1 / 26

Chapter 4. Partition

Consider n points in the Euclidean plane, each associated with a disk. Find a maximum independent set in an intersection disk graph. Implement a multi-layer partition method for efficient computation.

kchronister
Download Presentation

Chapter 4. Partition

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Chapter 4. Partition Ding-Zhu Du (2) Multi-layer Partition

  2. Intersection Disk Graph Consider n points in the Euclidean plane, each is associated with a disk. An edge exists between two points if and only if their associated disks have nonempty intersection.

  3. Maximum Independent Set in Intersection Disk Graph Given a intersection disk graph D, find a maximum Independent set opt(D).

  4. Multi-layer Suppose the largest disk has diameter 1-ε. Let dmin be The diameter of smallest disk. Fix an integer k > 0. Let Put all disks into m+1 layers. For 0 < j < m, layer j consists of all disks with diameter di,

  5. Partition P(0,0) in layer j (0,0)

  6. Partition P(0) in layer j and layer j+1

  7. Partition P(a,b) in layer j

  8. Layer j Layer j+1

  9. A disk hits a cut line. At each layer, a disk can hit at most one among Parallel lines apart each other with distance .

  10. D(a,b) In partition P(a,b), delete all disks each hitting a cut line in the same layer. The remaining disks form a collection D(a,b). Maximum Independent set opt(D(a,b)) can be computed in time Why use it? by dynamic programming.

  11. Dynamic Programming j-cell is a cell in layer j. For any j-cell S and a set I of independent disks in layers < j, intersecting S, Table(S,I) = maximum independent set of disks layers > j, contained in S, and disjoint from I. opt(D(a, b)) = US Table(S, Ǿ) where S is over all cells in layer 0.

  12. Recursive Relation For j-cell S and I,

  13. # of Table(S,I) # of S = too large How do we overcome this difficulty? Relevant cell: A j-cell is relevant if it contains a disk in layer j.

  14. Dynamic Programming j-cell is a cell in layer j. For any relevant j-cell S and a set I of independent disks in layers < j, intersecting S, Table(S,I) = maximum independent set of disks layers > j, contained in S, and disjoint from I. opt(D(a, b)) = USTable(S, Ǿ) where S is over all maximal relevant cells.

  15. Children of a relevant cell S S’’ S’

  16. Maximal relevant cell A relevant cell is maximal if it is not contained by Another relevant cell.

  17. Recursive Relation For j-cell S,

  18. # of Table(S,I) # of relevant S = n. # of I = # of Table(S,I) =

  19. # of I S # of I’s =

  20. Computation Time of Recursion # of S’ = # of J = Time = Running Time of dynamic programming

  21. # of J S

  22. (1+ε)-Approximation Choose k = ?. Compute opt(D(0,0)), opt(D(1,1)), …, opt(D(k-1,k-1). Choose maximum one among them.

  23. Analysis • Consider an optimal solution D*. • For each partition P(a,b), let H(a,b) be the collection of all disks hitting cut line in the same layer. • Estimate |H(0,0)|+|H(1,1)|+···+|H(k-1,k-1)|.

  24. |H(0,0)|+|H(1,1)|+···+|H(k-1,k-1)| Each disk appears in at most two terms in this sum. There exists i such that |H(2i,2i)| < 2|D*|/k.

  25. Performance ratio Opt/approx =1/(1-2/k) = 1 + 2/(k-4) Choose We obtain a (1+ε)-approximation With time

  26. Thanks, End

More Related