TITLE: Minimal combinatorial data needed to define a polytope
QUESTION [9 upvotes]: Suppose I give a list of vertices $(v_1, v_2, ..., v_n)$, and a list of "adjacencies", i.e. pairs of vertices $(v_i,v_j)$. Does it exists a unique polytope that has this vertices and realises the adjacencies with 1-dimensional faces? 
I do not know if that is important, but each and every vertex has exactly d adjacent vertices, and I'd expect the polytope to be simple.
I should add some information. The data I gave above is represented by a graph of which I know
1) it's connected 
2) every vertex has the same valence d.
I may come up with further constraints but these are the first I can think of.

REPLY [6 votes]: In fact, elaborating on Guillermo Pineda-Villavicencio's answer, Jürgen Richter-Gebert's universality theorem for 4-polytopes shows that even in four dimensions, deciding whether a graph is realized by the vertices and edges of a simple polytope is equivalent to the existential theory of the reals. 
I am fairly sure that the proof of this theorem can be extended to graphs of degree 4, so for an arbitrary semi-algebraic set defined by inequalities, you can in polynomial time find a graph so that the polytope is realizable if and only if this set is non-empty (although you should check this before you cite it in a paper).
See Realization spaces of 4-polytopes are universal by Richter-Gebert and Ziegler, although I suspect this theorem might have a better exposition in Richter-Gebert's monograph Realization Spaces of Polytopes.