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Problems related to graph drawing such as crossing numbers, layout designs, and intersection graphs.
3
votes
Accepted
Bounds on lengths of intervals in bounded-degree interval graphs
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 …
7
votes
Is it possible that every edge in a 1-planar drawing with minimum number of crossings is cro...
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 …
17
votes
Accepted
Can all crossings in a graph be moved to one point?
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 …
1
vote
Accepted
Given a vertex $u$ (of bounded degree $k$) and another vertex $v$ in a planar graph, what is...
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 …
11
votes
Accepted
Bounds on chromatic number of $k$-planar graphs
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 …
5
votes
Example to show pairwise crossing number is not equal to crossing number
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 …
15
votes
Is every graph an edge-crossing graph?
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 …