Extensions of the Koebe–Andreev–Thurston theorem to sphere packing? The Koebe–Andreev–Thurston theorem states that any planar graph can be represented 
"in such a way that its vertices correspond to disjoint disks, which touch if and only if
the corresponding vertices are adjacent" (to quote Günter Ziegler, Lectures on Polytopes, Springer, 1995 p.117.
(See also the Wikipedia article, "Circle packing theorem.")
         

(source: uci.edu)

(Image due to David Eppstein, here.)

What is the corresponding statement for spheres in $\mathbb{R}^3$?
  Every graph $G$ satisfying property $X$(?) can be represented by touching spheres.

This is surely known—Thanks for pointers!
 A: According to corollary 4.6 of "Representing Graphs by Disks and Balls (a survey of recognition-complexity results)" these graphs are NP hard to recognize.
A: Yes, certain restrictions are well known.  One reference is Kuperberg & Schramm here ("Average kissing numbers for non-congruent sphere packings", 1994) which says that such graphs would have to have average degree <15.  A more recent reference is Benjamini & Schramm here ("Lack of Sphere Packing of Graphs via Non-Linear Potential Theory", 2009) which shows that certain low degree infinite graphs are not realizable this way.  
A: I can make a pretty good case for sphere packing of a finite graph in some $\mathbb R^n.$ From http://en.wikipedia.org/wiki/Circle_packing_theorem#Generalizations_of_the_circle_packing_theorem 
we learn that a nonplanar graph still induces a circle packing on a compact orientable surface of larger genus, the surface having constant curvature. For example, for planar graphs we could realize them either as circle packings in the plane, then simply make those into spheres (thereby done), or we could make a circle packing on $\mathbb S^2$ and then ask whether we can blow up those circles into spheres with the same tangency relationships. The answer is yes, for each circle, take the sphere that intersects $\mathbb S^2$ orthogonally in precisely that circle.
For torus graphs, I am betting on 3-spheres in $\mathbb R^4,$ where we can take the flat Clifford torus. For larger genus, from Nash embedding we can take our compact surface with constant curvature $-1$ in some   $\mathbb R^n.$
Well, somebody just posted an answer, it says so at the top of the page, maybe they actually know something.  

The circle packing theorem generalizes
  to graphs that are not planar. If G is
  a graph that can be embedded on a
  surface S, then there is a constant
  curvature Riemannian metric d on S and
  a circle packing on (S, d) whose
  contacts graph is isomorphic to G. If
  S is closed (compact and without
  boundary) and G is a triangulation of
  S, then (S, d) and the packing are
  unique up to conformal equivalence. If
  S is the sphere, then this equivalence
  is up to Möbius transformations; if it
  is a torus, then the equivalence is up
  to scaling by a constant and
  isometries, while if S has genus at
  least 2, then the equivalence is up to
  isometries.

EDIT: I suspect it is worth trying to disprove $\mathbb R^3$ for, say, $K_7,$ which is a torus graph. See Topological Graph Theory by Jonathan L. Gross and Thomas W. Tucker. If $K_7$ works in  $\mathbb R^3$ try $K_8$ and $K_9,$ the dog graph.
EDIT TOO: some anecdotal evidence, we can always place $K_n$ as the regular simplex in $\mathbb R^{n-1}.$ So now it is a question of how to selectively erase edges in order to get the graph we are actually given...Note that the articles on ball-touching in $\mathbb R^3$ all seem to be about balls of fixed unit radius.
A: These are ball-touching graphs; according to some google sources they seem to be NP hard to recognize.
A: I don't know of any results on this. There are some reformulations in terms of Gram matrices, but I don't know if they help.
For $K_6$, there is no realization by touching spheres. Suppose you had such a realization. The property of having tangent spheres realizing a graph is invariant under Mobius transformations. In particular, one may perform a Mobius transformation sending the tangency between two spheres to infinity in $\mathbb{R}^3$. The two spheres are sent to parallel planes, and the other 4 spheres are tangent to both of these planes and to each other. In particular, the midplane between these two planes intersects the other 4 spheres in 4 tangent circles of equal radius. But 4 equal radius circles in the plane cannot be simultaneously tangent. So $K_6$ cannot be realized by tangent spheres.  
