Conflict-Free Coloring

Conflict-Free Coloring

Gila MorgensternCRI, Haifa University

Conflict-Free (CF) Coloring: Introduced by [Even, Lutker, Ron, Smorodinsky. 03], motivated by frequency assignment problems in radio network

Conflict-Free Coloring

CF-Coloring of Unit Disks with Respect to Points

Coloring of the disks s.t:For any point p, one of the disks covering

p has a unique color, supporting it.

How many colors we need?

Proper coloring

Is OK

But wasteful!

CF-Coloring of a Chain

1 2 3 4 5 6 7

All possible intervals

CF-Coloring of a chain is dual to CF-Coloring of points on the line w.r.t. intervals.

(Each interval must contain a unique colored point.)

1 2 3 4 5 6 7

SO…, how much can one save using colors ?

1 2 3 i n

[1,n]

(i), supporting of [1,n]

Independent, (i) is excluded

#c(n) ≥ 1 + maxi#c(i-1),#c (n-i) Ω(log n) colors required

CF-Coloring of a Chain cont.

1 2 3 4 5 6 7??

O(log n) colors are sufficient

CF-coloring unit disks

CF-coloring unit disks

1

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

7 8 9 7 8 9 7 8 9 7 8 9 7 8 9 7 8 9

4 5 6 4 5 6 4 5 6 4 5 6 4 5 6 4 5 6

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

7 8 9 7 8 9 7 8 9 7 8 9 7 8 9 7 8 9

4 5 6 4 5 6 4 5 6 4 5 6 4 5 6 4 5 6

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

7 8 9 7 8 9 7 8 9 7 8 9 7 8 9 7 8 9

4 5 6 4 5 6 4 5 6 4 5 6 4 5 6 4 5 6

CF-coloring unit disks

1

2

3

4

5

7

89

6

CF-coloring unit disksNot a chain, still

Θ(log ni) = Θ(log n) colors

Variants

• Squares, Pseudo disks, “Fat” objects, … - Θ(log n) colors.

• On-line CF-coloring - Θ(log2 n) colors.

• Many more ….

Matthew J. KatzNissan Lev-TovGila Morgenstern

Ben-Gurion University, Israel

Conflict-Free Coloring of Points on a Line W.R.T.

a Set of Intervals

What if we need not respect everyone ?

• Θ(log n) colors might be wasteful. Example: Proper interval graph: All intervals are non-

nested.

2 non-zero colors aresufficient

CF-Coloring of Points w.r.t. a Subset of

Intervals.• P = p1,…,pm - Set of points.• R = I1,…,In - Set of intervals.

• For each interval I R:– r(I) is the right endpoint of I.– r(I) P.

• Problem: Find coloring of P w.r.t. R, using minimum number of colors.

U l

U l

O(1)-Approximation

• 2-approximation algorithm:– Endpoints of all intervals are distinct.

• 4-approximation algorithm:– Intervals may share endpoints.

Algorithm CFCp1

Simple greedy approach:

Algorithm CFCp1:

For each I by increasing r(I):

χ(r(I)) = smallest color s.t. I is supported by some non-zero color.

Colors key : 531 420

Is this optimal?

No:

1 non-zero color

2 non-zero colors

Optimal CFCp1

CFCp1 computes a 2-approximation

Our goal is to prove the following Lemma:

If (p) = k, then CF-coloring of points leftward of p requires at least k/2 colors

Colors occuring in interval I.

• Observations:

If (r(I)) = k > 0 then:• k is the only unique color occurring in

I.• All colors smaller than k occur in I at.

least twice each.

Lemma: Each interval is supported by the

maximal color occurring in it.Proof: K’K’ KK

Supported by k<k’

All colors smaller than k’ occur at least twice

Maximal Color is unique

Optimal Sub-Coloring

•Lemma: If R = R1 U R2 U I s.t.:

Ul• Range(R1UR2) I

U

• Range(R1) Range(R2) = Ø R1 R2

I

R2R1

Cont.

I

2

?

1

Cont. …

• If 1 is an optimal CF-coloring w.r.t. R1,

and 2 is an optimal CF-coloring w.r.t. R2:

#colors required by an optimal CF-Coloring w.r.t. R is at least:

1 + || if |1|=|2|=||

max |1| , |2| else

CFCp1 Computes a2-Approximation

• Main Lemma:

If (p) = k, then CF-coloring of points leftward of p require at least k/2 colors.

Proof of Lemma

• Let p be such that (p)=k KK-1K-1 K-2K-2

(k-2/)2(k-2/)2

(k-2/)2 + 1 = k/2

Relieving Assumption

• So far, we assumed that endpoints of all intervals are distinct.=> only O(n) possible intervals (out of O(n2)).

• We now relieve this assumption, namely we allow intervals to share endpoints.

Algorithm CFCp2

Stage 1 (producing ’):

• for each p from left to right do:– If all intervals I with r(I)=p are already supported, then Χ’(p)=0.

– Otherwise, let Ip be the longest interval with r(Ip)=p not supported.

Χ’(p) = smallest color s.t. Ip is supported by some non-zero color.

Algorithm CFCp2

Stage 2 (producing ):

• for each color k, alternately change appearances of k into “dark k” and “light k”.

ObservationFor each point p and interval I with r(I)=p,

at least one of the following holds at the end of the first stage.

• (i) The color ’(p) is unique in I.• (ii) The color ’(p) occurs exactly twice in I.• (iii) There exists a color c ≠ ’(p) which is

unique in I.

• If ’(p)=0 then (iii) holds.• Otherwise:

Clearly, for Ip we have that (i) holds.Same is true for shorter intervals.Suppose I is longer than Ip.By the way we chose Ip, we have that I was supported by some color c just before CFCp2 colored the point p. if c=’(p), then (ii) holds, otherwise (iii).

Indeed,consider interval I with r(I)=p.

CFCp2 produces a CF-coloring

• Let I be an interval and put p=r(I).• (i) If ’(p) is unique in I, so does (p).• (ii) If ’(p) occurs exactly twice in I,

then “dark-’(p)” and “light-’(p)” occur in I once each.

• (iii) If there exists a color c ≠ ’(p) which is unique in I, then either “dark-c” or ”light-c” occur in I exactly once.

(p)

CFCp2 is 4-Appx

• Let R’ be the set of intervals Ip that we considered during CFCp2.

• Observation: Using CFCp1 to color P w.r.t R’, produces exactly ’.

• Let OPT and OPT’ be optimal CF-colorings of P w.r.t R and R’. We get that:

|| ≤ 2 |’| ≤ 4|OPT’| ≤ 4|OPT|

2 tints CFCp1 for R’

R’ R

THANKS!