Transcript
Page 1: First Fit Coloring of Interval Graphs

First Fit Coloring of Interval Graphs

William T. TrotterGeorgia Institute of Technology

October 14, 2005

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Interval Graphs

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First Fit with Left End Point Order Provides Optimal Coloring

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Interval Graphs are Perfect

Χ = ω = 4

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What Happens with Another Order?

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On-Line Coloring of Interval Graphs

Suppose the vertices of an interval graph are presented one at a time by a Graph Constructor. In turn, Graph Colorer must assign a legitimate color to the new vertex. Moves made by either player are irrevocable.

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Optimal On-Line Coloring

Theorem (Kierstead and Trotter, 1982) • There is an on-line algorithm that will use at most 3k-2 colors on an interval graph G for which the maximum clique size is at most k.• This result is best possible.• The algorithm does not need to know the value of k in advance.• The algorithm is not First Fit.• First Fit does worse when k is large.

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Dynamic Storage Allocation

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How Well Does First Fit Do?

For each positive integer k, let FF(k) denote the largest integer t for which First Fit can be forced to use t colors on an interval graph G for which the maximum clique size is at most k.

Woodall (1976) FF(k) = O(k log k).

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Upper Bounds on FF(k)

Theorem: Kierstead (1988) FF(k) ≤ 40k

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Upper Bounds on FF(k)

Theorem: Kierstead and Qin (1996) FF(k) ≤ 26.2k

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Upper Bounds on FF(k)

Theorem: Pemmaraju, Raman and Varadarajan(2003) FF(k) ≤ 10k

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Upper Bounds on FF(k)

Theorem: Brightwell, Kierstead and Trotter (2003) FF(k) ≤ 8k

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Upper Bounds on FF(k)

Theorem: Narayansamy and Babu (2004) FF(k) ≤ 8k - 3

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Analyzing First Fit Using Grids

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The Academic Algorithm

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Academic Algorithm - Rules A Belongs to an interval B Left neighbor is A C Right neighbor is A D Some terminal set of letters has more than 25% A’s F All else fails.

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A Pierced Interval

ABCCDBA

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The Piercing Lemma

Lemma: Every interval J is pierced by a column of passing grades.

Proof: We use a double induction. Suppose the interval J is at level j. We show that for every i = 1, 2, …, j, there is a column of grades passing at level i which is under interval J

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Double Induction

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Initial Segment Lemma

Lemma: In any initial segment of n letters all of which are passing,

a ≥ (n – b – c)/4

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A Column Surviving at the End

1. b ≤ n/42. c ≤ n/43. n ≥ h+34. h ≤ 8a - 3

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Lower Bounds on FF(k)

Theorem: Kierstead and Trotter (1982) There exists ε > 0 so that FF(k) ≥ (3 + ε)k when k is sufficiently large.

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Lower Bounds on FF(k)

Theorem: Chrobak and Slusarek (1988) FF(k) ≥ 4k - 9 when k ≥ 4.

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Lower Bounds on FF(k)

Theorem: Chrobak and Slusarek (1990) FF(k) ≥ 4.4 k when k is sufficiently large.

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Lower Bounds on FF(k)

Theorem: Kierstead and Trotter (2004) FF(k) ≥ 4.99 k when k is sufficiently large.

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A Likely Theorem

Our proof that FF(k) ≥ 4.99 k is computer assisted. However, there is good reason to believe that we can actually write out a proof to show:For every ε > 0, FF(k) ≥ (5 – ε) k when k is sufficiently large.

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Tree-Like Walls

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A Negative Result and a Conjecture

However, we have been able to show that the Tree-Like walls used by all authors to date in proving lower bounds will not give a performance ratio larger than 5. As a result it is natural to conjecture thatAs k tends to infinity, the ratio FF(k)/k tends to 5.


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