Convex Hulls. Helen Cameron. Helen Cameron Convex Hulls 1/101

Transcription

1 Convex Hulls Helen Cameron Helen Cameron Convex Hulls 1/101

2 What Is a Convex Hull?

3 Starting Point: Points in 2D y x Helen Cameron Convex Hulls 3/101

4 Convex Hull: Informally Imagine that the x, y-lane is a board and the oints are nails sticking out of the board If you stretch an elastic so that all the nails are inside it and then let go of the elastic, the elastic will tighten into the boundary of the convex hull of the oints y x Helen Cameron Convex Hulls 4/101

5 Starting Point: Simle Polygons y Polygon: A sequence of oints connected by edges into a closed curve Simle: Only consecutive edges intersect (at their common endoints) x Helen Cameron Convex Hulls 5/101

6 Starting Point: Simle Polygons y For comarison, here s a non-simle olygon x Helen Cameron Convex Hulls 6/101

7 Inside and Outside y A simle olygon has an inside and an outside The inside is always on your left if you walk the boundary in a counterclockwise direction x Helen Cameron Convex Hulls 7/101

8 Interior Angles y The interior angle at a oint is the angle inside the olygon between the oint s adjacent edges x Helen Cameron Convex Hulls 8/101

9 Convex Angles y A convex angle is an angle < 180 degrees (π in radians) not convex convex x Helen Cameron Convex Hulls 9/101

10 Convex Polygon y A convex olygon has only convex interior angles x Helen Cameron Convex Hulls 10/101

11 Convex Polygon y Equivalent definition: A olygon is convex if, for any two oints in or on the olygon, the line segment connecting the two oints is entirely contained in the olygon x Helen Cameron Convex Hulls 11/101

12 Convex Hull: Formally The convex hull of a set of 2D oints is the smallest convex olygon that contains the oints y x Helen Cameron Convex Hulls 12/101

13 Why Care About Convex Hulls? (Some alications!)

14 Convex Hulls Are Easier Collision avoidance is faster and easier using the convex hull of a robot or a car than working with its exact (more comlex) shae SevenTransarentjg By Brian Snelson (originally osted to Flickr as My car hotoshoed) [CC BY 20 (htts://creativecommons org/licenses/by/20)], via Wikimedia Commons Helen Cameron Convex Hulls 14/101

15 Removing Outliers Often, we want to estimate some arameter from a small, randomly-chosen samle of a oulation of oints This sort of estimation is used, for examle, in medical diagnostics and network intrusion detection File:Pankreas-Ca im CT - Coeliacusblockadeng By Hellerhoff [CC BY 20 (htts://creativecommonsorg/ licenses/by/20)], via Wikimedia Commons Helen Cameron Convex Hulls 15/101

16 Removing Outliers Examle outliers Some arameter estimations are very sensitive to outliers (oints that are far from most of the other oints in the oulation) Helen Cameron Convex Hulls 16/101

17 Removing Outliers Get rid of outlier oints in 2D: Reeatedly find and then eel off the convex hull, until no more than some chosen fraction of the original oints remains Then you can safely estimate the desired arameter Original oints Next hull Outermost hull Remaining oints Points to estimate from: Helen Cameron Convex Hulls 17/101

18 Finding the Convex Hull: Inut and Outut

19 The Inut (If You re Writing a Program) Inut: A list of 2D oints ( x, y ), in no articular order Examle inut: (98,106) (7,13) (128,87) (1035,121) (112, 145) (8,11) (75,98) (91,122) (78,73) (102,95) (117,96) (96,91) (95,78) (855,102) (107,101) (1255,83) (885,91) (84,138) (10,87) (122,107) (11, 104) (1055,83) Helen Cameron Convex Hulls 19/101

20 The Outut (If You re Writing a Program) Outut: The oints on the boundary of the convex hull of the inut oints, listed in the order of a counter-clockwise traversal of the boundary Examle outut: (7,13) (78,73) (1255,83) (128,87) (112, 145) (84,138) (7,13) Helen Cameron Convex Hulls 20/101

21 Left, Right, or Collinear?

22 Leftness Testing for a Line Segment Useful test when finding the convex hull: Is oint q to the left, right or collinear with a directed line segment a, b that is, if we re standing on endoint a and facing the other endoint b, is q to our left, to our right or somewhere on the infinite line through a and b? q 1 q 0 b q 2 a Helen Cameron Convex Hulls 22/101

23 Leftness Testing from a Line Segment A method that returns true if q is to the left of a, b (returns false otherwise): boolean isleft( Point q, Point a, Point b ) { return 0 < (bx-ax)*(qy-ay) - (by-ay)*(qx-ax); } A method that returns true if q is to the right of a, b (returns false otherwise): boolean isright( Point q, Point a, Point b ) { return 0 < (bx-ax)*(qy-ay) - (by-ay)*(qx-ax); } A method that returns true if q is collinear with a, b (returns false otherwise): boolean iscollinear( Point q, Point a, Point b ) { return 0 == (bx-ax)*(qy-ay) - (by-ay)*(qx-ax); } Helen Cameron Convex Hulls 23/101

24 Leftness Testing from a Line Segment The leftness test is using the right-hand rule : Lay the outer edge of your right hand along a, b (fingers ointing from a to b) with your thumb sticking u If your fingers naturally curl towards q, then q is to the left The right-hand rule has to do with the cross roduct of two vectors Helen Cameron Convex Hulls 24/101

25 A Vector A vector is a directed line segment starting at the origin and ending at oint Note: We ll work with a directed line segment starting at some oint a and ending at oint b we ll have to translate a to the origin by subtracting a from all oints (Really: subtract a x from all oints x-coordinates and subtract a y from all oints y-coordinates) Helen Cameron Convex Hulls 25/101

26 Area of a Triangle: Vector Cross Product The cross roduct q of two vectors = ( x, y, z ) and q = (q x, q y, q z ) only makes sense in 3D: it is the vector erendicular to both and q given by the right-hand rule and whose magnitude is given by the absolute value of x y z x y z q x q y q z = ( y q z z q y ) x +( z q x x q z ) y +( x q y y q x ) z Note: x is the unit vector in the ositive x-direction, y is the unit vector in the ositive y-direction, and z is the unit vector in the ositive z-direction Helen Cameron Convex Hulls 26/101

27 Area of a Triangle: Vector Cross Product For 2D vectors, the z-coordinates z = q z = 0 and the cross-roduct formula reduces to ( x q y y q x ) z If oint q is to the left of vector, then x q y y q x is ositive (follows right-hand rule) If oint q is to the right, then x q y y q x is negative (erendicular in the oosite direction) If oint q is collinear with vector, then x q y y q x is zero Helen Cameron Convex Hulls 27/101

28 Examles Examle 1: If = (3, 1) and q = (2, 4), then the calculation gives ( x q y y q x ) z = ( ) z = 10 z Conclusion: q is to the left of q (0, 0) Helen Cameron Convex Hulls 28/101

29 Examles Examle 2: If = ( 1, 4) and q = (2, 4), then the calculation is ( x q y y q x ) z = ( ) z = 12 z Conclusion: q is to the right of q (0, 0) Helen Cameron Convex Hulls 29/101

30 Examles Examle 3: If = (1, 1) and q = (2, 2), then the calculation gives ( x q y y q x ) z = ( ) z = 0 z Conclusion: q is collinear with q (0, 0) Helen Cameron Convex Hulls 30/101

31 Leftness Testing and Line Segments We re usually working not with a vector (tail at the origin and head at some oint ), but with a directed line segment (tail at some oint a and head at some other ont b) To test if oint q is to the left of directed line segment a, b, we take the cross roduct of vectors b a and q a Helen Cameron Convex Hulls 31/101

32 Leftness Testing and Line Segments Examle: If we want to know if oint q = (4, 5) is to the left of line segment a, b, where a = (2, 1) and b = (5, 2), we translate to asking if oint q a = (2, 4) is to the left of directed line segment a a, b a = vector b a = (3, 1) this translation is Examle 1 above: q a q b a b a (0, 0) Helen Cameron Convex Hulls 32/101

33 Graham s Scan

34 Graham s Scan: Overview Graham s scan: Finds a convex hull of a set S of two-dimensional oints Examle inut: Find the convex hull of (8, 8), (2, 4), (0, 11), (5, 6), (6, 10), (10, 5), (1, 6), (5, 12),(2, 8), and (9, 13): (0, 11) (5, 12) (6, 10) (9, 13) (2, 8) (8, 8) (1, 6) (2, 4) (5, 6) (10, 5) Helen Cameron Convex Hulls 34/101

35 Graham s Scan: Overview Graham s scan uses a list: Each oint is ushed onto the end of the list once, and eventually oed off the end of the list if the oint is not on the convex hull At the end of the algorithm, the list contains the counterclockwise ordering of oints on the hull, from the beginning to the end of the list Helen Cameron Convex Hulls 35/101

36 Graham s Scan: Overview Examle Outut: For the revious examle oints, Graham s scan returns (2, 4), (10, 5), (9, 13), (0, 11), and (1, 6) (in that order), because that s what its list contains (from beginning to end of the list) at the end of Graham s scan: (9, 13) (0, 11) (5, 12) (6, 10) (2, 8) (8, 8) (2, 4) (1, 6) (5, 6) (10, 5) Helen Cameron Convex Hulls 36/101

37 Graham s Scan Has Four Stes Ste 1: Find the oint 0 with the lowest y-coordinate (0, 11) (5, 12) (6, 10) (9, 13) (2, 8) (8, 8) (1, 6) 0 (2, 4) (5, 6) (10, 5) Helen Cameron Convex Hulls 37/101

38 Graham s Scan Has Four Stes Ste 2: Sort the oints by angle around 0 (that is, by olar coordinates ) discarded 1 0 Helen Cameron Convex Hulls 38/101

39 Graham s Scan Has Four Stes Ste 3: Initialize the list L to contain 0, 1, and 2 (in that order): discarded 1 0 Helen Cameron Convex Hulls 39/101

40 Graham s Scan Has Four Stes Ste 4: Process each of the other oints in order, 3, 4,, m 1 When we re rocessing oint i : (4a) Loo: Suose oints 2nd last and last are the last two oints on the end our list L Remove last from the end of L if i (the oint we re rocessing) is NOT to the left of directed line segment 2nd last last Reeat this until i is to the left of 2nd last last (4b) Add i to the end of L Helen Cameron Convex Hulls 40/101

41 Graham-Scan( S ) 1 Find 0, the oint in S with the lowest y-coordinate (break ties by choosing the leftmost such oint) 2 Sort the other oints in S by olar angle around 0 (if two or more oints have the same olar angle, remove all but the one farthest from 0 ) Let the sorted order be 1, 2,, m 1 3 L = emty list Add 0 to the end of L Add 1 to the end of L Add 2 to the end of L 4 For each oint i in order from 3 to m 1 do 4a While i is not left of directed line segment 2nd last last Remove the oint last from the end of L 4b Add i to the end of L

42 Ste 1: Finding Lowest Point 0 Question: How do we find the oint 0 with the lowest y-coordinate? (0, 11) (5, 12) (6, 10) (9, 13) (2, 8) (8, 8) (1, 6) 0 (2, 4) (5, 6) (10, 5) Answer: Simly loo through the oints, looking only at the y-coordinates of each oint (do a findlowest) Helen Cameron Convex Hulls 42/101

43 Ste 1: Finding Lowest Point 0 Question: How do we find the oint 0 with the lowest y-coordinate? findlowest( Point[] S ) returns int { int minindex = 0; // Assume S[0] is 0 for ( int i = 1; i < Slength; i++ ) { if (S[i]y < S[minIndex]y ) minindex = i; else if ((S[i]y == S[minIndex]y) && (S[i]x < S[minIndex]x)) minindex = i; } // end for return minindex; } // end findlowest At the end, S[minIndex] is the oint with the lowest y-coordinate (in case of ties, it s the leftmost of the lowest oints) Helen Cameron Convex Hulls 43/101

44 Ste 2: Sorting by Polar Coordinate Question: How do we sort by olar coordinate? discarded 1 0 Answer: Use any sorting algorithm such as merge sort or quick sort Helen Cameron Convex Hulls 44/101

45 Ste 2: Sorting by Polar Coordinate In the sorting algorithm, when comaring two oints i and j to decide which comes first in the sorted order ( Is i before j in the sorted order? ): Test if j is to the left of directed line segment 0, i using the cross roduct method If j is to the left, then i comes before j in the sorted order If j is to the right of 0, i, then i comes after j in the sorted order If the cross roduct gives 0, then 0, i and j are collinear Comute the distance of i and j from 0 and discard the closer one Helen Cameron Convex Hulls 45/101

46 Ste 2: Sorting by Polar Coordinate Further note: If the cross roduct gives 0, do not erform the square root in the calculation of the distance: Just discard i if is smaller than otherwise, discard j ( i x 0 x) 2 + ( i y 0 y) 2 ( j x 0 x) 2 + ( j y 0 y) 2 ; Comuting square roots is too exensive and we do not need to know the exact distance measurement, we just need to know which is closer to 0 Helen Cameron Convex Hulls 46/101

47 Graham s Scan Examle Examle: Given the following set of oints, comute the convex hull Helen Cameron Convex Hulls 47/101

48 Graham s Scan Examle Examle (continued): First choose 0 to be the leftmost oint among the two oints with the same lowest y-coordinate Then sort the remaining oints by olar angle: Helen Cameron Convex Hulls 48/101

49 Graham s Scan Examle Examle (continued): Now add 0, 1, 2 on the list L Helen Cameron Convex Hulls 49/101

50 Graham s Scan Examle Examle (continued): Consider oint 3 : Because 2, 3 is not a left turn from 1, 2, remove 2 from the list L Helen Cameron Convex Hulls 50/101

51 Graham s Scan Examle Examle (continued): Still considering oint 3 : Because 1, 3 is a left turn from 0, 1, add 3 onto the list L Helen Cameron Convex Hulls 51/101

52 Graham s Scan Examle Examle (continued): Consider oint 4 : Because 3, 4 is a left turn from 1, 3, add 4 onto the list L Helen Cameron Convex Hulls 52/101

53 Graham s Scan Examle Examle (continued): Consider oint 5 : Because 4, 5 is not a left turn from 3, 4, remove 4 from the list L Helen Cameron Convex Hulls 53/101

54 Graham s Scan Examle Examle (continued): Still considering oint 5 : Because 3, 5 is a left turn from 1, 3, add 5 onto the list L Helen Cameron Convex Hulls 54/101

55 Graham s Scan Examle Examle (continued): Consider oint 6 : Because 5, 6 is a left turn from 3, 5, add 6 onto the list L Helen Cameron Convex Hulls 55/101

56 Graham s Scan Examle Examle (continued): Consider oint 7 : Because 6, 7 is not a left turn from 5, 6, remove 6 from the list L Helen Cameron Convex Hulls 56/101

57 Graham s Scan Examle Examle (continued): Still considering oint 7 : Because 5, 7 is a left turn from 3, 5, add 7 onto the list L Helen Cameron Convex Hulls 57/101

58 Graham s Scan Examle Examle (continued): Consider oint 8 : Because 7, 8 is a left turn from 5, 7, add 8 onto the list L The convex hull is reresented by the counterclockwise ordering 0, 1, 3, 5, 7, 8 Helen Cameron Convex Hulls 58/101

59 What Is Ste 4(a) Doing? Idea: When Graham s scan is about to rocess i, list L contains the convex hull of the oints reviously rocessed 0,, i 1 Graham s scan tries to add the next oint i by simly adding triangle 0, i 1, i to the convex hull i i 1 CH( 0 i 1 ) 0 1 Helen Cameron Convex Hulls 59/101

60 What Is Ste 4(a) Doing? Idea: But the result is not a convex hull if i is not a left turn from the revious edge in CH( 0,, i 1 )) Graham s Scan fixes the convexity roblem by adding more triangles on the outside of the olygon ( filling in non-left turns to i with another triangle): i j k i 1 CH( 0 i 1 ) 0 1 Helen Cameron Convex Hulls 60/101

61 How Fast Is Graham s Scan? To find the convex hull of n oints using Graham s Scan: Running time = Time to find 0 + Time to sort by olar angle + Time for adds, removes, and left-turn tests = O(n) + O(n log n) + O(n) = O(n log n) Helen Cameron Convex Hulls 61/101

62 Divide and Conquer Convex Hull

63 Divide and Conquer Algorithm This Prearata and Hong (1977) algorithm can be extended to three or higher dimensions To kee things simle, assume: No three oints lie on a line (No three oints are collinear) No two oints lie on a vertical line These assumtions can be removed, but they make the exlanations simler, so we will use them Helen Cameron Convex Hulls 63/101

64 Divide and Conquer Algorithm: The Idea The algorithm sorts the oints by x-coordinate The algorithm then builds many small convex hulls of small grous of adjacent oints These small convex hulls will not intersect each other they re disjoint Then the algorithm merges (joins together) adjacent small convex hulls to make bigger convex hulls of bigger grous of adjacent oints The algorithm kees merging adjacent convex hulls until it has ONE convex hull of all the oints Helen Cameron Convex Hulls 64/101

65 Ste 1: Sort the Points Ste 1 gets as its inut an array of oints: (8, 3) ( 1, 4) (0, 1) (5, 2) (4, 4) (6, 0) (3, 6) Here s the inut oints lotted: [6] [1] [2] [4] [3] [5] [0] You can see that the oints are not in any articular order Helen Cameron Convex Hulls 65/101

66 Ste 1: Sort the Points Ste 1 sorts the oints into ascending order by x-coordinate: ( 1, 4) (0, 1) (3, 6) (4, 4) (5, 2) (6, 0) (8, 3) Here s the oints lotted, so you can see that they re now sorted [2] [0] [1] [3] [4] [5] [6] Helen Cameron Convex Hulls 66/101

67 Ste 1: Sort the Points Ste 1 can use any sorting method (eg, quick sort) when the sorting algorithm wants to comare two items (oints), it simly comares their x-coordinates Helen Cameron Convex Hulls 67/101

68 Ste 2: Pair U the Points Ste 2 airs adjacent oints to form edges (If there s an odd number of oints, grou the last three oints into a triangle) [2] [0] [1] [3] [4] [5] [6] Helen Cameron Convex Hulls 68/101

69 Ste 2: Pair U the Points An edge is a degenerate convex hull of the air of oints Consequence: Ste 2 is making the list of oints into a list of convex hulls of small grous of oints Helen Cameron Convex Hulls 69/101

70 Ste 2: Pair U the Points Note: Make sure you get a counter-clockwise ordering of the oints of a triangle When [i+2] is left of [i],[i+1] [i+2] When [i+2] is right of [i],[i+1] [i+1] [i+1] [i] Order: [i],[i+1],[i+2] [i+2] [i+2] [i] Order: [i],[i+2],[i+1] [i+1] [i] [i+1] [i+2] [i] Helen Cameron Convex Hulls 70/101

71 Ste 3: Merge Pairs of Hulls Ste 3 does the following Pair-and-Merge ste over and over again until we have ONE convex hull: Pair u adjacent small convex hulls and merge each air into one convex hull (more on merging in a minute) Before Pair-and-Merge 1: Helen Cameron Convex Hulls 71/101

72 Ste 3: Merge Pairs of Hulls Ste 3 airs u adjacent small convex hulls and merges each air into one convex hull (more on merging in a minute) After Pair-and-Merge 1: Helen Cameron Convex Hulls 72/101

73 Ste 3: Merge Pairs of Hulls Ste 3 airs u adjacent small convex hulls and merges each air into one convex hull (more on merging in a minute) After Pair-and-Merge 2: Helen Cameron Convex Hulls 73/101

74 Ste 3: Merge Pairs of Hulls Ste 3 airs u adjacent small convex hulls and merges each air into one convex hull (more on merging in a minute) After Pair-and-Merge 3: Helen Cameron Convex Hulls 74/101

75 Ste 3: Merge Pairs of Hulls Ste 3 airs u adjacent small convex hulls and merges each air into one convex hull (more on merging in a minute) After Pair-and-Merge 4: Helen Cameron Convex Hulls 75/101

76 Ste 3: Merge Pairs of Hulls Ste 3 Each time we air and merge, we are halving the number of convex hulls we have and doubling the number of oints that each convex hull contains If we have n oints, we can do that at most log 2 n times Helen Cameron Convex Hulls 76/101

77 Ste 3: Finding Tangent Lines to Merge Hulls The merge ste: How do we merge two non-overlaing convex hulls searated by an imaginary vertical line? We find two tangent lines, one suorting the two hulls below and the other resting on the two hulls above L R Helen Cameron Convex Hulls 77/101

78 Finding the Lower Tangent Line

79 Ste 3: Finding Tangent Lines to Merge Hulls What is the lower tangent line T? The two oints on L s hull immediately before and after T s endoint in L are both to the left of T (when T is directed from L to R) The two oints on R s hull immediately before and after T s endoint in R are both to the left of T (when T is directed from L to R) L R Helen Cameron Convex Hulls 79/101

80 Ste 3: Finding Tangent Lines to Merge Hulls To find the lower tangent line: Start by connecting the rightmost oint of L with the leftmost oint of R with a line segment T L T R Helen Cameron Convex Hulls 80/101

81 Ste 3: Finding Tangent Lines to Merge Hulls To find the lower tangent line: (continued) Then walk T downward Reeat while T is not tangent to both hulls at the same time: Move T s endoint down L s hull while T is not tangent to L s hull, then Move T s endoint down R s hull while T is not tangent to R s hull Helen Cameron Convex Hulls 81/101

82 Ste 3: Finding Tangent Lines to Merge Hulls To find the lower tangent line: (continued) Move T down L while T is not tangent to L: L R Helen Cameron Convex Hulls 82/101

83 Ste 3: Finding Tangent Lines to Merge Hulls To find the lower tangent line: (continued) Now move T down R while T is not tangent to R: L R Helen Cameron Convex Hulls 83/101

84 Ste 3: Finding Tangent Lines to Merge Hulls To find the lower tangent line: (continued) Move T down L again, while T is not tangent to L: L R Helen Cameron Convex Hulls 84/101

85 Ste 3: Finding Tangent Lines to Merge Hulls To find the lower tangent line: (continued) When we try to move T down R again, we see that T is already tangent to R Now T is the lower tangent line because it is tangent to both L and R at the same time! L R Helen Cameron Convex Hulls 85/101

86 Ste 3: Finding Tangent Lines to Merge Hulls To find the lower tangent line: (continued) Let l be the index of the endoint of T in L and r be the endoint of T in R L l T r R Helen Cameron Convex Hulls 86/101

87 Ste 3: Finding Tangent Lines to Merge Hulls To find the lower tangent line: (continued) Assume that, in each hull, the oints are indexed from 0 and that arithmetic on indices is done using modular arithmetic so that we can say l 1 and r + 1 and get the index of the next oint 7 L 6 5 l = 4 T 3 4 = r R 2 1 l 1 = = r Helen Cameron Convex Hulls 87/101

88 Ste 3: Finding Tangent Lines to Merge Hulls To find the lower tangent line: (continued) Notice that moving from l to l 1 is moving clockwise in L and moving from r to r + 1 is moving counterclockwise in R 7 L R 2 l = 4 T 4 = r 1 l 1 = = r Helen Cameron Convex Hulls 88/101

89 Ste 3: Finding Tangent Lines to Merge Hulls To find the lower tangent line: (continued) T will be tangent to both hulls when l 1 and l + 1 both lie above T (are to the left of lr) and when r 1 and r + 1 lie above T (to the left of lr) l 1 = 7 L R = r + 1 l = r 1 = = l + 1 T 0 = r Helen Cameron Convex Hulls 89/101

90 Ste 3: Finding Tangent Lines to Merge Hulls To find the lower tangent line: (continued) Algorithm LowerTangent 1 l = rightmost oint of L /* maximum x-coordinate */ 2 r = leftmost oint of R /* minimum x-coordinate */ 3 while (T =lr is not lower tangent to both L and R) { 4 while (T is not lower tangent to L) { 5 l=l-1; 6 } 7 while (T is not lower tangent to R) { 8 r=r+1; 9 } 10 } Helen Cameron Convex Hulls 90/101

91 Ste 3: Finding Tangent Lines to Merge Hulls To find the lower tangent line: L R Helen Cameron Convex Hulls 91/101

92 Finding the Uer Tangent Line

93 Ste 3: Finding Tangent Lines to Merge Hulls To find the uer tangent line: We find the uer tangent similarly we walk T u until it is uer tangent to both hulls Note: Instead of considering T to be the directed line segment from l to r, we consider T to be the directed line segment from r to l T is tangent if the before and after oints on L and R are to the left of rl Helen Cameron Convex Hulls 93/101

94 Ste 3: Finding Tangent Lines to Merge Hulls To find the uer tangent line: Again, start by connecting the rightmost oint of L with the leftmost oint of R with a line segment T L l T r R Helen Cameron Convex Hulls 94/101

95 Ste 3: Finding Tangent Lines to Merge Hulls To find the uer tangent line: Move T = rl u L while T is not tangent to L: l T L r R Helen Cameron Convex Hulls 95/101

96 Ste 3: Finding Tangent Lines to Merge Hulls To find the uer tangent line: Then move T = rl u R while T is not tangent to R: l T r L R Helen Cameron Convex Hulls 96/101

97 Ste 3: Finding Tangent Lines to Merge Hulls To find the uer tangent line: Since T is not yet tangent to both hulls at the same time, move T = rl u L while T is not tangent to L: l T r L R Helen Cameron Convex Hulls 97/101

98 Ste 3: Finding Tangent Lines to Merge Hulls To find the uer tangent line: When we try to move T u R again, we see that T is already tangent to R Now T is the uer tangent line because it is tangent both hulls at the same time, l T r L R Helen Cameron Convex Hulls 98/101

99 Ste 3: Finding Tangent Lines to Merge Hulls To find the uer tangent line: Algorithm UerTangent 1 l = rightmost oint of L /* maximum x-coordinate */ 2 r = leftmost oint of R /* minimum x-coordinate */ 3 while (T =rl is not uer tangent to both L and R) { 4 while (T is not uer tangent to L) { 5 l=l+1; 6 } 7 while (T is not uer tangent to R) { 8 r=r-1; 9 } 10 } Helen Cameron Convex Hulls 99/101

100 Merging the Two Convex Hulls

101 Ste 3: Merging Hulls with Tangent Lines The merged convex hull: Once we ve found the two tangents endoints, we simly coy the ieces of L s boundary and R s boundary and the tangent lines into the boundary of the convex hull of L and R merged L R Helen Cameron Convex Hulls 101/101