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

2. 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. x

3. 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 9 5 3 1 7 x

4. Motivation – Packet classificationB
Forward to Interface A
Forward to Interface B 1 2 3
block
block & report to Bill
This problem can be stated as follows:
The basic task of a router is to classify incoming packets into streams, and to handle each stream in a certain way.
A good way of doing that is to define intervals on the ip of the sender.
In this way, a packet with a sender which falls into a certain interval can be forwarded to an interface…
Or forwarded to another interface…
Or maybe blocked.
What happens when the packet falls into 2 intervals? We need priorities to decide.
IP address

5. Nested intervals, ip prefixes
190.0.*.*
190.1.*.* *
block
Forward to Interface A
Forward to Interface B 2 3
IP address

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

7. One dimensional data structure for nested intervals4 5 2 9 2 7 1

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

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

10. Hey, dynamic trees know how to do that4 5 2 9 2 7 1
We can use a dynamic tree to represent the containment tree. 2 7 1
Problem:
Updates => Many cuts & links 2 9 5 4

11. Insert

12. Binarization Node v => node v4 5 2 9 2 7 1
Node v => node v
Leftmost child of v => Left child of v 9 7 5 ∞
Adjust costs:
Left edge => priority of parent
Right edge => ∞ 2 7 1
Any other child of v => right child of its left sibling 2 9 5 4

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

14. Summary Containment tree C Represent C by binarized version B
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. How do you start the query ?4 2 9 5 2 7 1
Use a balanced search tree on the endpoints 7 1 9
Min(Mincost( ), pri( ))

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