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Can these 9 pieces be put together into a 3 x 3 x 3 cube?
Check out graphic of pieces:
http://i254.photobucket.com/albums/hh120/Scythian1...
On the left is a piece composed of 7 small cubes as shown, and on the right is 1 small cube. You have 3 of the pieces on the left, and 6 small cubes. Is it possible to assemble these 9 pieces together into a 3 x 3 x 3 cube? The 3 pieces that are composed of 7 small cubes are identical, no mirror copies are allowed. All small cubes are of the same size, so that:
3 pieces composed of 7 small cubes each = 21
21 + 6 small cubes = 27 = 3 x 3 x 3 cube.
Starrysky, if each of the 7-cube pieces take up 2 full edges of the 3 x 3 x 3 cube, or 3 corners, isn't that more than the 8 corners the cube has?
Some kind of sketch or diagram would be nice, otherwise other people will either 1) think it's not a solved puzzle, or 2) think it's a solved puzzle. And we wouldn't want that, right?
Falzoon, you've deduced that the 3 pieces can't each take up 3 corners of the 3 x 3 x 3 cube. We know this now. But does that mean a solution is impossible?
Quadrillerator, I don't have a convenient 3D drawing facility that can quickly put together an image of this problem, solved or not. What I can do is to post a wire frame of 27 nodes, the nodes which can be drawn over to indicate where the 7-cube pieces go, e.g., using squares, triangles, circles. Or you can go ahead and continue to use coordinates {x, y, z}, where x, y, z = -1, 0, 1.
I think the 3rd graphic above is probably best to use?
Quadrillerator, you're the only one here working towards a conclusive resolution, so certainly this will be left open for a while.
3 réponses
- ?Lv 6il y a 7 ansRéponse favorite
Let the 27 components of the cube be {-1,0,1}^3. Then the three 7-component pieces can be arranged this way:
(1): (-1, -1, -1), (-1, 0, -1), (-1, 0, 0), (-1, 0, 1), (0, 0, -1), (1, 0, -1 ), (1, -1,-1),
(2): (1, -1, 1), (1, -1, 0), (0, -1, 0), (-1, -1, 0), (1, 0, 0), (1, 1, 0 ), (1, 1, 1),
(3): (-1, 1, 1), (0, 1, 1), (0, 0, 1), (0, -1, 1), (0, 1, 0), (0, 1, -1), (-1, 1, -1).
In other words, if we take (1) to be the original copy, then we can obtain (2) and (3) via the orthonormal transforms
((0, 1, 0), (0, 0, -1), (-1, 0, 0)), and
((0, 0, -1), (1, 0, 0), (0, -1, 0),
respectively.
- falzoonLv 7il y a 7 ans
I say it's not possible.
First, the 3x3 L-shaped piece can only have it's middle cube in a corner.
The 3x3 L-piece takes up 3 corners.
So three 3x3 L-pieces would take up 9 corners.
But a cube has only 8 corners.
Ah! yes, I can see that at least one 7-cube piece can have the 3x3 piece as the
middle of two faces, but I'm unable to prove a cube can or cannot be formed,
as my 3D glasses are broken.
- QuadrilleratorLv 5il y a 7 ans
Consider the cube, M, of the 7-er that has the most number of cubes attached to it. It's clear that that can't occupy three corner position, because each time it occupies a corner position, it takes two other corners with it, and we don't have 9 corners. Also, it's clear that one of the unused singletons will be the center cube of the large cube.
So I don't have drawing software. If you can recommend a free website where I could conveniently assemble an image, I might give that a go. Otherwise, I could withdraw my entry. I'll take it that the centers of my cubes are at integer positions in (a,b,c) where each of a,b,c are integers ranging from -1 to 1, the center cube being at (0,0,0).
Let F be the cube on the 7-er that is farthest from M. If I give the positions of M and F, that will specify the layout completely:
7-er 1: M:(-1,0,-1), F:(1,-1,-1)
7-er 2: M:(0,1,1), F:(-1,1,-1)
7-er 3: M:(1,-1,0), F:(1,1,1)
Singletons: (0,0,0), (0,-1,-1), (-1,1,0), (1,1,-1), (1,0,1), (-1,-1,1)
Remaining cubes:
7-er 1: (-1,-1,-1), (-1,0,0), (-1,0,1), (0,0,-1), (1,0,-1)
7-er 2: (0,1,0), (0,1,-1), (-1,1,1), (0,0,1), (0,-1,1)
7-er 3: (-1,-1,0), (0,-1,0), (1,0,0), (1,1,0), (1,-1,1)
Update: It seems to me that putting M at a non-corner leaves only two possible corners for the other two Ms, but neither of them work. Ie. all three Ms must be an non corners. And given that, I think the solution is forced since that means each 7-er takes out an adjacent pair of center face cubes.
Scythian, thanks for that. I'm not sure I'll have the time today to color your cube in, but I hope to get back to it.
