Logical Progression
Solution
Layer 4 Layer 3 Layer 2 Layer 1
TOP BOTTOM
P h e f p H e l j h E l J b b l
k h e F p o O f i A o f j b o L
K a c c p a m C i a M c j B m m
k k d G D d d g i N g g I n n n
Bottom (layer 1) is the rightmost 4x4
Pins go left-right on layers 1,2,4
Pin go front-back on layer 3
Capital letters are the cell with the pin
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This puzzle was carefully designed to have a logical way of solving it. The pieces are such that the puzzle is rich in discovery during the solution process. With a little insight, the entire puzzle can be assembled piece by piece without any guesswork. The following gives a series of observations that allow you work your way to the solution from beginning to end. Note that this gives one path, while you might find some alternate steps along the way. The following might look a little long, but I'm trying to be a bit rigorous.
As with any puzzle, a good place to start is to analyze the pieces. What can you discover from them? What is in common between each piece and what is different?
There are 16 pieces, each consisting of four "cubes" or "cells" in an L-tetromino shape. The 16 pieces are put together into a 4x4x4 cube. 16 pieces of 4 cubes each provides 64 cubes and the finished cube has 64 cubes so there are no empty cells.
Each of the 16 pieces has a pin of length 4 -- sticking out three units, with an additional unit embedded in one cube of the piece. Each piece also has three holes, so no holes are unfilled in the finished cube.
Focusing on the pins, there are 16 pins of length 4, so every one of the 4x4x4 (= 64) cells in the finished cube has one pin going through it. Of course, no cell contains two pins; i.e., the pins don't intersect.
How many ways can you place the 16 pins in the 4x4x4 cube?
Is it possible to have pins running in all three directions (left-right, front-back, up-down)?
Answer: No. You might see this intuitively, but if you want a proof:
Identify two pins going in different directions and orient the cube so these two pins are both horizontal -- one, we'll call PinA going left-right and one, we'll call PinB going front-back. These two pins cannot intersect and must exist in different horizontal layers of the cube. Every cell above/below PinA that is in PinB's horizontal layer must have a pin that is parallel to PinB. Why? Because pins through these cells cannot be vertical or they will intersect PinA and they cannot be parallel to PinA or they will intersect PinB, so they must be parallel to PinB. This means all cells in PinB's layer will have pins that are parallel to PinB, making it impossible to have any pins that are vertical.
So, we can orient the finished cube so all pins are in horizontal layers and all pins in a given layer are parallel, either left-right or front-back.
Can we determine the exact pattern?
Examining the orientation of holes and pins in the 16 pieces, we see that some pieces must stand "vertically" in layers that alternate direction: for example, left-right then front-back then back to left-right. Some other pieces require two adjacent layers to have parallel pins.
Numbering the layers from 1 on the bottom to 4 on the top, we can therefore arbitrarily orient the finished puzzle so pins in layers 1 and 2 go left-right, pins in layer 3 go front-back, and pins in layer 4 go left-right. Each layer must have 16 cells and 4 pins.
Organize the pieces according to which layers they may, or must, be put in.
Pieces A, C, E, F, G and H must "stand vertically" in layers 2, 3 and 4.
Pieces L, M and O must "stand vertically" in layers 1, 2 and 3 -- with one end obviously in layer 1.
Pieces I and J must each exist in layers 1 and 2.
Pieces D and P must each exist in either layers 2 and 3 or in layers 3 and 4.
Pieces B and K must lay flat in a single layer.
Piece N can either lay flat in some layer, or it can exist in layers 1 and 2.
The bottom layer (layer 1) seems to be most constricted - not many pieces can fill it. Why not start there?
The pieces that must be in layer 1 are I, J, L, M, and O.
With these, contributions to layer 1 are:
L - two cells and one pin
M - two cells
O - one cell
J - either 3 cells and one pin or 1 cell and no pins
I - either 3 cells and no pins or 1 cell and one pin
With the layer 1 pins running left and right, pieces I, J and L must each be either on the left side or right side. However, it is easy to see that piece L, with two cells and one pin in layer 1, cannot be on the same side as either I or J, so L must be on one side and I and J must be "nested" on the other side.
Let's arbitrarily place I and J "nested" on the left side. Together they contribute 4 cells and either two or zero pins to the layer 1. Piece L is then placed on the right side, contributing two cells and one pin to layer 1.
Adding in contributions from pieces I, J, L, M and O, layer 1 now has a total of 9 cells and either one pin or three pins.
We require exactly 16 cells and 4 pins in this layer, so the remaining cells and pins must come from pieces B, K and N. Each, if laid flat in layer 1, would contribute 4 cells and one pin. Alternately, piece N could either provide 3 cells and no pins or one cell and one pin. Since we require an odd number of additional cells, piece N is needed to be in one of these two orientations.
If N is oriented to contribute one cell and one pin, that would bring the total so far to 10 cells in layer 1. Six more cells would be required and you can't get that from pieces B and/or K laying flat. So N must provide three cells and no pins in layer 1.
The total with pieces I, J, L, M, N and O is now 12 cells and either one or three pins. The remaining cells and pin(s) must come from piece B or K laying flat. Obviously piece B or K can only contribute one pin, so I and J must be oriented on the left side such that they contribute two pins to layer one (rather than zero).
Piece K cannot be used in layer 1, as it contributes 3 cells to a left or right edge. Pieces I/J take up all four cells on the left edge and piece L requires 2 cells on the right edge. Piece B must therefore lay flat in the bottom layer.
Now we know the complete makeup of the bottom layer:
I - 1 cell and one pin
J - 3 cells and one pin
L - 2 cells and one pin
M - 2 cells
N - 3 cells
O - 1 cell
B - 4 cells and one pin
I and J are on the left, with pins on the front edge and back edge
M has two cells in layer 1 and can only be oriented such that one cell is on the right side.
N has three cells in layer 1 and can only be oriented such that one cell is on the right side. Furthermore, piece N must be on either the front edge or the back edge in order to leave room for piece B somewhere in the middle.
If piece N is on the back edge, piece L must be in the middle of the right side and piece M must be on the front side. But this leaves no place for piece B in the bottom layer.
So piece N must be on the front edge.
Piece B must now lay flat with two cells along the back edge. It can only orient one way as the other doesn't allow a place for piece M to go.
Piece L can now only go one place now, right side and towards the back.
Piece M can only go one place.
Piece O now has only one place to go. Some disassembly/reassembly is required: pull off L and M, put M and O together, slide on, then replace L.
The bottom layer is now complete.
Nine pieces remain to be placed (in layers 2, 3 and 4). The following section gives one version of logic that can be used to place these nine pieces. Following that is an alternate approach.
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METHOD 1 SOLUTION LOGIC
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Piece K lays flat in one layer (2, 3, or 4).
Pieces D and P each exist in two layers (2 and 3 or 3 and 4)
Pieces A, C, E, F, G and H all "stand vertically" in layers 2, 3 and 4
For brevity, we'll call pieces D and P the "L2" pieces as they occupy two layers, and we'll call pieces A, C, E, F, G and H the "L3" pieces as they occupy three layers.
The first 7 pieces already placed occupy 8 cells in layer 2 and each of the 6 L3 pieces occupy at least one cell in layer 2. This leaves at most only two remaining cells, so, there is not enough room for piece K to lay flat in layer 2. Piece K must therefore lay in layer 3 or layer 4.
The first 7 pieces contribute 4 cells/one pin to layer 3. Pieces 6 L3 pieces collectively contribute exactly 6 cells and 2 pins to layer 3, so 10 cells and 3 pins are now accounted for in layer 3. 6 more cells and 1 pin are needed. Pieces D, K and P therefore collectively contribute 6 cells and 1 pin to layer 3.
If piece K were to lay flat in layer 3, then it would contribute 4 cells and one pin to layer 3, leaving 2 cells and no pins to be provided by pieces D and P. While D and P could contribute one cell each, that would put piece P's pin also in layer 3, exceeding the number of pins allowed. So, piece K cannot lay flat in layer 3 (or layer 2). Piece K must lay flat in layer 4.
The 6 cells and one pin still required in layer 3 must therefore come from pieces D and P alone. So, pieces D and P each contribute 3 cells to layer 3, with piece P contributing the remaining the pin.
For both pieces D and P, their 4th cell could go in layer 2 or layer 4. However, the pieces would each have only one place they could go if their 4th cell were in layer 2, and in each case this placement would leave cells in layer 2 that could not be filled by remaining (L3) pieces. So, D and P have three cells in layer 3 and one cell in layer 4.
Piece D has only 2 places it can be with one cell in layer 4, but one of these blocks a cell in layer 2 from being filled with remaining (L3) pieces. Piece D's position is now known (on the front side, shifted to the left).
Piece P now has only 2 places it can be with one cell in layer 4, but one of these blocks cells in layer 2 from being filled with remaining (L3) pieces. Piece P's position is now known (on the left side, shifted to the back).
Only piece G can fill the remaining layer 2 cell under piece D.
The remaining 5 L3 pieces must fill the remaining 6 cells and 2 pins of layer 2. Four L3 pieces will contribute one cell and one L3 piece will contribute 2 cells.
One of the 6 remaining layer 2 cells has an already-filled cell above it and pieces A and H could possibly fill that cell. However, only piece A would also supply a pin to that layer. Without that pin no other pieces could supply a pin in that location without also contributing two cells to layer 2. Piece A must therefore be in that position. Note the puzzle will have to be partially disassembled and reassembled to get that piece in place. Disassemble G - D - (P) - I.
Piece K lays flat in layer 4. It will currently fit in four positions, but two of those positions leave cells in layers 2 and 3 that cannot be filled with an L3 piece, and one position leaves a isolated cell in layer 4 that cannot be filled. Piece K must lay flat in layer 4 on the left edge with two cells on the front edge.
The remaining 4 L3 pieces each fill one cell in layer 2. One of these also supplys a pin. Only E can do this and it can go in only one position. Note the puzzle will have to be partially disassembled and reassembled to get that piece in place. Disassemble G - D - JP - L
With only one cell each in layer 2, the remaining pieces, C, F and H can each only go one place. The puzzle will have to be partially disassembled and reassembled to get them in place. Disassemble G - D - JP - L - E. Put E and H together and L, F, C and G together and assemble EH - JP - D - LFCG.
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ALTERNATE LOGIC FOR PLACING THE LAST NINE PIECES
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The observations below can be used to place the nine pieces remaining after the bottom layer is complete. Placements in a given section below can be in any order. After they are done, the remaining observations can be made.
E is the only piece to cover the back pin on the second layer
G has only one possible placement
D is the only piece to cover the leftmost pin in the third layer
P has only one possible placement
Also P is the only piece to cover the backmost pin on layer 4
C is the only piece to cover the righmost pin in layer 3
F has only one possible placement
Also, piece F is the only one to cover any of the locations or the pin it covers
A is the only piece to cover the remaining pin in layer 2
H has only one possible placement
Also, H is the only piece to cover the remaining pin in layer 3
K has only one possible placement
Also, K is the only piece to cover the remaining locations and pin