CMPS 6640/4040 Computational Geometry Spring 2016 Ham Sandwich Theorem Carola Wenk 3/8/16 1 CMPS 6640/4040 Computational Geometry
Ham-Sandwich Theorem Theorem: Let P and Q be two finite point sets in the plane Then there exists a line l such that on each side of l there are at most | P |/2 points of P and at most | Q |/2 points of Q . 3/8/16 2 CMPS 6640/4040 Computational Geometry
Ham-Sandwich Theorem Proof: Find a line l such that on each side of l there are at most | P |/2 points of P . Then rotate l counter-clockwise in such a way that there are always at most | P |/2 points of P on each side of l . 3/8/16 3 CMPS 6640/4040 Computational Geometry
Rotation Left: 4 Right: 4 3/8/16 4 CMPS 6640/4040 Computational Geometry
Rotation Left: 4 Right: 3 Rotate around this point now 3/8/16 5 CMPS 6640/4040 Computational Geometry
Rotation Left: 4 Right: 4 3/8/16 6 CMPS 6640/4040 Computational Geometry
Rotation Left: 3 Right: 4 rotate 3/8/16 7 CMPS 6640/4040 Computational Geometry
Rotation Left: 3 Right: 4 rotate 3/8/16 8 CMPS 6640/4040 Computational Geometry
Rotation Left: 4 Right: 3 rotate 3/8/16 9 CMPS 6640/4040 Computational Geometry
Rotation Left: 3 Right: 4 rotate 3/8/16 10 CMPS 6640/4040 Computational Geometry
Rotation Left: 4 Right: 3 rotate 3/8/16 11 CMPS 6640/4040 Computational Geometry
Rotation Left: 2 Right: 4 Rotate around this point now 3/8/16 12 CMPS 6640/4040 Computational Geometry
Rotation Left: 4 Right: 4 Rotate around this point now 3/8/16 13 CMPS 6640/4040 Computational Geometry
Rotation Left: 4 rotate Right: 3 3/8/16 14 CMPS 6640/4040 Computational Geometry
Rotation Left: 3 Right: 4 rotate 3/8/16 15 CMPS 6640/4040 Computational Geometry
Rotation Left: 4 Right: 3 rotate 3/8/16 16 CMPS 6640/4040 Computational Geometry
Rotation Left: 4 Right: 4 3/8/16 17 CMPS 6640/4040 Computational Geometry
Proof Continued In general, choose the rotation point such that the number of points on each side of l does not change. k points m points rotate rotate m points k points 3/8/16 18 CMPS 6640/4040 Computational Geometry
Proof Continued Throughout the rotation, there are at most |P|/2 points on each side of l . After 180 rotation, we get the line which is l but directed in the other direction. Let t be the number of blue points to the left of l at the beginning. In the end, t points are on the right side of l , so | Q |- t -1 are on the left side. Therefore, there must have been one orientation of l such that there were t most | Q |/2 points on each side of l . 3/8/16 19 CMPS 6640/4040 Computational Geometry
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