Double-Hole Pin Cube Solution
Exchange puzzle for International
Puzzle Party 20
Los Angeles, August 2000
The following is my version of the
solution logic. Read these sequentially for increasing levels of help.
There are 9 pins of length 3 among
the 9 pieces, and every "cell" (a.k.a. "cube") has a pin passing through
it. There are only a three ways that 9 pins can be oriented in a 3x3x3
The three ways are A) all pins in the
same orientation, B) 3 pins in one plane (let's say the top layer) are
all in one direction, while the other 6 pins are in the middle and bottom
layers in a perpendicular direction, and C) pins in top and bottom layers
are in one direction, while pins in the middle layer are in a perpendicular
direction (may need to rotate the cube).
There exist pieces with pins and holes
in perpendicular directions, so method A is not possible.
Method B would require at least 3 pieces
to have all pins/holes parallel. (I.e., these pieces fill the bottom layer).
Only one piece, piece A, has the pin and holes parallel,
so Method C must be the one; i.e., the pins alternate directions in each
Piece A must lay flat
in a layer because its pin and holes are in the same direction.
All pieces other than piece A
utilize two directions for pins/holes, so these pieces must reside in two
layers, either top and middle or bottom and middle.
Piece A cannot reside
in the middle layer, because that would only leave 6 cells in the middle
layer for the other 8 pieces. Assume piece A lays flat in
the bottom layer (or rotate the cube to make it so.)
Because 8 pieces share the 9 cells
in the middle layer, exactly one of the pieces has two cells in the middle
layer, while the rest have one.
The bottom layer, which already has
piece A, must also include three more pieces, each with 2
cells in the bottom layer and one in the middle. (Only one piece will contribute
a single cell to the top/bottom layers - all others contribute two).
Piece A must have its
pin in the middle of the bottom layer, because if it were on an edge, there
would not be room for 2 cells of some other piece to fill its "inside"
There are only two possibilities for
the placement of piece A in the bottom layer (or you can
rotate the cube to get one of these):
The pin/hole directions of each piece
restricts the piece to only a few possible orientations.
With the possibility at the right above,
only pieces B, C and G can go
at position a, however placing either B or C
here leaves no valid piece for position b. Therefore only piece
can go at position a. Note however, that only piece
can go at position c. Since piece G cannot be in both
places, the possibility at the right above will not work and the possibility
at the left must be used.
With the possibility at the left, pieces
and C cannot go in position c because, as before,
no valid piece could then be placed in position b. Only pieces
and H can go in position c, and also, only pieces
and H can go in position a. Only pieces
and E will then go at position
b. I.e., we have the
following four possibilities: abc =
Only abc = HDF
8 cells in the top layer are occupied
with 4 pieces, each with 2 cells in the top layer and one in the middle.
I'll add more later here when I have
© 2000, Richard Eason
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