Conflict-Free Coloring
Conflict-Free Coloring
Gila Morgenstern
CRI, Haifa University
Conflict-Free Coloring
Conflict-Free (CF) Coloring: Introduced
by [Even, Lutker, Ron, Smorodinsky. 03],
motivated by frequency assignment
problems in radio network
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
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.)
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All
possible
intervals
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SO…, how much can one
save using colors ?
(i), supporting of [1,n]
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[1,n]
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i
n
#c(n) ≥ 1 + maxi{#c(i-1),#c (n-i)}
 Ω(log n) colors required
Independent,
(i) is excluded
CF-Coloring of a Chain cont.
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?2
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5?
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O(log n) colors are sufficient
CF-coloring unit disks
CF-coloring unit disks
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CF-coloring unit disks
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95
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CF-coloring unit disks
Not a chain, still
Θ(logni) = Θ(log n)
colors
Variants
• Squares, Pseudo disks, “Fat” objects, …
- Θ(logn) colors.
• On-line CF-coloring - Θ(log2 n) colors.
• Many more ….
Conflict-Free Coloring of
Points on a Line W.R.T.
a Set of Intervals
Matthew J. Katz
Nissan Lev-Tov
Gila Morgenstern
Ben-Gurion University, Israel
What if we need not respect
everyone ?
• Θ(logn) colors might be wasteful.
Example: Proper interval graph:
All intervals are non-nested.
2 non-zero colors are
sufficient
CF-Coloring of Points w.r.t.
a Subset of Intervals.
• P = {p1,…,pm} - Set of points.
• R = {I1,…,In} - Set of intervals.
l
U
• For each interval I
R:
l
U
– 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.
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:
0
1
2
3
4
5
Is this optimal?
No:
Optimal
1 non-zero color
CFCp1
2 non-zero colors
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.
Maximal Color is unique
Lemma:
Each interval is supported by the
maximal color occurring in it.
Proof:
K’
K
K K’
Supported by k<k’
All colors smaller than k’ occur at least twice
Optimal Sub-Coloring
•Lemma: If
R = R1 U R2 U {I} s.t.:
Ul
• Range(R1UR2)
I
• Range(R1) Range(R2) = Ø
U
R1
{I}
R2
Cont.
?

R1
{I}
1
2
R2
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 + | |
max{ |1| , |2| }
if |1|=|2|=||
else
CFCp1 Computes a
2-Approximation
• Main Lemma:
If (p) = k, then CF-coloring of points
leftward of p require at least k/2
colors.
Proof of Lemma
= k/2
• Let p be such (k-2)/2
that+1(p)=k
K-2
K-1
(k-2)/2
K-2 K-1
(k-2)/2
K
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”.
Observation
For 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.
Indeed,consider
interval I with r(I)=p.
• 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).
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).
(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.
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!