18
$\begingroup$

Consider the following tetracube (made from 4 unit cubes):

three unit cubes attached to adjacent faces of a central cube, which functions as a corner

It is easy to pack two of these tetracubes to form a 2x2x2 rectangular block. And from that simple packing it is easy to pack any rectangular block where all three side lengths are positive even integers. But is it possible to pack a rectangular block with these tetracubes where at least one of the side lengths is an odd number?

Improve this question
$\endgroup$
0

2 Answers 2

Reset to default
17
$\begingroup$

It is

possible

because

Here is a solution for a $4\times4\times5$ cuboid.
There are two $2\times2\times2$ sub-blocks which I've denoted by 2's, and the other pieces are denoted by letters. I don't know if this is the smallest example.

 A A B B   A E E B   G E K H   M K K N   M M N N
 A C C B   G E C F   G G H H   L O K H   M O O N
 2 2 C D   2 2 F F   L I J F   L L 2 2   P O 2 2
 2 2 D D   2 2 J D   I I J J   P I 2 2   P P 2 2

Thanks to mousetail in the comments, here is a picture:

enter image description here

Improve this answer
$\endgroup$
3
9
$\begingroup$

Jaap already gave the answer, here is a complement.

Indeed, the problem is ...

... solvable.

If my small program is correct, the smallest possible sizes are:
(4, 4, 5), (4, 5, 6), (2, 5, 8), (2, 5, 12), ...
plus those you get trivially by extending it with 2x2x2 cubes. Jaap nailed the smallest.

You might have fun solving the (2, 5, 8) and (2, 5, 12) cases. Solution in the next spoiler block.

The following picture represents the two superposed layers of the 2x5x8 case. The delimited shapes can easily be filled with tetracubes.

enter image description here

Next is the 2x5x12 case showing how it extends but non-trivially:

enter image description here

And a followup when 2x2x2 cuboids are forbidden.

Incidentally this constraint removes the triviality of the all-even cases. So these cases become intersting again.

Here are the number of possibilities I found for sizes up to 200 cubes.
(4, 4, 5): 8 solutions
(4, 4, 6): 8 solutions
(4, 5, 6): 80 solutions
(4, 6, 6): 136 solutions
(4, 4, 7): 24 solutions
(4, 6, 7): 640 solutions
(4, 4, 8): 48 solutions
(2, 5, 8): 8 solutions
(2, 6, 8): 32 solutions
(2, 7, 8): 8 solutions
(2, 8, 8): 224 solutions

It seems the 4x4x5 case has a unique solution up to symmetry.
The 2x5x8 case has 2 related solutions. They can be found easily by deduction.

The no-2x2x2-cuboid solution for the 2x5x8 problem is given by the following image. Just fill the paths with tetracubes.

enter image description here

And the solution to the 4x4x5 one is below:

A A N N A M M N C H M L B H H I B B I I
A P P N C P M O C C L L D H G L B G G I
R P S O Q Q O O E Q K T D D K K D F G J
R R S S R Q S T E E T T E F K J F F J J

Improve this answer
$\endgroup$
4
  • 2
    $\begingroup$ Nice. Out of curiosity: Does every solution contain 2x2x2-cubes? Do you think there are any solutions with two odd sides? $\endgroup$
    – loopy walt
    Commented May 29, 2023 at 3:11
  • 6
    $\begingroup$ @loopywalt Regarding your second question: Consider checkerboard coloring all cells from a single layer of a cuboid, and then coloring all other layers exactly the same (i.e. not alternating across layers). Then every placement of the tile takes up equal amounts of light and dark cells, which should imply a cuboid with two or more odd sides is unsolvable. $\endgroup$
    – user78949
    Commented May 29, 2023 at 3:57
  • 3
    $\begingroup$ I was also wondering about the two-odd case and couldn't find one. Thanks for the explanation. $\endgroup$
    – Florian F
    Commented May 29, 2023 at 8:39
  • $\begingroup$ I came up with a four-color coloring that also disproves the two-odd case. $\endgroup$
    – mathlander
    Commented Aug 24 at 21:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.