for this cast I try scramble followed Burgo, first and second time the puzzles not block, but the third blocked.
Yes, the hiden layer was fliped

very interesting case, I show picture that can turn R and r

Why it happen, if base on my 3x4x5 cuz' the hidden edges Flipped, cuz' when first move Dd the hidden move together.
so if turn by hidden layer not move this blocked cast will not happen.

<<this is nomally hidden edge can turn R

<<and this flipped hidden edge can't turn R
Hope I can improve my mechanism for avoid this case, but look impossible for my mechanism

Thanks to you and Tom for your replies,

What you have said about trying it a few times before finding it is what strikes me, because the only thing not symetrical about the MF8 345 is the single bandaged hidden edge. I assume that is a contributor? I have found another position that involves different pieces, maybe it will help track down a cause (or confound us).

Try [4x5 F, 3x5 U] Now do Uu2 L: that works right? Take account of the pieces on Ll.
Now from the start position do: Rr2 Uu Rr2 Uu2 L.
Note: that will all work on one side, just do y2 and try it on the other side.
Now for the side that didn't work: if you took account of the pieces on Ll when you did Uu2 L, unless I'm mistaken, they were the same when it was blocked .

EDIT: I think it's got do do with turning the 2 outer layers onto a `4x5 face centre` when the 4x5 centre moves around, which I think is dictated by the single bandaged hidden edge.

PS, Although it is a problem with the mechanism, I don't want to make people think that it is a catastrophic problem with the puzzle, (using these sequences as an alg could even undo the problem if you encounter it). And it is mostly unnoticable.

So if you had a blocked position and you wanted to do the R2 turn you could do something in the effect of Uu2 Ll2 Uu Ll2 Uu2 and then do your R2.

Cheers,
Burgo.

I got your 3x4x5 one and a half years ago and have seen this locking behaviour very recently only.
Nobody who is interested in buying either Tom's or the mf8 version should be pushed back by this little locking problem. I find it interesting that even different designs (Tom's, mf8 and traiphum's) have got the same problem.
Nevertheless, I would be interested in your redesigned pieces, if they are not too expensive. (You'll remember that I have got the V3 prototype. I hope this will make no difference to the internal edges.) I would order them together with my next Shapeways puzzle to avoid additional shipping costs. (Probably Shapeways would charge € 11 for shipping them the standard UPS way )

@Andrea: This locking problem should not discourage you to buy the mf8 3x4x5.
You will have seen unfriendly reviews about its turning and popping even exploding in the other thread.
As always , I was the lucky one and got a sample that is turning almost perfectly after breaking in.
Others have reported that it was OK after adjusting the screws (I have never touched them, because mine is so good out of the box.)

And I'll just add that after getting the tensioning right and lubricating, it now turns (and has consistently turned) as well as pretty much every other puzzle I have. Definitely worth getting, despite my initial frustration with it.

Thanks everyone for the interesting information on the 3x4x5 blocked moves: Burgo for your clear explanation on how to demonstrate it and get around it, and TomZ and traiphum for your feedback. I have a TomZ V4 (final version, I think) 3x4x5 and I never experienced a blocked outer 3x4 face move until I followed Burgo's instructions.

@TomZ: I would definitely be interested in some Shapeways pieces to get around the problem, thanks.

@Everyone: The 3x4x5 is an amazingly fun puzzle and very well worth buying! The 3x4x5 is like two puzzles in one. First I did only scrambles with 180 degree turns to always keep the cuboid shape, and figured out how to solve it every time that way, then I started doing shape-shifting scrambles, which was like a whole new puzzle. If you have never played with a cuboid before, you could be in for a particularly brain-melting challenge.

Hi Julian, have you got both versions too? I support everything you have said about this puzzle.
I would highly recommend it too. As a first cuboid it would be hard, but with all the sequences explained in this thread it is doable.

@rline: It is good to hear that your puzzle is turning OK now.

I have now tried various methods. (It is fun too trying different things and finding out what suits you best).
My current favourite is this:
1. Bring it back to cuboid shape AND solve the eight outer corners in this first step.
2. Pair the 4x5 AND 3x5 wedges. (This is a different kind of reduction as described by Burgo. I solve the outer wedges pairs only.) Surprisingly, very often this can be done intuitively. In far less than 50% of the cases I do need any 3-cycle move sequences.
3. Solve the 4x5 centre pieces - the innermost 3x3x3 domino - with move sequences not disturbing any outer paired pieces. e.g. (r2 u2)x2 If I recognize a necessary 3-cycle (Some will call it a ´parity´ of two outer centres.) I leave this for the last step.
4. Solve all remaining pieces by 180 degree turns. I have two work horses to do this: (Ll)2 U2 F2 l2 F2 U2 (Ll)2 (or in shorter notation [(Ll)2 U2 F2, l2] ) and [R2 U2]x3 (or [R2 (Uu)]x3) where I hold the puzzle in varying positions - F= {4x5, 3x5, 3x4}.
5. If necessary I do a 3-cycle of 4x5 outer centres in the last step. I leave this for last because I think I can avoid the locking problem. At least I came across the locking thing in step 3 but never in step 5.

Tanner Frisby

Tanner, I interpret these two pictures saying that your own custom built 3x4x5 shows the same behaviour, right?

I just have TomZ's version. I think that with 180 degree turns only, the 3x4x5 is confusing but not too evil (I solve the corners intuitively and then the rest of the puzzle using fairly simple 6- or 8-move algos, inefficiently but reliably). However, shape-shifting scrambles seem to require at least knowledge of solving a Domino/2x3x3 and the edges of higher cubes, or discovering equivalent algorithms that are at least as hard to find, with the added confusing recognition given the strange shapes the puzzle can get into. I suppose my ranking for the 3x4x5 as a first cuboid would be:

(scrambles of no more than 3 moves -- "irritatingly easy" )

180 degree scrambles only -- fairly difficult

shape-shifting scrambles having first mastered 180 degree scrambles -- difficult

jumping straight into shape-shifting scrambles -- very difficult

I find this puzzle very difficult. Indeed, I am even not able to return to cube shape !!!
Any hints ? I can make the middle layer, but I am not able to oriente the pieces on the exterior layers.

Thank you

Hi Alacoume,
The principle is similar to how you would solve the shape of the Rubik's 2x2x4, if you've solved one of those. I don't know what you mean by `solved the middle layer`, because there's more than one way to view the puzzle. But nevertheless, I have posted ^^ how to view the puzzle as a 3x3x4 [4X5 F, 3X5 U], just use your 3x3x4 domino sequences on the `internal 3x3x4 part`, but concentrate on moving the extended shapes rather than colours. Just remember, it's only possible to do a 90* turn on one axis.
Cheers,
Burgo.

Hi Burgo,

What I meant is that extended pieces are on their correct location (the colour are still scrambled), but not correctly rotated. When using 3x3x4 algorithms, all you could do is permute, because pieces are always correctly rotated, so seeing this puzzle as a 3x3x4 cube does not help me for instance. Where am I wrong ? What do I not understand here ?

Yes, but how ?

I don't know what you mean by `rotated` (orientate?). The easiest thing to do in this situation is to provide a photo and I can give you a sequence.

Although.. all that you need to do is permute because the other 2 axis only turn by 180*, you need 2 axis to turn 90* to orientate.

Cheers,
Burgo.

Here is a picture to illustrate what I meant. I just want to rotate extended pieces (on U face in the picture) to restore cuboid shape.
What is frustating is that I am pretty sure this is not difficult, but I cannot just see how to do it!

Although, maybe the problem is not rotating these pieces, but that they do not belong to the right face ? Problem is therefore more permuting pieces than rotating them ?

Please have a look at my diagrams above for how to view it as a 334 [F=4x5, U=3X5].

This is the result of (Rr2 Uu Rr2 Uu' Rr2) y' (Rr2 Uu' Rr2 Uu Rr2) Dd'

Applying that to the various layers (and possibly some edge swapping too) should be what you need.
The edge swapping can be done with (Rr2 U)X2 (Rr2 U2)X2 (Rr2 U Rr2 U' Rr2).

Cheers,
Burgo.

Thank you Burgo - I will try this this week end !

edit : I have returned to the cuboid shape. I have messed all because I have mixed up y' and y!! Thus, I write your first alg this way :

1) To exchange and rotate corner RF and RB on both layers U and u :
(Rr2 Uu Rr2 Uu' Rr2) Uu' Dd (Rr2 Uu' Rr2 Uu Rr2) Dd' Uu
where I assumed thet Uu' = (Uu)' = U'u'
You can replace Uu by U (resp. u) if you want to rotate only the corners RF and RB on U layer (resp. u).

2) To exchange FU and RU edge :
(Rr2 U)×2 (Rr2 U2)×2 (Rr2 U Rr2 U' Rr2)
You could replace U by u or Uu if needed.

Thank you Burgo for your help. Next step for me : reduce to a 3×3×4 (with crazy Jupiter or Uranus method) and finish the solve. I think I can finish it now.

Next cube after to solve : Burr cube!! I have to read all the post Andrea and Burgo have written, because my cube is scrambled since a long time and I have no idea of how to solve it!!

Hi Alacoume,

Sorry, I missed your response because it was an edit.

Yes, this corner permutation just pretends to be a swap, of course it must be a 3cycle, so to treat it as a `swap` you are putting a 90* twist into the centre piece also. If you are using it where centre orientation matters (like the wall/burr) then it pays to be aware of that.

Sorry for the confusion about the y turn, if you are using it as a domino alg, you just need to do y', but (as you found out) if you are doing it above a section as with a 3x3x4 or a 3x3x3 then you need to do U'D instead (and if you want to see the corner 3cycle do a UD' turn at the end instead of just a D').

Good luck with the Wall4 Cube, it's a lot of fun.
Cheers,
Burgo.

Hi,
I use 2 sequences with variants.

1) (2L2 U 2R2 U' ) x2

this is a 3 cycle of 3 corner pairs. LUB -> RFB -> LFB
variant
F2 (1) F2 -> F2 (2L2 U 2R2 U' ) x2 F2

exchanges BL -> FL -> Fr in upper layer.

2) 2R2 U 2R2 U' 2R2 U2 2R2 U2 2R2 U' 2R2 U 2R2 U2 2R2 U2

This exchanges edgegroups ( edges on 3x3x4) in upper layer.
b - > f -> r

And the mirrored:
2R2 U' 2R2 U 2R2 U2 2R2 U2 2R2 U 2R2 U' 2R2 U2 2R2 U2

b - > r -> f

With this sequences I 'm able to transform the fully scrambled puzzle to a 3x4x5 cuboid shape.

For the rest , I had no idea. Pairing the pieces that it works like a 3x3x4 , i cannot do this.

Cheers, Andrea

Hi Andrea,

I know English is a second language so I know going through all the first posts is difficult, there are some useful diagrams there though too. Once you are sure of how to orientate the cube and what it looks like when it is reduced to a 334. Pair edges first and then corners.. edges are fairly easy, and we did discuss corner pairing 3cycles and how to generate them.. I found an adaption to Julian's sequence suited me the best in the end (but there are other ways):
The goal here is to place your desired pieces to exchange on the Rr layer and then perform an R2 turn to make the exchanges. A corner 3cycle is only needed occasionally.

I found reducing it was a lot like reducing the Jupiter 3x3x3. If this is not helpful just ask more questions.

Cheers,
Burgo.

Actually and surprisingly, I am facing same problems than Andrea !
The problem is just placing the desired piece to be able to perform a R2 turn to pair edges or corners.
For example to exchange 2 opposed diagonals edges, the only 3x3x4 alg I know involve U or D turn and therefore destroy the cuboid shape.

Am I clear ?

Hi alacoume

If memory serves me, I'm fairly sure that destroying the cuboid shape is not the end of the world, here. In case it helps, I made a tutorial for this puzzle, which will perhaps give you an alternative way of placing things. It certainly incorporates some, but not all, of the things in this thread. I hope it might help. You can find it at http://rubiksultimatesolution.blogspot. ... uboid.html

Hi Alacoume,

It is not necessary to completely maintain the cuboid shape, particularly when you are reducing corners (only the cuboid shape on your desired working side -like the RHS- needs to be kept intact). The reason I make the cuboid shape is that it is easier to reduce the edges like that. When it comes time to solve the reduced 3x3x4 you will necessarily break the cuboid shape too.

The edge algos I use are:
(R2 U2)X3 and
(R2 U)X2 (R2 U2)X2 R2 U R2 U' R2

The process for reducing edges and corners is slightly different, first solve the edges. Then set up the `same coloured edges` on the RHS so that a R2 corner exchange does not affect the edges. Eg on a solved cube with [4x5 F, 3x5 U] perform Ll2 B2 U2, now in this position you would cycle in your corners for an R2 exchange.

If I did not hit your problem, include a picture and I will provide some sequences.

Cheers,
Burgo.

Thank you Guys !!

Arnaud

Hi Andrea, maybe, building a complete 3x3x4 first and doing the rest in a second step makes things easier for you?

Notation: (Rr)2 and Rw and 2R are all synonyms.

1. Build the 3x3x4 like in this diagram and do not care about the grey pieces yet:

The outer pieces define the locations where the still unpaired `corners` and `edges` should go to.
Corners surrounded by purple lines, edges in the u and d layer by light green and edges on the U and D layer by light yellow.

2. Pair as many of the pieces in the grey slices by intuition or by move sequences as shown on page 1 of this thread:


The 3-cycle of the last diagram is explained in more detail below.

3. Very often you will need 3-cycles as shown here:
If you do your move sequence (1) first inverse, you'll get the result as in the upper left diagram (I show front and back view similar to Gelatinbrain diagrams), you can recognize that the right hand part of the puzzle allows r2 or R2 turns of a single layer. Reverse the move sequence and you have created a pure 3-cycle in 22 moves. (I have included R2 based commutators for completeness).


By the same logic you can build commutators for the u layer:


You can click onto the pictures to enlarge them.

EDIT:
I made a complete scramble and solve and took pictures of important phases. I hope this helps. We have discussed several different methods to solve a 3x4x5. Solving the outer 3x3x4 first seems to be a visually easy way :

Hi Konrad and all together,

Thanks for the pictures. Which software do you use ?
I didn't found a 3x4x5 simulator. The problem is, I cannot explore sequences, beacause my 345 is scrambled.
I tried to program a simulator ( expand my self programmed circlecube simmulator), but I don't know exactly the mechanics.

I thought for a single outer face turning all pieces must exist. But it's not so.
One ore two inner cubies can be away.
What are the restrictions for twisting a single plane? Perhaps the 10 cubies around the two middle cubies must exist ? I explored the blocking situation posted by Burgo.

Is there a simulator ?

Cheers,
Andrea

Hi Andrea, I do not know a software simulator. I made the pictures with my Canon and Adobe Photoshop It's the real mf8 3x4x5 you see here and the diagrams are made with Coreldraw.

You can explore sequences on a Gelatinbrain 5x5x5, though.
Imagine the 3x5 face as U and the 3x4 face as F.
Just turn Ff instead of F and imagine the E layer as non existent.
It is a bit hard to see the F face of the 3x4x5, but for the sequences shown above it is OK.
(Cubies in the F layer of the 5x5x5 do exist only, if they would stick out of the 3x4x5).
Still using a Gelatinbrain 5x5x5 is much easier than programming a 3x4x5

The blocking situation described by Burgo is not so hard to overcome.
Whenever I got this (not very often indeed), I just reversed the intended sequence and used the other 3x5 face as U.

You are right, that it is possible to turn an outer 3x4 layer, if all 10 outer cubies exist, even if the centres are `hollow`.

Which cuboids have you solved so far? If you can solve an ordinary 3x3x4 and can get the 3x4x5 back to cuboid shape, the first step (the `outer` 3x3x4) is not so hard, I think.

Good luck!

Konrad

EDIT: Here is an example for a 5x5x5 simulation in the case of my move sequence above
r2 (Ff)2 [U (Rr)2 U´ (Ll)2]x2 (Ff)2 r2 (Ff)2 [(Ll)2 U (Rr)2 U´]x2 (Ff)2



Just imagine that the F and B layer have a width of zero and view the stickers on the F and B face to be on the f and b slice.
The E layer (between the u and d slice) does not exist on the 3x4x5.

The Gelatinbrain move sequence is:
[R2&2, F2, F2&2, U, R2, R2&2, U', L2, L2&2, U, R2, R2&2, U', L2, L2&2, F2, F2&2, R2&2, F2, F2&2, L2, L2&2, U, R2, R2&2, U', L2, L2&2, U, R2, R2&2, U', F2, F2&2]
Because Gelatinbrain counts things like L2, L2&2, as two moves there are more than 22 moves. I count 22.

Edit2: The obvious restrictions are that all vertical layers can turn by 180 degrees only and that single R2, r2, L2, l2 turns are possible only, if L or R are a `complete` layer on the 3x4x5. For many of the sequences above this is no problem because they never shapeshift the 3x4x5.
For the mentioned 3-cycles it is not so hard to see that the rule applies.
For the unexpected blocking situation mentioned by Burgo, I have no means to predict it for the 5x5x5 simulation of a 3x4x5. It will not be a problem for exploring move sequences, though.

Hi Konrad,



I'm able to solve my crazy 3x3x2 and my Witeden Super 3x3x4. I don't own a regular 3x3x4. But the Witeden Super 3x3x4 is not very different to a regular 3x3x4, I think.

Cheers,

Andrea

Hi Andrea, with this experience you should have no problems to solve it to a cuboid 3x3x4 with `extended` corners and edges as in my diagram:

Is this assumption correct?

Hi Konrad,



Yes. This was my first idea. But I cannot separate outer from inner pieces in the 3x3x4 blocks.

I programmed a little simulator.( sorry the programm is simple , I'm not a good programmer)


After Konrad gave the tip with gelatin-brain 5x5x5 cube I had more than an halve ready.

I own no photoshop software, and haven't enough money to buy that. So I wrote a little program in C++ and Qt4.

This program shows a solved cube. You can input some easy sequences and look what they change.
In past I used it while waiting for my crazy cubes from Calvin.

The progam shows the 3x4x5 and 8 crazy planets and crazy 4x4x4.

It's not the best. No opengl, no mouse input, no Animation, no parser for special (commutator) input, no shading, etc.
But it was usefull for me to find some sequences for the crazy planets.

Here some pictures and the installer.


(edit)
the 3x4x5 has a bug, so I deleted it.

Hi Andrea,

can it be that we have an essential difference with the notation used?
Your program uses `r` as in the SIGN notation representing a turn of two layers?
This thread uses the WCA notation so far where `r` represents the inner slice below face R.
(Rr)2 in all sequences above means turn the outer layer R and the slice below by 180 degrees.
In your notation my 3-cycle of inner edges would read
R r f U r U' l U r U' l f R r f l U r U' l U r U' f

EDIT: fixed a typo in the sequence above

Your program is not so bad at all , but it needs an essential bug fix!

Unfortunately your program does not accept a legal R move after the first part of the sequence r R f U r U' l U r U' l f


In this situation an R move is legal.
Your program does nothing!
You can cut and paste r R f U r U' l U r U' l f and try R.

Hi Andrea, since you've used Qt4 it should be portable enough to compile on Linux. Are you willing to share your source? I'd like to take a few pokes at it

HI Konrad,


yes. I tried some sequences and a 'R' move was not possible.
In the readme file I explained the multislice notation. ( r = R and the adjacent slice)
Some Websites and programs use this. I'm not sure that this is not wca conform.

But I cannot find the bug. Perhaps this takes a while.

I try to delete the programs in attachment.

Cheers,
Andrea

Hi Brandon,

yes , this works under linux and windows. I put the source in unix/linux textformat.

under linux :

qmake circlecube.pro
make

Cheers,
Andrea

It compiles warning free, thanks

Hi,

I found the bug and hope that no more bugs are in the program.
To install click the CircleCubeSetup.exe program in the zip file. The first zip file is only for programmers.

I found 2 bugs. File cube5.cpp in Line 131 and 133. (Internal check of possible turn is different to make permutations)

I add the source in windows textformat. It's possible to compile it under linux,too.
Perhaps now I found solutions for the 3x4x5 cuboid.

Much fun with the program.

Cheers,
Andrea

Here the new Version for windows without the bug.

Yesss! You have fixed the bug and I fixed a little but essential typo:
The 3-cycle in your notation should read R r f U r U' l U r U' l f R r f l U r U' l U r U' f
This is the result on a solved 3x4x5 using your program:

Regarding WCA notation:quoted from http://www.worldcubeassociation.org/reg ... /#notation

Luke mentioned in this thread above that he does not like lower case notation like `r`, because he remembers the odd meaning of r being a turn of two layers.
Anyway, everybody can use the notation he/she likes for his / her purposes.
For the purpose of exchanging sequences we need a common notation.
For this thread I propose to continue using WCA notation.
r = single inner slice
r2 R2 = Rw = 2R = (Rr)2 a 180 degree turn of two layers

Andrea, where are you now with the physical 3x4x5?
I understand that you can make the 3x3x4 as in this diagram:

At least your reply seems to indicate this:When you can build a cuboid shape but with the edges and centres not yet correct (the grey slices in my diagram) you can the do cuboid sequences where each single turn preserves the cuboid shape,

e.g. (Ll)2 U2 F2 l2 F2 U2 (Ll)2 does a double swap of 3x5 edges
or (r2 u2)x2 swaps centres around.

In many cases you will find 3-cycles necessary at the end.
In this case you need to break the cuboid shape temporarily.

Do you fully understand the logic behind my 3-cycle sequences?
The main part is a move sequence you have mentioned yourself.
The complete sequences do a pure 3-cycle of 3x5 edges or 4x5 centres (in the r and l slices).

If there are remaining questions, we can switch to German in PM, which will make things easier for both uf us

I want to say nothing against Burgo's method pairing pieces before solving the 3x3x4.
Personally, I find it easier when I see a bigger part solved already.

Hi Konrad,



Yes. Very Happy.


I would like understand it. Cubing is not memorizing seqences. Cubing is understanding!!!!

Some words to my notation:
I thought while invent the program, that you don't like my notation.
The 3x4x5 is a 5x5x4 cube with 2 invisible faces. After U three cubies are in the invisible front face.
After y the front face is turnable.

The rR (double) face in your notation is always turnable.
Important !
The idea of my notation is not only to have less work.
The idea was to minimize the blockable moves. The program doesn't make a blockable move and continue with the following moves. So the result is not very deterministic.

The problem is minimized if the blockable moves are minimized. So only R L F U are the outer faces are blockable.
The slicees r ( im my notation rR) are not always possible.
In past I had a scramble timer program,there was a possibilitie to turn on or off the multislice.
I your notation the always possible moves have two letters.This is an disadvantage.
In multislice all outer faces turn with the slice.

Solution:
Find out a sequence, then translate it to your notation.
replace r with rR and rR with r.

The program replaces all numbers after r R F f l L B b with 1 . So R2 = R1 ( the only possible turn)
( Same idea , minimize blocking moves)

The wca notation was invented for regular cubes and for speedcubers. Look to the definition of moves M E S, no logic there. ( All other cubes in my program use WCA notation)

Now I have enough programmed. I want solve my 3x4x5.
Making the cuboid shape is no problem for me. But the many sequences. I don't memorize sequences. I only use sequences that I understand. Perhaps it's easier first pair blocks, and then make the 3x4x5 (like a 3x3x4 cuboid) shape. I don't know.

I cannot solve my 3x4x5

Its more difficult than Burr Cube 4x4x4 !

Cheers,
Andrea

Hi Konrad,




Thanks for sharing this. Great work. ( can you explain how it works ?)

With this 3 cycle and a parity sequence I get this :



(edit)
With your sequence from
Sat Feb 04, 2012 5:14 pm

OmG I solved the 3x4x5 !
The solution was accidentaly. I cannot say, that I am able to solve this puzzle.

Thanks for your help

Congratulation! I think you can easily understand my 3-cycles because they are based on the very same sequence you came up with yourself, your variant `F2(1).` (see the quote below)

I translate the WCA notation to the variant I'm using in my pictures:
F2(1): F2 [(Ll)2 U (Rr)2 U']x 2 F2
In my example above I use the inverse of this: F2 [U (Rr)2 U' (Ll)2]x2 F2 as the first part.
In my picture sequence I had tried to explain how it works and why it works:

After these first 10 moves you have a cuboid like the one in the upper right picture.
The important thing is that the right hand side remains in cuboid shape.
The cubie UrF is the only one that has changed in the r layer.
Doing r2 and inverse the first part (which becomes now exactly your `F2(1)`) is what we need to get a 3-cycle commutator.

Notation: I do not argue that WCA is better than yours (SIGN is the official name for it). Notation is like the basic words in a language and to be able to communicate we have to decide which language we will use.
The 3x4x5 is a derivative of a 5x5x5 in a sense.
WCA is a wide spread common language for describing move sequences.
Why not use it?
BTW, in your language it would be hard to express what I describe as the r layer or r slice.
An `R r slice` is a strange wording.

Andrea


I can say without doubt the 345 is much easier than the 444 wall! Did you work through my tutorial given in a link a few posts back?

Hi rline,


Today I viewed you videos and read your text one more time. Execellent work. Now I understand how the 3 cycle works.

Ok, the 3x4x5 is easier than the Burr-Cube.

Helpfully is to remember which opposite colors belong together. white <-> yellow
orange <-> red blue <-> green. This allows one configuration that is not solveable with only 180 degrees turns. I knew the basic corner 3-cycle, too. The combination with F2f2 R2r2 is the key.

Konrad: In past I had readed a translated tutorial about crazy 4x4x4. It was in multislice notation. This was no problem for me. I don't know that the notation is such a problem, sorry. In my postings I use wca conform notation. In the program I can integrate a conversion if people find it usefull. Ok ?


Cheers,
Andrea

Thanks for the simulator Andrea.

Konrad, is this now your preferred method? I have been following along but haven't felt the need to contribute, thanks for your work I like to see the different methods.

Perhaps what people might not see in the way I would do things with reduction is that once the edges are made, most of the time I can just twist the corners in without sequences and then just use edge sequences to setup for an R2 turn. Sometimes at the end I might need a corner 3cycle, but not every time.

Cheers,
Burgo.

You really have not understood my text and pictures above, before you read rline's text? This is frustrating to me Where was I not clear enough? I will certainly think twice before I put so much effort into pictures and text that cannot be understood. As the puzzle comes unstickered, there are many variants possible how the six colours can be arranged. You may have seen that my 3x4x5 uses the old traditional scheme white opposite blue. My first TomZ 3x4x5 was stickered by Tom in this way. I have not said that the notation in your program is a problem, have I?
We are certainly in agreement that it is necessary to explain which notation we use. In your program you do it on your help screen. No problem at all.
I was just afraid that you would not interpret the notation used in this thread, correctly.
As long as we all agree that WCA is a reasonable default when applicable, I do not have the slightest problem. If somebody wants to use something different, I would expect an explanation.

I remember some posts in the Bermuda thread where you used lower case letters in a way that was surprising to me.
For sure it was different from the interpretation in SIGN (which is the notation in your program).

@Burgo: Actually I thought that building the outer 3x3x4 first is easier to explain and to understand. (Same assumtion as of Superantonivivaldi! Obviously I was wrong in this case )

My own method is
1. cuboid with 8 correct corners

Depending on the resulting random edge and centre configuration, I decide how to proceed.
Very often I go on like this:

2. pair edges (four on the 3x5 face; four on the 3x4 face) by 180 degree turns
3. do 3-cycles IF necessary. (I will not need them more often than in you reduction method)
4. place the edge groups correctly with 180 degree turns only

Hi together, hi Konrad,



This was not my intention. Sorry that you feel so.I didn't write that. I wrote to you: Thu Mar 22, 2012 9:19 am :



You, rline, Burgo and many more users do an excellent work for all interested people.
Your work for the dodecahedron notation is very good, too. Thanks! I understand your pictures and rline's videos about 3-cycle. Often my problem is my bad english.

Cheers,
Andrea

Hi Andrea, your English is probably better than the German of most English native speakers here
Yeah, for both of us a conversation in German would be easier.
Ciao Konrad

Helle Konrad,

Thank you for sharing your methodology and for your clear pictures.
I have been able to solve mine using one of your last post. I was thinking first reducing the cube to a 3x4x4, but I cannot do it. Starting solving the 3x4x4 part is easier I think. The difficult point was managing the 3-cycle, but I used your algs. After that parity is as usual.