# Given a vertex $u$ (of bounded degree $k$) and another vertex $v$ in a planar graph, what is the smallest number of “curves”?

Given a vertex $$u$$ (of bounded degree $$k$$) and another vertex $$v$$ in a planar graph $$G$$, what is the smallest number of "curves" in the plane drawn from $$u$$ to $$v$$ such that no $$u$$--$$v$$ path in $$G$$ intersects each curve at a point other than $$u$$ or $$v$$?

(any finite bound is good as well)

Also, can anyone recommend a reference for graph drawings? Sorry if this is a stupid question.

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 degree-$$4$$ vertex in the 'right half' of the grid. Let $$\mathcal{I}$$ be a family of $$u$$--$$v$$ curves such that there is no $$u$$--$$v$$ path in $$G$$ which intersects each $$I \in \mathcal I$$ at a point other than $$u$$ or $$v$$.
Let $$C$$ be the cycle in $$G$$ consisting of the 'middle' vertical path $$P$$ of $$G$$ and the part of the boundary of $$G$$ to the left of $$P$$. Since $$C$$ topologically separates $$u$$ from $$v$$, every curve $$I \in \mathcal I$$ must intersect an edge of $$C$$ (for this proof, an edge contains its endpoints). Choose such an edge $$e(I)$$ for each $$I \in \mathcal I$$. Note that $$E(P) \subseteq \{e(I) \mid I \in \mathcal I\}$$, because for each $$e \in E(P)$$, there is a $$u$$--$$v$$ path in $$G$$ using all edges of $$E(C) \setminus \{e\}$$. Thus, $$|\mathcal I| \geq |E(P)|$$, which can be arbitrarily large if we increase the size of the grid.