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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

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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.