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Published byKaela Weymouth Modified over 2 years ago

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**Why Graham-Scan Needs to Sort Vertices Before Scanning**

Bill Thies November 12, 2004 6.046 Recitation Supplement

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**Graham Scan: Sorting Step**

The first stage of Graham-Scan sorts the points by their polar angle from the bottom-left vertex, p0: In recitation, we asked: would the scanning phase of Graham-Scan work on any simple polygon? That is, can you omit this sorting phase if you start from a simple polygon? The answer is NO. p0

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**Graham Scan: Sorting Step**

The first stage of Graham-Scan sorts the points by their polar angle from the bottom-left vertex, p0: In recitation, we asked: would the scanning phase of Graham-Scan work on any simple polygon? That is, can you omit this sorting phase if you start from a simple polygon? The answer is NO. p0

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**Graham Scan: Sorting Step**

The first stage of Graham-Scan sorts the points by their polar angle from the bottom-left vertex, p0: In recitation, we asked: would the scanning phase of Graham-Scan work on any simple polygon? That is, can you omit this sorting phase if you start from a simple polygon? The answer is NO. p6 p3 p4 p5 p2 p1 p0

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**Graham Scan: Sorting Step**

The first stage of Graham-Scan sorts the points by their polar angle from the bottom-left vertex, p0: In recitation, we asked: would the scanning phase of Graham-Scan work on any simple polygon? That is, can you omit this sorting phase if you start from a simple polygon? The answer is NO. p6 p3 p4 p5 Simple polygon (not convex) p2 p1 p0

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**Graham Scan on Simple Polygons**

The first stage of Graham-Scan sorts the points by their polar angle from the bottom-left vertex, p0: In recitation, we asked: would the scanning phase of Graham-Scan work on any simple polygon? That is, can you omit this sorting phase if you start from a simple polygon? The answer is NO. p6 p3 p4 p5 Simple polygon (not convex) p2 p1 p0

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Counter-Example Consider the same set of points with the following simple polygon:

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Counter-Example Consider the same set of points with the following simple polygon: Run Graham’s Scan starting from p0 p3 p2 Stack p4 p6 p5 p1 p0

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Counter-Example Consider the same set of points with the following simple polygon: Run Graham’s Scan starting from p0 p3 p2 p2 p1 p0 Stack p4 p6 p5 p1 p0

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Counter-Example Consider the same set of points with the following simple polygon: Run Graham’s Scan starting from p0 p3 p2 p2 p1 p0 Stack p4 p6 p5 p1 p0 p0

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Counter-Example Consider the same set of points with the following simple polygon: Run Graham’s Scan starting from p0 p3 p2 p3 p2 p1 p0 Stack p4 p6 p5 p1 p0 p0

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Counter-Example Consider the same set of points with the following simple polygon: Run Graham’s Scan starting from p0 p3 p2 p3 p2 p1 p0 Stack p4 p6 p5 p1 p0 p0

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Counter-Example Consider the same set of points with the following simple polygon: Run Graham’s Scan starting from p0 p3 p2 p4 p3 p2 p1 p0 Stack p4 p6 p5 p1 p0 p0

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Counter-Example Consider the same set of points with the following simple polygon: Run Graham’s Scan starting from p0 p3 p2 p4 p3 p2 p1 p0 Stack p4 p6 p5 p1 p0 p0

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Counter-Example Consider the same set of points with the following simple polygon: Run Graham’s Scan starting from p0 p3 p2 p4 p3 p2 p1 p0 Stack p4 p6 p5 p1 p0 p0

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Counter-Example Consider the same set of points with the following simple polygon: Run Graham’s Scan starting from p0 p3 p2 p3 p2 p1 p0 Stack p4 p6 p5 p1 p0 p0

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Counter-Example Consider the same set of points with the following simple polygon: Run Graham’s Scan starting from p0 p3 p2 p3 p2 p1 p0 Stack p4 p6 p5 p1 p0 p0

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Counter-Example Consider the same set of points with the following simple polygon: Run Graham’s Scan starting from p0 p3 p2 p3 p2 p1 p0 Stack p4 p6 p5 p1 p0 p0

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Counter-Example Consider the same set of points with the following simple polygon: Run Graham’s Scan starting from p0 p3 p2 p5 p3 p2 p1 p0 Stack p4 p6 p5 p1 p0 p0

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Counter-Example Consider the same set of points with the following simple polygon: Run Graham’s Scan starting from p0 p3 p2 p5 p3 p2 p1 p0 Stack p4 p6 p5 p1 p0 p0

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Counter-Example Consider the same set of points with the following simple polygon: Run Graham’s Scan starting from p0 p3 p2 p6 p5 p3 p2 p1 p0 Stack p4 p6 p5 p1 p0 p0

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Counter-Example Consider the same set of points with the following simple polygon: Run Graham’s Scan starting from p0 p3 p2 p6 p5 p3 p2 p1 p0 Stack p4 p6 p5 p1 p0 p0 At end, get convex hull = <p0, p1, p2, p3, p5, p6> But actually, convex hull = <p0, p1, p2, p3>

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Counter-Example Consider the same set of points with the following simple polygon: p3 p2 p6 p5 p3 p2 p1 p0 Stack p4 p6 p5 original p1 p0 p0 What went wrong? Vertex p3 was not “visible” from p0, so the line p3->p5 crossed p6->p0 when building the convex hull. That is, the polygon became non-simple during course of the algorithm.

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**Graham-Scan Builds Star-Shaped Polygons**

When vertices are sorted by polar angle from p0, all other vertices are visible from p0 in resulting polygon: A polygon with a point visible from each vertex is called star-shaped (CLRS p. 957, Ex ). Graham-Scan works for all star-shaped polygons, but not for all simple ones p6 p3 p4 p5 p2 p1 p0

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For More Information There do exist linear-time algorithms for building the convex hull of a simple polygon. Many of the first algorithms proposed were actually incorrect! See here for an interesting history: Is the question we considered a million-dollar question? Probably not, but it can be worth up to $200! Click here for a good time:

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# of sides # of triangles Sum of measures of interior angles 31 1(180)=180 42 2(180)=360 5 33(180)=540 644(180)=720 n n-2(n-2) 180.

# of sides # of triangles Sum of measures of interior angles 31 1(180)=180 42 2(180)=360 5 33(180)=540 644(180)=720 n n-2(n-2) 180.

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