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**CS16: Introduction to Data Structures & Algorithms**

Thursday March 12, 2015 Convex Hull CS16: Introduction to Data Structures & Algorithms

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**Outline Overview Convex Hull Overview Graham Scan Algorithm**

Thursday March 12, 2015 Outline Overview Convex Hull Overview Graham Scan Algorithm Incremental Algorithm

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Thursday March 12, 2015 Convex Hull The convex hull of a set of points is the smallest convex polygon containing the points A convex polygon is a nonintersecting polygon whose internal angles are all convex (i.e., less than 180 degrees) In a convex polygon, a segment joining any two lies entirely inside the polygon segment not contained! convex nonconvex

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**Convex Hull (2) Think of a rubber band snapping around the points**

Thursday March 12, 2015 Convex Hull (2) Think of a rubber band snapping around the points Remember to think of special cases Colinearity: A point that lies on a segment of the convex is not included in the convex hull

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**Applications Motion planning**

Thursday March 12, 2015 Applications Motion planning Find an optimal route that avoids obstacles for a robot Bounding Box Obtain a closer bounding box in computer graphics Pattern Matching Compare two objects using their convex hulls obstacle start end

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Thursday March 12, 2015 Finding a Convex Hull One algorithm for determining a convex polygon is one which, when following the points in counterclockwise order, always produces left-turns

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**Calculating Orientation**

Thursday March 12, 2015 Calculating Orientation a b c The orientation of three points (a, b, c) in the plane is clockwise, counterclockwise, or collinear Clockwise (CW): right turn Counterclockwise (CCW): left turn Collinear (COLL): no turn The orientation of three points is characterized by the sign of the determinant Δ(a, b, c) CW c b CCW a c b a COLL function isLeftTurn(a, b, c): return (b.x - a.x)*(c.y - a.y)– (b.y - a.y)*(c.x - a.x) > 0

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**Calculating Orientation (2)**

Thursday March 12, 2015 Calculating Orientation (2) Using the isLeftTurn() method: (0.5-0) * ( ) - (1-0) * (1-0) = (CW) (1-0) * (1-0) - (0.5-0) * (0.5-0) = 0.75 (CCW) (1-0) * (2-0) - (1-0) * (2-0) = 0 (COLL) a (0,0) b (0.5,1) c (1, 0.5) CW c (0.5,1) b (1, 0.5) CCW a (0,0) c(2,2) b(1,1) a(0,0) COLL

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**Calculating Orientation (3)**

Thursday March 12, 2015 Calculating Orientation (3) Let a, b, c be three consecutive vertices of a polygon, in counterclockwise order b’ is not included in the hull if a’, b,’ c’ is non-convex, i.e. orientation(a’, b’, c’) = CW or COLL b is included in the hull if a, b, c is convex, i.e. orientation(a, b, c) = CCW, and all other non-hull points have been removed c’ c b’ a’ b a

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Thursday March 12, 2015 Graham Scan Algorithm Find the anchor point (the point with the smallest y value) Sort points in CCW order around the anchor You can sort points by comparing the angle between the anchor and the point you’re looking at (the smaller the angle, the closer the point) 4 7 5 6 3 8 2 Anchor 1

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Thursday March 12, 2015 Graham Scan Algorithm The polygon is traversed in sorted order and a sequence H of vertices in the hull is maintained For each point a, add a to H While the last turn is a right turn, remove the second to last point from H In the image below, p, q, r forms a right turn, and thus q is removed from H. Similarly, o, p, r forms a right turn, and thus p is removed from H. p q r H o p r H n o r H

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**Graham Scan: Pseudocode**

Thursday March 12, 2015 Graham Scan: Pseudocode function graham_scan(pts): // Input: Set of points pts // Output: Hull of points find anchor point sort other points in CCW order around anchor hull = [] for p in pts: add p to hull while last turn is a “right turn” remove 2nd to last point add anchor to hull return hull Work on a diagram on the board B D C A P Hull: A few more pieces “sort in counter-clockwise order” X coordinate of normalized vector from P to any Q works for the sort it is the cos of angle with x axis (Side adjacent over hypotenuse (1)) V = (A – P) / | A-P | Sort according to V.x Right or left turn: Given points A B C take cross product of BA and BC answer is perpendicular to plane and follows right/left hand rule for in vs. out of plane so size of the z coordinate Sign of (Ax-Bx)(Cy-By) – (Ay-By)(Cx-Bx) gives turn direction if positive, left turn if negative, right turn (or maybe vice versa; depends on handedness of coordinate system) Complexity O(n) O(n log n) Loop: O(n) while: O(1) times amortized, since max n points can be removed Note: this is very high-level pseudocode. There are many special cases to consider!

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**Overall run time: O(nlogn)**

Thursday March 12, 2015 Graham Scan: Run Time function graham_scan(pts): // Input: Set of points pts // Output: Hull of points find anchor point // O(n) sort other points in CCW order around anchor // O(nlogn) create hull (empty list representing ordered points) for p in pts: // O(n) add p to hull while last turn is a “right turn” // O(1) amortized remove 2nd to last point add anchor to hull return hull Overall run time: O(nlogn)

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**Incremental Algorithm**

Thursday March 12, 2015 Incremental Algorithm What if we already have a convex hull, and we just want to add one point q? This is what you’ll be doing on the Java project after heap! Get the angle from the anchor to q and find points p and r, the hull points on either side of q If p, q, r forms a left turn, add q to the hull Check if adding q creates a concave shape If you have right turns on either side of q, remove vertices until the shape becomes convex This is done in the same way as the static Graham Scan

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**Incremental Algorithm**

Thursday March 12, 2015 Incremental Algorithm Original Hull: Want to add point q: Find p and r: q H q H H p r p, q, r forms a left turn, so add q: o, p, q forms a right turn, so remove p n, o, q forms a right turn, so remove o p r q H r q H o n r q H n n o s s s

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**Incremental Algorithm**

Thursday March 12, 2015 Incremental Algorithm Since m, n, q is a left turn, we’re done with that side. Now we look at the other side: Since q, r, s is a left turn, we’re done! r q H q H r q H n m s s Remember that you can have right turns on either or both sides, so make sure to check in both directions and remove concave points!

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Thursday March 12, 2015 Incremental Analysis Let n be the current size of the convex hull, stored in a binary search tree for efficiency How are they sorted? Around the anchor To check if a point q should be in the hull, we insert it into the tree and get its neighbors (p and r on the prior slides) in O(log n) time We then traverse the ring, possibly deleting d points from the convex hull in O((1 + d) log n) time Therefore, incremental insertion is O(d log n) where d is the number of points removed by the insertion

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