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

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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

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

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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

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

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

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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.

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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

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

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

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

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

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Answer: 4 colors

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

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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)

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set-system representation disks cells coalesce cells with identical neighbors diskscells cell connected to disks in N(cell)

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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

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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

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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.

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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.

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coloring chain with O(log n) colors

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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.

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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)

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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

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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

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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

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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.

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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.

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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.

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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

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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!

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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

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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…

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