TITLE: Can finite metric space embedded into planar graph?
QUESTION [9 upvotes]: Suppose $(X,d)$ is a finite metric, i.e. $|X|<\infty$.
Does there always exist a weighted planar graph $G=(V,E,w)$ and an injection $f:X\rightarrow V$ such that the following is satisfied? $$\forall x,y\in X, d(x,y)=\min\{\sum_{e\in P}w(e)\mid P\text{ is a path from }f(x)\text{ to }f(y)\}$$ 
Note that there could be some transfer point, i.e. $|V|\geq |X|$.
If not, what is the necessary and sufficient condition? I'm quite interested in this problem, and I could not find many relevant results. Most of the articles I found consider the opposite cases, say embedding graph metric to some normed space with distortion. I started to think about this question when I learnt tree metric.

REPLY [3 votes]: In general, the answer is negative. Consider the set $X$ of vertices of the bipartite graph $K_{3,3}$  as a metric space. Define the distance on the set $X$ by putting $d(x,y)=3$ if the vertices $x$ and $y$ belong to one part of $K_{3,3}$ and $d(x,y)=2$ if the vertices $x$ and $y$ belong to different parts of $K_{3,3}$. Since the graph $K_{3,3}$ is not planar, there exists different vertices $x_1, y_1, x_2, y_2\in X$ such that the vertices $x_1$ and $x_2$ belong to one part of the graph $K_{3,3}$, the vertices $y_1$ and $y_2$ belong to the other part the graph $K_{3,3}$, and in the drawing of the graph $G(X, E, w)$ the shortest weighted paths connecting $f(x_1)$ with $f(y_1)$  and $f(x_2)$ with $f(y_2)$  intersects in a point $p$. Then one of the paths $f(x_1)-p-f(x_2)$ or $f(y_1)-p-f(y_2)$ has the weighted length at most $2$, but $d(x_1,x_2)=d(y_1,y_2)=3$.