Multicoloring bounded tree
width graphs and planar graphs
Guy Kortsarz, Rutgers University,
Camden
Joint work with M.M Halldorsson.
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Scheduling dependent jobs
Jobs compete on resources
Create a graph. Each job is a vertex.
Two vertices are adjacent if dependent
Two possibilities:
Single unit jobs
Jobs that require more than one unit of
processing
Two conflicting jobs can not be executed
at the same time unit
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Relation of single units jobs to
graph coloring
•
•
Given a graph G, find
a mapping :VN so
that if (v)=(u) then
uvE
Objective, usually, is
to minimize the
number of colors
used.
Classic applications:
Timetabling, frequency allocation
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Multicoloring bipartite graphs
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4
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2
1
4
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Multicoloring bipartite graphs
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4
2
2
1
4
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Formal definition of the
multicoloring problem
• Input: Graph G=(V,E), with lengths x(v) on
the vertices
• Output: An assignment :V 2N such that
adjacent vertices do not receive
overlapping times. (v)(u)   implies
that uvE
• Goal: minimize the maximum integer:
• Minimize MaxuV { f(u) }
• f(u)- The largest color of u.
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Multicoloring is often easy
6-clique
More generally, chordal graphs, even perfect,
are no harder to multicolor than to color.
Maybe one reason why multicolorings are not
more common.
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Some other objectives than
makespan (= number of colors)
•
•
•
•
Sum of completion times of jobs
– For each vertex, count the last time unit
assigned, and sum these values up
Weighted sum of completion times
– Vertices additionally have importance value
attached
Total lateness
– Assumes deadline for each task
Sum of flow times
– Assumes release time of each job
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“Sum of completion times” in Graph
coloring?
•
•
•
Recall we color with numbers, :VN
Sum of a coloring = Sum of the values assigned to
the vertices
Sum coloring problem:
– Find a coloring that minimizes the chromatic
sum,
 (v)
vV
•
This measure is more favorable to the users (as
a whole), while the makespan is desired by the
system (machines)
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The Sum Multicoloring problem
 Each vertex v V requires x(v)  1 distinct colors.
 A proper coloring of G is a function Ψ: V → 2N, such that
adjacent vertices are assigned distinct sets of numbers
(colors). Let fΨ(v) denote the maximum color assigned to v
by Ψ.
 The sum multicoloring (SMC) problem: find a multicoloring
Ψ that minimizes vV fΨ(v).
• The preemptive model (pSMC): each vertex can get
any set of colors.
• The no-preemption model (npSMC): assign to each
vertex a contiguous set of colors.
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Is “Sum coloring” any different
from ordinary coloring? YES!
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1
2
3
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Ok, isn’t at least the sum of an
ordinary coloring “good enough”?
Factor k for kchromatic
graphs.
Any 3-coloring has sum of 2n...
... while a certain 4-coloring has sum of n+6
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Multicoloring p max demand
bounded tree width graphs
• Sum multicoloring non preemptively of trees is
in P. [Halldórsson, K, Proskurowski, Salman,
Shachnai and Telle]
• Preemptive sum multicoloring of Paths is in P.
An ICALP paper (!) [Kovács]
• Preemptive coloring of trees is NPC (!) [Marx]
• See the collection of papers maintained by
Daniel Marx for more results:
http://www2.informatik.huberlin.de/~dmarx/sum.php
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An (important?) breakpoint lemma
Consider the collection of breakpoints
r,r+1,…..,s
There exists r  j  s so that:
|{ ai | ai  jµ }| n/(j ln(s/r))
Probably not written in any place that you can
find, but known.
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Breakpoint Lemma
•
For any q, there is a sequence of breakpoints bi
satisfying
bi
q
q
bi  1
such that the total breakpoint overhead is at
most
1
ln q
b1
b2
  xi 
i
1
ln q
b3
 OPT
b4
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np-MC bounded tree width graphs
• This can be solved by DP in what I call the standard
way. Running time O(n · (kp log n)k+1)
• Two difficulties: bad time for large p
• Also, we need a sum that uses few colors
• This is because we are going to color the
classes in the breakpoint lemma one after
the other.
• How much they wait depends on the
makespann not the sum.
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Sum coloring with few colors BTW
graphs
• Ignore all bits but roughly log n for every
x(v)
• This means that the numbers can be
divided by a large number q.
• Dividing by q we get an instance with small
ratio p/pmin
• Can use the previous solution and
duplications and get only 1+ penalty and
get roughly (n/)k running time.
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Application for np-SMC planar
graphs
•
•
It is shown that np-SMC planar graphs admits a
PTAS in time roughly (log n)d *n with d =p/pmin
We make d=log n/loglog n by methods discussed.
• Baker decomposition into k2-outerplanar G1
and outerplanar and small GO.
• Biased round robin: color mainly G1 . Color GO
only after k rounds of coloring G1.
• This is OK as GO is negligible.
• We can assume d is small due to the
breakpoint lemma.
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The color sum
• G1 has size at most n/k2 (one of the
choices). We also get very small d.
• Solve optimally by dp, but truncate the
coloring after (1/)b/a colors.
 Cost ≤
OPT. And poly time due to value of d.
• Finish by 4-coloring remaining vertices
• The color starts with (1/)b/a
 Cost of O(n/k2)* (1/)b/a ≤  n when k >> -2
(b/a).
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Preemptive multicoloring of graphs
• A general tool for O(1) makspan coloring:
preemptive scaling.
We are able to reduce job lengths to log n paying
only (1+), even if start with p exponential.
• Claim 1: Say that all x(v) are divisible by q. Then
then if x(v)/qc ln n for every v then:
(I) q(I/q)(1+)(I)
• Was not a problem in non-pree’. Here use random
proof and its non trivial.
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Reduce to log n
• Split jobs to (roughly) log n most significant
bits and the rest
• The large parts of numbers are all divisible by some
large q
• Reduce these numbers to log n bits (small loss). The
O(log n) is derived
• Color first by round robin the small part of the
numbers non-preemptively
• Small delay because constant colors and small
numbers
• Then take the solution to the O(log n) instance
• The coloring for x’/q is repeated q times. Uses the
randomized proof.
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My most surprising claim ever
• Say you want to preemptively multicolor a BTW
graph
• There is a family of colorings so that for every
instance there exists a coloring in the family that
approximates the SMC by (1+) factor
• The key property: the number of different
colorings of v in the family is polynomial in n
• This allows finding the best solution via DP if
treewidth constant
• The existence is proven by modifying OPT
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What is the surprise?
• The colorings in our family depend only on a
specific k-coloring of the graph, on n and
on p
• In particular, it does not depend on the
actual connections in the graph, nor on the
distribution of color requirements
• Hence the name universal
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The universal family
• Split the colors of every vertex in powers
of (1+)
• Segment i: colors (1+)i to (1+)i+1
• In every segment, treat the coloring as a
makespann instance
• Thus in every segment O(log n) colors
• Make the number of segments in which a
vertex is colored, constant
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How to bound # of segments
• For every vertex v, its colors that are
smaller than x(v) or larger than (2/)x(v)
are removed from OPT
• Instead they are replace by round robins
that are performed every (roughly) 1/
rounds
• The key: the number of “non-standard”
segments for every v is O(log 1+ (2 ) )
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The number of preemption for v
In every segment v has O(log n)
preemptions
• # colors per segment clog n. Choose c’ log
n of them for v
• The number of possibilities for v is
O((2clog n) f() ) hence polynomial in n
• The round robin, executed only every 1/
rounds and adds O( opt)
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Open problems
• What can we say about np-MC of planar
graphs?
• Of course 4/3 lower bound.
• I have no idea how to do 4/3.
• But giving 2 ratio is easy.
• A planar graph can be decomposed into two
bipartite graphs.
• Optimally color first the one with smallest
p. Then the other. Ratio 2. 4/3 anyone?
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Quotes on knowledge
. Fiery the angels fell. Deep thunder rolled
around their shoulders... burning with the
fires of Orc. Blade Runner & Paradise Lost.
• I've seen things you people wouldn't
believe. Attack ships on fire off the
shoulder of Orion. I watched c-beams
glitter in the dark near Tannhäuser Gate.
All those moments will be lost in time, like
tears in rain. Time to die. Blade Runner
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What did I not know in 1999?
• De-vos et al. 2004; Demaine,
Hajiaghayi, Kawarabayashi 2005
• H-minor-free graphs can have their
vertices partitioned into k pieces such
that deleting any one piece results in
bounded treewidth.
• Paradise lost: if I knew then, I would
write the paper for H-minor free
graphs. Also bounded genus graphs, etc.
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