Introduction to Algorithms 6.046J/18.401J/SMA5503 Lecture 12 Prof. Erik Demaine Introduction to Algorithms Day 21 L12.2? 2001 by Erik D. Demaine Computational geometry Algorithms for solving “geometric problems” in 2D and higher. Fundamental objects: point line segment line Basic structures: polygonpoint set Introduction to Algorithms Day 21 L12.3? 2001 by Erik D. Demaine Computational geometry Algorithms for solving “geometric problems” in 2D and higher. Fundamental objects: point line segment line Basic structures: convex hulltriangulation Introduction to Algorithms Day 21 L12.4? 2001 by Erik D. Demaine Orthogonal range searching Input: n points in d dimensions ? E.g., representing a database of n records each with d numeric fields Query: Axis-aligned box (in 2D, a rectangle) ? Report on the points inside the box: ? Are there any points? ? How many are there? ? List the points. Introduction to Algorithms Day 21 L12.5? 2001 by Erik D. Demaine Orthogonal range searching Input: n points in d dimensions Query: Axis-aligned box (in 2D, a rectangle) ? Report on the points inside the box Goal: Preprocess points into a data structure to support fast queries ? Primary goal: Static data structure ? In 1D, we will also obtain a dynamic data structure supporting insert and delete Introduction to Algorithms Day 21 L12.6? 2001 by Erik D. Demaine 1D range searching In 1D, the query is an interval: First solution using ideas we know: ? Interval trees ? Represent each point x by the interval [x, x]. ? Obtain a dynamic structure that can list k answers in a query in O(k lg n) time. Introduction to Algorithms Day 21 L12.7? 2001 by Erik D. Demaine 1D range searching In 1D, the query is an interval: Second solution using ideas we know: ? Sort the points and store them in an array ? Solve query by binary search on endpoints. ? Obtain a static structure that can list k answers in a query in O(k + lg n) time. Goal: Obtain a dynamic structure that can list k answers in a query in O(k + lg n) time. Introduction to Algorithms Day 21 L12.8? 2001 by Erik D. Demaine 1D range searching In 1D, the query is an interval: New solution that extends to higher dimensions: ? Balanced binary search tree ? New organization principle: Store points in the leaves of the tree. ? Internal nodes store copies of the leaves to satisfy binary search property: ? Node x stores in key[x] the maximum key of any leaf in the left subtree of x. Introduction to Algorithms Day 21 L12.9? 2001 by Erik D. Demaine Example of a 1D range tree 1 1 6 6 8 8 12 12 14 14 17 17 26 26 35 35 41 41 42 42 43 43 59 59 61 61 Introduction to Algorithms Day 21 L12.10? 2001 by Erik D. Demaine Example of a 1D range tree 12 12 1 1 6 6 8 8 12 12 14 14 17 17 26 26 35 35 41 41 42 42 43 43 59 59 61 61 6 6 26 26 41 41 59 59 1 1 14 14 35 35 43 43 42 42 8 8 17 17 x x ≤ x > x Introduction to Algorithms Day 21 L12.11? 2001 by Erik D. Demaine 12 12 8 8 12 12 14 14 17 17 26 26 35 35 41 41 26 26 14 14 Example of a 1D range query 1 1 6 6 42 42 43 43 59 59 61 61 6 6 41 41 59 59 1 1 35 35 43 43 42 42 8 8 17 17 RANGE-QUERY([7, 41]) x x ≤ x > x Introduction to Algorithms Day 21 L12.12? 2001 by Erik D. Demaine General 1D range query root split node Introduction to Algorithms Day 21 L12.13? 2001 by Erik D. Demaine Pseudocode, part 1: Find the split node 1D-RANGE-QUERY(T, [x 1 , x 2 ]) w ← root[T] while w is not a leaf and (x 2 ≤ key[w] or key[w] < x 1 ) do if x 2 ≤ key[w] then w ← left[w] else w ← right[w] ? w is now the split node [traverse left and right from w and report relevant subtrees] Introduction to Algorithms Day 21 L12.14? 2001 by Erik D. Demaine Pseudocode, part 2: Traverse left and right from split node 1D-RANGE-QUERY(T, [x 1 , x 2 ]) [find the split node] ? w is now the split node if w is a leaf then output the leaf w if x 1 ≤ key[w] ≤ x 2 else v ← left[w] ? Left traversal while v is not a leaf do if x 1 ≤ key[v] then output the subtree rooted at right[v] v ← left[v] else v ← right[v] output the leaf v if x 1 ≤ key[v] ≤ x 2 [symmetrically for right traversal] Introduction to Algorithms Day 21 L12.15? 2001 by Erik D. Demaine Analysis of 1D-RANGE-QUERY Query time: Answer to range query represented by O(lg n) subtrees found in O(lg n) time. Thus: ? Can test for points in interval in O(lg n) time. ? Can count points in interval in O(lg n) time if we augment the tree with subtree sizes. ? Can report the first k points in interval in O(k + lg n) time. Space: O(n) Preprocessing time: O(n lg n) Introduction to Algorithms Day 21 L12.16? 2001 by Erik D. Demaine 2D range trees Store a primary 1D range tree for all the points based on x-coordinate. Thus in O(lg n) time we can find O(lg n) subtrees representing the points with proper x-coordinate. How to restrict to points with proper y-coordinate? Introduction to Algorithms Day 21 L12.17? 2001 by Erik D. Demaine 2D range trees Idea: In primary 1D range tree of x-coordinate, every node stores a secondary 1D range tree based on y-coordinate for all points in the subtree of the node. Recursively search within each. Introduction to Algorithms Day 21 L12.18? 2001 by Erik D. Demaine Analysis of 2D range trees Query time: In O(lg 2 n) = O((lg n) 2 ) time, we can represent answer to range query by O(lg 2 n) subtrees. Total cost for reporting k points: O(k + (lg n) 2 ). Preprocessing time: O(n lg n) Space: The secondary trees at each level of the primary tree together store a copy of the points. Also, each point is present in each secondary tree along the path from the leaf to the root. Either way, we obtain that the space is O(n lg n). Introduction to Algorithms Day 21 L12.19? 2001 by Erik D. Demaine d-dimensional range trees Query time: O(k + lg d n) to report k points. Space: O(n lg d –1 n) Preprocessing time: O(n lg d –1 n) Each node of the secondary y-structure stores a tertiary z-structure representing the points in the subtree rooted at the node, etc. Best data structure to date: Query time: O(k + lg d –1 n) to report k points. Space: O(n (lg n / lg lg n) d –1 ) Preprocessing time: O(n lg d –1 n) Introduction to Algorithms Day 21 L12.20? 2001 by Erik D. Demaine Primitive operations: Crossproduct Given two vectors v 1 = (x 1 , y 1 ) and v 2 = (x 2 , y 2 ), is their counterclockwise angle θ ? convex (< 180o), ? reflex (> 180o), or ? borderline (0 or 180o)? v 1 v 2 θ v 2 v 1 θ convex reflex Crossproduct v 1 × v 2 = x 1 y 2 – y 1 x 2 = |v 1 | |v 2 | sin θ . Thus, sign(v 1 × v 2 ) = sign(sin θ) > 0 if θ convex, < 0 if θ reflex, = 0 if θ borderline. Introduction to Algorithms Day 21 L12.21? 2001 by Erik D. Demaine Primitive operations: Orientation test Given three points p 1 , p 2 , p 3 are they ? in clockwise (cw) order, ? in counterclockwise (ccw) order, or ? collinear? (p 2 – p 1 ) × (p 3 – p 1 ) > 0 if ccw < 0 if cw = 0 if collinear p 1 p 3 p 2 cw p 1 p 2 p 3 ccw p 1 p 2 p 3 collinear Introduction to Algorithms Day 21 L12.22? 2001 by Erik D. Demaine Primitive operations: Sidedness test Given three points p 1 , p 2 , p 3 are they ? in clockwise (cw) order, ? in counterclockwise (ccw) order, or ? collinear? Let L be the oriented line from p 1 to p 2 . Equivalently, is the point p 3 ? right of L, ? left of L, or ? on L? p 1 p 3 p 2 cw p 1 p 2 p 3 ccw p 1 p 2 p 3 collinear Introduction to Algorithms Day 21 L12.23? 2001 by Erik D. Demaine Line-segment intersection Given n line segments, does any pair intersect? Obvious algorithm: O(n 2 ). a b c d e f Introduction to Algorithms Day 21 L12.24? 2001 by Erik D. Demaine Sweep-line algorithm ? Sweep a vertical line from left to right (conceptually replacing x-coordinate with time). ? Maintain dynamic set S of segments that intersect the sweep line, ordered (tentatively) by y-coordinate of intersection. ? Order changes when ? new segment is encountered, ? existing segment finishes, or ? two segments cross ? Key event points are therefore segment endpoints. segment endpoints Introduction to Algorithms Day 21 L12.25? 2001 by Erik D. Demaine a b c d e f a a bbb b b bffff c a c a d d edbee d c c dbddd eeeb Introduction to Algorithms Day 21 L12.26? 2001 by Erik D. Demaine Sweep-line algorithm Process event points in order by sorting segment endpoints by x-coordinate and looping through: ? For a left endpoint of segment s: ? Add segment s to dynamic set S. ? Check for intersection between s and its neighbors in S. ? For a right endpoint of segment s: ? Remove segment s from dynamic set S. ? Check for intersection between the neighbors of s in S. Introduction to Algorithms Day 21 L12.27? 2001 by Erik D. Demaine Analysis Use red-black tree to store dynamic set S. Total running time: O(n lg n). Introduction to Algorithms Day 21 L12.28? 2001 by Erik D. Demaine Correctness Theorem: If there is an intersection, the algorithm finds it. Proof: Let X be the leftmost intersection point. Assume for simplicity that ? only two segments s 1 , s 2 pass through X, and ? no two points have the same x-coordinate. At some point before we reach X, s 1 and s 2 become consecutive in the order of S. Either initially consecutive when s 1 or s 2 inserted, or became consecutive when another deleted.