1. Guy Even, Zvi Lotker, Dana Ron, Shakhar Smorodinsky Tel Aviv University Conflict-free colorings of simple geometric regions with applications to frequency assignment in cellular networks

2. Guy Even, Zvi Lotker, Dana Ron, Shakhar Smorodinsky Tel-Aviv University Conflict-free colorings of simple geometric regions with applications to frequency assignment in cellular networks Now that’s a pretty LONG title!!!Guy, are you sure you you didn’t forget to add something to the title?

3. outline motivation definition of problem results primal & dual versions reductions algorithm for unit disks open problems

4. r=range every client within range can communicate with base station cellular networks – a base-station

5. more antennas  increase covered region cellular networks – multiple base-stations backbone network: between base-stations radio link: client  base-station mobile clients: dynamically create links with base-stations

6. interfering base-stations base-stations using same frequency  interference in intersection of regions

7. non-interfering base-stations base-stations use different frequencies  no interference!

8. base-station frequency assignment Coloring: intersecting base-stations must use different frequencies Coloring is too restrictive: every base-station can serve region of intersection. but, one is enough! Most models deal with interference between pairs of base-stations, 3rd base-station can´t resolve an interference.

9. Def: Conflict-free coloring Input: set of regions (e.g. disks). Output: a Coloring of the regions regions that cover a point P: N(P) = {regions d: P  d} point P is served by region d, if CF-coloring: all covered points are served. 1 2

10. What is the min #colors needed in a CF-coloring ?

11. What is the minimum number of colors we need ? every 2 “adjacent” disks must have different colors

12. Answer: 3 colors What is the minimum number of colors we need ?

13. What is the min #colors needed in a CF-coloring?

14. Answer: 4 colors

15. Hardness: Min CF-coloring of unit disks NP hardness – reduction similar to [CCJ90] vertex coloring of planar graph  Vertex coloring of intersection graphs of unit disks Reduction implies also that: (4/3-  )-approximation is NP-hard. Extends also to squares.

16. Results: CF-coloring of disks Poly algorithm: uses O(log(n)) colors. Tightness:  arrangement of disks that requires  (log(n)) colors. Locality: if radiuses  [1,const] then O(log(local density)) colors. Uniform coloring of congruent disks: given centers of congruent disks, same coloring is good for all radiuses. models clients with different reception ranges

17. Results: CF-coloring of “fat” rectangles O(1)-approx algorithm for CF-coloring arrangements of axis-parallel rectangles provided that: Similar result also for hexagons.

18. Results: CF-coloring “ellipses” Uniform CF-coloring of congruent copies of centrally symmetric convex regions using O(log(n)) colors. Uniform = centers are given, scaling factor is not given Centrally symmetric convex shapes: regular polygons with even #edges, …

19. arrangements of unit disks Topological arrangement: sub-division of plane into cells. a cell

20. examples of arrangements 7 cells : all non-empty subsets 6 cells : missing red-blue cell 7 cells: missing red-blue cell but brown cell appears twice. (view it as a single cell  combinatorially equiv. to previous arrangement)

21. set-system representation 1 2 3 4 5 6 7 1 2 3 4 5 6 disks 1 2 3 4 5 7 6 cells coalesce cells with identical neighbors 1 2 3 4 5 7 6 diskscells 12345 12345 6 7 disk-cell edge if cell in disk

22. primal/dual set-systems primal: sets elements dual: elements sets

23. indexed arrangements (skip) (skip) assign indexes to disks (not arbitrary!). represent set system by diagram (i.e. is cell covered by disk?) cells disks 245 7 89 N(cell) is an interval N(cell) is not an interval

24. Interval property of arrangements Full interval property: interval property and, for every interval [i,j], there exists a cell such that N(v) = [i,j]. Indexed arrangement: every disk has an index. Interval property: if, for every cell v, there exist i  j such that: N(v) = [i,j]. Chain: an indexed arrangement that satisfies the full interval property Equivalent DEF: dual set system representation isomorphic to the set system ({1,…,n}, {[i,j]} )

25. chains Claim: for every n, there exists a chain C(n) of n unit circles. Proof: index circles from left to right same proof works with axis-parallel squares, hexagons, etc.

26. CF-colorings of chains Claim: every CF-coloring of C(n) requires  (log n) colors. proof: “query”: which disk serves cell v: N(v)=[1,n]? color of this disk appears once (unique color). -red disk partitions chain into 2 disjoint chains. -pick larger part, and continue “queries” recursively.

27. coloring chain with O(log n) colors

28. duality in arrangements of unit disks arrangement corresponding to dual set system:

29. self-duality A collection of set-systems A is self-dual if (X,R)  A implies that (R,X*)  A. Consider set systems of “points & unit disks”: X – set of points in the plane R – set of ranges induced by intersection with unit disks. Claim: set systems of “points & unit disks” are self-dual. More general: “points & regions”: Claim: set system of “points & regions” is self-dual if regions are translations of a centrally-symmetric body (e.g. square, hexagon, rectangle). “points & arbitrary disks” NOT self-dual

30. Dual Problem: CF-coloring of points wrt ranges Input: set of points and set of ranges. Output: Coloring Require: for every range d, there exists a color i, such that {P  d:  (P)=i} contains a single point. Compare with: coloring regions so that every point is served… Simply means: CF-coloring of the dual set system.

31. CF-coloring: primal vs. dual primal: CF-color unit disks every point is served dual: CF-color points every disk contains a point with unique color reduction: CF-color unit disks  CF-color points wrt unit disks disk centers  points cells (representatives)  centers of disks extended reduction: uniform CF-color congruent disks  CF-color points wrt all disks disk centers  points cells (representatives)  centers of disks

32. CF-color points in plane wrt all disks using O(log n) colors Intuition: ALG(X,i) pick “good” set of points S  X color every point x  S with color i recurse: ALG(X-S,i+1) “Good” choice: one node at a time – correct but expensive. if every disk d satisfies: d  X  S  |d  S|  1. Q: can we guarantee |S| always a constant fraction of |X| ? if yes, then O(log n) colors. Q: how is S computed?

33. CF-color points in plane wrt all disks using O(log n) colors Trivial: empty disks & disks with single point Remaining: ranges (disk  points) with at least 2 points. observations: pick a pair (P,Q) in every minimal range, and draw an edge (P,Q). independent set S in this graph can not contain a range. this graph is the Delaunay graph of X.Delaunay graph Delauney graph is planar  independent set  |X|/4 ALG (X,i) : find an independent set IND  X in DG(X), color every point x  IND with color i recurse: ALG(X-IND, i+1) IND  |X|/4 implies O(log n) colors!

34. pick independent set IND in DG(X) color IND red draw DG(X) remove IND DEMO

35. pick independent set IND in DG(X) color IND red draw DG(X) remove IND

36. pick independent set IND in DG(X-RED) color IND green draw DG(X-Red)

37. pick independent set IND in DG(X-RED) color IND green draw DG(X-Red) remove IND

38. draw DG(X-(Red+green)) pick independent set IND in DG(X-RED) color IND blue remove IND

39. draw DG(X-(Red+green)) pick independent set IND in DG(X-RED) color IND blue remove IND

40. color remaining 2 points

41. final coloring

42. bi-criteria algorithms for unit-disks THM: Inflate radius by . Poly-time algorithm for coloring “inflated” disks using O(log (1/  )) colors so that all points in unit disks are served.  =1/2 O(opt)  opt colors! THM: Poly-time algorithm for coloring unit disks using O(log (1/  )) colors so that all but  - fraction of points in unit disks are served.

43. Open questions O(1)-approximation algorithm for disks (have one for case of intersecting unit disks). CF-coloring of arrangements of regions similar to coverage areas of antennas: 60º sectors… Capacitated versions: center may serve a limited #clients

45. indexed arrangements assign indexes to disks (not arbitrary!). represent set system by diagram (i.e. is cell covered by disk?) cells disks 245 7 89 N(cell) is an interval N(cell) is not an interval

46. Interval property of arrangements Full interval property: interval property and, for every interval [i,j], there exists a cell such that N(v) = [i,j]. Indexed arrangement: every disk has an index. Interval property: if, for every cell v, there exist i  j such that: N(v) = [i,j]. Chain: an indexed arrangement that satisfies the full interval property Equivalent DEF: dual set system representation isomorphic to the set system ({1,…,n}, {[i,j]} )

47. chains Claim: for every n, there exists a chain C(n) of n unit circles. Proof: index circles from left to right same proof works with axis-parallel squares, hexagons, etc.

48. CF-colorings of chains Claim: every CF-coloring of C(n) requires  (log n) colors. proof: “query”: which disk serves cell v: N(v)=[1,n]? color of this disk appears once (unique color). -red disk partitions chain into 2 disjoint chains. -pick larger part, and continue “queries” recursively.

49. coloring chain with O(log n) colors

50. theorem for unit disks a tile: a square of unit diameter. local density  (A(C)) of arrangement A(C): max #disk centers in tile. Theorem: There exists a poly-time algorithm: Input: a collection C of unit disks Output: a CF-coloring of C Number of colors: O(log  (A(C))) Tightness: see chains… [BY] every set-system can be CF-colored using O(log 2 C) colors

51. reduction to case: all disks centers in the same tile - Tile the plane: diameter(tile) = 1. center(unit disk)  tile  tile  unit disk -Assign a palette to each tile (periodically to blocks of 4  4 tiles), so disks from different tiles with same palette do not intersect. suffices now to CF-color disks with centers in the same tile. (in particular, intersection of all disks contains the tile)

52. reduction to case: all disks in the same tile have a boundary arc boundary disk: disk with a boundary arc. Reduction based on lemma:  boundary disks=  disks.  need to consider only boundary disks in tile. boundary arc non-boundary arc

53. boundary arcs set of disks C: - all centers in same tile - all disks have a boundary arc Lemma: every disk in C has at most two boundary arcs. distance(centers)  1 angle of intersection at least 2  /3

54. decomposition of boundary disks: disks on one side of a line - all the disks cut r twice -  two disks intersect once -  boundary disk WRT H has one boundary arc in H - no nesting of boundary disks - boundary disks WRT H are a chain r H This is where proof fails for non-identical disks

55. decomposition of boundary disks: (assume that all the disks have precisely one boundary arc) pick 4 disks (that intersect extensions of vert sides) color 4 circles with 4 new distinct colors remaining disks: 4 disjoint chains. color each chain.

56. decompositions of boundary disks (disks that have 2 boundary arcs) previous method gives 2 colors per disk. 4 chains & each disk in 2 chains. partition disks into parts. 2 chains in each part.

57. decompositions of boundary disks (disks that have 2 boundary arcs) Lemma: pairs of chains have the same “orders”. use 1 indexing for both chains. colors of disk in 2 chains agree.

58. summary of CF-coloring algorithm Tiling: 16 palettes Decomposing boundary disks: 4 disks 4 chains of disks with 1 boundary arc: 4  log (#boundary disks in tile) chains of disks with 2 boundary arcs: 6  log (#boundary disks in tile)  O(log(max (#boundary disks in tile))) colors. Observation: if all disks belong to same tile, then ALG uses at most 10  OPT + 4 colors

59. applications: a bi-criteria algorithm C – set of unit disks with  C non-empty CF * (C) – min #colors in CF-coloring of C C  = {Disk(x,1+  ): x center of unit disk in C} Serve  C with a coloring of C . CORO: exists coloring of C  that serves  (C) using O(log 1/  ) colors. Proof: dilute centers so that d min  . CORO:  =1/2 O(CF*(C))  CF*(C) colors!

62. far from optimal ALG uses log n colors but, OPT uses only 4 colors… reason: ALG ignores “help” from disks centered in other tiles. local OPT  global OPT

63. Outline cellular networks – Frequency Assignment Problem conflict-free coloring – Model of FAP primal/dual range spaces results more results open problems

64. More results Arrangements of squares: constant approximation algorithm. Arrangements of regular polygons: constant approximation algorithm. (also for case of constant #”angle types”. Open problems: constant approximation for unit disks, non-identical disks… OPEN: NP-completeness…