conflict-free coloring
DESCRIPTION
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. - PowerPoint PPT PresentationTRANSCRIPT
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Conflict-Free Coloring
Conflict-Free Coloring
Gila MorgensternCRI, Haifa University
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Conflict-Free (CF) Coloring: Introduced by [Even, Lutker, Ron, Smorodinsky. 03], motivated by frequency assignment problems in radio network
Conflict-Free Coloring
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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.
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How many colors we need?
Proper coloring
Is OK
But wasteful!
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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
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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
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CF-Coloring of a Chain cont.
1 2 3 4 5 6 7??
O(log n) colors are sufficient
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CF-coloring unit disks
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CF-coloring unit disks
1
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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
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CF-coloring unit disksNot a chain, still
Θ(log ni) = Θ(log n) colors
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Variants
• Squares, Pseudo disks, “Fat” objects, … - Θ(log n) colors.
• On-line CF-coloring - Θ(log2 n) colors.
• Many more ….
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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
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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
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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
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O(1)-Approximation
• 2-approximation algorithm:– Endpoints of all intervals are distinct.
• 4-approximation algorithm:– Intervals may share endpoints.
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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.
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Colors key : 531 420
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Is this optimal?
No:
1 non-zero color
2 non-zero colors
Optimal CFCp1
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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
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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.
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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
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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
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R2R1
Cont.
I
2
?
1
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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
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CFCp1 Computes a2-Approximation
• Main Lemma:
If (p) = k, then CF-coloring of points leftward of p require at least k/2 colors.
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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
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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.
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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.
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Algorithm CFCp2
Stage 2 (producing ):
• for each color k, alternately change appearances of k into “dark k” and “light k”.
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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.
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• 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.
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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)
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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
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THANKS!