Abstract

We show that any planar straight line graph (PSLG) with $n$ vertices has a conforming triangulation by $O(n^{2.5})$ nonobtuse triangles (all angles $\leq 90^\circ$), answering the question of whether any polynomial bound exists. A nonobtuse triangulation is Delaunay, so this result also improves a previous $O(n^3)$ bound of Eldesbrunner and Tan for conforming Delaunay triangulations of PSLGs. In the special case that the PSLG is the triangulation of a simple polygon, we will show that only $O(n^2)$ triangles are needed, improving an $O(n^4)$ bound of Bern and Eppstein. We also show that for any $\epsilon >0$, every PSLG has a conforming triangulation with $O(n^2/\epsilon^2)$ elements and with all angles bounded above by $90^\circ + \epsilon$. This improves a result of S. Mitchell when $\epsilon = 3 \pi /8 = 67.5^\circ $ and Tan when $\epsilon = 7\pi/30 =42^\circ$.

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