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Disclaimer: I'm not a mathematican. Please answer in a way a non-mathematican can understand. Thank you.

I'm building kind of a wooden puzzle and got stuck. My problem is: I have squares whose 4 edges have x different key-and-slot-patterns. Each square looks the same. Now I can join different squares to each other (edge-to-edge) as long as the edges don't share the same key-and-slot-pattern. I prefer to see the keys and slots as a "color". That way each square has x colors whose edges can be joined to each other as long as their color differentiates. Joining may happen planar or perpendicular. In a first step I want to build a cube whose 6 faces consist out of 6 squares. I want to know how many different edge colors I need when building a) an ordinary cube b) a cube in cube system like the rubic's cube (3x3x3). Can anybody give me a tipp where to start?

Here's a picture of the "keys-and-slots" and the resulting cube:

enter image description here

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    Please consider that the squares (tiles) can be flipped which results in a mirrored "key-and-slot-pattern". They also can be rotated.2012-01-07
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    I thought I understood what you're saying but the sentence "That way each square has $x$ colors whose edges can be joined to each other as long as their color differentiates." threw me off. It seems to refer to colors of edges of colors. Did you mean "That way each square has $x$ colors and the edges of the squares can be joined to each other as long as their color differs"? Also, how should we interpret "I have squares whose $4$ edges have $x$ different key-and-slot-patterns." and "each square has $x$ colors"? Is it the edges or the squares that have $x$ colours each?2012-01-07
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    I think you'll also need to say more about the flipping and rotating. If I flip a square and get a mirror image of its key/slot pattern, will that correspond to one of the other patterns? Also, if the squares can be joined perpendicularly and their patterns can be rotated and flipped, how do you decide which of them differ? This no longer seems to gel with the colour paradigm.2012-01-07

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Maybe I'm misunderstanding your problem, but it seems to me that you only need 2 colors to build a cube from 6 squares. If each square has top and bottom edges blue, left and right edges red, then you can fit 6 of them together to form a cube in such a way that wherever two squares meet, one has a blue edge, the other, red.

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    Right. But does this hold true when building a rubic's cube, where several boxes share tiles?2012-01-07
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    I don't know, but it should be easy enough for you to check. Oh, and it's "Rubik", with a "k".2012-01-07
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    Independent of whether this particular scheme of colouring the squares works, $2$ colours will always suffice, since you can take each edge in the finished cube and colour the two edges that will be forming it differently.2012-01-07
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    @joriki: Not sure if I explained it clear enough. I'm not talking about physical colouring the edges of the finished cube. I talk about the positions/alignment of the "nozzles" at the tiles edges. E.g. if I'd had only one nozzle in the middle of the tiles edges, it would not be possible to join two tiles together, as they would collide.2012-01-07
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    @prinzdezibel: Yes, that was clear enough. If you put your cube together (without nozzles at first) and for each cube edge mark one of the square edges forming it "$1$" and the other one "$2$", then take the cube apart, then put $1$ nozzle in the middle of every square edge marked "$1$" and two nozzles to both sides of the middle of every square edge marked "$2$" and then put the cube back together, everything will fit, and you've only used two different patterns a.k.a. colours.2012-01-07
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    @joriki: Correct. But if you start to append tiles to the existing cube in x, y and z direction you find that two patterns won't suffice any more, because each edge can have up to 4 tiles participating. I'm thinking about a 3 x 3 x 3 "super-cube" which is made of single cubes and share tiles in the inner of it.2012-01-07
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    @prinzdezibel: Judging from your image, it seems that the $4$ tile edges that meet in one cube edge will all have to have different colours. Thus in that case the required number of colours will be $4$, and you can distribute them according to the same principle, arbitrarily marking the $4$ tile edges that meet in one cube edge with "$1$" through "$4$".2012-01-07
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    @joriki. Yes, that is what I've found as well. At the moment I have 8 colors. But as you said, it should be sufficient to have 4 colors. They should stack up nicely. Thank you for your help!2012-01-07