Abstract

Let $K$ be a closed polygonal curve in $\RR^3$ consisting of $n$ line segments. Assume that $K$ is unknotted, so that it is the boundary of an embedded disk in $\RR^3$. This paper considers the question: How many triangles are needed to triangulate a Piecewise-Linear spanning disk of $K$? The main result exhibits a family of unknotted polygons with $n$ edges, $n \to \infty$, such that the minimal number of triangles needed in any triangulated spanning disk grows exponentially with $n$. For each integer $n \ge 0$, there is a closed, unknotted, polygonal curve $K_n$ in $R^3$ having less than $10n+9$ edges, with the property that any Piecewise-Linear triangulated disk spanning the curve contains at least $2^{n-1}$ triangles.