1. Guy EvenZvi LotkerDana Ron Tel Aviv University Conflict-free colorings of unit disks, squares, & hexagons

2. Outline cellular networks – frequency assignment problem (FAP) Conflict-Free coloring – a model for FAP chains – special arrangements of unit disks CF-coloring of unit disks CF-coloring of squares & regular hexagons

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

4. 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

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

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

7. base-station frequency assignment Coloring: intersecting base-stations must use different frequencies too restrictive: every base can serve region of intersection. but, one is enough! Most models deal with interference between pairs of base-stations, 3rd base-station can not resolve an interference.

8. Def: Conflict-free coloring Coloring: Disks that cover a point P: N(P) = {disks d: P d} point P is served by disk d, if CF-coloring: all covered points are served. 1 2

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

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

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

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

13. Answer: 4 colors

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

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

16. 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 cell connected to disks in N(cell)

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

18. Interval property of arrangements Full interval property: interval property and, for every 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

19. 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.

20. 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.

21. coloring chain with O(log n) colors

22. 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 Multi-CF-colored using O(log 2 C) colors O(1) approx. algorithm for CF-coloring disks in one tile.

23. 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)

24. 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

25. 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

26. 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

27. 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.

28. 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.

29. 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.

30. 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

31. 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!

34. 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

35. More results Arrangements of squares: constant approximation algorithm. Arrangements of regular polygons: constant approximation algorithm. Open problems: constant approximation for unit disks, non-identical disks… OPEN: NP-completeness…