Using a second recursive construction, we are able to improve still
further on this path, thus obtaining the best known results on the
notorious "middle levels" problem (to show that the graph formed by
the middle two levels of B_2k+1 is hamiltonian for all k). We show
in fact that for every epsilon > 0, there is an h >= 1 so that
if a hamilton cycle exists in the middle two levels of B_2k+1 for
1 >= k <= h, then there is a cycle of length at least (1-epsilon)N(k)
for ALL k >= 1, where N(k) = 2(k+1 choose k). Using the fact that
hamilton cycles are currently known to exist for 1 <= k <= 11,
the construction guarantees a cycle of length at least .839 N(k) in
the middle two levels of B_2k+1 for all k.
Paper available at:
ftp://dimacs.rutgers.edu/pub/dimacs/TechnicalReports/TechReports/1993/93-05.ps