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First of all …. Thanks to Janos

First of all …. Thanks to Janos. Conflict-free coloring problems. Shakhar Smorodinsky Tel Aviv University. Part of this work is joint with Guy Even , Zvi Lotker and Dana Ron. Definition of Conflict-Free Coloring. Any (non-empty) disc contains a unique color. A Coloring of pts.

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First of all …. Thanks to Janos

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  1. First of all …. Thanks to Janos

  2. Conflict-free coloring problems Shakhar Smorodinsky Tel Aviv University Part of this work is joint with Guy Even, Zvi Lotker and Dana Ron.

  3. Definition of Conflict-Free Coloring Any (non-empty) disc contains a unique color A Coloring of pts is Conflict Free if: 4 1 3 2 4 3 3 2 1

  4. What the … is Conflict-Free Coloring? A Coloring of pts 1 is Conflict Free if: 1 3 2 4 3 3 2 1

  5. Problem Statement What is the smallest number f(n) s.t. any n points can be CF-colored with only f(n) colors? Remark: We can define a CF-coloring for a general set system (X,A) where AP(X) i.e., a coloring of the elements of X s.t. each set sA contains an element with a unique color

  6. Motivation from Frequency Assignment in cellular networks 1 mobile clients: create links with base-stations within reception radius

  7. If 2 is unique Frequency assignment 1 2 1 Power and location of clients’ cellular may vary

  8. log n colors n pts n/2  n/4 Problem Statement (cont) What is the minimum number f(n) s.t. any n points can be CF-colored with f(n) colors? Thm: f(n) > log n Easy: n pts on a line! Discs = Intervals 3 1 2

  9. Points on a line (cont) log n colors suffice (in this special case) Divide & Conquer 1 3 3 2 1 3 2 3 Color median with 1 Recurse on right and left Reusing colors!

  10. CF-coloring in general case Divide & Conquer doesn’t work! Thm: f(n) = O(log n) n pts

  11. Proof of: f(n) = O(log n) Consider the Delauney Graph i.e., the “empty pairs” graph n pts It is planar. Hence, By the four colors Thm  “large” independent set

  12. 1 1 1 Proof of: f(n) = O(log n) (cont) • ISP s.t. |IS| n/4 and IS isindependent |P|=n 1. Color IS with 1 2. Remove IS

  13. 2 2 Proof of: f(n) = O(log n) (cont) • ISP s.t. |IS| n/4 and IS isindependent! |P|=n 1.Color IS with 1 2. remove IS 3. Construct the new Delauney graph … anditerate (O(log n) times)on remaining pts

  14. Proof of: f(n) = O(log n) (cont) • ISP s.t. |IS| n/4 and IS isindependent! |P|=n 1.Color IS with 1 2. remove IS 5 4 3. Iterate (O(log n) times)on remaining pts 3

  15. Proof of: f(n) = O(log n) (cont) Algorithm is correct Consider a non-empty disc 2 “maximal” color is unique 5 1 1 4 3 1 2 “maximal” color 3

  16. Proof of: f(n) = O(log n) (cont) “maximal” color i is unique Proof: Assume i is not unique and ignore colors < i i “maximal” color i

  17. Proof of: f(n) = O(log n) (cont) “maximal” color i is unique Assume i is not unique and ignore colors < i i i “maximal” color i

  18. Proof: maximal color iis unique Consider the i’th iteration independent i i i A third point exists

  19. Proof: maximal color iis unique Consider the i’th iteration Contradiction! i i i

  20. Remarks: Algorithm is very easy to implement

  21. How about axis-parallel rectangles? What about other ranges? CF-coloring pts w/ respect to other ranges? Previous proof works for homothetic copies of a convex body Thm: O(sqrt (n)) colors always suffice

  22. CF-coloring pts w.r.t axis-parallel rectangles Thm: O(sqrt (n)) colors always suffice How small is an independent set in the “Delauney” graph ? I DON’T KNOW!

  23. CF-coloring pts w.r.t axis-parallel rectangles Note:

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