Klee’s Measure Problem Made Easy

Timothy ChanU of Waterloo & HKUST

Beginning of Story…

• Victor Klee, “Can the measure of U1n [ai,bi] be

computed in less than O(n log n) steps?”, Amer. Math. Monthly, 84:284-285, 1977

The Problem

• Given n rectangular boxes in d dimensions, compute the measure of the union

Known Result in 2D

• Union may have Ω(n2) boundary edges

• Bentley'77: O(n log n) time

Bentley’s Algorithm in 2D

• Idea: sweep!

dynamic data structure problem in 1D

Known Results in d ≥ 3 Dimensions

• Union may have Ω(nd) boundary faces

• Bentley'77: O(nd-1 log n) time• Van Leeuwen,Wood'81: O(nd-1) time

• Overmars,Yap [FOCS'88]: O(nd/2 log n) time• C.'08: O(nd/2 loglog n)

time

Known Results in d ≥ 3 Dimensions

• Union may have Ω(nd) boundary faces

• Bentley'77: O(nd-1 log n) time• Van Leeuwen,Wood'81: O(nd-1) time

• Overmars,Yap [FOCS'88]: O(nd/2 log n) time• C.'08: O(nd/2 logloglog n)

Known Results in d ≥ 3 Dimensions

• Union may have Ω(nd) boundary faces

• Bentley'77: O(nd-1 log n) time• Van Leeuwen,Wood'81: O(nd-1) time

• Overmars,Yap [FOCS'88]: O(nd/2 log n) time• C.'08: O(nd/2 loglogloglog

n)

Known Results in d ≥ 3 Dimensions

• Union may have Ω(nd) boundary faces

• Bentley'77: O(nd-1 log n) time• Van Leeuwen,Wood'81: O(nd-1) time

• Overmars,Yap [FOCS'88]: O(nd/2 log n) time• C.'08: O(nd/2 2O(log*n)) time• Today: O(nd/2) time

New Algorithm: Warm-Up in 2D

• Idea: good old divide&conquer!

New Algorithm in 2D: First Attempt

• Given set B of n rectangles inside cell C:

1. divide C into Cleft & Cright

2. recurse inside Cleft & inside Cright

New Algorithm in 2D: How to Divide?

• … by median-x

• Problem: “bad” rectangles!

C

Cleft Cright

New Algorithm in 2D

• Given set B of n rectangles inside cell C: 0. simplify B new twist!

1. divide C into Cleft & Cright

2. recurse inside Cleft & inside Cright

New Algorithm in 2D: How to Simplify?

• … by removing all “bad” rectangles that don't have vertices inside C

New Algorithm in 2D: Summary & Analysis

• Given set B of n rectangles inside cell C: 0. simplify B

… by removing all “bad” rectangles that don't have vertices inside C

1. divide C into Cleft & Cright

… by median-x

2. recurse inside Cleft & inside Cright

• Let N = # vertices inside C

T(N) = 2 T(N/2) + O(N) O(N log N)

Extending New Algorithm to 3D

• Given set B of n boxes inside cell C: 0. simplify B

… by removing all “bad” boxes that don't have edges intersecting C

1. divide C into Cleft & Cright

… by weighted median-x

2. recurse inside Cleft & inside Cright

New Algorithm in 3D: How to Simplify?

• … by removing all “bad” boxes that don't have edges intersecting C

New Algorithm in 3D:How to Divide?

• … by weighted median-xx

y

z

New Algorithm in 3D:How to Divide?

• … by weighted median-x

• Define weight of an edge:

• Total weight of edges intersecting Cleft (or Cright) =

… after factoring out 2-2/3

x

y

z

1 21/322/3

1 21/3 /2 22/3 /22-2/3 2-1/322/321/31

New Algorithm in 3D: Analysis

• Let N = total weight of edges intersecting C

Nleft, Nright ≤ 2-2/3 N after renaming axes

T(N) = 2 T(N / 22/3) + O(N) O(N3/2)

Extending New Algorithm to d Dimensions

• Define weight of a (d-2)-dimensional face f= 2(i+j)/d if f is orthogonal to axis i & j

T(N) = 2 T(N / 22/d) + O(N) O(Nd/2)

FINAL REMARKS

Other Applications

• Cover problem: given n boxes in d dimensions, does their union completely cover C?

• Depth problem: given n boxes in d dimensions, find a point q minimizing # of boxes containing q

• Weighted depth problem: given n weighted boxes in d dimensions, find a point q minimizing total weight of boxes containing q

• Etc.

Lower Bounds?

• C'08: Finding cliques of size d in a graph with V=n1/2 vertices reduces to the cover/depth/ measure problem

hard to beat O(Vd) = O(nd/2) time (ignoring log factors) by “purely combinatorial”

algorithms

• Open question: can matrix multiplication help??

Log Factor Speedups

• Cover/depth problem: O**(nd/2 / logd/2 n)by table lookup

• Weighted depth problem: O**(nd/2 / logd/2 - 5 n) by linear decision trees [Meyer auf der

Heide’84,Meiser’93]

• Measure problem: O**(nd/2 / logd/2 - 2 n) by Chinese remainder theorem for integer

input

• Open question: speedup for measure problem for real input?

Special Case: Hypercubes (or “Fat” Boxes)

• Union may have Ω(nd/2) boundary faces

• Agarwal,Kaplan,Sharir'07: O(n4/3 log2 n) time (d=3)

• Agarwal'10: O(n log4 n) time (d=3)

• Bringman'10: O(n(d+2)/3) • New: O(n(d+1)/3 polylog

n)by more complicated additional ideas…

• Open question: faster algorithms for hypercubes?