Guy Even Zvi Lotker Dana Ron

Tel Aviv University

Conflict-free colorings of unit disks, squares, & hexagons

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

r=range

every client within range cancommunicate with base station

cellular networks – a base-station

more antennas

increase covered region

cellular networks – multiple base-stations

backbone network:between base-stations

radio link:client base-station

mobile clients: dynamicallycreate links with base-stations

interfering base-stations

base-stations using same frequency

interference in intersection of regions

non-interfering base-stations

base-stations use different frequencies

no interference!

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.

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.

)()'(:)(' dddPNd

dP

Ndisks :

1

2

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

What is the minimum number of colors we need ?

every 2 “adjacent” disks must have different colors

Answer: 3 colors

What is the minimum number of colors we need ?

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

Answer: 4 colors

arrangements of unit disks

arrangement: sub-division of plane into cells.

a cell

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)

set-system representation 1

23

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

disks cells

1 2 3 4 51 2 3 4 5

6

7

cell connected todisks in N(cell)

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

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

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.

CF-colorings of chainsClaim: 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.

).(log)()1(),(max 1

:equation recurrence

i nnfinfif f(n)

coloring chain with O(log n) colors

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(log2 C) colors

O(1) approx. algorithm for CF-coloring disks in one tile.

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 44 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)

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

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

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

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.

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.

2

4

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.

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.

2

4

Observation: if all disks belong to same tile,

then ALG uses at most 10OPT + 4 colors

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 dmin .

• CORO: =1/2O(CF*(C)) CF*(C) colors!

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

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…