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 Post subject: The Complex NxNxN PuzzlesPosted: Sun Aug 22, 2010 4:44 pm

Joined: Thu Dec 02, 2004 12:09 pm
Location: Missouri
Before you jump into this thread you may want to read these:

http://twistypuzzles.com/forum/viewtopic.php?f=1&t=15667
http://twistypuzzles.com/forum/viewtopic.php?f=1&p=226290

Let's start with the Complex 3x3x3 as it is the first Complex NxNxN that has been pretty well defined. See first link above.

Order = The number of linearly independant twists available per axis of rotation. Can also be thought of as the number of cuts per axis of rotation. This means the order of a 3x3x3 is 2.

MultiCube = An NxNxN with a (N-2)x(N-2)x(N-2) inside. Inside that is a (N-4)x(N-4)x(N-4), etc. Until you get to a 1x1x1 for N=odd or a 2x2x2 for N=even.

Holding Point = A point that is unaffected by a complete set of linearly independant twists for a given puzzle. Example, the core of a 3x3x3 never moves with this set of 6 twists [RLUDFB].

Imaginary Piece = A piece of a puzzle that can not serve as a holding point.

Note: the 3 layers along a given axis aren't all linearly independant. A slice turn can be thought of as two face turns. Or a face turn can be thought of as a turn of the opposite face and the slice layer.

Now on to the complex 3x3x3. It looks like a 5x5x5 MultiCube with its face turns defined like this:

R maps to http://www.randelshofer.ch/cube/professor/?RMR
L maps to http://www.randelshofer.ch/cube/professor/?LML
U maps to http://www.randelshofer.ch/cube/professor/?UMU
D maps to http://www.randelshofer.ch/cube/professor/?DMD
F maps to http://www.randelshofer.ch/cube/professor/?FMF
B maps to http://www.randelshofer.ch/cube/professor/?BMB

This produces the 10 piece types found in the complex 3x3x3:

The 10 defining patterns:
( 1) UDLRFB => inverted core; 1 moved in every turn
( 2) UDLRF => inverted faces; 5 moved per turn
( 3) UDLR => not yet explained; inverted version of UD
( 4) UDLF => inverted edges; 8 moved per turn
( 5) ULF => normal corners; 4 moved per turn; impossible to be inverted
( 6) UDL => not yet explained; impossible to be inverted
( 7) UD => not yet explained;
( 8) UL => normal edges; 4 moved per turn
( 9) U => normal faces; 1 moved per turn
(10) [] => normal core; 0 moved per turn

The central core is type (1)
The inner 3x3x3 corners are collectively one piece, type (10)
The inner 3x3x3 face centers in groups of 2 are type (3)
The inner 3x3x3 edges move in groups of 4 pieces and are type (7)
The 5x5x5 face centers are type (2)
The 5x5x5 X-centers move in groups of 4 pieces at a time which are the type (9) pieces.
The 5x5x5 T-centers are our elusive type (6) pieces in groups of 2 pieces.
The 5x5x5 middle edges are our type (4) pieces
The 5x5x5 corners are our type (5) pieces
The 5x5x5 wings are our type (8) pieces again in groups of 2 pieces

And so far in this analysis we've only looked at one holding point, the core. That makes things easy as there is only one piece of that type. However if we are going to define Complex NxNxN puzzles for N=even we must be able to deal with other holding points. So instead of picking the core as a holding point let's pick a corner. There are 8 corners so mathematically we must look at all 8 choices together.

When we pick a corner we are in effect picking one face turn and one slice turn as being linearly independant along each axis of rotation. The other face turn is a linear combination of these two.

I'm not sure if this is a standard notation or not but I want to use numbers to represent the slice turns so they are distinct from face turns. So I'll define my notation.

R = Right Face Turn (along x-axis)
L = Left Face Turn (along x-axis)
1 = Slice Turn along x-axis

U = Up Face Turn (along y-axis)
D = Down Face Turn (along y-axis)
2 = Slice Turn along y-axis

F = Front Face Turn (along z-axis)
B = Back Face Turn (along z-axis)
3 = Slice Turn along z-axis

So what does a slice turn look like on a Complex 3x3x3? Remember 1 must be a linear combination of R and L. Here is a nice choice:

1 maps to http://www.randelshofer.ch/cube/professor/?WRMR
2 maps to http://www.randelshofer.ch/cube/professor/?WUMU
3 maps to http://www.randelshofer.ch/cube/professor/?WFMF

Above we picked the 6 face turns (or RLUDFB) as our choice of linearly independant turns. This is equivalent to picking the core as the holding point. Now we want to look at what happens if we pick a corner as a holding point. Note the puzzle should still be the same puzzle with the same 10 types of pieces in the same positions. How you hold the puzzle doesn't change the nature of the puzzle, its still the same puzzle.

So here are the 8 choices of linearly independant turns that match each of the 8 corners:

RUF123
RUB123
RDF123
RDB123
LUF123
LUB123
LDF123
LDB123

And in this table you can see you get the same 64 pieces and its easy to identify the type of each piece.
Attachment:

C3x3x3Table.png [ 102.67 KiB | Viewed 4366 times ]

Using the same method we can now define the Complex 2x2x2.
Attachment:

C2x2x2Table.png [ 14.72 KiB | Viewed 4366 times ]

And we see its just the normal 2x2x2. The Complex 2x2x2 only has 8 pieces and none of them are imaginary. Going the other way we can now define the Complex 4x4x4 and as making the above table by hand took me several hours I'm hoping Andreas or someone can help me as I'm sure a program would be much faster. But I believe the Complex 4x4x4 will have 64 real pieces, 448 imaginary pieces, and look like an 8x8x8 MultiCube with the 3 independant layers per axis of rotation defined something like this.
Attachment:

Complex4.png [ 6.95 KiB | Viewed 4366 times ]

Carl

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 Post subject: Re: The Complex NxNxN PuzzlesPosted: Sun Aug 22, 2010 5:17 pm

Joined: Mon Mar 30, 2009 5:13 pm
I read through this entire post 3 times (and also read the previous threads). But I still can't work out if there's a point to it.

What are you trying to achieve, exactly?

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 Post subject: Re: The Complex NxNxN PuzzlesPosted: Sun Aug 22, 2010 5:41 pm

Joined: Thu Dec 02, 2004 12:09 pm
Location: Missouri
Kelvin Stott wrote:
What are you trying to achieve, exactly?

We came up with the idea of a Complex 3x3x3 some time ago. Matt did most of the work coming up with the 10 piece types and I found a way to fit them all into one puzzle. However at the time the Complex 2x2x2 and Complex 4x4x4 weren't well defined. I believe Matt had an idea what the Complex 5x5x5 might be but until now a general definition of a Complex NxNxN didn't exist. I've proposed a set of definitions that I believe will define what these puzzles are and a way to classify their piece types.

For example the Complex 2x2x2 has 1 piece type, a real corner. The Complex 3x3x3 has 10 piece types, seen above. And until I go though the math I still don't know how many piece types there are in the Complex 4x4x4, but we now have a way to find out.

That "way" is the achievement. Its a new tool even if I haven't finished using it myself yet.

Carl

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 Post subject: Re: The Complex NxNxN PuzzlesPosted: Sun Aug 22, 2010 6:00 pm

Joined: Sat Mar 29, 2008 12:55 am
Location: WA, USA
is there a way of showing all these pieces? maybe in an animation or 3d rendering?

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 Post subject: Re: The Complex NxNxN PuzzlesPosted: Sun Aug 22, 2010 8:16 pm

Joined: Fri Jan 29, 2010 2:34 pm
Location: Scotland, UK
Interesting stuff . I'm in the process of making a table for the complex 5x5x5 and I'm at 188 pieces and counting. I noticed that some of the pieces are chiral, the same way as the two mirror image pieces on a 6x6x6 or 7x7x7, and I chose to count each pair as two pieces. Incidentally, providing I haven't made a mistake, the number of pieces which aren't by two layers of the same face is 729 (57 unique), which you might recognise as 3^6.
I have also started looking at the complex 3x3x3x3 and it has 15 unique pieces, which matches the number of unique pieces on a 5x5x5x5 multicube. I will look tomorrow to see if the obvious emulation works, but I think it will. It has a total of 256 (2^8) pieces compared to the 64 (2^6) of the complex 3x3x3.
Have you checked the 8x8x8 multicube emulation for the complex 4x4x4?

Matt

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 Post subject: Re: The Complex NxNxN PuzzlesPosted: Sun Aug 22, 2010 10:25 pm

Joined: Sat Apr 10, 2010 8:07 pm
Ok first off I'm am totally lost on this thread. Is the complex 3x3x3 the cube in your avatar?

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 Post subject: Re: The Complex NxNxN PuzzlesPosted: Mon Aug 23, 2010 12:00 am

Joined: Thu Dec 02, 2004 12:09 pm
Location: Missouri
Beans wrote:
Ok first off I'm am totally lost on this thread. Is the complex 3x3x3 the cube in your avatar?

No. That is the Mixup Circle Cube. You can read more about that puzzle here:
http://twistypuzzles.com/forum/viewtopic.php?f=9&t=17890

Carl

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 Post subject: Re: The Complex NxNxN PuzzlesPosted: Mon Aug 23, 2010 12:03 am

Joined: Thu Dec 02, 2004 12:09 pm
Location: Missouri
elijah wrote:
is there a way of showing all these pieces? maybe in an animation or 3d rendering?

Yes and Yes and I will try to make some. Just have very little free time at the moment. Better yet it should be possible to make a version of this that could be play with via gelatinbrain or a similiar site. I would envision the Complex 3x3x3 looking like a 5x5x5 Circle Cube.

Carl

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 Post subject: Re: The Complex NxNxN PuzzlesPosted: Mon Aug 23, 2010 12:10 am

Joined: Sat Mar 29, 2008 12:55 am
Location: WA, USA
wwwmwww wrote:
Beans wrote:
Ok first off I'm am totally lost on this thread. Is the complex 3x3x3 the cube in your avatar?

No. That is the Mixup Circle Cube. You can read more about that puzzle here:
http://twistypuzzles.com/forum/viewtopic.php?f=9&t=17890

Carl

wasn't the final conclusion that the mixup circle cube is impossible due to infinite cuts similar to unbandaging a jumbling puzzle?

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 Post subject: Re: The Complex NxNxN PuzzlesPosted: Mon Aug 23, 2010 12:12 am

Joined: Thu Dec 02, 2004 12:09 pm
Location: Missouri
bobthegiraffemonkey wrote:
Interesting stuff

I couldn't agree more. I eat this stuff up. I'll need to think a bit about the 4D stuff.
bobthegiraffemonkey wrote:
Have you checked the 8x8x8 multicube emulation for the complex 4x4x4?

Not fully but I'm about 99% sure its correct. The Complex 4x4x4 will have 512 pieces and that is just a bit much to do by hand with my current schedule. I'm hoping Andreas might have a bit of code that could spit out its full table in a matter of minutes. If not I may take a shot at making some code myself.

Carl

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 Post subject: Re: The Complex NxNxN PuzzlesPosted: Mon Aug 23, 2010 12:16 am

Joined: Thu Dec 02, 2004 12:09 pm
Location: Missouri
elijah wrote:
wasn't the final conclusion that the mixup circle cube is impossible due to infinite cuts similar to unbandaging a jumbling puzzle?

No... the Mixup Cube isn't a pure twisty puzzle for the reason you mention. I call it a twisty/slidey puzzle. But just as its possible to make a Mixup Cube you should be able to make a functional Mixup Circle Cube. See the other thread for a design that I believe will work. I still hope to get it on Shapeways someday...

Carl

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 Post subject: Re: The Complex NxNxN PuzzlesPosted: Mon Aug 23, 2010 1:49 am

Joined: Wed May 13, 2009 4:58 pm
Location: Vancouver, Washington
wwwmwww wrote:
Is there any special reason you put them in this order?

As a computer scientist, I've taken what seems like a dozen logic coarse where I've made lots of truth tables. I also work with binary on a semi-regular basis so I would have put them in this order. I'm just curious if there's some special meaning or not.
Attachment:

complex-cube.png [ 4.15 KiB | Viewed 4156 times ]

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 Post subject: Re: The Complex NxNxN PuzzlesPosted: Mon Aug 23, 2010 9:17 am

Joined: Thu Dec 02, 2004 12:09 pm
Location: Missouri
GuiltyBystander wrote:
Is there any special reason you put them in this order?

No. If you go back to the tread I linked to above where we first came up with the Complex 3x3x3 you'll see I mention 3 different ways of packing the pieces into a 5x5x5 MultiCube. Your method would have the holding point being one of the 8x8x8 corners and mine would have had the holding point as one of the inner 2x2x2 corners but you should still get all the same pieces. Some would be in different positions though.

Of the 8 columns on a face I believe all you need is the following:

(1) One that doesn't turn with any of the turable layers.
(2) One that turns with just layer 1.
(3) One that turns with just layer 2.
(4) One that turns with just layer 3.
(5) One that turns with layers 1 and 2.
(6) One that turns with layers 1 and 3.
(7) One that turns with layers 2 and 3.
(8) One that turns with all 3 layers.

Which ever order you choose should give you the same set of pieces. Here is another one that I like:
Attachment:

Complex4B.png [ 7.98 KiB | Viewed 4113 times ]

Assuming we are looking at the x-axis we could call the red layer the Left layer (L), the green layer the Right layer (R), and the blue layer could be called the Slice layer (X). To do a proper analysis of this Complex 4x4x4 we also need to define the 4th layer. A turn of this layer (1) will be a linear combination of L, R, and X.

Carl

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 Post subject: Re: The Complex NxNxN PuzzlesPosted: Tue Aug 24, 2010 12:59 pm

Joined: Mon Aug 02, 2004 7:03 am
Location: Koblenz, Germany
Urgh!
I calculated the number of permutations for the Complex3x3x3 and now this!
I will try to see what I can do but not now and not tomorrow.

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 Post subject: Re: The Complex NxNxN PuzzlesPosted: Tue Aug 24, 2010 5:27 pm

Joined: Fri Jan 29, 2010 2:34 pm
Location: Scotland, UK
I've been doing some work on this over the last couple of days, and I've made some good progress (got distracted from calculating the number of positions on the complex 3x3x3, Andreas, but I will get back to that soon). In fact, I have calculated the number of unique pieces in the complex 5x5x5, how many are chiral, and the total number of pieces. So what? Well, I found a general approach, so I have the same figures for the 7x7x7, 9x9x9, 11x11x11 and 13x13x13, and I can easily extend this to any odd order by changing a single cell on an Excel sheet. I have only looked at odd orders so far, and I'm less concerned with the even ones, but anyone who is interested can probably adapt my approach. The idea is as follows:

First, I made a list of all possible configurations of distinct faces. For example, the UR and the UDRF pieces have the same arrangement of distinct faces, so this is one configuration. There are 23 such configurations, 5 of which are chiral. Then, I worked out how many pieces were associated with each (call this p); for example there are 12 each of the UR and UDRF pieces. You also need to know how many different combinations of layers you can turn for each face. For the 3x3x3 this is easy, there are 2. Either the face affects the piece, or it doesn't. Call these 1 and 0 respectively. For the 5x5x5 we have 2 layers per face, and assigning a 1 or 0 to each we get: 00, 01, 10 or 11. Thus there are four possibilities. It is easy to realise that this number is simply 2^(number of layers per face), and I called this m.

It took three attempts to get a working model from there, but I managed eventually. First realise that for the configurations of distinct faces, there will be 1, 2, 3, 4, 5 or 6 groups of distinct faces, and call this number f for each configuration. For the 3x3x3, there are only 2 possible choices for a face and so only configurations involving all faces being the same or two distinct faces are relevant (f=1,2). For the 5x5x5, we have up to 4 available (f=1,2,3,4), and for anything bigger we have all configurations to play with. The number of ways of choosing f different combinations of layers from the m possible is simply mCf (where C is a combinatorial=m!/(f!(m-f!))). If the number of ways of distributing f possibilities over the f distinct groups of faces is then calculated (call this n), then to find out the number of distinct pieces for each face configuration the calculation is (mCf)*n. To find out the total number of pieces for each configuration we simply multiply this by p. To find the total for any number of layers, we add these numbers up for each relevant configuration. Anyone still with me? Descriptions are not my strong point! Just try and stick with it, the table below should help.

The left column is the combinations of distinct faces (with a * for chiral pieces), followed by f, p and n respectively. Obviously there is more than one way to write most of the entries in the left column, but one representative entry is all that is needed.

Code:
f   p   n   chiral
UDRLFB             1   1   1
UDRLF, B           2   6   2
UDRL, FB           2   3   2
UDRF, LB           2   12  2
UDR, LFB           2   12  1
URF, DLB           2   8   1
UDRL, F, B         3   6   3
UDRF, L, B         3   24  3
UDR, FB, L         3   12  6
UDR, LF, B         3   24  6
URF, DL, B         3   24  6
UD, RL, FB         3   6   1
UD, RF, LB         3   12  3
UR, DF, LB         3   48  1   *
UDR, L, F, B       4   24  12
URF, D, L, B       4   48  4   *
UD, RL, F, B       4   12  6
UD, RF, L, B       4   24  12
UR, DL, F, B       4   24  6
UR, DF, L, B       4   48  12  *
UD, R, L, F, B     5   24  15
UR, D, L, F, B     5   48  30  *
U, D, R, L, F, B   6   48  15  *

(I had to put this in code for the spaces to work. Anyone know why that is/a better way to fix it?)

And there you have it, all the info needed for any odd order nxnxn complex cube.

Edit: To add the number of distinct pieces for some of the complex cubes, and to correct me being an idiot.

3x3x3: 10 distinct pieces, 0 chiral, 64 total
5x5x5: 220, 20, 4096
7x7x7: 8436, 3276, 262144
9x9x9: 428536, 280840, 16777216

I commented on how I thought the trend for the total number of pieces always worked, but I wasn't sure. It is actually obvious, I was just being slow! There are clearly m^6 pieces for each complex cube, since there are m choices for each of the 6 faces.

I also mentioned the complex 3x3x3x3 (I love 4D puzzles), which has a total of 256 (or 2^8) pieces, so a similar trend probably exists in higher dimensions too. I also verified that the complex 3x3x3x3 can be emulated on a 5x5x5x5 multicube, and I think that this works in all higher dimensions too. I may work on a similar result to all this for 4D, and if that works, maybe even 5D.

Now to get back to the number of possible positions on the complex 3x3x3 . Hope this post makes enough sense! Please feel free to ask for any clarifications.

Matt

Last edited by bobthegiraffemonkey on Sat Aug 28, 2010 6:40 am, edited 1 time in total.

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 Post subject: Re: The Complex NxNxN PuzzlesPosted: Fri Aug 27, 2010 7:37 am

Joined: Fri Jan 29, 2010 2:34 pm
Location: Scotland, UK
This is going to be two long posts in a row!

Anyway, I realised that the even order complex nxnxn puzzles are actually pretty interesting, so I have had a look at them over the last few days. First, let me point out that there are two problems (sort of) with Carl's approach to even order complex cubes. First, when dealing with orders larger than the 2x2x2 it is hard to realise how many pieces there are. When allowing U, R and F twists you get four types of pieces: URF, UR, R and []. For the 2x2x2, this is easy to sort through by hand, but I'm not sure how easy it would be for, say the complex 6x6x6. The second problem is more interesting. I have found two other ways to define the even order complex nxnxn's, which I think are no less valid. I first thought that one of them would yeild the same total number of pieces but would calculate how many distinct pieces there were easily, but it actually gives a different result.

The first way stems from a way to turn (for example) a 3x3x3 into a 2x2x2. Consider the 3x3x3 to have 1 layer per face, such that for one axis we have, for example, R and L slices. If we remove all the pieces which are unaffected by all these twists and do the same for the other 3 axes, we get a 2x2x2. So, one way to define an even order complex nxnxn's is to take the corresponding (n+1)x(n+1)x(n+1) complex cube and eliminate any pieces which are unaffected by all layers of any pair of opposite faces. For the complex 2x2x2, this would give: URF (standard corners, and the only piece defined by Carl's model), UDRF, UDRLF and UDRLFB. This is a total of 4 distinct pieces, and 27 pieces in total. To calculate this, I used the same setup as before for odd orders, which some tweaks. First, there are (m-1)Cf*n distinct pieces for each configuration which involve each piece being turned by at least one layer per face. Then there are mCf-(m-1)Cf*e possibilities where a piece is not affected by one or more layers, where e<=n and is the number of ways of arranging the f blocks of distinct faces bearing in mind that one of them cannot occupy two opposite faces. I have tried to explain that as best as possible, but it probably isn't too clear. I will edit my last post to include the values of e, as well as including the number of distinct pieces on the complex 5x5x5 (did anyone even notice I forgot to include that ).

The second way was meant to be a workaround for calculating the number of distinct pieces in Carl's model, but it seems to give different results. I have yet to properly figure out why, so any ideas would be welcome. There are 8 blocks of pieces on an even nxnxn, one for each corner. The idea was to calculate how many pieces there were in one such block which should give the number of distinct pieces, then the total number of pieces is obtained by multiplying the number of pieces in the corner by 8. This gives different figures for the total number of pieces from both other approaches. For both this approach and calculating numbers from Carl's approach I used a similar table to before, but for only 3 faces instead of 6, which makes things considerably easier!

So here is a table (which I should have had in my last post, oh well) for the various figures for various cubes. Format is: number of distinct pieces, number of which are chiral, total number of pieces (what this is equal to (showing trend)). First row is the 3x3x3->2x2x2 approach, second is using Carl's approach, and third is my variation on Carl's approach which turned out to not be a variation of it but actually completely different.

2x2x2:
4, 0, 27 (3^3)
4, 0, 8, (2^3)
1, 0, 8 (2^3)

4x4x4:
165, 20, 3375 (15^3)
120, 56, 512 (2^9)
10, 1, 216 (6^3)

6x6x6:
7770, 3276, 250047 (63^3)
5984, 4960, 32768 (2^15)
84, 35, 2744 (14^3)

Again, not found any obvous trend for the number of distinct pieces, but I did for total pieces. Let l=number of layers per face, and we have:

(2^(2l)-1)^3
2^(6l-3)
(2^(l+1)-2)^3

I have run out of time just now, I will edit my last post later. Also, work on the 4D version is ongoing. I would be interested to hear other people's opinions about the various definitions for even complex nxnxn's.

Matt

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