Abstract
Proof: Let T be any spanning tree in Qn. For any vertex v Of Qn there is an antipodal vertex v in Qn. There is a unique path in T from v to v; orient its first edge towards v and repeat this process for every vertex v. Since the tree T has fewer edges than vertices, the pigeonhole principle implies there is an edge (u, v) which has received two orientations. The distance between u and -u (and between v and v) is n in Qn and hence is at least n in T. Finally, if (u, v) is an edge in Qn then (-u, v) also is an edge in Qn. Hence (u, v) and (-u, v) and the edges of the u -u and v v paths yield the cycle sought. Q
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