Let G be a finite, planar, disconnected graph with two components.
Is it always possible to add an edge to the graph to make it connected and still planar?
Let G be a finite, planar, disconnected graph with two components.
Is it always possible to add an edge to the graph to make it connected and still planar?
On
Let $C,C'$ the two connected components of $G$. By Fáry's theorem, there is an embedding into the plane that maps edges to line segments. We will identify $G$ with its image under such an embedding. WLOG, the convex hull $H$ of $C$ and the convex hull $H'$ of $C'$ have empty intersection (just translate $C'$ out of $H$). Let's call $d$ the Euclidean distance.
Both $H$ and $H'$ are polygons whose edges are edges of $C$ and $C'$ respectively. Take vertices $v, v'$ in the boundaries of $H,H'$ respectively such that $d(v, v')$ is minimal.
Clearly, if you add the edge $(v,v')$ to your graph, it is connected and still planar.
The answer is yes.
Rigorous proofs regarding planarity of a graph usually use non-trivial theorems such as Jordan curve theorem, and theorems about embedding, which I will avoid, and therefore will give intuition, and not a full answer.
Constructive Proof
Denote the two disconnect planar graphs by $G,H$. Since they are both planar, one could embed them on the plane $\mathbb R^2$ with straight edges (and not just curves) This is equivalent to stating:
Now Denote by $F_1,F_2$ thier outer faces, with bouderies $A\subset V(G),B\subset V(H)$.
Proof:
Pick $v\in G$ whose $x$ coordinate is minimal (maybe not unique). Pick $u\in H$ whose $x$ coordinate is maximal (maybe not unique). Connect them by a straight edge.
By the choice of $v,u$, and the fact all previous edges were segments, we can conclude the new graph is also planar, by the new obtained embedded.
Existence Proof
One a second thought, I think it might be easier to use Kuratowski's theorem and Wagner theorem and prooving:
A proof sketch would be: $$\forall v\in G\ \ \deg_H(v)=0\\ \forall u\in H\ \ \deg_G(v)=0\\ $$