There seems to be no solution for packing a 3xMxN-box with copies of this tetracubetetracube.

Does anybody know of a proof that this is impossible or is this still an open question?


1 Answer 1


Is it possible:

No. The reasoning is below:

First of all, $M$x$N$ must be divisible by 4 in a $3$x$M$x$N$ box because you're using shapes made up of 4 cubes. If we paint all the unit cubes like a 3D checkerboard, half of the "middle cubes" of the tetracubes must be of the opposite color of the rest. For this to be possible, $M$x$N$ must be divisible by 8.

Let's say there are $t_{bottom}$ tetracubes with three cubes at the bottom on the box's base, and $o_{bottom}$ with one. The base's area is $3t_{bottom}+o_{bottom}$. If $3t_{bottom}+o_{bottom} = MN$, then $t_{bottom}>o_{bottom}$ and $t_{top}>o_{top}$.

$3t_{bottom}+o_{bottom} = MN$
$3t_{top}+o_{top} = MN$
$t_{bottom}+t_{top}+3o_{top}+3o_{bottom} = MN$
$3(MN-3t_{bottom}) + 3(MN-3t_{top}) + t_{bottom}+t_{top} = MN$
$6MN-8t_{bottom}-8t_{top} = MN$
$5MN = 8t_{bottom}+8t_{top}$
$t_{top} = 5MN/8 - t_{bottom}$

Either the top or bottom layer has $5MN/16$ or more, and each has at least $5MN/8 - MN/3 = 7MN/24$. Both layers have a combined $MN/8$ lone legs of the tetracubes.

If two tetracubes combine into a cube, the latter can't be at a corner.

Let's look at the edges. There are $4*(N+1)*M + 4*N*(M+1) + 3*(N+1)*(M+1) = 11MN + 7N + 7M + 3$ in/on our prism.

Each tetracube has $48$ edges including the ones counted multiple times within the tetracube itself and there are $3MN/4$ tetracubes, so the product is $36MN$.

We also know that there are $5MN/8$ top- and bottom-centered ("justified") tetracubes in total and two edges at the top of each overlap with the surface.

Also, at least the 3 "deep" edges on every tetracube are counted only twice as well as the top/bottom 6 edges on each justified tetracube (the latter can overlap though). Assuming the rest are counted four times, the total count is:

4 times: $11MN + 7N + 7M + 3 - (23MN/4) = 21MN/4 + 7N + 7M + 3$
Twice: $18MN/4$
Once: $5MN/4$.

The total is at most $41MN/4 + 21MN + 28N + 28M + 12$.

$41MN/4 + 21MN + 28N + 28M + 12 >= 36MN$
$MN/4 + 28N + 28M + 12 >= 5MN$
$28N + 28M + 12 >= 19MN/4$
$112N + 112M + 48 >= 19MN$

Even assuming it could potentially work for sufficiently small $N$ and $M$ values, the fact that it's not applicable for any larger value (not even multiples) means it's impossible for any $N$ and $M$.

  • $\begingroup$ "Each tetracube has 48 edges": Here, several edges are counted more than once, e.g., the "deep" edges are counted three times because they are shared by three cubes. Each tetracube has only 36 edges, so $3MN/4$ tetracubes contribute a total of only 27MN edges. $\endgroup$ Aug 31, 2023 at 4:52

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.