By Godfried T Toussaint

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**Example text**

2) From q it can be determined in constant time whether, in case a last vertex pg of P with respect to H exists, pg is a leaf of Tq. Simple On-Line Algorithms for Convex 33 Polygons Proof: We will prove the first part of this lemma. The proof of the second part is similar. If g is a leaf oi T, then the vertex v of P corresponding to q is the first point with respect to H iff v is in the interior of H or on the line L , but its counterclockwise predecessor, w, does not. previous. Now assume 9 is a non-leaf and non-root node ot T.

E. it is O (logn), where n is the number of vertices of P . This bound relies heavily on the fact that the test CONTAINS_LTP can be performed in constant time. 3 D. Avis, H. ElGindy and R. Seidel 28 So far we have shown how the left and right tangent points for some point p can be determined quickly. Now it remains to show bow the tree T representing P has to be modified to a tree T' that represents P1 . Let us assume that the point p lies outside the polygon P and let Up and rip denote the left and right tangent point of P for p , respectively.

If the straight line L intersects the closed line segment between a and s , then pf is in Cq if s lies in H and 0 is not in the interior of H, or in case both 8 and a lie on L , if b lies in / / . This covers all the cases where L intersects the closed line segment between a and 8 . Now assume L does not intersect this line segment. We want to decide whether Tq is the subtree to be checked for an intersection between L and the convex polygon P . Using the convexity of this chain we can check for intersection as follows: Let z be a point on L , let La be the straight line parallel to L through point a , and let Ls be the line parallel to L through point 8 .