# Binary search trees ppt

###### optimal binary search trees using dynamic programming ppt and binary search tree insertion and deletion ppt
Dr.BenjaminClark,United States,Teacher
Published Date:21-07-2017
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ROBERT SEDGEWICK KEVIN WAYNE Algorithms GEOMETRIC APPLICATIONS OF BSTS 1d range search ‣ line segment intersection ‣ kd trees ‣ Algorithms FOUR TH EDITION interval search trees ‣ rectangle intersection ‣ ROBERT SEDGEWICK KEVIN WAYNE http://algs4.cs.princeton.eduOverview This lecture. Intersections among geometric objects. 2d orthogonal range search orthogonal rectangle intersection Applications. CAD, games, movies, virtual reality, databases, GIS, .… Efficient solutions. Binary search trees (and extensions). 2GEOMETRIC APPLICATIONS OF BSTS 1d range search ‣ line segment intersection ‣ kd trees ‣ Algorithms interval search trees ‣ rectangle intersection ‣ ROBERT SEDGEWICK KEVIN WAYNE http://algs4.cs.princeton.edu1d range search Extension of ordered symbol table. Insert key-value pair. Search for key k. Delete key k. Range search: find all keys between k and k . 1 2 Range count: number of keys between k and k . 1 2 Application. Database queries. insert B B insert D B D insert A A B D Geometric interpretation. insert I A B D I insert H A B D H I Keys are point on a line. insert F A B D F H I Find/count points in a given 1d interval. insert P A B D F H I P search G to K H I count G to K 2 41d range search: elementary implementations Unordered list. Fast insert, slow range search. Ordered array. Slow insert, binary search for k and k to do range search. 1 2 order of growth of running time for 1d range search data structure insert range count range search unordered list 1 N N ordered array N log N R + log N goal log N log N R + log N N = number of keys R = number of keys that match 51d range count: BST implementation 1d range count. How many keys between lo and hi ? 6 rank S 2 7 E X 5 0 A R 3 1 C H 4 M public int size(Key lo, Key hi) if (contains(hi)) return rank(hi) - rank(lo) + 1; else return rank(hi) - rank(lo); number of keys hi Proposition. Running time proportional to log N. Pf. Nodes examined = search path to lo + search path to hi. 61d range search: BST implementation 1d range search. Find all keys between lo and hi. Recursively find all keys in left subtree (if any could fall in range). Check key in current node. Recursively find all keys in right subtree (if any could fall in range). searching in the range F..T red keys are used in compares but are not in the range S E X A R H C M L P black keys are in the range Range search in a BST Proposition. Running time proportional to R + log N. Pf. Nodes examined = search path to lo + search path to hi + matches. 7GEOMETRIC APPLICATIONS OF BSTS 1d range search ‣ line segment intersection ‣ kd trees ‣ Algorithms interval search trees ‣ rectangle intersection ‣ ROBERT SEDGEWICK KEVIN WAYNE http://algs4.cs.princeton.eduOrthogonal line segment intersection Given N horizontal and vertical line segments, find all intersections. Quadratic algorithm. Check all pairs of line segments for intersection. Nondegeneracy assumption. All x- and y-coordinates are distinct. 9Orthogonal line segment intersection: sweep-line algorithm Sweep vertical line from left to right. x-coordinates define events. h-segment (left endpoint): insert y-coordinate into BST. 3 3 4 2 2 1 1 0 0 y-coordinates 10Orthogonal line segment intersection: sweep-line algorithm Sweep vertical line from left to right. x-coordinates define events. h-segment (left endpoint): insert y-coordinate into BST. h-segment (right endpoint): remove y-coordinate from BST. 3 3 4 2 1 1 0 0 y-coordinates 11Orthogonal line segment intersection: sweep-line algorithm Sweep vertical line from left to right. x-coordinates define events. h-segment (left endpoint): insert y-coordinate into BST. h-segment (right endpoint): remove y-coordinate from BST. v-segment: range search for interval of y-endpoints. 3 3 1d range 4 search 2 1 1 0 0 y-coordinates 12Orthogonal line segment intersection: sweep-line analysis Proposition. The sweep-line algorithm takes time proportional to N log N + R to find all R intersections among N orthogonal line segments. Pf. N log N Put x-coordinates on a PQ (or sort). N log N Insert y-coordinates into BST. Delete y-coordinates from BST. N log N Range searches in BST. N log N + R Bottom line. Sweep line reduces 2d orthogonal line segment intersection search to 1d range search. 13GEOMETRIC APPLICATIONS OF BSTS 1d range search ‣ line segment intersection ‣ kd trees ‣ Algorithms interval search trees ‣ rectangle intersection ‣ ROBERT SEDGEWICK KEVIN WAYNE http://algs4.cs.princeton.edu2-d orthogonal range search Extension of ordered symbol-table to 2d keys. Insert a 2d key. Delete a 2d key. Search for a 2d key. Range search: find all keys that lie in a 2d range. Range count: number of keys that lie in a 2d range. Applications. Networking, circuit design, databases, ... Geometric interpretation. Keys are point in the plane. Find/count points in a given h-v rectangle rectangle is axis-aligned 152d orthogonal range search: grid implementation Grid implementation. Divide space into M-by-M grid of squares. Create list of points contained in each square. Use 2d array to directly index relevant square. Insert: add (x, y) to list for corresponding square. Range search: examine only squares that intersect 2d range query. RT LB 162d orthogonal range search: grid implementation analysis Space-time tradeoff. 2 Space: M + N. 2 Time: 1 + N / M per square examined, on average. Choose grid square size to tune performance. Too small: wastes space. Too large: too many points per square. Rule of thumb: √N-by-√N grid. RT Running time. if points are evenly distributed Initialize data structure: N. choose M √N Insert point: 1. Range search: 1 per point in range. LB 17Clustering Grid implementation. Fast, simple solution for evenly-distributed points. Problem. Clustering a well-known phenomenon in geometric data. Lists are too long, even though average length is short. Need data structure that adapts gracefully to data. 18Clustering Grid implementation. Fast, simple solution for evenly-distributed points. Problem. Clustering a well-known phenomenon in geometric data. Ex. USA map data. 13,000 points, 1000 grid squares half the squares are empty half the points are in 10% of the squares 19Space-partitioning trees Use a tree to represent a recursive subdivision of 2d space. Grid. Divide space uniformly into squares. 2d tree. Recursively divide space into two halfplanes. Quadtree. Recursively divide space into four quadrants. BSP tree. Recursively divide space into two regions. Grid 2d tree Quadtree BSP tree 20
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