First Fit Coloring of Interval Graphs William T. Trotter Georgia Institute of Technology October 14, 2005

Interval Graphs

First Fit with Left End Point Order Provides Optimal Coloring

Interval Graphs are Perfect Χ = ω = 4

What Happens with Another Order?

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.

Dynamic Storage Allocation

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

Analyzing First Fit Using Grids

The Academic Algorithm

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 D

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

Double Induction

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 - 9 when 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.

Tree-Like Walls

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.