I've managed to solve C4 in a way similar to those mentioned, but using a
different transform.  I solve C3 a little differently than most folk because I
use only one (!) transform to do ALL of the edge work.  On C3 the result of
this transform is (looking at the edges only):
               a                            a
from	    b     c             to      ~d     b
	       d                           ~c

Using it and its inverse and a few simple conjugates of it I can do all the
work that is necessary for the edges. For me, at least, it has the twin
advantages of 1) being easy to remember (since it is fairly short and there is
only one of it), and 2) it has such bounded consequences that it is easy to
fix a cube without requiring a lot of planning (in the picture above, nothing
that is not shown changes: no other edges, no corners).

Anyhow, since that little move is a favorite of mine, I tried it on C4.  On C3
it comes in two flavors (the move and its inverse, or as it turns out, the
right-handed and left-handed versions).  On C4 it comes in four versions: the
move and its inverse, but each in a `left central slice' and `right central
slice' version.  Now for the fun part:  First off I started ignoring the
centers and I noticed that the move (lets call it `M') only moves around edges
in a single plane.  As I tried to figure out what the damn thing did I
discovered that it is a move of order 5!!!.  I find it truly hard to plan out
what happens when five cubes move around a little orbit, but I'm getting better
at predicting it.  The result is:
from:                       to:
	a   b                           a   b
    c           d                  ~d           e
    e           f                  ~f           g
	g   h                           c   h

With some pain I have been able to use ONLY M (and conjugates and powers of it,
of course) to get all the edges in place. Then I looked at M a little more to
see what it did with centers. This one is NOT planar, unfortunately, but is
simple enough to be useful: only two sets of centers are affected. If you are
doing M on the top (to get the above edge transform), only the top centers and
the rear centers are affected. And what happens is that there are two disjoint
three-cycles each involving two of the top center cubes and one of the rear
center cubes. thus, you can easily use conjugates of M to move cubes, one or
two at a time, into place on all of the faces around any particular top.
Voila: done!

An amusing thing about M: since its edge permutation is of order 5 and the
center permutation is of order 3, the damn thing is an order 15 move.

/Bernie