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A polycube is a solid three-dimensional connected figure formed by joining one or more unit cubes face to face. Polycubes can be thought of as the three-dimensional generalization of polyominoes.

Can a 3x3x3 cube be packed with three congruent polycubes (rotation allowed but not mirroring)? The answer is yes and can easily be done with the following polycube. The fact that the unit cubes are of two colors is simply a visual aid and can be ignored.

1x3x3 block made of nine unit cubes

My question is, “Can a 3x3x3 cube be packed with three congruent polycubes (rotation allowed but not mirroring) using any other polycube?”

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  • $\begingroup$ I don't understand. Why can't you stack the 3x3x1 polycubes to form a 3x3x3? No rotation or mirroring is needed. Or do they need to be all different, but congruent? $\endgroup$ Dec 21, 2023 at 3:04
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    $\begingroup$ @DmitryKamenetsky "using any other polycube" $\endgroup$
    – RobPratt
    Dec 21, 2023 at 3:17
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    $\begingroup$ @DmitryKamenetsky You are right, 3x3x1 polycubes can pack a 3x3x3 cube. But can the cube be packed using three congruent polycubes that aren’t 3x3x1 polycubes? $\endgroup$ Dec 21, 2023 at 5:03
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    $\begingroup$ Thanks, I understand now. $\endgroup$ Dec 21, 2023 at 5:30
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    $\begingroup$ @TheFootprint I am not aware of any simpler solution to my puzzle. $\endgroup$ Dec 29, 2023 at 3:24

3 Answers 3

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It is

impossible.

Let's call a boxing of a polycube the smallest parallelepiped that contains it. As ulutku observed, it is impossible for our polycube to have a $3 \times 3 \times 3$ boxing, because its two clones can differ only by rotation, and rotation doesn't change the position of the $(2,2,2)$ unit cube (so it's either empty or overlaps 3 times). It clearly can't have a $2 \times 2 \times 2$ boxing, and one can easily convince oneself that a $2 \times 2 \times 3$ boxing is also too small for three of them to cover the whole cube: there are two ways (up to rotation) to cover a face with $\le 3$ boxings, and they will leave empty space. (Obviously, no boxing can have a dimension equal to $1$, else the polycube would be a subset of the forbidden 3x3x1 polycube.)

So let's now focus on polycubes with a $3 \times 3 \times 2$ boxing. That is, they are contained in a structure like this:

enter image description here

Let's narrow down further the possible polycubes. In general, the vertices of the big cube must be vertices of the constituent polycubes (a vertex of a polycube is a unit cube that touches a vertex of its boxing). I claim that this implies that there must be exactly $3$ vertices on the green face and $2$ vertices on the blue face (without loss of generality).

Suppose there were $4$ vertices on the green face. Then the three green faces would have to be parallel. Therefore, two of them must be facing each other, forming a "sandwich", i.e. one polycube and a clone with a 180° rotation about one axis parallel to its green face. The third one would face in the same direction as one of the other two. Well, this situation is impossible: if the sandwich is not full, there is no way the third polycube can fill it (its boxing can't fill it), and if it is full, the third polycube will have holes in its green side (because only a sandwich can be full then).

If there were $2$ vertices on the green face and $2$ vertices on the blue face, it would be impossible to fill a cube's 8 vertices with just 3 polycubes, because only the vertices from one face per polycube can be vertices of the big cube. Note that $3$ vertices on the green face and $3$ on the blue face is also impossible, because then we wouldn't be able to fill exactly 8 vertices. Similar arguments allow us to discard lower vertices combinations, leaving us with $3$ vertices on the green face and $2$ vertices on the blue face. By playing with these polycubes, one can convince oneself that there are only two possible vertex configurations (up to rotation).

One is the following (the gray unit cubes are just for grouping vertices in a polycube, they are not the only possible way to connect vertices):

enter image description here

There can be no real configuration with these characteristics. The green side of C is either parallel to the green side of A, therefore overlapping with the blue side of A, or the green side of B, therefore overlapping with the blue side of B. In both cases, on a single 3x3x1 parallelepiped there would have to be 5 vertices!

The other is this:

enter image description here

Why can't I rotate B on its green face? Because then C's blue face vertices won't be on a 3x1x1 polycube, and they must be on a 3x1x1 polycube or else there will be at least 5 unit cubes on a blue side, and this is impossible because there are always at least 5 cubes on a green side. Now back to our image, notice that A's and B's blue sides must overlap, this means that the four vertices of the 3x3x1 parallelepiped between A's and B's green sides are all filled. One of these is the unit cube that must be part of C's blue side (per the previous argument), a contradiction.

Note that this argument

doesn't use the fact that reflection is not allowed. So it seems that this would be impossible even if reflection were allowed.

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It is not possible.

First of all the polycube’s we are looking for can’t have length 3 in all of x, y, z axes because then either all of them controls (2,2,2) or non of them. Then they to have min. length 2 in all axes because when length is 1, they would be same as given 3x3x1 polycube.

Let’s assume that there is a possible solution polycube. First let’s select one of it‘s length 2 axes. Polycube consists of 9 unit cubes, so it‘s broken down in this 2 layers in unequal numbers. Without losing the generality let‘s name the layer with more unit cubes bottom layer & less unit cubes top layer. Not three of the polycubes lie horizontally (so not all of them has only length 2 in z-axis) (or vertically but we can rotate the end cube so we can assume that they are horizontal) because 1. & 3. Polycubes top layers lie on z=2 plane, 2. Polycubes top layer lies on z=1 or z=3 plane, no polycube’s top layer lies on either z=1 or z=3 plane. So without loss of generality 2 polycubes bottom layer lie horizontally & 1 polycubes bottom layer lie vertically.

Bottom layers can’t have 7 or 8 unit cubes because then horizontal one can’t fit in 3x3x3 cube.

Bottom layers can’t have 6 unit cubes because if you fit 6 unit cubes in 3x3 grid then you will find 3 unit cubes long line in bottom layer like y=3 ( only other possibility is [0,1,1,1,1,0,1,0,1] which makes putting the third layer vertically impossible). If this lines lie in the middle (like y=2) then who controls 2x2x2 is problem. If one of horizontal ones control, then where does vertical ones bulge in middle lie?

So in bottom layer 3 units long lines lie in sides, only possible via occupying 2 sides, which makes it impossible to fit vertical polycube in the end cube.

Last option is 5 unit cubes in bottom layer, we don’t want to occupy middle line with 3 units. So possible solutions are enter image description here

In 1,2,3,5,6,7,8,9 who controls the 2x2x2 is problem. In 4,10 fitting the vertical one is impossible.

So there is no such solution!

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Looking at parity with the middle square black we have 13 black and 14 white squares.
That means the polycube must be 5-4, with a 5-black polycube in the center, and 2 5-white polycubes with their 'center' 1 square shifted from it.
(note that 3 tiles shifted is not possible since 9 squares do not fit in 2x2x2 or 1x2x3)

Possibility 1: opposite shifts (i.e. in the same dimension) . - This means the original must fit in 1x3x3 i.e. leads to the trivial solution.

Possibility 2: the shifts are in different dimensions. Then the polycubes are within a 2x2x3 area, and can cover at most 2 corners each/6 corners total.
So there is no nontrivial solution

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  • $\begingroup$ Why do you only consider shifting? There are ways to change the color parity by a combination of shifting and rotating $\endgroup$ Dec 26, 2023 at 16:51
  • $\begingroup$ I consider the displacement of the 'center square', the rotation (of the entire poly-cubes) does not matter for the (theoretical) maximum amount of corners reached. $\endgroup$
    – Retudin
    Dec 26, 2023 at 17:13
  • $\begingroup$ It seems that only by assuming shifting only can you make those conclusions in possibility 1 and 2 $\endgroup$ Dec 26, 2023 at 17:32

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