TITLE: Is every graph an edge-crossing graph?
QUESTION [12 upvotes]: Consider a circular drawing of a simple (in particular, loopless) graph $G$ in which edges are drawn as straight lines inside the circle. The crossing graph for such a drawing is the simple graph whose nodes correspond to the edges of $G$ and in which two nodes are adjacent if and only if the corresponding edges cross.
Example. The graph $G$ has four vertices (1–4) and three edges (a–c) where $a = 12$, $b = 13$, $c = 24$. In the circular drawing, $b$ and $c$ cross, so the crossing graph has three nodes and a single edge $bc$.

Here are my questions:

Is every simple graph the crossing graph of some circular graph drawing?
If not, how does a counterexample look like?
If yes, how can such a graph drawing be constructed?

REPLY [8 votes]: No, and you can see this from just a counting argument.
For determining which of the $ n $ chords of the circle intersect, it is enough to know the order of the $ 2n $ endpoints on the circle.  (You can assume that no two endpoints coincide.)  There are at most $ (2n)!/2^n $ such orders (the two endpoints of a chord aren't distinguishable).  On the other hand, there are exactly $ 2^{n(n-1)/2} $ simple graphs on $ n $ nodes.  The latter grows faster.  
You can even get an explicit bound from this: there are more simple graphs on 16 nodes than arrangement of 13 chords in the circle, so at least one of those graphs can't be generated this way.