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Klee’s Measure Problem Made Easy

Klee’s Measure Problem Made Easy. Timothy Chan U of Waterloo & HKUST. Beginning of Story…. Victor Klee, “Can the measure of U 1 n [ a i ,b i ] be computed in less than O(n log n) steps?”, Amer. Math. Monthly, 84:284-285, 1977. The Problem.

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Klee’s Measure Problem Made Easy

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  1. Klee’s Measure Problem Made Easy Timothy Chan U of Waterloo & HKUST

  2. 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

  3. The Problem • Given n rectangular boxes in d dimensions, compute the measure of the union

  4. Known Result in 2D • Union may have Ω(n2) boundary edges • Bentley'77: O(n log n) time

  5. Bentley’s Algorithm in 2D • Idea: sweep!  dynamicdata structure problem in 1D

  6. 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/2loglogn) time

  7. 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/2logloglog n)

  8. 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/2loglogloglog n)

  9. 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/22O(log*n)) time • Today:O(nd/2) time

  10. New Algorithm: Warm-Up in 2D • Idea: good old divide&conquer!

  11. New Algorithm in 2D: First Attempt • Given set B of n rectangles inside cell C: 1. divide C into Cleft & Cright 2. recurseinside Cleft & inside Cright

  12. New Algorithm in 2D: How to Divide? • … by median-x • Problem: “bad”rectangles! C Cright Cleft

  13. 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. recurseinside Cleft & inside Cright

  14. New Algorithm in 2D: How to Simplify? • … by removing all “bad” rectangles that don't have vertices inside C 

  15. 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)

  16. 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. recurseinside Cleft & inside Cright

  17. New Algorithm in 3D: How to Simplify? • … by removing all “bad” boxes that don't have edges intersecting C 

  18. New Algorithm in 3D:How to Divide? y • … by weightedmedian-x x z

  19. New Algorithm in 3D:How to Divide? y • … by weightedmedian-x • Define weight of an edge: • Total weight of edges intersecting Cleft(or Cright) = … after factoring out 2-2/3 x z 21/3 1 22/3 21/3 /2 1 22/3 /2 2-2/3 22/3 2-1/3 1 21/3

  20. 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)

  21. 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)

  22. FINAL REMARKS

  23. 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.

  24. 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??

  25. Log Factor Speedups • Cover/depth problem: O**(nd/2/ logd/2 n) by table lookup • Weighted depth problem: O**(nd/2/ logd/2 - 5n) by linear decision trees [Meyer auf der Heide’84,Meiser’93] • Measure problem: O**(nd/2 / logd/2 - 2n) by Chinese remainder theorem for integer input • Open question: speedup for measure problem for real input?

  26. Special Case: Hypercubes(or “Fat” Boxes) • Union may have Ω(nd/2) boundary faces • Agarwal,Kaplan,Sharir'07: O(n4/3 log2n) time (d=3) • Agarwal'10: O(n log4n) 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?

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