By Jurek Czyzowicz, David Ilcinkas, Arnaud Labourel (auth.), Haim Kaplan (eds.)

ISBN-10: 3642137318

ISBN-13: 9783642137310

This ebook constitutes the court cases of the twelfth foreign Scandinavian Workshop on set of rules thought, held in Bergen, Norway in June 2010.

**Read Online or Download Algorithm Theory - SWAT 2010: 12th Scandinavian Symposium and Workshops on Algorithm Theory, Bergen, Norway, June 21-23, 2010. Proceedings PDF**

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**Additional info for Algorithm Theory - SWAT 2010: 12th Scandinavian Symposium and Workshops on Algorithm Theory, Bergen, Norway, June 21-23, 2010. Proceedings**

**Example text**

From before we also know that they both lie left of the oriented line vi vj . We now get to the central lemma that essentially states that the existence of a triangle witness is necessary and suﬃcient for a pair of vertices to see each other. Lemma 4. Let vi , vj ∈ V with |chain(vi , vj )| > 2. There is a triangle witness vl for (vi , vj ), if and only if {vi , vj } ∈ Evis . 22 Y. Disser, M. Mihalák, and P. Widmayer Fig. 8. Sketch of the deﬁnitions in the proof of Lemma 4 Proof. If {vi , vj } ∈ Evis , because any polygon can be triangulated, there must be a vertex vl ∈ chain(vi+1 , vj−1 ) for which both edges {vi , vl } and {vl , vj } are in Evis .

If and only if this is the case, we update Evis , F, B with the edge {vi , vi+k }. For a listing of the triangle witness algorithm see Algorithm 1. In the following we prove the correctness of the triangle witness algorithm. For this we mainly have to show that having a triangle witness is necessary and suﬃcient for a pair of vertices to see each other. To show this, we will need the notion of blockers and shortest paths in polygons. Fig. 6. Illustration of the generalized angle sum condition of Deﬁnition 1.

As no two vertices can be further apart than n2 steps along the boundary, this immediately implies that Reconstructing a Simple Polygon from Its Angles 23 Evis eventually contains exactly the edges of the visibility graph. More precisely, we inductively show that after step k of the iteration, F [vi ] contains the vertices of chainvi (vi+1 , vi+k ) and B[vi ] contains the vertices of chainvi (vi−k , vi−1 ) for all vi ∈ V . For sake sake of simplicity we write F [vi ] = chainvi (vi+1 , vi+k ) and B[vi ] = chainvi (vi−k , vi−1 ).

### Algorithm Theory - SWAT 2010: 12th Scandinavian Symposium and Workshops on Algorithm Theory, Bergen, Norway, June 21-23, 2010. Proceedings by Jurek Czyzowicz, David Ilcinkas, Arnaud Labourel (auth.), Haim Kaplan (eds.)

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