Conflict-Free Coloring

1. Conflict-Free Coloring Conflict-Free Coloring Gila Morgenstern CRI, Haifa University

2. Conflict-Free Coloring Conflict-Free (CF) Coloring: Introduced by [Even, Lutker, Ron, Smorodinsky. 03], motivated by frequency assignment problems in radio network

3. CF-Coloring of Unit Disks with Respect to Points Coloring of the disks s.t: For any point p, one of the disks covering p has a unique color, supporting it.

4. How many colors we need? Proper coloring Is OK But wasteful!

5. 1 2 3 4 5 6 7 All possible intervals 1 2 3 4 5 6 7 CF-Coloring of a Chain CF-Coloring of a chain is dual to CF-Coloring of points on the line w.r.t. intervals. (Each interval must contain a unique colored point.)

6. 1 2 3 i n SO…, how much can one save using colors ? (i), supporting of [1,n] #c(n) ≥ 1 + maxi{#c(i-1),#c (n-i)}  Ω(log n) colors required [1,n] Independent, (i) is excluded

7. 1 2 3 4 5 6 7 CF-Coloring of a Chain cont. ? ? O(log n) colors are sufficient

8. CF-coloring unit disks

9. CF-coloring unit disks 1

10. CF-coloring unit disks 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 6 4 5 6 4 5 6 4 5 6 4 5 6 4 5 6 4 5 6 7 8 9 7 8 9 7 8 9 7 8 9 7 8 9 7 8 9 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 4 5 6 4 5 6 4 5 6 4 5 6 4 5 6 4 5 6 5 7 7 8 9 7 8 9 7 8 9 7 8 9 7 8 9 7 8 9 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 3 4 4 5 6 4 5 6 4 5 6 4 5 6 4 5 6 4 5 6 9 8 2 7 8 9 7 8 9 7 8 9 7 8 9 7 8 9 7 8 9

11. CF-coloring unit disks Not a chain, still Θ(logni)= Θ(log n) colors

12. Variants • Squares, Pseudo disks, “Fat” objects, … - Θ(logn)colors. • On-line CF-coloring - Θ(log2n)colors. • Many more ….

13. Matthew J. Katz Nissan Lev-Tov Gila Morgenstern Ben-Gurion University, Israel Conflict-Free Coloring of Points on a Line W.R.T. a Set of Intervals

14. What if we need not respect everyone ? • Θ(logn)colors might bewasteful. Example: Proper interval graph: All intervals are non-nested. 2 non-zero colors are sufficient

15. l l U U CF-Coloring of Points w.r.t. a Subset of Intervals. • P = {p1,…,pm} - Set of points. • R = {I1,…,In} - Set of intervals. • For each interval I R: • r(I) is the right endpoint of I. • r(I) P. • Problem: Find coloring  of P w.r.t. R, using minimum number of colors.

16. O(1)-Approximation • 2-approximation algorithm: • Endpoints of all intervals are distinct. • 4-approximation algorithm: • Intervals may share endpoints.

17. Algorithm CFCp1 Simple greedy approach: Algorithm CFCp1: For each I by increasing r(I): χ(r(I)) = smallest color s.t. I is supported by some non-zero color.

18. 1 3 5 Colors key: 0 2 4

19. CFCp1 Optimal 2 non-zero colors 1 non-zero color Is this optimal? No:

20. CFCp1 computes a 2-approximation Our goal is to prove the following Lemma: If (p) = k, then CF-coloring of points leftward of p requires at least k/2 colors

21. Colors occuring in interval I. • Observations: If (r(I)) = k > 0 then: • k is the only unique color occurring in I. • All colors smaller than k occur in I at. least twice each.

22. K’ K’ K K Maximal Color is unique Lemma: Each interval is supported by the maximal color occurring in it. Proof: All colors smaller than k’ occur at least twice Supported by k<k’

23. U l • Range(R1UR2)I • Range(R1) Range(R2)= Ø U Optimal Sub-Coloring • Lemma: If R = R1U R2U {I} s.t.: R1 R2 {I}

24. Cont. ?  R1 R2 2 1 {I}

25. Cont. … • If 1 is an optimal CF-coloring w.r.t. R1,and 2 is an optimal CF-coloring w.r.t. R2: #colors required by an optimal CF-Coloring w.r.t. R is at least: 1 + || if |1|=|2|=|| max{ |1| , |2| } else

26. CFCp1 Computes a2-Approximation • Main Lemma: If (p) = k, then CF-coloring of points leftward of p require at least k/2 colors.

27. K K-2 K-2 K-1 K-1 Proof of Lemma • Let p be such that (p)=k (k-2)/2 + 1 = k/2 (k-2)/2 (k-2)/2

28. Relieving Assumption • So far, we assumed that endpoints of all intervals are distinct. => only O(n) possible intervals (out of O(n2)). • We now relieve this assumption, namely we allow intervals to share endpoints.

29. Algorithm CFCp2 Stage 1 (producing ’): • foreach p from left to right do: • If all intervals Iwith r(I)=p are already supported, then Χ’(p)=0. • Otherwise, let Ip be the longest interval with r(Ip)=p not supported. Χ’(p) = smallest color s.t. Ip is supported by some non-zero color.

30. Algorithm CFCp2 Stage 2 (producing ): • for each color k, alternately change appearances of k into “dark k” and “light k”.

31. Observation For each point p and interval I with r(I)=p, at least one of the following holds at the end of the first stage. • (i) The color ’(p) is unique in I. • (ii) The color ’(p) occurs exactly twice in I. • (iii) There exists a color c ≠ ’(p) which is unique in I.

32. Indeed,consider interval I with r(I)=p. • If ’(p)=0 then (iii) holds. • Otherwise: Clearly, for Ip we have that (i) holds. Same is true for shorter intervals. Suppose I is longer than Ip. By the way we chose Ip, we have that I was supported by some color c just before CFCp2 colored the point p. if c=’(p), then (ii) holds, otherwise (iii).

33. CFCp2 produces a CF-coloring • Let I be an interval and put p=r(I). • (i) If ’(p) is unique in I, so does (p). • (ii) If ’(p) occurs exactly twice in I, then “dark-’(p)” and “light-’(p)” occur in I once each. • (iii) If there exists a color c ≠ ’(p) which is unique in I, then either “dark-c” or ”light-c” occur in I exactly once. (p)

34. CFCp2 is 4-Appx • Let R’ be the set of intervals Ip that we considered during CFCp2. • Observation: Using CFCp1 to color P w.r.t R’, produces exactly ’. • Let OPT and OPT’ be optimal CF-colorings of P w.r.t R and R’. We get that: || ≤ 2 |’| ≤ 4|OPT’| ≤ 4|OPT| 2 tints CFCp1 for R’ R’ R

35. THANKS!