Lecture 7: Computational Geometry

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1 Lecture 7: Computational Geometry CS 491 CAP Uttam Thakore Friday, October 7 th, 2016 Credit for many of the slides on solving geometry problems goes to the Stanford CS 97SI course lecture on computational geometry ( CS 491 CAP - Intro to Competitive Algorithmic Programming 1/66

2 Table of Contents Basic HS Geometry Review Trigonometry Vectors Solving Geometry Problems Line-line intersection Sweep line algorithm Area of a polygon Binary search Ternary search Convex Hull Other Geometry Concepts More Resources CS 491 CAP - Intro to Competitive Algorithmic Programming 2/66

3 Trigonometry Table of Contents Basic HS Geometry Review Trigonometry Vectors Solving Geometry Problems Line-line intersection Sweep line algorithm Area of a polygon Binary search Ternary search Convex Hull Other Geometry Concepts More Resources CS 491 CAP - Intro to Competitive Algorithmic Programming 3/66

4 Trigonometry sin, cos, & tan B sin θ = cos θ = opposite hypotenuse adjacent hypotenuse tan θ = opposite adjacent hypotenuse A θ h b a C opposite adjacent CS 491 CAP - Intro to Competitive Algorithmic Programming 4/66

5 Trigonometry sin, cos, & tan B sin θ = cos θ = opposite hypotenuse adjacent hypotenuse tan θ = opposite adjacent hypotenuse A θ h b a C opposite adjacent CS 491 CAP - Intro to Competitive Algorithmic Programming 4/66

6 Trigonometry sin, cos, & tan B sin θ = cos θ = opposite hypotenuse adjacent hypotenuse tan θ = opposite adjacent hypotenuse A θ h b a C opposite adjacent CS 491 CAP - Intro to Competitive Algorithmic Programming 4/66

7 Trigonometry sin, cos, & tan B sin θ = cos θ = opposite hypotenuse adjacent hypotenuse tan θ = opposite adjacent hypotenuse A θ h b a C opposite adjacent CS 491 CAP - Intro to Competitive Algorithmic Programming 4/66

8 Trigonometry Law of sines & law of cosines = sin(β) b = sin(γ) c Law of sines: sin(α) a Law of cosines: a 2 = b 2 + c 2 2bc cos(α) Put another way: cos(α) = a 2 +b 2 +c 2 Similar for b and c 2bc CS 491 CAP - Intro to Competitive Algorithmic Programming 5/66

9 Trigonometry Law of sines & law of cosines = sin(β) b = sin(γ) c Law of sines: sin(α) a Law of cosines: a 2 = b 2 + c 2 2bc cos(α) Put another way: cos(α) = a 2 +b 2 +c 2 Similar for b and c 2bc CS 491 CAP - Intro to Competitive Algorithmic Programming 5/66

10 Trigonometry Law of sines & law of cosines = sin(β) b = sin(γ) c Law of sines: sin(α) a Law of cosines: a 2 = b 2 + c 2 2bc cos(α) Put another way: cos(α) = a 2 +b 2 +c 2 Similar for b and c 2bc CS 491 CAP - Intro to Competitive Algorithmic Programming 5/66

11 Trigonometry Law of sines & law of cosines = sin(β) b = sin(γ) c Law of sines: sin(α) a Law of cosines: a 2 = b 2 + c 2 2bc cos(α) Put another way: cos(α) = a 2 +b 2 +c 2 Similar for b and c 2bc CS 491 CAP - Intro to Competitive Algorithmic Programming 5/66

12 Trigonometry Law of sines & law of cosines = sin(β) b = sin(γ) c Law of sines: sin(α) a Law of cosines: a 2 = b 2 + c 2 2bc cos(α) Put another way: cos(α) = a 2 +b 2 +c 2 Similar for b and c 2bc CS 491 CAP - Intro to Competitive Algorithmic Programming 5/66

13 Trigonometry Heron s formula B a C c b A Area of triangle = s (s a) (s b) (s c) s = semiperimeter = a+b+c 2 But there is an easier way to find the area of a triangle... CS 491 CAP - Intro to Competitive Algorithmic Programming 6/66

14 Trigonometry Heron s formula B a C c b A Area of triangle = s (s a) (s b) (s c) s = semiperimeter = a+b+c 2 But there is an easier way to find the area of a triangle... CS 491 CAP - Intro to Competitive Algorithmic Programming 6/66

15 Trigonometry Heron s formula B a C c b A Area of triangle = s (s a) (s b) (s c) s = semiperimeter = a+b+c 2 But there is an easier way to find the area of a triangle... CS 491 CAP - Intro to Competitive Algorithmic Programming 6/66

16 Trigonometry Heron s formula B a C c b A Area of triangle = s (s a) (s b) (s c) s = semiperimeter = a+b+c 2 But there is an easier way to find the area of a triangle... CS 491 CAP - Intro to Competitive Algorithmic Programming 6/66

17 Vectors Table of Contents Basic HS Geometry Review Trigonometry Vectors Solving Geometry Problems Line-line intersection Sweep line algorithm Area of a polygon Binary search Ternary search Convex Hull Other Geometry Concepts More Resources CS 491 CAP - Intro to Competitive Algorithmic Programming 7/66

18 Vectors What is a vector? Direction Magnitude Alternatively, magnitude & angle from positive x-axis A a B CS 491 CAP - Intro to Competitive Algorithmic Programming 8/66

19 Vectors What is a vector? Direction Magnitude Alternatively, magnitude & angle from positive x-axis A a B CS 491 CAP - Intro to Competitive Algorithmic Programming 8/66

20 Vectors What is a vector? Direction Magnitude Alternatively, magnitude & angle from positive x-axis A a B CS 491 CAP - Intro to Competitive Algorithmic Programming 8/66

21 Vectors What is a vector? Direction Magnitude Alternatively, magnitude & angle from positive x-axis A a B CS 491 CAP - Intro to Competitive Algorithmic Programming 8/66

22 Vectors Why are vectors important? Basis for large number of geometry problems Basic operations on vectors provide very useful information CS 491 CAP - Intro to Competitive Algorithmic Programming 9/66

23 Vectors Why are vectors important? Basis for large number of geometry problems Basic operations on vectors provide very useful information CS 491 CAP - Intro to Competitive Algorithmic Programming 9/66

24 Vectors Why are vectors important? Basis for large number of geometry problems Basic operations on vectors provide very useful information CS 491 CAP - Intro to Competitive Algorithmic Programming 9/66

25 Vectors Vector Addition Place beginning of one vector at end of other New vector from beginning of first vector to end of last CS 491 CAP - Intro to Competitive Algorithmic Programming 10/66

26 Vectors Vector Addition Place beginning of one vector at end of other New vector from beginning of first vector to end of last CS 491 CAP - Intro to Competitive Algorithmic Programming 10/66

27 Vectors Vector Addition Place beginning of one vector at end of other New vector from beginning of first vector to end of last + CS 491 CAP - Intro to Competitive Algorithmic Programming 10/66

28 Vectors Vector Dot Product A B = x 1 x 2 + y 1 y 2 Why is this important? A B = A B cos θ Can use to find cos θ CS 491 CAP - Intro to Competitive Algorithmic Programming 11/66

29 Vectors Vector Dot Product A B = x 1 x 2 + y 1 y 2 Why is this important? A B = A B cos θ Can use to find cos θ CS 491 CAP - Intro to Competitive Algorithmic Programming 11/66

30 Vectors Vector Dot Product A B = x 1 x 2 + y 1 y 2 Why is this important? A B = A B cos θ Can use to find cos θ CS 491 CAP - Intro to Competitive Algorithmic Programming 11/66

31 Vectors Vector Dot Product A B = x 1 x 2 + y 1 y 2 Why is this important? A B = A B cos θ Can use to find cos θ CS 491 CAP - Intro to Competitive Algorithmic Programming 11/66

32 Vectors Vector Dot Product A B = x 1 x 2 + y 1 y 2 Why is this important? A B = A B cos θ Can use to find cos θ CS 491 CAP - Intro to Competitive Algorithmic Programming 11/66

33 Vectors Vector Cross Product A B = x 1 y 2 y 1 x 2 Why is this important? A B = A B sin θ Can use to find sin θ Significance: For θ < 0, sin θ < 0 For θ > 0, sin θ > 0 Also, A B = area of parallelogram created by A and B A B 2 is the area of the triangle CS 491 CAP - Intro to Competitive Algorithmic Programming 12/66

34 Vectors Vector Cross Product A B = x 1 y 2 y 1 x 2 Why is this important? A B = A B sin θ Can use to find sin θ Significance: For θ < 0, sin θ < 0 For θ > 0, sin θ > 0 Also, A B = area of parallelogram created by A and B A B 2 is the area of the triangle CS 491 CAP - Intro to Competitive Algorithmic Programming 12/66

35 Vectors Vector Cross Product A B = x 1 y 2 y 1 x 2 Why is this important? A B = A B sin θ Can use to find sin θ Significance: For θ < 0, sin θ < 0 For θ > 0, sin θ > 0 Also, A B = area of parallelogram created by A and B A B 2 is the area of the triangle CS 491 CAP - Intro to Competitive Algorithmic Programming 12/66

36 Vectors Vector Cross Product A B = x 1 y 2 y 1 x 2 Why is this important? A B = A B sin θ Can use to find sin θ Significance: For θ < 0, sin θ < 0 For θ > 0, sin θ > 0 Also, A B = area of parallelogram created by A and B A B 2 is the area of the triangle CS 491 CAP - Intro to Competitive Algorithmic Programming 12/66

37 Vectors Vector Cross Product A B = x 1 y 2 y 1 x 2 Why is this important? A B = A B sin θ Can use to find sin θ Significance: For θ < 0, sin θ < 0 For θ > 0, sin θ > 0 Also, A B = area of parallelogram created by A and B A B 2 is the area of the triangle CS 491 CAP - Intro to Competitive Algorithmic Programming 12/66

38 Vectors Vector Cross Product A B = x 1 y 2 y 1 x 2 Why is this important? A B = A B sin θ Can use to find sin θ Significance: For θ < 0, sin θ < 0 For θ > 0, sin θ > 0 Also, A B = area of parallelogram created by A and B A B 2 is the area of the triangle CS 491 CAP - Intro to Competitive Algorithmic Programming 12/66

39 Vectors Vector Cross Product A B = x 1 y 2 y 1 x 2 Why is this important? A B = A B sin θ Can use to find sin θ Significance: For θ < 0, sin θ < 0 For θ > 0, sin θ > 0 Also, A B = area of parallelogram created by A and B A B 2 is the area of the triangle CS 491 CAP - Intro to Competitive Algorithmic Programming 12/66

40 Vectors Vector Cross Product A B = x 1 y 2 y 1 x 2 Why is this important? A B = A B sin θ Can use to find sin θ Significance: For θ < 0, sin θ < 0 For θ > 0, sin θ > 0 Also, A B = area of parallelogram created by A and B A B 2 is the area of the triangle CS 491 CAP - Intro to Competitive Algorithmic Programming 12/66

41 Vectors Vector Cross Product A B = x 1 y 2 y 1 x 2 Why is this important? A B = A B sin θ Can use to find sin θ Significance: For θ < 0, sin θ < 0 For θ > 0, sin θ > 0 Also, A B = area of parallelogram created by A and B A B 2 is the area of the triangle CS 491 CAP - Intro to Competitive Algorithmic Programming 12/66

42 Vectors Vector Cross Product A B = x 1 y 2 y 1 x 2 Why is this important? A B = A B sin θ Can use to find sin θ Significance: For θ < 0, sin θ < 0 For θ > 0, sin θ > 0 Also, A B = area of parallelogram created by A and B A B 2 is the area of the triangle CS 491 CAP - Intro to Competitive Algorithmic Programming 12/66

43 Table of Contents Basic HS Geometry Review Trigonometry Vectors Solving Geometry Problems Line-line intersection Sweep line algorithm Area of a polygon Binary search Ternary search Convex Hull Other Geometry Concepts More Resources CS 491 CAP - Intro to Competitive Algorithmic Programming 13/66

44 Line-line intersection Table of Contents Basic HS Geometry Review Trigonometry Vectors Solving Geometry Problems Line-line intersection Sweep line algorithm Area of a polygon Binary search Ternary search Convex Hull Other Geometry Concepts More Resources CS 491 CAP - Intro to Competitive Algorithmic Programming 14/66

45 Cross Product Define ccw A, B, C = B A C A ccw A, B, C > 0 A B C is a left turn B A ccw A, B, C < 0 A B C is a right turn

46 Segment-Segment Intersection Test Given two segments AB and CD Want to determine if they intersect properly: two segments meet at a single point that are strictly inside both segments Non-intersecting Proper Not Proper

47 Segment-Segment Intersection Test Assume that the segments intersect From A s point of view, looking straight to B, C and D must lie on different sides Holds true for the other segment as well The intersection exists and is proper if: ccw A, B, C ccw A, B, D < 0 AND ccw C, D, A ccw C, D, B < 0

48 Segment-Segment Intersection Test Determining non-proper intersections We need more special cases to consider! e.g. If ccw A, B, C, ccw A, B, D, ccw C, D, A, ccw C, D, B are all zeros, then two segments are collinear Very careful implementation is required

49 Sweep line algorithm Table of Contents Basic HS Geometry Review Trigonometry Vectors Solving Geometry Problems Line-line intersection Sweep line algorithm Area of a polygon Binary search Ternary search Convex Hull Other Geometry Concepts More Resources CS 491 CAP - Intro to Competitive Algorithmic Programming 19/66

50 Sweep Line Algorithm A problem solving strategy for geometry problems The main idea is to maintain a line (with some auxiliary data structure) that sweeps through the entire plane and solve the problem locally We can t simulate a continuous process, (e.g. sweeping a line) so we define events that causes certain changes in our data structure And process the events in the order of occurrence We ll cover one sweep line algorithm

51 Sweep Line Algorithm Problem: Given n axis-aligned rectangles, find the area of the union of them We will sweep the plane from left to right Events: left and right edges of the rectangles The main idea is to maintain the set of active rectangles in order It suffices to store the y-coordinates of the rectangles

52 Example Sweep line

53 Example Blue interval is added to the data structure

54 Example green area = blue segment length (distance that the sweep line traveled)

55 Example

56 Example

57 Example

58 Example

59 Example

60 Example

61 Example One of the segments is removed

62 Example

63 Pseudopseudocode If the sweep line hits the left edge of a rectangle Insert it to the data structure Right edge? Remove it Move to the next event, and add the area(s) of the green rectangle(s) Finding the length of the union of the blue segments is the hardest step There is an easy O n method for this step

64 Notes on Sweep Line Algorithms Sweep line algorithm is a generic concept Come up with the right set of events and data structures for each problem Exercise problems Finding the perimeter of the union of rectangles Finding all k intersections of n line segments in O n + k log n time

65 Area of a polygon Table of Contents Basic HS Geometry Review Trigonometry Vectors Solving Geometry Problems Line-line intersection Sweep line algorithm Area of a polygon Binary search Ternary search Convex Hull Other Geometry Concepts More Resources CS 491 CAP - Intro to Competitive Algorithmic Programming 35/66

66 Area of a polygon Triangulation! 1. Start at any point (let s call it A) 2. Go through all other points, summing areas of triangles made by vectors from A (using cross product) F A E L G J K I H D C B CS 491 CAP - Intro to Competitive Algorithmic Programming 36/66

67 Area of a polygon Triangulation! 1. Start at any point (let s call it A) 2. Go through all other points, summing areas of triangles made by vectors from A (using cross product) F A E L G J K I H D C B CS 491 CAP - Intro to Competitive Algorithmic Programming 36/66

68 Area of a polygon Triangulation! 1. Start at any point (let s call it A) 2. Go through all other points, summing areas of triangles made by vectors from A (using cross product) F A E L G J K I H D C B CS 491 CAP - Intro to Competitive Algorithmic Programming 36/66

69 Area of a polygon Triangulation! 1. Start at any point (let s call it A) 2. Go through all other points, summing areas of triangles made by vectors from A (using cross product) F A E L G J K I H D C B CS 491 CAP - Intro to Competitive Algorithmic Programming 36/66

70 Area of a polygon Why does this work? For convex polygons, obvious For concave polygons, when we turn inwards, area is negative Sum of the negative and positive triangles equals area of polygon CS 491 CAP - Intro to Competitive Algorithmic Programming 37/66

71 Area of a polygon Why does this work? For convex polygons, obvious For concave polygons, when we turn inwards, area is negative Sum of the negative and positive triangles equals area of polygon CS 491 CAP - Intro to Competitive Algorithmic Programming 37/66

72 Area of a polygon Why does this work? For convex polygons, obvious For concave polygons, when we turn inwards, area is negative Sum of the negative and positive triangles equals area of polygon CS 491 CAP - Intro to Competitive Algorithmic Programming 37/66

73 Area of a polygon Why does this work? For convex polygons, obvious For concave polygons, when we turn inwards, area is negative Sum of the negative and positive triangles equals area of polygon CS 491 CAP - Intro to Competitive Algorithmic Programming 37/66

74 Binary search Table of Contents Basic HS Geometry Review Trigonometry Vectors Solving Geometry Problems Line-line intersection Sweep line algorithm Area of a polygon Binary search Ternary search Convex Hull Other Geometry Concepts More Resources CS 491 CAP - Intro to Competitive Algorithmic Programming 38/66

75 Note on Binary Search Usually, binary search is used to find an item of interest in a sorted array There is a nice application of binary search, often used in geometry problems Example: finding the largest circle that fits into a given polygon Don t try to find a closed form solution or anything like that! Instead, binary search on the answer

76 Ternary search Table of Contents Basic HS Geometry Review Trigonometry Vectors Solving Geometry Problems Line-line intersection Sweep line algorithm Area of a polygon Binary search Ternary search Convex Hull Other Geometry Concepts More Resources CS 491 CAP - Intro to Competitive Algorithmic Programming 40/66

77 Ternary Search Another useful method in many geometry problems Finds the minimum point of a convex function f Not exactly convex, but let s use this word anyway Initialize the search interval [s, e] Until e s becomes small: m 1 = s + (e s)/3, m 2 = e (e s)/3 If f m 1 f(m 2 ), then set e to m 2 Otherwise, set s to m 1

78 Convex Hull Table of Contents Basic HS Geometry Review Trigonometry Vectors Solving Geometry Problems Line-line intersection Sweep line algorithm Area of a polygon Binary search Ternary search Convex Hull Other Geometry Concepts More Resources CS 491 CAP - Intro to Competitive Algorithmic Programming 42/66

79 Convex Hull Problem Given n points on the plane, find the smallest convex polygon that contains all the given points For simplicity, assume that no three points are collinear

80 Simple O n 3 algorithm AB is an edge of the convex hull iff ccw A, B, C have the same sign for all other given points C This gives us a simple algorithm For each A and B: If ccw A, B, C > 0 for all C A, B: Record the edge A B Walk along the recorded edges to recover the convex hull

81 Faster Algorithm: Graham Scan We know that the leftmost given point has to be in the convex hull We assume that there is a unique leftmost point Make the leftmost point the origin So that all other points have positive x coordinates Sort the points in increasing order of y/x Increasing order of angle, whatever you like to call it Incrementally construct the convex hull using a stack

82 Incremental Construction We maintain a convex chain of the given points For each i, we do the following: Append point i to the current chain If the new point causes a concave corner, remove the bad vertex from the chain that causes it Repeat until the new chain becomes convex

83 Example Points are numbered in increasing order of y/x new origin

84 Example Add the first two points in the chain new origin

85 Example Adding point 3 causes a concave corner Remove new origin

86 Example That s better new origin

87 Example Adding 4 to the chain causes a problem Remove new origin

88 Example Continue adding points new origin

89 Example Continue adding points new origin

90 Example Continue adding points new origin

91 Example Bad corner! new origin

92 Example Bad corner again! new origin 5 4 1

93 Example Continue adding points new origin

94 Example Continue adding points new origin 4 1

95 Example Continue adding points new origin 4 1

96 Example Done! new origin 4 1

97 Pseudocode Set the leftmost point 0,0, and sort the rest of the points in increasing order of y/x Initialize stack S For i = 1 n: Let A be the second topmost element of S, B be the topmost element of S, C be the ith point If ccw A, B, C < 0, pop S and go back Push C to S Points in S form the convex hull

98 Convex Hull Graham Scan Complexity Sorting of points takes O(n log n) time Construction of the hull requires only traversal of all points once, with removal from the stack at most once, so O(n) time Total runtime complexity: O(n log n) CS 491 CAP - Intro to Competitive Algorithmic Programming 62/66

99 Convex Hull Graham Scan Complexity Sorting of points takes O(n log n) time Construction of the hull requires only traversal of all points once, with removal from the stack at most once, so O(n) time Total runtime complexity: O(n log n) CS 491 CAP - Intro to Competitive Algorithmic Programming 62/66

100 Convex Hull Graham Scan Complexity Sorting of points takes O(n log n) time Construction of the hull requires only traversal of all points once, with removal from the stack at most once, so O(n) time Total runtime complexity: O(n log n) CS 491 CAP - Intro to Competitive Algorithmic Programming 62/66

101 Convex Hull Graham Scan Complexity Sorting of points takes O(n log n) time Construction of the hull requires only traversal of all points once, with removal from the stack at most once, so O(n) time Total runtime complexity: O(n log n) CS 491 CAP - Intro to Competitive Algorithmic Programming 62/66

102 Table of Contents Basic HS Geometry Review Trigonometry Vectors Solving Geometry Problems Line-line intersection Sweep line algorithm Area of a polygon Binary search Ternary search Convex Hull Other Geometry Concepts More Resources CS 491 CAP - Intro to Competitive Algorithmic Programming 63/66

103 There are more geometry concepts to know! Circle concepts chord length, inscribed and circumscribed polygons, circles and intersecting tangent/secant lines Sphere concepts great circle distance Polygons checking if point is inside polygon (extension of concepts covered today) For formulas, algorithms, and implementations, see Chapter 7 of Competitive Programming by Steven Halim Thankfully, most of these concepts do not appear in our regionals However, you should know these for World Finals! CS 491 CAP - Intro to Competitive Algorithmic Programming 64/66

104 Table of Contents Basic HS Geometry Review Trigonometry Vectors Solving Geometry Problems Line-line intersection Sweep line algorithm Area of a polygon Binary search Ternary search Convex Hull Other Geometry Concepts More Resources CS 491 CAP - Intro to Competitive Algorithmic Programming 65/66

105 Some geometry resources TopCoder Algorithm Tutorials - Basic Geometry Concepts TopCoder Algorithm Tutorials - Line Intersection and its Applications TopCoder Algorithm Tutorials - Practice TopCoder geometry problems Ahmed-Aly list of geometry problems Stanford Introduction to ICPC Course Geometry Lecture Slides Stanford ICPC Notebook (contains geometry algorithm implementations in C++) Chapter 7 of Competitive Programming by Steven Halim CS 491 CAP - Intro to Competitive Algorithmic Programming 66/66

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