An Introduction To Range Searching

Transcription

1 An Introduction To Range Searching Jan Vahrenhold eartment of Comuter Science Westfälische Wilhelms-Universität Münster, Germany. Overview 1. Introduction: Problem Statement, Lower Bounds 2. Range Searching in 1 and 1.5 imensions 3. Range Searching in 2 imensions 4. Summary and Outlook Jan Vahrenhold Range Searching 1

2 Problem Setting Given: Collection S of n oints in d dimensions (S IR d ). Wanted: Algorithm for efficiently reorting all k oints in S falling into a given axis-arallel query range IR d. Alications: Geograhic Information Systems; atabases having relations in which the keys can be totally ordered. Gehalt Mayer Maier Eintrittsdatum Meier 1985 Ma Me Name Jan Vahrenhold Range Searching 2 A First Aroach Assume that S = { 0,..., n 1 } is stored in an array. Scan though the array and test for each i whether i Need to scan the whole array, regardless of how many oints are reorted. Comlexity: Θ (n) time and sace. Jan Vahrenhold Range Searching 3

3 Lower (and Uer) Bounds Change the model to also include k (the number of oints reorted) as a arameter. Algorithm on revious slide has comlexity O (n + k) = O (n). Time comlexity: rerocessing time query time Can disregard rerocessing time for many alications (one-time oeration). Query time comosed of two comonents: Search time: Time to locate the first element to be reorted. Retrieval time: Time to fetch and reort all k elements to be reorted. Sace requirement (lower bound for rerocessing time). Jan Vahrenhold Range Searching 4 Lower Bounds [Bentley & Maurer, 1980] Parameters: n oints, k oints reorted, d dimensions. Sace requirement: Ω (n). Retrieval time: Ω (k). Search time: Using binary decision tree ( sorting lower bound). Lower bound construction: (n =) 2ad oints, each with exactly one unique non-zero integer coordinate taken from [ a, a] \ {0}. = [b 1,..., b d ] [c 1,..., c d ], with b i [ a, 1], c i [1, a], 1 i d. Query ranges not-emty, each roduces a different answer. Overall: a 2d = (n/(2d)) 2d different answers. ( eth of decision tree: Ω log (n/(2d)) 2d) = Ω (d log n). Lower bound not tight for all d. Jan Vahrenhold Range Searching 5

4 Overview 1. Introduction: Problem Statement, Lower Bounds 2. Range Searching in 1 and 1.5 imensions 3. Range Searching in 2 imensions 4. Summary and Outlook Jan Vahrenhold Range Searching 6 One-imensional Range Searching Point set S = { 0,..., n 1 } IR, stored in an array. Query range = [x 1, x 2 ]. Scanning is sub-otimal; lower bound: Ω(1 log 2 n + k). Prerocessing: Sort the oints, e.g., using heasort in O (n log 2 n) time Query: Binary search for smallest i x 1... O (log 2 n)... scan forward until first i < x 2 (or end of array). O (k + 1) Jan Vahrenhold Range Searching 7

5 oes Sorting Hel in Two imensions? There is no total order on oints in two dimensions sorting according to which guarantees Θ (2 log 2 n + k) query time for range searching. Jan Vahrenhold Range Searching 8 Reca: One-imensional Range Searching Key ingredient: binary search (bisection). Relace (sorted) array by binary search tree Time Comlexity: Prerocessing time: O (n log n) Query time: O (log n + k) Sace Comlexity: O (n). Inserts/eletes ossible Jan Vahrenhold Range Searching 9

6 Three-sided (1.5-dim.) Range Searching Given: Point set S = { 0,..., n 1 } IR 2, stored in an array. Wanted: Method to efficiently retrieve all S that, for given (x 1, x 2, y), fall into [x 1, x 2 ] ], y]. Look at two subroblems: Reort all oints in [x 1, x 2 ] IR using, e.g., a threaded binary search tree. Reort all oints in IR ], y] using, e.g., a hea: Almost comlete binary tree. key(v) min{key(lson(v)), key(rson(v))} Jan Vahrenhold Range Searching 10 Combining the best of both worlds(?) Binary search tree with hea roerty: Binary search tree unique w.r.t. inorder -traversal. No (direct) way of incororating hea roerty. Hea with search tree roerty: Hea not unique. More recisely: Children of a node may be switched. Priority Search Tree: Binary tree H storing a two-dimensional oint at each node s.t. the hea roerty w.r.t. the y-coordinates is fulfilled. Additional requirement: v H : x v IR : l x v < r l LSUBTREE(v), r RSUBTREE(v). Jan Vahrenhold Range Searching 11

7 Building a riority search tree Use recursive definition [McCreight, 1985]: Build riority search tree H(S) for a given set S of oints in the lane. Assume w.l.o.g. that all coordinates are airwise distinct. If S =, construct H(S) as an (emty) leaf. Else let min be the oint in S having the minimum y-coordinate. Let x mid be the median of the x-coordinates in S \ { min }. Partition S \ { min }: S left := { S \ { min }.x x mid } S right := { S \ { min }.x > x mid } Construct search tree node v storing x mid and set (v) := min. Recursively comute v s children H(S left ) and H(S right ). Comlexity: O (n) sace; O (n log n) time (why?). Jan Vahrenhold Range Searching 12 Querying a riority search tree Query range [x 1, x 2 ] [, y]: Queries for x 1 and x 2 result in two search aths in H. Check all oints on these aths. All subtrees embraced by these aths contain oints in [x 1, x 2 ] IR. Query these subtrees a follows: x 1 x 2 SearchInSubtree(v, y) if v not a leaf and (v).y y then Reort (v); SearchInSubtree(LSON(v), y); SearchInSubtree(RSON(v), y); Query time: O (1 + k v ). Examle for y = 5. Jan Vahrenhold Range Searching 13

8 Summary Missing Comonents: A more detailed descrition of the query algorithm. Proof of correctness. [de Berg et al., 2000] Theorem 2.1 Priority search trees allow for answering three-sided range queries on oints in IR 2 with time and sace comlexities as follows: Prerocessing time: Θ (n log n) Query time: O (log n + k) Sace requirement: Θ (n) Jan Vahrenhold Range Searching 14 Overview 1. Introduction: Problem Statement, Lower Bounds 2. Range Searching in 1 and 1.5 imensions 3. Range Searching in 2 imensions 4. Summary and Outlook Jan Vahrenhold Range Searching 15

9 Multidimensional Binary Search Tree Extend the concet of binary search by bisection to higher dimensions. Instead of intervals, artition (hyer-)rectangles; do the artitioning alternating arallel to the coordinate axes. R i is artitioned into R j and R k R j R k 1 2 R i. Structure corresonding to artitioning: balanced binary tree (k-tree [Bentley, 1975]). Node v corresonds to hyerrectangle R(v), R(root) = IR d ; children corresond to sub-hyerrectangles. Each node v is augmented to store: S(v): oints contained in R(v) (imlicitly). l(v): reresentation of slit axis. P (v): median of S(v) w.r.t. l(v). Jan Vahrenhold Range Searching 16 Examle Alternating artitioning along the coordinate axes. Jan Vahrenhold Range Searching 17

10 Querying a 2-tree void search(node v, rectangle, list oint & result) double left, median, right; if v.tye == vertical then left =.x1; right =.x2; median = v.p.x; else left =.y1; right =.y2; median = v.p.y; if left median right and.contains(v.p) then result.aend(v.p); if!isleaf(v) then if left < median then search(leftson(v),, result); if median < right then search(rightson(v),, result); return; Jan Vahrenhold Range Searching 18 Comlexity of a 2-tree Sace requirement: R(v) = P (v) R(q) for any descendant q of v. O (1) sace requirement er node, exactly one oint stored at each node O (n) overall sace requirement. Construction time (rerocessing): Linear-time median finding er artitioning ste, i.e., recurrence: T (n) = 2 T ( n/2 ) + O (n) O (n log n) Alternative: Relace median-finding by re-sorting (coies of) the oint by their x- and y-coordinates, resectively. Can find median w.r.t. x-coordinate in O (1) time. Can construct sorted y-arrays to be assed to the children in linear time. Jan Vahrenhold Range Searching 19

11 Analysis of worst-case query time Query time roortional to number of nodes visited. v roductive P (v). Nodes visited: roductive and unroductive nodes. efinition 3.1 Let R(v) be a rectangle and let 0 i 4. and R(v) form a tyei situation i sides of R(v) intersect the interior of. R(v) R(v) R(v) R(v) Tye 0 Tye 1 Tye 2 R(v) Tye 3 Tye 4 Tye-4 situation always roductive, all other situations may be unroductive. Jan Vahrenhold Range Searching 20 Constructing a worst-case situation I Use self-relicating tye-2/tye-3 situations [Lee & Wong, 1977]. A E B C B Y A G H F I C E F G Y H I T (h-2) T(h-2) T (h-3) h Recurrence for worst-case query time: T (h) = }{{} 1 + }{{} 1 + }{{} 1 + T (h 2) + T (h 2) + }{{}}{{}}{{} 1 A B C G F + T (h 3) }{{} H Jan Vahrenhold Range Searching 21

12 Constructing a worst-case situation II A closer look at situation subtree rooted at node. E B Y A G H F I C T (h-2) T (h-2) Y h Recurrence for this situation: T (h) = }{{} 1 + }{{} 1 + }{{} 1 Y + 2 T (h 2) }{{} Children of and Y Jan Vahrenhold Range Searching 22 Constructing a worst-case situation III The following recurrence holds for T (h): Y T (h) = 2 T (h 2) + 3 h with T (0) = 0 and T (1) = 1. T (h-2) T (h-2) Solve recurrence for T (h), w.l.o.g. h = 2 i, i IN. T ( 2 i ) = T ( 2(i 1) ) = ( T ( 2(i 2) )) = i 1 j=0 Similarly: T (2 i + 1) = 4 2 i j = 3 2 i 3 Jan Vahrenhold Range Searching 23

13 Constructing a worst-case situation IV The following recurrence holds for T (h): T (h) = T (h 2) + T (h 2) + T (h 3) + 4 T (h) = 4 2 i 3 for h = 2 i i 3 for h = 2 i with T (0) = T (0) = 0 and T (1) = T (1) = 1. A B C E F G Y H I T (h-2) T(h-2) T (h-3) h Solve recurrence for T (h), w.l.o.g. h = 2 i, i IN. T ( 2 i ) = 4 + T ( 2(i 1) ) i i 2 3 = T ( 2(i 1) ) i 1 2 = 5 (2 h/2 1 ) h Similarly: T (2 i + 1) = 7 (2 h/2 1 ) h + 2. Overall (for n 2 h 1): T (n) O ( 2 n 1/2). Jan Vahrenhold Range Searching 24 Summary Worst-case query time indeendent of the number of oints reorted. k-tree very relevant in ractice! Extension to higher dimensions (oints in IR d ): o artitioning in a round-robin manner of the coordinate axes x 1 x 2... x d x 1... Theorem 3.2 Multidimensional search trees (k-trees) allow for answering foursided range queries on oints in IR d, d 2 with time and sace comlexities as follows: Prerocessing time: Θ (d n log n) Query time: O ( d n 1 1/d + k ) Sace requirement: Θ (n) Jan Vahrenhold Range Searching 25

14 Overview 1. Introduction: Problem Statement, Lower Bounds 2. Range Searching in 1 and 1.5 imensions 3. Range Searching in 2 imensions 4. Summary and Outlook Jan Vahrenhold Range Searching 26 Summary Lower bounds: Ω (d log 2 n + k) time, Ω (n) sace. Results: One dimension: otimal O (log 2 n + k) algorithm, Θ (n) sace. 1.5 dimensions: otimal O (log 2 n + k) algorithm, Θ (n) sace. Two dimensions: sub-otimal O ( n + k ) algorithm, Θ (n) sace. d dimensions: sub-otimal O ( n 1 1/d + k ) algorithm, Θ (n) sace. Outlook: Otimal query time ossible of one is willing to send suerlinear sace [Chazelle, 1990]. Beware: choosing the adequate model of comutation is crucial. Jan Vahrenhold Range Searching 27

15 Bibliograhy [Bentley & Maurer, 1980] J. L. Bentley and H. A. Maurer. Efficient worst-case data structures for range searching. Acta Informatica, 13: , [Bentley, 1975] J. L. Bentley. Multidimensional binary search trees used for associative searching. Communications of the ACM, 18(9): , Setember [Chazelle, 1990] B. M. Chazelle. Lower bounds for orthogonal range searching. I: The reorting case. Journal of the ACM, 37(2): , Aril [de Berg et al., 2000] M. de Berg, M. J. van Kreveld, M. H. Overmars, and O. Schwarzkof. Comutational Geometry: Algorithms and Alications. Sringer, Berlin, second edition, [Lee & Wong, 1977].-T. Lee and C. K. Wong. Worst-case analysis for region and artial region searches in multidimensional binary search trees and balanced quad trees. Acta Informatica, 9:23 29, [McCreight, 1985] E. M. McCreight. Priority search trees. SIAM Journal on Comuting, 14(2): , May 1985.