An Optimal Dynamic Interval Stabbing-Max Data Structure?

1
An Optimal Dynamic Interval Stabbing-Max Data
Structure?

• Pankaj K. Agarwal, Lars Arge and Ke Yi
• Department of Computer Science
• Duke University

2
Problem Definition

• S n intervals in 1D
• w(s) weight of s?S w(s)?R
• Stabbing-max query q?R maxw(s) s?S, q?s

• Three operations
• Query
• Insert an interval
• Delete an interval

3
Previous Results

• Trivial if no deletions
• A binary search tree O(n) space, O(log n) query
  insert
• Interval tree
• Size O(n)
• Query O(log2 n)
• Insert O(log n)
• Delete O(log n)
• Kaplan, Molad and Tarjan STOC03
• The best known result
• Size O(n)
• Query O(log n)
• Insert O(log n)
• Delete O(log n loglog n)

q
4
4
3
2
4
3
1
4
Our Results

• Trivial if no deletions
• A binary search tree O(n) space, O(log n) query
  insert
• Interval tree
• Size O(n)
• Query O(log2 n)
• Insert O(log n)
• Delete O(log n)
• Kaplan, Molad and Tarjan STOC03
• The best known result
• Size O(n)
• Query O(log n)
• Insert O(log n)
• Delete O(log n loglog n)

q
4
4
3
2
4
3
1
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 / log3/2 n) internal nodes, O(n / log n)
  nodes in total

O( ) slabs
slab
slab
multislab
multislab
multislab
O(log n) multislabs
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
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 Mv is affected
• Total time O(log n)
• Size O(S(v) log3/2 n)
• O(n / log3/2 n) internal nodes
• Total size O(n)

• A secondary structure Mv for each internal node v
  of the base tree

S(v)
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Secondary Structure Mv

• 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 3 4 5
1
2
3
4
5
Size O(log3/2n)
2-4
3-3
2-5
4-5
1-2
Size O(S(v))
9
Left Segments A Static Structure

• L(v) segments in ?iS(ui) with left endpoints in
  the slab of v
• Basic idea answer query inL(v) - L(w4) when
  visiting v
• f(wj) maximum interval inL(v) with left
  endpoint in the slab of wj
• Organize in a tournament tree
• Can find the maximumof the red segmentsin time
  O(loglog n)
• Total O(log n)

q
f(w2)
u2
u1
v
w1 w2 w3 w4 w5
f(w1)
f(w2)
f(w3)
f(w4)
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Updating f

• Each segment at u2 affects the f values at O(log
  n / loglog n) descendents
• Each f can be updatedin O(loglog n) time (using
  additional structures)
• Total update time O(log n)
• Other issues (omitted)
• Rebalancing of the base tree

u2
u1
v
w1 w2 w3 w4 w5
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(logd-1n) ? O(logdn)

12
Thank you!
13
Putting Everything Together
y(u, v) maxy(u, w), for all of vs children
w f(w) maxy(u, w), for all vs ancestors u
execpt p(w)
Ta
Base tree
y(a,b)
a
y(a,c)
y(a,d)
y(a,e)
b
Tb
c
d
y(b,c)
y(b,d)
Consider f(e)
y(b,e)
y(b,f)
y(b,g)
g
f
e
h
i
j
Td
Tc
y(d,e)
y(c,e)
y(c,f)
y(c,g)
y(d,f)
y(d,g)
14
Putting Everything Together
Update O(log n / loglog n loglog n) O(log n)
Only one Tv is affected
Ta
Base tree
y(a,b)
a
y(a,c)
y(a,d)
y(a,e)
b
Tb
c
d
Total size is still linear
y(b,c)
y(b,d)
y(b,e)
g
f
e
h
i
j
Td
Tc
y(d,e)
y(c,e)