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How do I show that a graph created by adding a single edge between antipodal vertices of a 3-dimensional cube is not hypercube embeddable?

Section 2.1 in this paper states that

A connected graph $G$ can be embedded into $Q_n$ (hyper cube of $n$ dimensions) if and only if it is possible to label the edges of $G$ with the integers $\{1,\dots, n\}$ such that

  1. Edges incident with a common node are of different labels;
  2. In each path of G there is some label that appears an odd number of times; and
  3. In each cycle of G no label appears an odd number of times.

Each such labelling gives rise to a (not necessarily unique) embedding in which the label of an edge is the dimension along which its endpoints differ.

Which condition does it fail to satisfy?

Benjamin Wang
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Ning Bao
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1 Answers1

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Label the vertices of $Q_3$ as $\{000,001,010,011,100,101,110,111\}$, with adjacencies between vertices that differ in one position, and let's say that $000$ and $111$ are joined by an edge in the modified graph.

Suppose that the edge between $000$ and $111$ is given label $k$ in some labeling that satisfies all these conditions. Then property (3) demands that label $k$ occurs an even number of times on any cycle using that edge; in other words, every path from $000$ to $111$ must use label $k$ an odd number of times.

Consider the path $000, 001, 011, 111$. The edge from $000$ to $001$ cannot have label $k$: there can only be one edge with label $k$ out of $000$. Similarly, the edge from $011$ to $111$ cannot have label $k$. So the only possible edge with label $k$ is the edge from $001$ to $011$.

But now consider the path $000, 001, 101, 111$. Vertices $000$, $001$, and $111$ already have a label-$k$ edge that's not part of this path. So there is nowhere to put the label-$k$ edge that this path needs; contradiction.

Misha Lavrov
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