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No, this is not always possible. Lemma. Let $G$ be an $n$-vertex graph with at least $3n-2$ edges. Then $G$ cannot be drawn in the plane so that all crossings occur at the same point. Proof. We make …

The answer to 1 is no. To see this, note that every edge-crossing graph is a string graph. A string graph is a graph which is the intersection graph of arbitrary curves in the plane. However, there …

Pach and Tóth proved that if $G$ is a $k$-planar graph (with $k \geq 1$), then $|E(G)| \leq 4.108 \sqrt{k} |V(G)|$. Thus, every $k$-planar graph has a vertex of degree at most $\lfloor8.216 \sqrt{k …

It is not possible that in an optimal drawing of a 1-planar graph, every edge is crossed. Here is a proof. Suppose not and let $G$ be a smallest counterexample. I claim that $G$ is 2-connected. If …

As far as I know this is still an open problem. It is listed as an open problem in the paper Which crossing-number is it anyway? by Pach and Tóth, and also in the introduction of this more recent sur …

Yes, we may take the function to be $2\Delta$. Lemma. Every interval graph $G$ has an interval representation where all intervals have length between $1$ and $2\Delta$, where $\Delta$ is the maximum d …

Unfortunately, there is no finite bound, even if both vertices have bounded degree. To see this, consider a large grid graph $G$ with $u$ a degree-$4$ vertex in the 'left half' of the grid and $v$ a …