  Consider the Hales-Jewett theorem. The $k$-dimensional version of it tells us
that the combinatorial space $\mathcal{U}_{M, \Lambda} = \{ \eta \mid \eta: M
\to \Lambda \}$ has, under suitable assumptions, monochromatic $k$-dimensional
subspaces, where by a $k$-dimensional subspace we mean there exist a partition
$\langle N_0, N_1, \cdots, N_k \rangle$ of $M$ such that $N_1, \cdots, N_k \neq
\emptyset$ (but we allow $N_0$ to be empty) and some $\rho_0: N_0 \to \Lambda$,
such that the subspace consists of those $\rho \in \mathcal{U}_{M, \Lambda}$
such that for $0<l<k+1, \rho \restriction N_l$ is constant and $\rho
\restriction N_0= \rho_0.$
  It seems natural to think it is better to have each $N_{l}, 0<l<k+1$ a
singleton. However it is then impossible to always find monochromatic
$k$-dimensional subspaces (for example color $\eta$ by $0$ if
$|\eta^{-1}\{\alpha \}|$ is an even number and by $1$ otherwise). But modulo
restricting the sign of each $|\eta^{-1}\{\alpha \}|$, we prove the parallel
theorem -- whose proof is not related to the Hales-Jewett theorem. We then
connect the two numbers by showing that the Hales-Jewett numbers are not too
much above the present ones. This gives an alternative proof of the
Hales-Jewett theorem.
