first of all …. thanks to janos shakhar smorodinsky tel aviv university conflict-free coloring...

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First of all …. Thanks to Janos

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First of all …. Thanks to Janos

Shakhar Smorodinsky

Tel Aviv University

Conflict-free coloring problems

Part of this work is joint with

Guy Even, Zvi Lotker and Dana Ron.

1

A Coloring of pts

Definition of Conflict-Free Coloring

2 1

23

3

3

4

is Conflict Free if:

4

Any (non-empty) disc contains a unique color

1

A Coloring of pts

What the … is Conflict-Free Coloring?

2 1

23

3

3

4

is Conflict Free if:1

Problem Statement

What is the smallest number f(n) s.t.

any n points can be CF-colored with only f(n) colors?

Remark: We can define a CF-coloring for

a general set system (X,A) where AP(X)

i.e., a coloring of the elements of X s.t. each set sA

contains an element with a unique color

Motivation from Frequency Assignment

in cellular networks

mobile clients:create links with base-stations

within reception radius

1

Frequency assignment 1

1

2

If 2 is unique

Power and location of

clients’ cellular may

vary

Problem Statement (cont)

Thm: f(n) > log n

What is the minimum number f(n) s.t.

any n points can be CF-colored with f(n) colors?

Easy:

n pts on a line! Discs = Intervals

1 3 2

log n colors

n ptsn/2 n/4

Points on a line (cont)

log n colors suffice (in this special case)

Divide & Conquer

1 32

Color median with 1

Recurse on right and left

Reusing colors!

32 33

1

CF-coloring in general case

Thm:

f(n) = O(log n)

Divide & Conquer doesn’t work!

n pts

Proof of: f(n) = O(log n)

Consider the Delauney Graph

i.e., the “empty pairs” graph

It is planar.

Hence, By the four colors Thm

“large” independent set

n pts

Proof of: f(n) = O(log n) (cont)

IS P s.t. |IS| n/4 and

IS is independent |P|=n

1. Color IS with 1

2. Remove IS

1

1

1

Proof of: f(n) = O(log n) (cont)

IS P s.t. |IS| n/4 and

IS is independent!

|P|=n

1.Color IS with 1

2. remove IS

3. Construct the new Delauney graph … and iterate (O(log n) times) on remaining pts

2

2

Proof of: f(n) = O(log n) (cont)

IS P s.t. |IS| n/4 and

IS is independent!

|P|=n

1.Color IS with 1

2. remove IS

3. Iterate (O(log n) times) on remaining pts

5

34

Algorithm is correct

Proof of: f(n) = O(log n) (cont)

1

1 15

34

2

2

Consider a non-empty disc

“maximal” color 3

“maximal” color is unique

Proof of: f(n) = O(log n) (cont)

“maximal” color i is unique

Proof:

Assume i is not unique and

ignore colors < i

“maximal” color i

i

Proof of: f(n) = O(log n) (cont)

“maximal” color i is unique

Assume i is not unique and

ignore colors < i

“maximal” color i

ii

Proof: maximal color i is unique

Consider the i’th iteration

independent

ii

A third point exists

i

Proof: maximal color i is unique

Consider the i’th iteration

ii

Contradiction!

i

Remarks:

Algorithm is very easy to implement

What about other ranges?

CF-coloring pts w/ respect to other ranges?

Previous proof works for homothetic copies of a convex body

Thm: O(sqrt (n)) colors always suffice

How about axis-parallel rectangles?

Thm: O(sqrt (n)) colors always suffice

CF-coloring pts w.r.t axis-parallel rectangles

How small is an independent set in the “Delauney” graph ?

I DON’T KNOW!

Note:

CF-coloring pts w.r.t axis-parallel rectangles