Double-Hole Pin Cube Solution Logic
Exchange puzzle for International Puzzle Party 20
Los Angeles, August 2000
Rick Eason

The following is my version of the solution logic. Read these sequentially for increasing levels of help.

1. 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 array.
2. 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).
3. There exist pieces with pins and holes in perpendicular directions, so method A is not possible.
4. 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 layer.
5. Piece A must lay flat in a layer because its pin and holes are in the same direction.
6. 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.
7. 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.)
8. 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.
9. 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).
10. 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" corner.
11. 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):
1. The pin/hole directions of each piece restricts the piece to only a few possible orientations.
2. 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 G can go at position a. Note however, that only piece G 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.
3. With the possibility at the left, pieces B and C cannot go in position c because, as before, no valid piece could then be placed in position b. Only pieces F and H can go in position c, and also, only pieces F and H can go in position a. Only pieces D and E will then go at position b. I.e., we have the following four possibilities: abc = FDH, FEH, HDF, or HEF.
4. Only abc = HDF will work.
5. 8 cells in the top layer are occupied with 4 pieces, each with 2 cells in the top layer and one in the middle.
6. I'll add more later here when I have time.