We got reduction graph to planar bounded treewidth graph,
but this is unlikely to be true.


Let $H$, the planarizing gadget, be planar graph with four
distinguished vertices $u,u',v,v'$ on the outer faces.


Take graph $G$ drawn on the plane. Add new vertex $S$,
adjacent to all vertices of $G$. So far the diameter is
at most two.


Replace each pair of crossing edges $(u,u'),(v,v')$ by
new copy of the gadget $H$.


The resulting graph $G'$ is planar with diameter $D = 2\max(d(u,u'),d(v,v'))$
where $d$ is the distance in $H$.


The treewidth of $G'$ is $O(D)$, which is constant for fixed $H$.


Similar reduction with specially chosen $H$ is used to show
NP-hardness of problems for planar graphs.



> 
> What is wrong with this reduction?
> 
> 
> 


Correctness of the reduction is unlikely, because for
bounded treewidth graphs a lot of graph invariants are
computable in polynomial time and choosing suitable
gadget $H$ might give relation between invariants
of $G$ and $G'$, implying $P=NP$.


**Added** Example of 3-Coloring planarizing gadget is in [this lecture p.1](https://courses.cs.washington.edu/courses/cse431/14sp/scribes/lec15.pdf).


We cannot use it directly as gadget $H$ since the $S$ vertex increases
the chromatic number, but there is potential approach to use it
and then subdivide all edges $(S,t)$ twice, having the property
that $S$ can be any color in a valid 3-coloring by adjusting
the colors of the degree 2 vertices in the subdivision.



Reduction graph to planar bounded treewidth graph
If gadgets are applied to pairs of crossing edges as described then I'm not sure such gadgets exist as they probably preserve the presence of forbidden minors.


If gadgets are applied around crossing points then two vertices at distance $d$ in $G$ are not necessarily at distance $O(d)$ in $G'$, as a path of length $d$ in $G$ corresponds to paths of length $O(d) + O(\text{number of crossings})$ in $G'$.


