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Suppose we take the set of all cross sections of a cube and construct from them a set $A$ whose elements are sets of vertices of the cube as follows. If there exists a cross section of the cube which keeps those vertices on the same side of the plane (used during the cross section) then the set of those vertices is an element of $A$. Clearly, the set of a single vertex is an element of $A$, so is the set of two vertices connected by an edge. Is there a simple way to figure out if a set of vertices of the cube is an element of $A$ (maybe some way of labeling vertices based on cartesian coordinates might help)? Can we do this in every dimension?

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For example, using the picture above, the set composed out of vertices $B,F,H,E,A,D$ is an element of $A$ and the set composed out of vertices $E,A,C,G$ is definitely not an element of $A$, but I cannot tell whether the set composed out of vertices $E,D,C,A$ is an element of $A$. Thanks.

Yitzchak Shmalo
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1 Answers1

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I have an answer that most likely does not work well in higher dimensions, but here it goes anyways:

First perform a reduction step, that is, if you are testing a set $S$ with $n>4$ vertices, then instead test the complement $S$ which has $8-n$ vertices and negate the boolean result.

Now, to know if a set matches your criteria, first consider a graph $G$ of the vertices, where every vertex is connected with every other vertex, except with it's antipode. Essentially, it is identical to the graph you have in the question, except it also has an edge at every face diagonal, connecting both pairs of opposite vertices of each face.

Your set fulfills your criteria if $S\cap G$ is a complete graph (all vertices are connected with each other).

vero
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  • Ya, this seems to work, thanks (and E,D,C,A is not an element of A). This also trivially works in 2 dimensions (with a line partitioning the vertices and with n greater than 2). Obviously, in higher dimensions we are talking about the partition being done with a hyper plane. – Yitzchak Shmalo Apr 04 '18 at 23:48