Thoughts on solving the Magic 3D Hyperbolic Tile {6,3,3}

[Djair Maynart]

Hello, hypercubing group. I’m Djair and I have recently solved all of the seven Magic 3D Hyperbolic Tile {6,3,3} versions. It has been a fun journey and I’m here to talk a little bit about it.

At first, the puzzle might look very hard, but to me, it’s not really that different from the others like Magic Cube 4d. It has 4 types of pieces: fixed 1-color pieces, 2-color pieces, 3-color pieces and 4-color pieces which almost look analogous to those in the 3^4. I used a piece-by-piece method, where I would first place all 2-color pieces on their spots, then use algorithms and setup moves to cycle 3-color pieces without moving the already placed ones and finally place all the 4-color pieces. It probably looks difficult because of the infinitely repeating cells and because each turn moves multiple cells at the same time, but if you focus on solving it piece by piece without caring about its intimidating infinite nature, you will probably manage to get around it.

As I said before, I used algorithms to cycle each type of piece and also used these algorithms to rotate pieces that were already on their proper spots. Interestingly, it was actually trickier to make these algorithms in the 8 colors version because the repeating cells are too close to each other and there isn't much space to use commutators to isolate the desired pieces.

But there is one case where these algorithms won’t work, which is in some kind of parity where a 3-color piece or a 4-color piece is flipped in a specific way. The only way I know to solve this case is to click on this piece and flip it manually, and then solve again everything that got out of its place because of this flip. I already made a post about this case here.

That’s pretty much what’s needed to know. Apparently, the 8 colors version has 344 pieces, while the 52 colors one has 7852. These puzzles are very similar, they just get increasingly more repetitive. I don’t mind this and actually like to work through it while listening to something else. I don’t think I would recommend going through all of them if you don’t enjoy monotonous solves, but I certainly recommend trying at least the simple ones, or the largest one if you are interested in a bigger challenge. If you liked it, you could probably try the other ones sometime.

Lastly, I would like to point out a possible bug I ran into while solving the 52 colors version. I even restarted the solution when I noticed it but the problem persisted. It apparently doesn’t exist in the default solved version of the puzzle but happened some time during my solution. Apart from the periodically repeating cells, it seems that there are more than one cell with the same color. And this can’t be just a misunderstanding based on similar shades, because even when I edit the color of a cell manually, more than one cell actually changes color with it. This confused me because I would finish solving one cell and would later find a similar one with the same coloration but still unsolved. Because of this, there were interesting situations like finding different pieces with the same coloration which could be placed interchangeably. I would shift + click both pieces and they both would show that their proper place was really the same. Sometimes I would look for a specific piece only to find it was already in its right spot. I tried to paint every single cell in this 52 colors file black, but could manage to turn everything black by editing the colors of only 35 cells. It seems that the shades of some cells got mixed up and turned into the same coloration. I don’t know if it happened because of some macro I applied, or during the loading of the file or it naturally hapenned during the solution, because as I said, this is not seen in the solved default puzzle.

Honestly, this actually made the solution more interesting since all the previous puzzles were pretty much the same. I ran into an interesting case where I had to swap only 2 pieces, which shouldn’t be possible. But just like a Penrose cube or similar puzzles, this is actually possible because there were different pieces with similar colors which could be placed interchangeably, so it is be possible to swap 2 pairs of 2 pieces. This recoloring of a puzzle with fewer colors than faces is something I’ve seen before and it makes solving it a little bit more challenging. It was surprising to see this, even if it wasn’t on purpose.

I have some examples of this happening below. Here there are some pictures of two pieces that belong in the same spot and a vídeo of me searching for a blue and pink piece which is already in its place (because there were actually two blue and two pink cells with the same shades). Also, here is the file of the solution I stopped halfway through when i realized this was happening.

     

Hopefully more people will be interested in trying to solve it, because it still doesn’t have many solvers. Hyperbolic 3D geometry is a very interesting idea for a puzzle and I would love to see a hyperbolic 4D one in the future, or other unusual geometries.

Best regards,

Djair

[Raymond Zhao]

Hi Djair,

Congratulations again for solving all of the MHT633 puzzles, and thanks for the notes!

To add to your notes, the 8c version is compact, but one advantage is that the pieces seem close by when you're looking for them. I ended up spamming commutators for that one. For the larger puzzles, pieces seem farther away, but based on my experience so far with the 12c puzzle, I speculate that one can blockbuild on those puzzles if they really wanted to.

As for the 52c puzzle bug, did you try using the Recalculate file menu entry in the program mid-solve to fix the issue?

I also looked over your post on 3c and 4c parity and realized that I didn't hit 3c parity.

I don't know if there is a better way to solve those parity cases other than what you wrote in that post, but I believe the reason those parity cases occur in the first place has to do with modular arithmetic. For example, if clicking a 3c of a cell flips an odd number of 3c's, but the usual flip algorithm flips two 3c's, then it's possible to have all the 2c's solved but have a single 3c flipped.

Kind regards,

Raymond

[Djair Maynart]

Hi, Raymond,

 

Thank you and congratulations for you recent solution too!

Yes, the 8c version may be easier to solve by commutators and by blockbuilding than other larger versions. I’m not used to these methods so I prefer to generalize my solutions with just a few algorithms for every type of piece. Even if that’s not very efficient in terms of number of moves, it’s faster for me because I’m already used to it. I just thought it was interesting that being a smaller puzzle could actually make these algorithms a little bit longer than in larger ones.

It looks like using Recalculate really fixed it, but since it’s a big puzzle, it actually takes quite some time to load every move it and it would be annoying to do it every time I wanted to continue my solve. Thank you nonetheless!

I think that flipping the parity piece directly might be the only way to solve that case. It reminds me of puzzles like Master Skewb where you need to place the corners correctly from the start or else you won’t be able to permute them in the end. When this happens, you basically solved the corner pieces in a wrong way so you need to fix them and solve everything that got scrambled in the process again. The problem is that you can only know if the corners are placed correctly or not at the end, when everything is almost solved. As for why it happens, I was just curious to know why it occurs in hyperbolic space but not in usual Euclidean space. It probably really has to do with the number of pieces it affects. Thank you for the insight.

 

Best regards,

Djair

[Djair Maynart]

Hello again, hypercubing group. I have some new information to share about this puzzle.

 

I actually discovered what caused this bug to happen. When a file is saved, each color is being represented by a character. The 52 colors version has colors corresponding to the digits 1 through 9, uppercase letters A to Z and lowercase letters ‘a’ to ‘q’. When you select the 52 colors puzzle, all of the colors are loaded, but the bug happens when you save a file and then load it. The software will actually interpret lowercase and uppercase letters as being the same, and will save all lowercase letters as if they were uppercase. That’s why there are only 35 colors in this bugged version, each correspond to the 9 digits plus 26 uppercase letters of the alphabet, since all lowercase letters were turned into uppercase ones. Fortunately, the file saves all movements that were done, so it’s possible to restore the puzzle to its original coloration by clicking “Recalculate”, as Raymond pointed out. This issue doesn’t occur in the 32 and fewer colors versions because they don’t need to use lowercase letters to represent every color. Here’s an example:

File after being saved for the first time

File after being loaded and saved again

Also, I previously said that the 8 colors version had 344 pieces, while the 52 colors one had 7852. I said that based on the number of characters in the second line of the saved file, but it turns out that’s actually the number of stickers, not pieces. If anyone has any idea on how to calculate how many pieces each version has, it would be very interesting to know.

 

Best regards,

Djair

[Djair Maynart]

If anyone is interested, here are the number of pieces for each puzzle: 

8 colors:

8 1c pieces, 28 2c pieces, 56 3c pieces and 28 4c pieces
This results in 120 pieces and 344 stickers

12 colors:

12 1c pieces, 54 2c pieces, 108 3c pieces and 54 4c pieces
This results in 228 pieces and 660 stickers

14 colors:

14 1c pieces, 91 2c pieces, 182 3c pieces and 91 4c pieces
This results in 378 pieces and 1106 stickers

20 colors A:

20 1c pieces, 160 2c pieces, 320 3c pieces and 160 4c pieces
This results in 660 pieces and 1940 stickers

20 colors B:

20 1c pieces, 120 2c pieces, 240 3c pieces and 120 4c pieces
This results in 500 pieces and 1460 stickers

32 colors:

32 1c pieces, 336 2c pieces, 672 3c pieces and 336 4c pieces
This results in 1376 pieces and 4064 stickers

52 colors:

52 1c pieces, 650 2c pieces, 1300 3c pieces and 650 4c pieces
This results in 2652 pieces and 7852 stickers

It's interesting to notice that every puzzle has the same number of 2c and 4c pieces, which is half the number of 3c pieces. Also, the 20 colors A version actually has less pieces and stickers than 20 colors B. 

I'm confident these are correct because the number of stickers these calculations predict match up perfectly with the number of characters that represent stickers in the saved files of each puzzle.

Best regards,

Djair

[Raymond Zhao]

Hi Djair,

I find it interesting that the numbers work out so well. It turns out the formulas cancel out in a way such that for the MHT633 puzzles, the numbers of 2c and 4c pieces are always the same, and the number of 3c pieces is always double the number of 2c pieces.

Let N>=7 be the number of colours in an orientable {6,3} shallow-cut face-turning puzzle, like the Torus Rubik or 16-colour {6,3} puzzles in MagicTile. Then, there are N 1c pieces. There are 6 edges per hexagon, so there are 6N/2 edges in total. We divided by two there to not double-count the edges. Therefore, there are 6N/2=3N edges, or 2c pieces (result 1). Similarly, there are 6N/3=2N corners, or 3c pieces (result 2).

Moving up a dimension, let C>=8 be the number of colours in the MHT633 puzzle, and let N>=7 be the number of neighbours of each cell. It turns out that the neighbours of a cell form an orientable {6,3} tiling, so we can use the earlier results.

The number of 2c pieces per cell is N. There are C cells, so there are NC/2 2c pieces. Again, we divided by two to avoid double-counting. There are 3N 3c pieces per cell by result 1, and three 3c pieces meet at an edge of the MHT633 puzzle, so there are 3NC/3=NC 3c pieces in total. Similarly, by result 2, there are 2NC/4=NC/2 4c pieces in total.

Therefore, in an MHT633 puzzle with C colours and N neighbours per cell, there are NC/2 2c pieces, NC 3c pieces, and NC/2 4c pieces, adding up to 2NC pieces in total. If we include the 1c pieces, there are 2NC+C=(2N+1)C pieces in total.

With your total piece count numbers, we obtain N=12 for the 20-colour pattern B puzzle, N=21 for the 32-colour puzzle, and N=25 for the 52-colour puzzle. Combining that information with my post on determining MHT633 parity cases, we conclude that the 20-colour pattern B puzzle cannot get any orientation parity cases! Meanwhile, the 32-colour puzzle can get 3c orientation parity, and the 52-colour puzzle can get both 3c and 4c orientation parity.

Thanks,
Raymond

[Roice Nelson]

I like seeing the activity and posts on MHT633. 

On the topic of parity possibilities, I haven't looked into the 3-dimensional case enough but still wanted to comment... I suspect the behavior relates to the topology of the puzzle rather than modular arithmetic.

For example, in the 2-dimensional case with MagicTile, you can get single unsolved 3C pieces for Klein bottle puzzles because they are non-orientable. I suppose modular arithmetic could map onto orientability in some subtle way, and maybe it does for the supported set of puzzles, but that doesn't feel general.

A dimension up, perhaps both the orientability of the cells and how those cells connect up comes into play. To start digging, a topology concept to study could be Betti numbers (a higher dimensional version of genus. See also Euler characteristic).

In any case, when it comes to solving puzzles, I bet fixing parity requires what Djair found. In 2-dimensions, it is not possible to see if you are going to get them ahead of time, and you have to undo work for the final fix. I think this relates to moving pieces "around the horn", kind of like moving something along a Möbius strip to an original spot but mirrored.

Best,

Roice

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