is the following true?

  • for $n\in\mathbb{N}$ every Hamilton cycle in an $n$-dimensional hypercube $Q_n$ there exist $2^{n-1}$ edges that are mutually parallel
  • $Q_2$ is the only case in which every Hamilton cycle contains two such sets of edges, in all other cases the set of edges is unique.
  • $\begingroup$ yes to the first question (I'm not sure I get the statement of the second one). imagine $Q_n$ as having vertex set $\{0,1\}^n$. if I take an edge that flips the $i$-th bit, I must have another edge flipping this bit to close the cycle. so you can pair up the edges so that each pair "flips the same bit" (i.e., they are parallel). $\endgroup$ Commented Jun 20 at 17:07
  • $\begingroup$ or maybe more evocative is to imagine $Q_n$ as the Cayley graph of $\mathbb{F}_2^n$ generated by the basis vectors $e_1,\dots,e_n$. if $l_1,\dots,l_{2^n}$ are the labels of the edges in a cycle, we must have $\sum_i l_i = 0_{\mathbb{F}_2^n}$ (since you return to the start), whence each label is used an even number of times. $\endgroup$ Commented Jun 20 at 17:10
  • $\begingroup$ @ZachHunter: I think the question asks for one set of $2^{n-1}$ edges all of which are parallel, whereas your argument gives $2^{n-1}$ sets of 2 edges each. $\endgroup$ Commented Jun 20 at 19:50

1 Answer 1


If there is a set of $2^{n-1}$ mutually parallel edges, then it is unique by the pigeonhole principle: a Hamilton cycle has $2^n$ edges, and contains at least one edge in each "direction".

However, for $n\geq 4$ it is possible that there is no such set. Below is a sketch for $n=4$ which is easy to generalise (I did not include all edges in the matching between the two 3-cubes so the drawing wouldn't get too messy).



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