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Transcript of 1 Multicoloring bounded tree width graphs and planar graphs Guy Kortsarz, Rutgers University, Camden...
![Page 1: 1 Multicoloring bounded tree width graphs and planar graphs Guy Kortsarz, Rutgers University, Camden Joint work with M.M Halldorsson.](https://reader035.fdocuments.us/reader035/viewer/2022070403/56649f2e5503460f94c48373/html5/thumbnails/1.jpg)
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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 resourcesCreate a graph. Each job is a vertex.Two vertices are adjacent if dependentTwo possibilities:
Single unit jobsJobs that require more than one unit of processingTwo 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 :VN so that if (v)=(u) then uvE
• 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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Multicoloring bipartite graphs
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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 uvE
• Goal: minimize the maximum integer:• Minimize MaxuV { 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, :VN • Sum of a coloring = Sum of the values
assigned to the vertices• Sum coloring problem:
– Find a coloring that minimizes the chromatic sum,
• This measure is more favorable to the users (as a whole), while the makespan is desired by the system (machines)
Vv
v)(
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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 → 2, 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 vV 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?
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YES!
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Ok, isn’t at least the sum of an ordinary coloring “good
enough”?
Any 3-coloring has sum of 2n...
... while a certain 4-coloring has sum of n+6
Factor k for k-chromatic graphs.
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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.hu-berlin.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 jµ }| 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
such that the total breakpoint overhead is at most
OPTq
xq i
i ln
1
ln
1
qb
bq
i
i
1
b1 b2 b3 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)/qc 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, De-vos et al. 2004; Demaine, Hajiaghayi, Kawarabayashi 2005Hajiaghayi, 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.