Math-1345 · mathematics · Graph theory

Math-1345

Obstacle number of planar graphs

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Statement

Definitions and mathematical statement

For a finite simple graph $G$, an obstacle representation consists of distinct points $p_v\in\mathbb R^2$, one for each vertex, together with finitely many closed polygonal obstacles avoiding all these points. For distinct vertices $u,v$, require \[ uv\in E(G)\quad\Longleftrightarrow\quad (p_up_v)\cap\bigcup_{O\in\mathcal O}O=\varnothing, \] where $(p_up_v)$ is the open straight segment between their representing points. Thus obstacles block every nonedge and no edge. A polygonal obstacle is a filled simple polygon; arbitrarily many sides are permitted. Vertex points may be chosen in general position to eliminate collinearity degeneracies.

The obstacle number $\operatorname{obs}(G)$ is the minimum number of obstacles in such a representation, with zero allowed. A graph is planar if it has some drawing in the plane with no crossings between edge interiors. This planarity condition is a property of the abstract graph; an obstacle representation need not be a crossing-free planar drawing.

The source asks two questions: does a planar graph with obstacle number greater than one exist, and is there an absolute finite constant $C$ such that $\operatorname{obs}(G)\le C$ for every planar graph $G$? The second question concerns arbitrary graph order and arbitrary planar graphs, not a fixed embedding. The first asks for a single example and is logically distinct from proving that planar obstacle numbers are unbounded. The definition retains both original questions without treating their historical status as identical.

Short English statement

Determine whether planar graphs ever need more than one obstacle, and whether all planar graphs admit a uniformly bounded number of obstacles.

Sources

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