  A 1-factorization $\mathcal{M} = \{M_1,M_2,\ldots,M_n\}$ of a graph $G$ is
called perfect if the union of any pair of 1-factors $M_i, M_j$ with $i \ne j$
is a Hamilton cycle. It is called $k$-semi-perfect if the union of any pair of
1-factors $M_i, M_j$ with $1 \le i \le k$ and $k+1 \le j \le n$ is a Hamilton
cycle.
  We consider 1-factorizations of the discrete cube $Q_d$. There is no perfect
1-factorization of $Q_d$, but it was previously shown that there is a
1-semi-perfect 1-factorization of $Q_d$ for all $d$. Our main result is to
prove that there is a $k$-semi-perfect 1-factorization of $Q_d$ for all $k$ and
all $d$, except for one possible exception when $k=3$ and $d=6$. This is, in
some sense, best possible.
  We conclude with some questions concerning other generalisations of perfect
1-factorizations.
