1. An Optimal Dynamic Interval Stabbing-Max Data Structure? Pankaj K. Agarwal, Lars Arge and Ke Yi Department of Computer Science Duke University

2. 2 Problem Definition S : n intervals in 1D w(s) : weight of s  S; w(s)  R Stabbing-max query q  R : max{w(s) | s  S, q  s} q 2 3 4 5 1 Three operations: Query Insert an interval Delete an interval

3. 3 Trivial if no deletions A binary search tree: O(n) space, O(log n) query & insert Interval tree Size: O(n) Query: O(log 2 n) Insert: O(log n) Delete: O(log n) Kaplan, Molad and Tarjan [STOC ’ 03] The best known result Size: O(n) Query: O(log n) Insert: O(log n) Delete: O(log n loglog n) Previous Results 1 3 4 2 3 4 4 q

4. 4 Trivial if no deletions A binary search tree: O(n) space, O(log n) query & insert Interval tree Size: O(n) Query: O(log 2 n) Insert: O(log n) Delete: O(log n) Kaplan, Molad and Tarjan [STOC ’ 03] The best known result Size: O(n) Query: O(log n) Insert: O(log n) Delete: O(log n loglog n) Our Results 1 3 4 2 3 4 4 q

5. 5 Base Tree T Main idea: use a base interval tree with fan-out f = Height of tree: O(log n / loglog n) Each leaf stores Q (log n) endpoints O(n / log 3/2 n) internal nodes, O(n / log n) nodes in total slab multislab slab multislab O( ) slabs O(log n) multislabs

6. 6 Base Tree T v S(v): Intervals associated with node v Each interval is broken into a left, a right and a middle segment

7. 7 Middle Segments: Overview Answers query within S(v) in O(loglog n) time O(log n / loglog n) nodes on the path O(log n) time in total Insert or delete a segment in O(log n) time Only one M v is affected Total time: O(log n) Size: O(|S(v)| + log 3/2 n) O(n / log 3/2 n) internal nodes Total size: O(n) S(v)S(v) A secondary structure M v for each internal node v of the base tree

8. 8 Secondary Structure M v Build a multislab heap for each multislab Stores all segments spanning exactly this multislab Build a slab heap for each slab Stores the top elements of the relevant multislab heaps Query: O(loglog n) Find the slab heap Update: O(log n) Update the multislab heap: O(log n) Update the slab heaps: 1-2 2-43-3 2-5 123 1 2 3 4 5 Size: O(|S(v)|) Size: O(log 3/2 n) 45 4-5

9. 9 Left Segments: A Static Structure L(v) : segments in  i S(u i ) with left endpoints in the slab of v Basic idea: answer query in L(v) − L( w 4 ) when visiting v f (w j ) : maximum interval in L(v) with left endpoint in the slab of w j Organize in a tournament tree Can find the maximum of the red segments in time O(loglog n) Total: O(log n) w 1 w 2 w 3 w 4 w 5 v u1u1 f(w2)f(w2) u2u2 f(w1)f(w1) f(w2)f(w2) f(w3)f(w3) f(w4)f(w4) q

10. 10 Updating f Each segment at u 2 affects the f values at O(log n / loglog n) descendents Each f can be updated in O(loglog n) time (using additional structures) Total update time: O(log n) Other issues (omitted) Rebalancing of the base tree w 1 w 2 w 3 w 4 w 5 v u1u1 u2u2

11. 11 Is It Really Optimal? Is the O(log n) deletion bound really optimal? The O(log n) query bound is optimal in comparison model Can show O(log n) query → W (log n) insert O(log n) query & insert → W (log n) delete ? Probably yes Errata (Theorem 4.2 & 4.3) Query & update time: O(log d-1 n) → O(log d n)

12. Thank you!

13. 13 Putting Everything Together y (u, v) = max{ y (u, w), for all of v ’s children w} f (w) = max{ y (u, w), for all v ’s ancestors u execpt p(w)} a b cd e gf hij Base tree y (a,b) y (a,c) y (a,d) y (a,e) TaTa y (b,c) y (b,d) y (b,e) TbTb y (c,e) TcTc y (d,e) TdTd Consider f (e) y (b,f) y (b,g) y (c,f) y (c,g) y (d,f) y (d,g)

14. 14 Putting Everything Together Update: O(log n / loglog n · loglog n) = O(log n) a b cd e gf hij Base tree y (a,b) y (a,c) y (a,d) y (a,e) TaTa y (b,c) y (b,d) y (b,e) TbTb y (c,e) TcTc y (d,e) TdTd Only one T v is affected Total size is still linear