necessary.

@Vykeria: Nr 8 is a 3-cycle of 3x4 middle edges. I guess that was your last problem.

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Last edited by Konrad on Mon Mar 19, 2012 8:03 am, edited 2 times in total.

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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Sun Feb 05, 2012 10:17 am 
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Konrad wrote:
Here are a bunch of move sequences handling edges (they should be sufficient if beginners scramble the 3x4x5 by 180 degree turns only)... I hope I have not made any typos. Please correct me if necessary.

@Vykeria: Nr 8 is a 3-cycle of 3x4 middle edges. I guess that was you last problem.
I couldn't get the 8th algo to work. I do that cycle with [F² R² U² R²]*2. Your 9th/last algo is really clever!


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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Mon Feb 06, 2012 1:37 am 
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Hi Julian,
I couldn't get your sequence to work {[F = 4x5, U = 3x4] (B² R B²) [(Uu)² L (Uu)² L' (Uu)², R'] (B² R' B²),} I put a picture of what I got.
I look forward to seeing what you think of reduction.
Konrad, yes your 8th sequence moves 4 corners.
I tried on both to find a small fix but my searching was in vain :( .

Konrad wrote:
BTW has anybody thought about naming the pieces?
Using the paradigm of an inner 3x3x4 and an outer 3x3x4 we could name the slice pieces:
- inner corner (a corner of the inner 3x3x4
- inner edges (an edge of the inner 3x3x4) equal to a corner of the innermost 3x3x2 Domino.
- inner centres (only four of them are visible)
Respectively we would talk about outer corners, edges, centres.
Still, we would have sometimes the need to differentiate between different edge types.
Are we going with this then, so we say inner edge 3cycle FUR>BUL>FDL and such, and refer to the puzzle in 4x5 F, 3x5 U as the standard position?

Cheers,
Burgo.


Attachments:
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345 b.jpg [ 66.01 KiB | Viewed 7604 times ]

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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Mon Feb 06, 2012 4:12 am 
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Burgo wrote:
Hi Julian,
I couldn't get your sequence to work {[F = 4x5, U = 3x4] (B² R B²) [(Uu)² L (Uu)² L' (Uu)², R'] (B² R' B²)} I put a picture of what I got.
I used commutator notation for the middle part to save space, because it's 5 moves, a single move, then the inverse of the 5 moves, then the inverse of the single move. I probably shouldn't do that outside the Gelatinbrain thread without an explanation, so sorry for any confusion. In full, writing out every move, the sequence is:

{[F = 4x5, U = 3x4] (B² R B²) (Uu)² L (Uu)² L' (Uu)² R' (Uu)² L (Uu)² L' (Uu)² R (B² R' B²)

Burgo wrote:
Konrad, yes your 8th sequence moves 4 corners.
I tried on both to find a small fix but my searching was in vain :( .
[F² R² U² R²]*2 works for the 8th pattern.


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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Mon Feb 06, 2012 4:58 am 
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Thanks for the clarification Julian, at least now I know what the comma means, I thought it was a typo :) .

[3x4 U, 3x5 F][F² R² U² R²]*2 is much shorter than what I had^^ [3x4 U, 3x5 F][(R2 B2 U2 R2 U2 B2 R2 U2) F2]X2
Cheers,
Burgo.

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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Mon Feb 06, 2012 5:34 am 
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Julian wrote:
...I couldn't get the 8th algo to work. I do that cycle with [F² R² U² R²]*2. Your 9th/last algo is really clever!
Sorry for the typo :oops: and thanks for the hint :) : the x2 needed to be x3 (corrected above).
I reused some trivial move sequence, yours is much shorter.

EDIT:
Julian wrote:
...[F = 4x5, U = 3x4] (B² R B²) [(Uu)² L (Uu)² L' (Uu)², R'] (B² R' B²)

Or in shorter notation, [B² : R] : [(Uu)² L : (Uu)², R']

...
If we hold the puzzle with [F =4x5; U =3x4] the sequence becomes:
B2 D B2 [(Rr)2 U (Rr)2 U’ (Rr)2 D’ (Rr)2 U (Rr)2 U’ (Rr)2 D] B2 D’ B2
or in shorter notation
[B2 : D] : [[ (Rr)2 U : (Rr)2 ], D’]
The result is:
Image
EDIT2: Regarding names:
Obviously, my naming proposal was a bit confusing.
Therefore I would like to propose something that is closer to the physical appearance of the puzzle.
We perceive F= 4x5; U=3x5 as the standard position.
• As on other puzzles we name pieces with one sticker centres.
• Pieces with two stickers are edges .
• Pieces with three stickers are corners
All piece types are enumerated by the following picture. Each piece type is shown once with 1,2,3 letters / numbers on its stickers.
Image
We call the slice layer between l and r M and the inner layer between F and B we call S.
There are 5 centre types:
• 1 = 4x5 middle centre (FMu)
• 2 = 4x5 x-centre (Fru)
• 3 = 3x5 middle centre (UMS)
• 4 = 3x5 outer centre (UrS)
• 5 = 3x4 centre (RdS)
There are 4 edge types
• 6 = 4x5 middle edge (FUM)
• 7 = 3x5 wedge (wider edge) (FUr)
• 8 = 3x4 middle edge (URS)
• 9 = 3x4 wedge (FRu)
and obviously one corner type c (URF).

I do not assume that we need such names very often if we go on in this conversation, but the names could be used if necessary.

Still unsolved: Who do we name a position in space that does not exist on the standard cuboid form.
E.g. where does piece 8 arrive after a 90 degree U turn?
We could use f instead of F corresponding to a 5x5x5 and name the position in former empty space FUm.

Let's assume we do not this latest level of detail at all.

EDIT3: I have corrected some mistakes. Please, find an improved text below.

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Last edited by Konrad on Tue Feb 07, 2012 10:38 am, edited 2 times in total.

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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Tue Feb 07, 2012 2:01 am 
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Konrad wrote:
Still unsolved: Who do we name a position in space that does not exist on the standard cuboid form.
And even better, what do we call piece 8 when it is not there?? :shock: :lol: Come to think of it more seriously though, if you used the original idea of the 3x3x4 analogy it could account for the outer slices being out of shape on F,R,B or L. You could just refer to any 3x3x4 inner slice piece in the lower case (f,r,b,l) and any 3x3x4 outer slice in capitals (F,R,B,L), U&u and D&d layers still as normal? Then you could name any piece even if it was shape shifted.

With a system like that you could even talk about turning f instead of F (or with 3x4 F, 3x5U R would really be r), that's realistically what it would be, but I don't know if that would catch on (although you could use it and it wouldn't be misinterpreted, I don't think, because if we use the M&S layers terminology, what else could it be?).

I think for consistency you should probably use `S` (standing layer) instead of `e` which is confusing because it is like (E) equatorial layer (especially since you used M).

Cheers,
Burgo.

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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Tue Feb 07, 2012 5:11 am 
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Burgo wrote:
Konrad wrote:
Still unsolved: Who do we name a position in space that does not exist on the standard cuboid form.
And even better, what do we call piece 8 when it is not there?? :shock: :lol: Come to think of it more seriously though, if you used the original idea of the 3x3x4 analogy it could account for the outer slices being out of shape on F,R,B or L. You could just refer to any 3x3x4 inner slice piece in the lower case (f,r,b,l) and any 3x3x4 outer slice in capitals (F,R,B,L), U&u and D&d layers still as normal? Then you could name any piece even if it was shape shifted.

With a system like that you could even talk about turning f instead of F (or with 3x4 F, 3x5U R would really be r), that's realistically what it would be, but I don't know if that would catch on (although you could use it and it wouldn't be misinterpreted, I don't think, because if we use the M&S layers terminology, what else could it be?).

I think for consistency you should probably use `S` (standing layer) instead of `e` which is confusing because it is like (E) equatorial layer (especially since you used M).

Cheers,
Burgo.
The problem with the inner 3x3x4 analogy is the contradiction between logical 'corners` (piece type 7) and physical edges. Even you have talked about edges. I think the idea with f instead of F or r instead of R (for a 4x5 face) makes the naming consistent because it reflects the underlying 5x5x5.
The e for naming the S layer was a plain mistake :oops:

I rewrite the text above:

We look at the puzzle as if it were a 5x5x5 puzzle with some missing layers!
We will use WCA notation.
We look at the front 4x5 face and have the 3x5 face on top and call this the `standard position`.
We imagine the `original` 5x5x5 and see that the F and B faces and additionally the layer between u and d (in WCA called `E`) are missing.
We recognize that after a 90 degree U turn several pieces arrive in the former empty F layer.

We continue to use the letters F and B for face turns and make them synonyms for f and b.
(BTW, if you ever simulate move sequences for the 3x4x5 on a 5x5x5, be careful to turn F and f together.)

• As on other puzzles we name pieces with one sticker centres.
• Pieces with two stickers are edges .
• Pieces with three stickers are corners
All piece types are enumerated by the following picture. Each piece type is shown once with 1,2,3 letters / numbers on its stickers.

Image

We call the slice layer between l and r M and the inner layer between f and b we call S.
There are 5 centre types:
• 1 = 4x5 middle centre (fMu)
• 2 = 4x5 x-centre (fru)
• 3 = 3x5 middle centre (UMS)
• 4 = 3x5 outer centre (UrS)
• 5 = 3x4 centre (RdS)
There are 4 edge types
• 6 = 4x5 middle edge (fUM)
• 7 = 3x5 wedge (wider edge) (fUr)
• 8 = 3x4 middle edge (URS)
• 9 = 3x4 wedge (fRu)
and obviously one corner type c (URf).

I do not assume that we need such names very often if we go on in this conversation, but the names could be used if necessary.

Where does piece 8 arrive after a 90 degree U turn?
Corresponding to a 5x5x5 we will name the position in former empty space FUM.

We could use the term `corners of the inner 3x3x4` for piece type 7 as well (synonym for 3x5 wedges).

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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Tue Feb 07, 2012 4:03 pm 
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Julian wrote:
...1. Restore the cuboid shape. ...
2. Solve the corners and the 3x5 edges adjacent to the corners relative to the center pieces of the 3x5 faces. ...
3. Solve the rest of the puzzle using 180 degree moves only. ...

Hi Julian, I think there is some step missing in your outline.
This picture shows a legal state after your step 2, right?
Image

Done by l2 F2 [u‘ Ll)2 u (Rr)2]x2 F2 l2
F2 [(Rr)2 u‘ (Ll)2 u]x2 F2 on a solved 3x4x5.

We have the cuboid shape, all 8 corners are solved and all `3x5 edges adjacent to the corners`.
I cannot see how this pattern could be solved in step 3 by 180 degree turns only.

I assume you want to add to step 2 that the inner 3x3x2 Domino should be solved (or should be in a pattern solvable by 180 degree turns)?
I seem to remember something like this in your original post.

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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Tue Feb 07, 2012 7:19 pm 
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Konrad wrote:
I assume you want to add to step 2 that the inner 3x3x2 Domino should be solved (or should be in a pattern solvable by 180 degree turns)?
I seem to remember something like this in your original post.
Yes, you're right. Both piece types 2 and 9 in your earlier diagram can get into situations where 90 degree turns are needed to solve them, and the extra step you mention above takes care of that. I am starting to like my method less and less, and reduction to 3x3x4 is seeming more and more attractive!


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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Wed Feb 08, 2012 1:47 am 
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Julian wrote:
reduction to 3x3x4 is seeming more and more attractive!
The more I solve it, the more I am starting to like this:

[F = 4x5, U = 3x5][(Rr2 U Rr2 U' Rr2) U' D (Rr2 U' Rr2 U Rr2) D' U]X2

for the wedge-corner-group pure 3cycle [UFL>UBR>ULB] it's 28 moves, but it's using my domino & 3x3x4 method and it returns edges and therefore sets up for the R2 turn better. Most of the time you can spot the `type of parity- when odd corner swapping is required` at the 5 corner stage and just do a [swap of 4 unsolved corners that results in only one paired] back there- and that's what I was doing before and tried to show^^, but `this` works well as an alternative and is probably more attractive.

A reverse sequence would require an edge fix and that is fairly long as well, because it's the adjacent 2 edge one [(Rr2 U)X2 (Rr2 U2)X2 Rr2 U Rr2 U' Rr2], so I'm finding a double application OK in practice.

Julian's equivalent [DFL>UFR>UBL] in the standard position: [F = 4x5, U = 3x5]
applied as: [F = 4x5, U = 3x4][(B² R B²) (Uu)² L (Uu)² L' (Uu)² R' (Uu)² L (Uu)² L' (Uu)² R (B² R' B²)]
is shorter at 18 moves.

EDIT: Come to think of it, I just looked at Julian's sequence more closely, and it suits my purposes with a different setup:
[F = 4x5, U = 3x5][ (Ll2 D2) (Rr2 U Rr2 U' Rr2) D' (Rr2 U Rr2 U' Rr2) D' D (D2 Ll2) ] = [ UBL>UFR>UFL] 15 moves, 13 moves without (Ll2) :D

or (D2) [(Rr2 U Rr2 U' Rr2) D']X2

Cheers,
Burgo.

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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Wed Feb 08, 2012 8:13 am 
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Burgo wrote:
....I am starting to like this:

[F = 4x5, U = 3x5][(Rr2 U Rr2 U' Rr2) U' D (Rr2 U' Rr2 U Rr2) D' U]X2

for the wedge-corner-group pure 3cycle [UFL>UBR>ULB] it's 28 moves, but it's using my domino & 3x3x4 method and it returns edges and therefore sets up for the R2 turn better....
As you said this is a (clockwise) 3-cycle of corner-wedge pairs.
This is the result on a solved 3x4x5 with F = 4x5 white U= 3x5 green R= orange B = blue L = red D = yellow.
I'll try to show this in a shapeshifted net diagram:
Image
You could achieve the same result by these 10 moves:
D2 [(Ll)2 U' (Rr)2 U]x2 D2
Usually I would turn the whole puzzle by 180 degrees and have the D face as F face:
F2 [(Rr)2 U' (Ll)2 U]x2 F2
I find this move sequence extremely useful on all my cuboids.
It's logic is the base of all my pure 3-cycles for single pieces (shown above).

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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Wed Feb 08, 2012 9:06 am 
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Hi Konrad,

*all algs using standard position [F=4X5, U=3X5]

Thanks for showing that.
To work with reduction I'm looking to keep (one 3x4 outer layer) `say`the Rr face/slice non-shape shifted.
D2 [(Ll)2 U' (Rr)2 U]x2 D2 with a different setup this is achieveable:
So something like F2 D (Rr2 U Ll2 U')X2 D' F2 or U' F2 (Rr2 U' Ll2 U)x2 F2 U would be more like it.

Did you see what I did with Julian's alg in my edit, that's quite efficient and quite nice (ergonomic- I like using the Rs & Us):
(D2) [(Rr2 U Rr2 U' Rr2) D']X2
It's using the commutator on D that is a bit flexible.

That post ended up as a sort of `think out` loud post with the final result at the end. I decided to leave it all in tact because you never know what someone else might like.

EDIT: Here's the thing with reduction though: it's nice to use the `switch 2 corners and repair edges` most of the time, because it doesn't matter if 1/2 of the cube begins to shape shift, but you need the other 1/2 intact. With these 3cycles, they are requiring the cube to start in a non-shape shifted form, so you have to potentially repair the shape and then apply it.. so a shorter alg can end up longer anyway.

I'm yet to decide on the most efficient way.

Cheers,
Burgo.

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PB 3x3 55sec Jan 2011 (When I was a kid 1:30 was speedcubing so I'm stoked).
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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Wed Feb 08, 2012 4:18 pm 
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It should be noted that SuperAntonioVivaldi has revisted his tutorials for the 3x4x5 since it has been mass produced.

Part 1 (Basic 3x3x4 parts): http://youtu.be/KOIhylQUNtg

Part 2 (Final edges): http://youtu.be/AXnL-7aMN7o

His second part is a bit of a long method, but it works with one algorithm and a slight variation. If you're looking for quicker algorithms, Konrad seems to have it down.

Jack

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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Thu Feb 09, 2012 4:48 am 
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I was just scrambling mine and I got this situation. The outer 3X4 layer is locked and will not turn. The puzzle may have started with the orientation the other way, but I still can't see why this turn is not possible.. It should be possible, right? Nothing seems to be out of alignment in the mechanism, and all other turns are possible :? .

EDIT: I just had a look at the mechanism and it is physically blocked, (2nd photo added).. it seems that it is not as `fully functional` as I thought.

If you want to try it you can do [4x5 F, 3x5 U] Dd Rr2 Uu2 R (With one 3X5 face as U it is blocked, with the other it is functional :o ).
Cheers,
Burgo.
Attachment:
345.jpg
345.jpg [ 135.62 KiB | Viewed 7217 times ]
Attachment:
345 2.jpg
345 2.jpg [ 75.31 KiB | Viewed 7201 times ]

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PB 3x3 55sec Jan 2011 (When I was a kid 1:30 was speedcubing so I'm stoked).
1st 3x3 Earth (nemesis) solve Jan 2011 My You Tube (Now has ALLCrazy 3X3 Planets with Reduction)


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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Thu Feb 09, 2012 7:42 am 
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Burgo wrote:
I was just scrambling mine and I got this situation. The outer 3X4 layer is locked and will not turn. ...
Now this interesting indeed. TomZ's has exactly the same behaviour :roll: I had such blocking only once and thought that it was a temporary thing.
As the 3x4x5 from Tom and from mf8 have a different mechanism, there must be a common underlying problem.
It would be interesting to get feedback from the designers (I recollect that Steryne, Traiphum and Rouricht have built there own custom cuboids.)

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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Fri Feb 10, 2012 2:14 am 
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What's really bizzare to me is that the whole puzzle mechanism in general appears symetrical, but as I have shown with that little sequence above: if you flip the puzzle `upside down` there is no blocking! If you open the centre caps as I did in that photo you will see that the mechanism clashes and doesn't permit the turn.

It's not a problem that I would likely get solving, that's why I think I only came across the situation during a scramble.

It's very interesting to know that Tom's puzzle has it too, I presume they all will.

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PB 3x3 55sec Jan 2011 (When I was a kid 1:30 was speedcubing so I'm stoked).
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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Fri Feb 10, 2012 4:26 am 
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Hi together,

Burgo, I saw your pictures. Perhaps I want to order this puzzle. Hmm. What about the lockings. Is it possible to solve each scrambled cube ( block) without disassembling ?

Cheers,
Andrea


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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Fri Feb 10, 2012 4:39 am 
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My puzzle had that too. When it locks up, the hidden edges are the wrong way, bandaging the outer layer.
If the hidden edges were symmetrically cut, would it be possible for them to get out of alignment, or get to awkward positions inside the puzzle?

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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Fri Feb 10, 2012 4:59 am 
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Andrea wrote:
Perhaps I want to order this puzzle. Hmm. What about the lockings. Is it possible to solve each scrambled cube ( block) without disassembling ?
Hi Andrea,
It is a very unusual situation for the puzzle and I only found it by a fluke. I have solved the puzzle many times and haven't come across it before. It is not a reason to boycott the puzzle at all, if you got the situation you could get around it with a few simple twists. Konrad said Tom's version has it as well. It's likely you might not even notice it unless you try to find it.

Here is a closer look at the situation, and the inversion of the internal edges:


Attachments:
345 3.jpg
345 3.jpg [ 109.69 KiB | Viewed 7084 times ]

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1st 3x3 solve Oct 2010 (Even though I lived through the 80s).
PB 3x3 55sec Jan 2011 (When I was a kid 1:30 was speedcubing so I'm stoked).
1st 3x3 Earth (nemesis) solve Jan 2011 My You Tube (Now has ALLCrazy 3X3 Planets with Reduction)
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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Sat Feb 11, 2012 11:10 am 
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Burgo wrote:
....Did you see what I did with Julian's alg in my edit, that's quite efficient and quite nice (ergonomic- I like using the Rs & Us):
(D2) [(Rr2 U Rr2 U' Rr2) D']X2
It's using the commutator on D that is a bit flexible.

That post ended up as a sort of `think out` loud post with the final result at the end. I decided to leave it all in tact because you never know what someone else might like.....

[F = 4x5, U = 3x5][ (Ll2 D2) (Rr2 U Rr2 U' Rr2) D' (Rr2 U Rr2 U' Rr2) D' D (D2 Ll2) ] = [ UBL>UFR>UFL] 15 moves
Yes I had seen this, but I did not have the time to respond earlier. (I wrote in my PM why :) )
It is interesting how relatively small adaptations can produce interesting results.
E.g. I'm using R2 U R2 D' R2 U R2 U' R2 U' D R2 U' (13 moves) since ages for my cuboids.
Now I've seen from you R2 U R2 U' R2 U' D R2 U' R2 U R2 D' which is exactly the same move sequence just with a different starting point (my turn # 5 is your turn # 1).
The result is similar, but the edges move differently.

My 10 moves F2 [U (Rr)2 U' (Ll)2 ]x2 F2 would produce exactly the same result as the mentioned 15 moves
[ (Ll2 D2) (Rr2 U Rr2 U' Rr2) D' (Rr2 U Rr2 U' Rr2) D' D (D2 Ll2) ]
and leave (Rr) in cuboid shape.
Image

As the main part [U (Rr)2 U' (Ll)2 ]x2 does a pure 3-cycle of corner- wedge pairs DFR -> ULB -> DLF -DFR you can get variations too e. g. by U (Rr)2 U2 [U (Rr)2 U' (Ll)2 ]x2 U2 (Rr)2 U’ (13 moves because the first U2 and the U in the brackets melt together to U') where you retain the cuboid shape completely as in Julian's original move sequence.

Image

Have you ever tried to use a `parity` move sequence from 4x4x4 / 5x5x5 on the 3x4x5?
This one is quite interesting:
u2 (Ll)2 F2 d F2 u´ F2 u F2 (Rr)2 u (R2)2 d’ (Ll)2 u2
Image
I used it when I solved TomZ's cuboid one and a half years ago.
Now I usually prefer my 3-cycles.
But you never know who will like it.
It is a very well known move sequence, but probably you have the swapped edge pair in the U layer and you will usually not do the R and L moves but r and l on there own.

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 Post subject: Re: How to solve a 3x4x5?
PostPosted: Sat Feb 11, 2012 5:25 pm 
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This is interesting. I assumed (and so did MF8) that internal edges would never get flipped but apparently there's a rare case where they can be.
I can probably design some additional parts (for the Shapeways version) that would solve this problem. Is there any interest in those?

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