First Fit Coloring of Interval Graphs

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## First Fit Coloring of Interval Graphs

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**First Fit Coloring of Interval Graphs**William T. Trotter Georgia Institute of Technology October 14, 2005**First Fit with Left End Point Order Provides Optimal**Coloring**Interval Graphs are Perfect**Χ = ω = 4**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.**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.**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).**Upper Bounds on FF(k)**Theorem: Kierstead (1988) FF(k) ≤ 40k**Upper Bounds on FF(k)**Theorem: Kierstead and Qin (1996) FF(k) ≤ 26.2k**Upper Bounds on FF(k)**Theorem: Pemmaraju, Raman and Varadarajan(2003) FF(k) ≤ 10k**Upper Bounds on FF(k)**Theorem: Brightwell, Kierstead and Trotter (2003) FF(k) ≤ 8k**Upper Bounds on FF(k)**Theorem: Narayansamy and Babu (2004) FF(k) ≤ 8k - 3**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.**A Pierced Interval**A B C C D B A**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**Initial Segment Lemma**Lemma: In any initial segment of n letters all of which are passing, a ≥ (n – b – c)/4**A Column Surviving at the End**• b ≤ n/4 • c ≤ n/4 • n ≥ h+3 • h ≤ 8a - 3**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.**Lower Bounds on FF(k)**Theorem: Chrobak and Slusarek (1988) FF(k) ≥ 4k - 9when k ≥ 4.**Lower Bounds on FF(k)**Theorem: Chrobak and Slusarek (1990) FF(k) ≥ 4.4 k when k is sufficiently large.**Lower Bounds on FF(k)**Theorem: Kierstead and Trotter (2004) FF(k) ≥ 4.99 k when k is sufficiently large.**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.**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 that As k tends to infinity, the ratio FF(k)/k tends to 5.