Recently I received my Face turning octahedron. I have tried to solve it myself but I can't seem to figure it out. Please post your solutions. Thanks

Once again, a wonderfull website , not in english but pics speak by themselves.

http://tw.myblog.yahoo.com/jclei0411/article?mid=1088

Hope that's what you want.

It's great! Thanks!

Hello,

the solution from the link from William works. I used it.
But it uses many many commutators. For this solution you must spent much time.
Solution from that link:

1) Solve the corners intuitive
2) Flip 2 corners with R L' R' L
3) solve the centers with commutator
4) solve the edges with commutator

the commutator is: ( R U R' U ) x 2

This makes a 3 cycle of 3 centers AND a 3cycle of 3 edges on one face.
After solving the centers you can only permute edges on 1 face, because the centers permute too.

Perhaps someone can post better algorithm.
Perhaps it's effective to solve corners at last.
Bu I don't find corner-permutations that don't destroy centers and edges.

Cheers,
Andrea

The solution from Jaap solves first the edges like a Dino Cube then the corners.
At last the centers.

The sequence ( r U r' U' ) x 5 solves the centers.
There are 24 centers. Assume that 4 centers are solved automaticaly we need 400 moves to solve the centers.

You can solve the centers pure with a [[1:1],1] commutated conjugate (8 moves).

Here you go in Gelatinbrain notation:

[ULB, URF, ULB'], URF&2, [ULB, URF', ULB'], URF'&2

There are a couple minor variations on the above concept that will cycle different pieces. Combined with setup moves you should be able to solve at least two, often three triangles at a time.

Supposing you average 2 setup moves to solve 3 triangles at a time then you can solve them all in 96 moves.

Hello bmenrigh,


I don't know this notation. ( Gelatinbrain notation ?)

In a video I saw an other sequence. That's shorter.

r' F' U F U' r U F' U' F


Notation like Jaap's website.



If it's possible in easy way permute 3 centers then it uses 24/3 * ( 2+10+2)
= 112
I think it's not possible permute every 3 centers with 2 setup moves.
But I am not sure. 3 setup-moves is to much.
You must memorize 1-2 setup moves for each 3 cycle.
One mistake and you must begin from start. I like more easy solutions.

Cheers, Andrea

bmenrigh's [3,1] commutator is just 8 moves. Your algorithm is 10. Gelatinbrain is a wonderful collection of virtual puzzles. Here is the link of the FTO: http://users.skynet.be/gelatinbrain/App ... cta_f1.htm
The notation is a bit strange in this case. You can cut and paste bmenrigh's algorithm and see what it does.
Here is the index page http://users.skynet.be/gelatinbrain/App ... /index.htm

Hello Konrad,



I prefer easy sequences. They are easy to memorize. 1 or 2 moves more, no problem.


the first link doesn't work. It shows a white screen.
The program gelatinbrain, I found a zip file. This program doesn't show notation input. The program has not a input-field for notation. The only input interface is the mouse.
To find out moves for real puzzles it's better to do the input with notation and copy and paste. My self coded simulator for crazy-cubes was very simple (no animation) and accept only notation input. I had not found an explanation for the notation.

I hope, that I must not read the complete gelatinbrain thread.

Hi Andrea,

Without the ability to copy-and-past the notation into Gelatinbrain's applet I agree the notation isn't as useful.

I did decompose the sequence into its constituent building blocks: [[1:1],1]

If you are not familiar with that notation, the : and , operators are conjugation and commutation respectively.

So X:Y expands to X Y X'
And X,Y expands to X Y X' Y'

[] are just for grouping so that you can nest.

Using those rules you can see that [[A:B],C] expands to A B A' C A B' A' C'

The numbers in the notation refer to the length of the operand sequences. [1,1] means you are building a 4 move sequence out of two 1 move sequences.

I could have also notated the sequence as [3,1] but using [1:1] in place of the 3 is a good way of saying the 3 moves are actually composed of 2 face turns of the form X Y X'

The sequence you posted is [1,[1,1]] or [1,4] which is just the inverse of a [4,1].

As you can see, the only difference between the sequence I created and the one you posted is that instead of commutating a [1,1] I used a [1:1] which is a careful observation that X:Y is a truncation of X,Y.

That is, X:Y is X Y X' and X,Y is X,Y,X',Y' which includes an unnecessary Y'. If you leave off a one-move Y' you save two moves because the move is used in a conjugation operation which means it will get applied once as Y' and another time as Y, saving 2 moves overall.

Regarding memorization, I think of a [[1:1],1] sequence as needing to memorize 3 moves and how they are put together (the operators) rather than having to memorize 8 moves. Your [1,[1,1]] is basically the same amount of memorization (three moves put together in a different way).

Regarding setup moves, the triangles in a FTO are in two orbits. Usually orbits reduce the number of setup moves because a piece that goes in a particular spot has fewer locations that it could be. You can solve by orbit ignoring the triangles in the other orbit. The 96 moves I suggested is really a worst-case upper bound. In reality you should be able to solve the entire puzzle in ~50 moves if you are very careful.

The downloadable program has no cut and paste input. The online applet has.
To give you another example of a (1,3) commutator that is very easy to memorize, I introduce first a very common notation:

If the FTO is flat on a table with one tip pointing at you, R is a clockwise turn of he face to your right and r is an inner slice move of the layer below R. L is to your left.
The sequence r (L R' L') r' (L R L') producces this result on a solved FTO:

Mirrored and inverse sequences are easy to see.

The (X,Y) notation would be (r,L R' L); X= r; Y = L R' L and that retranslates to X, Y,X',Y' = r (L R' L) r' (L R L')
The construction logic helps to memorize this or similar samples quite easily.

The right part of the diagram shows the unchanged other four faces.
The diagram has been produced with the Gelatinbrain applet. The Gelatinbrain notation of the very same sequence is: UFL&2,UBR,UFL',UBR',UFL'&2,UBR,UFL,UBR',
The &2 denotes an inner slice move under the face described by the first three letters. I have no idea why Gelatinbrain has chosen a three letter notation for a face turning puzzle.
I would be glad myself, if somebody can explain the logic behind this. My theory is that it is using the three corners that are in a specific face. But why UFL stands for the right hand face, I have no idea.

I had problems with the applet myself, there are dependencies on the browser, the operating system, the driver, whatever ...
bmenrigh can give you certainly tips what to check. He has helped me some time ago, as well.

Brandon, I hope you can explain the logic behind the Gelatinbrain notation.
Andrea don't worry, for most Gelatinbrain puzzles the notation is easier.

EDIT: @AndreaThis not a commutator! As bmenrigh has described it above, a commutator is described by two sequences X and Y and [X,Y] translates to X, Y, X', Y'.
In R U R' U R U R' U you cannot identify X and Y.

Hello Konrad, bmenrigh

thanks for your explanation.
The shown sequence is shorter as mine.
If I must memorize more than 2 setup moves I make mistakes.
So I must make more moves.




I thought that every sequence can be a commutator.
Example A = R L U
B = D2 L

B A B' = D2L R L U L' D2 uses A as commutator
I know it from mathematics.

The form A B A' B' I know it as conjugate.


I hope that's ok ? Perhaps I don't understand what's meant here.
Sorry.
Solving a puzzle and translate this "language / notation" is difficult.
But to analyse a puzzle this is fine.

Cheers, Andrea

Any sequence of moves can be called a permutation. Conjugates and Commutators are defined specifically.B A B' is a conjugation.Actually, this is a commutator. Conjugation (or conjugacy) is defined e.g. here.
bmenrigh describes the terms correctly, maybe a bit short for somebody who is not so familiar with them.X:Y is just a notation describing "conjugation". X:Y := X Y X'. If you have algorithm Y, X is your setup sequence!Again just a short cut for "commutators". by Ryan HeiseA good site describing the stuff related to the Rubik's Cube is Ryan Heise's site
You do not need it solving a puzzle but the stuff is useful understandig algorithms, finding new ones and memorizing them.
E.g. you can solve a Rubik's Cube without memorizing any algorithms, if you understand the concepts of conjugation and commutation, thoroughly. Not an easy way
I hope this helps

Hello Konrad,

oh, I exchanged commutator and conjugate. Thanks for explaining it.
I know both, but my definition was vice versa.
Yes, its possible to solve the Rubiks Cube without memorize algorithm.
The Ryan Heise method can be modifyed. Building blocks like petrus method. Then solve edges with keyhole technique. Then permute corners with "conjugators" and orient corners with "conjugators".
Very interesting.
Sorry that I exchanged the names of concepts.

Cheers,
Andrea

Aaah, I see, you are familiar with the concepts, you have just mixed up the terms. Don't worry

The Gelatinbrain notation makes a lot of sense when you break it down. It isn't always convenient to read and write manually but if you have the Jar working it doesn't matter.

Gelatinbrain's notation is based on the face-turning tetrahedra / cubes / dodecahedra. The other platonic solids are the duals of these three and their notation is based off of their face-turning dual counterpart. So for the FTO, the face-turning is equivalent to vertex turning on a cube and the vertices of the cube are labeled by the three faces that intersect at that vertex.

So face-turning dodecahedra and vertex turning icosahedra share the same notation. Vertex turning dodecahedra and face turning icosahedra share the same notation and so forth.

The only "strange" notation is the tetrahedra because they are self-dual. If you turn a vertex you get vertex notation and if you turn a face you get face notation.


The other tricky part is slices and turning a face more than a "quarter" turn. Slices are a bit-mask where the face is the 1 bit, the second slice is the 2 bit, and if there is a deeper slice it is the 4 bit.

So if you want to turn a face you could specify A or A&1. If you want to turn just the slice then you can do A&2. If you want to turn both you can do A&3. Inverse turns look like A', A'&2, etc. Turning a face twice looks like A2, A'2, etc. Finally, you can turn a slice twice too with A2&2, A'2&2, etc.

I like Gelatinbrain's notation because it is complete and unambiguous and there is a program you can copy and paste into. If you specify routines in a different notation you have to use a physical puzzle or translate which is error-prone.

Thanks Brandon for your explanation. Still I'm missing something. I understand the part that the Octahedron is dual to the Hexahedron (Cube). I understand that the face turns are named as the corresponding cube vertexes.
Still I do not see the logic behind using R = Right L = Left F = Front and B = Back in the Gelatinbrain notation.
This is the Gelatinbrain notation for the FTO:

I have no intuitive way to understand why UFL is at my righthand side and ULB is at the front to me.

I guess you have to imagine yourself standing at the top right corner of the left view.

Maybe Gelatinbrain uses this weird POV because the code is more compatible with that for the cube. Like, in Gelatinbrain's cube, the F face is not directly in front, but in the left bottom corner of the left view, and the URF corner is in the center. Maybe the convention for the octahedron has something to do with that.

Yeah it's a bit odd but I think the strange locations for the vertex labels are an artifact of the view snapping for the octahedra. With the octahedra you can only view the shape with a vertex pointing straight out at you. It would make sense to make the vertex pointing at you the U face and then it's really a random choice if you choose the upper right or lower right vertex as the F face.

Here are the vertex labels:

It doesn't seem like there are any natural views of an octahedron that are also natural views for the cube. I think the choice of labels that Gelatinbrain made are just as good the few other obvious choices.

This choice is arbitrary. I had to choose. That's all.
With my fixed orientation, there's no way to keep the L-R symmetry.
If one of 3 pairs of opposite corners(U-D,L-R,F-B) is brought to center(U-D, in this case),
other 2 pairs comme to diagonals.

The most intuitive notation will be like this:



But to realize this, I have to rotate the entire puzzle by 45º.
Or maybe I should at least bring the F-B pair to center. It's more intuitive.

If it's confusing at the first time, you can be easily accustomed.
Just like in the universe, there's no fixed axis on a twistypuzzle.
You are right, Conrad! Einstein must love twistypuzzles, if he were still living today.

I think given the notations for the cube, a natural notations for the octahedron should be as follows:


But I don't mind the current notation, because when I'm looking for algorithms, I twist the puzzle by clicking the mouse and then copy/paste the list of the moves. I never have to think which face is called what.

Sometimes I need to "mirror" an algorithm. When I do that I just swap F<->R, B<->L, clockwise <-> counterclockwise. This rule works for both the cube and the octahedron.

Yeah the current labels don't bother me either. I never have to think about the labels.

Regarding mirroring and inverting, etc, I have a series of perl scripts to help me:


I have been planning on creating a simple web interface to these tools for people who want to mirror/rotate/invert/whatever the macros/notation.

Nan, when you're solving a puzzle with macros, do you pre-make the sequence, its inverse, its mirror, and its mirrored inverse? I only make the sequence and the inverse and sometimes I have to spend a bunch of time doing setup moves to make the sequence I need fit. Do you create a bunch of variants for each macro sequence?

I recently solved 1.2.28 and 1.2.29 and they took me ~50 minutes with macros and they only took you ~13 minutes. The majority of my time was with finding setup moves to cycle the center triangles. I was pretty amazed how much faster you were than me. I wasn't trying to be efficient. Did you have a lot more macro variants made for the triangles cycle?

Brandon, I replied your post in the Gelatinbrain thread here:

viewtopic.php?p=262054#p262054

because I think that's a more relevant place.