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I have got this task at high-school math-contest seminar.

The theme is graphs.

Let us have $n \in \mathbb{N}$ and the cube $ n \times n \times n$. An ant can go over a diagonal of little cubes, but it can't turn at the intersection of two little diagonals.

Is it true that if $n$ is odd then an ant can't visit every little facet (all of $6 n \times n$) exactly once? I didn't find the path for $n = 1,3$

I don't see a particular method to solve this task, because you can't treat facets as vertices. You can't treat them as edges also. It isn't an Euler or Hamilton cycle. I don't see an invariant that should be saved as well. The graph is planar.

Jean Marie
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Lada Dudnikova
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2 Answers2

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Hint: can you see why the graph the ant is moving on is actually disconnected, with two components? Each facet is covered by two edges, but these edges are in different components. So the ant wants to cover all the edges of the component it is in.

Especially Lime
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Another explanation in the same vein as @Especially Lime placing in evidence an invariant.

This invariant is the parity of the sum of coordinates of the two endpoints of a given diagonal edge.

Case $n=1$ : Think to the cube as being $\{0,1\}^3$.

A diagonal path connects a vertex to another vertex by changing two coordinates (a change equivalent to a boolean complementation) ; this induces a partitionning of the 8 vertices into 2 classes of 4 that cannot be connected between themselves :

$$\begin{matrix}(0,0,0) \ \ &\|& \ \ (1,1,1)\\ (1,1,0) \ \ &\|& \ \ (0,0,1)\\ (0,1,1) \ \ &\|& \ \ (1,0,0)\\ (1,0,1) \ \ &\|& \ \ (0,1,0) \end{matrix}$$

This vertices' partitionning induces a partitionning of diagonals.

Another way to say this is that the left set has a parity-check (sum of coordinates) $0$ whereas it is $1$ for the second.

Case $n=2$ : This time, the vertices a subset of $\{0,1,2\}^3$ (we must take out interior point $(1,1,1)$). We have the same principle : if a diagonal in one face connects two points it means that

  • one coordinate remains fixed

  • the two others move by one unit.

Let us consider the cases where the coordinate with a fixed value is $z=z_0$ (with $z_0=0$ or $z_0=2$); the only possible "moves" are of type

$$(x,y,z_0) \leftrightarrow (x + e_1,y+e_2,z_0) \ \ \text{where} \ e_1,e_2=\pm1.$$

In all cases, the sum of coordinates of connected points either staies the same or is changed by $2$ units.

(same reasoning if the fixed coordinate is $y$ or $x$).

Therefore the $3^3-1=26$ vertices can be separated in the same way as for the case $n=1$ into two subsets that cannot "communicate", according to the parity of the sum of their coordinates. As a consequence, we have an induced partionning of the diagonal segments.

What we have done for these two cases extends immediately to the general case.

I am indebted to @Misha Lavrov who has spotted a default in my presentation.

Jean Marie
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  • In the $n=1$ case, isn't our goal to visit all $6$ faces of the cube rather than all $8$ corners? – Misha Lavrov Aug 30 '19 at 14:50
  • @Misha Lavrov : you are right : I haven't clearly said that the partition of vertices induces an impossibility for diagonal edges. I will do it. – Jean Marie Aug 30 '19 at 18:43