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Dynamic rectangular intersection with priorities

Dynamic rectangular intersection with priorities. K, Molad, Tarjan. 0. x. Related classical problem: Range reporting. Given a set of intervals S on the line, preprocess them to build a structure that allows efficient queries of the from: Given a point x find all intervals containing it. 3.

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Dynamic rectangular intersection with priorities

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  1. Dynamic rectangular intersection with priorities K, Molad, Tarjan

  2. 0 x Related classical problem: Range reporting Given a set of intervals S on the line, preprocess them to build a structure that allows efficient queries of the from: Given a point x find all intervals containing it.

  3. 3 x Range reporting + priorities Given a set of intervals S on the line, each with priority assigned to it, build a structure that allows efficient queries of the from: Given a point x find interval with minimum priority containing it. Updates – insert or delete an interval 5 9 1 7 0

  4. B A 3 3 2 1 190.0.1.0 190.0.7.0 190.0.2.0 190.0.3.0 190.0.4.0 190.0.5.0 190.0.6.0 190.0.8.0 190.0.9.0 190.0.10.0 190.0.11.0 190.0.12.0 190.0.13.0 190.0.14.0 Motivation – Packet classification Forward to Interface A Forward to Interface B block block & report to Bill IP address

  5. 190.0.*.* 190.1.*.* Forward to Interface A Forward to Interface B 3 3 190.0.1.* block 2 Nested intervals, ip prefixes IP address 190.1.0.0 190.0.0.0 190.0.1.0 190.0.2.255 190.0.255.255 190.1.255.255

  6. Extension to 2D • Query = point in R2 • (Sender IP, receiver IP) • interval = rectangle with priority 5 9 7

  7. One dimensional data structure for nested intervals 4 5 9 2 2 7 1

  8. Nested Intervals 4 5 9 2 2 7 1 Containment tree: The parent of interval v is the smallest interval containing v 7 2 1 2 9 5 4

  9. Nested Intervals 4 5 9 2 2 7 1 7 2 Query: Starting node s = smallest interval containing the query point Relevant priorities are on the path from s to the root. 1 2 9 5 Problem: path may be long… 4

  10. Hey, dynamic trees know how to do that 4 5 9 2 2 7 1 We can use a dynamic tree to represent the containment tree. 7 2 1 Problem: Updates => Many cuts & links 2 9 5 4

  11. Insert

  12. Node v => node v Leftmost child of v => Left child of v ∞ ∞ 7 ∞ Any other child of v => right child of its left sibling 9 Adjust costs: Left edge => priority of parent Right edge => ∞ 5 Binarization 4 5 9 2 2 7 1 7 2 1 2 9 5 4

  13. Insert (Cont.) Constant number of links and cuts

  14. Summary • Containment tree C • Query = min cost on path from starting point to root • Represent C by binarized version B • Represent B by dynamic tree D • How do you find the point to start the query ? • How do you find the edges to cut ?

  15. Min(Mincost( ), pri( )) How do you start the query ? 4 2 9 5 2 7 1 Use a balanced search tree on the endpoints 7 1 9

  16. Mincost( ) query (cont) 4 2 9 5 2 7 1 7 1 9

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