An Analysis of First-Fit
N.S. Narayanaswamy (IITM)
Work with R. Subhash Babu
Interval Graphs
5
4
3
1
2
1
2
Clique number ω:
3
4
5
The maximum no. of
intervals that share a
point.
Coloring Intervals
5
color
4
3
1
2
Resource Allocation:
• Each interval ~ Request for a resource
for a period of time
• Color
~ Resource
time
Coloring

Offline coloring
Optimal coloring : Consider intervals in non-decreasing
order of their left end points, assign the least feasible color.
 chromatic number = clique number


Online Coloring

Requests/intervals are presented in a sequences one
after another.
Example
Competitive Ratio of First-Fit
No of colors used by the online A
Competitive
= max
Ratio of A
G,s No of colors used by the optimal offline
algorithm
= max
G,s
No of colors used by First-Fit
ω
First fit




Principle: Assign the least feasible color to the
incoming interval.
Currently proved to use at most 8ω-3 colors (our
work).
First 8ω by Kierstead, Brightwell, Trotter
improving 10ω by Pemmaraju, Raman,
Varadarajan.
There exists an instance on which First fit uses at least
4.4ω colors.
First fit: Example
• Clique size is 2, therefore offline uses 2.
• No of colors used is 4
Properties of First fit
• Property: If an interval I is colored j, then there exists
an interval I’ in each color i, 1 ≤ i ≤ j such that I
intersects I’.
• Wall like structure, height=number of colors used
Analysis of First-Fit



Goal : Find a lower bound on the clique size in terms
of the height of the wall.
Column construction procedure : a counting technique
Idea




Consider the First-Fit wall as a grid.
Assign one of three symbols to each cell in the grid.
Count relative occurrences of symbols.
Lower bound the clique size
Column Construction Procedure
m
C4
C3
C2
C1
• Elementary Columns/Intervals
• Assign symbols to each cell – R, $, F
( ,, )
• Find relations between , and  to find a lower bound
on the clique number.
Column Construction
C4
C3
C2
C1
R
R
R
R
R
R
R
R
R
• First Step: Consider C1 = set of cells colored 1, they
get the label R.
• Ending Condition: For i = 2,3,… stop when Ci
obtained from Ci-1 becomes empty.
Rule - R
C4
C3
C2
C1
R
R
R
R
R
R
R
R
R
R
R
R
R
R1: For each cell e in Ci-1, if e is occupied by an
interval I colored i, then add e to Ci with the symbol R.
Rule - $
C4
C3
C2
C1
$
R
R
$
R
R
R
R
R
$
R
R
$
R
R
R
R
R2: For each remaining cell e є Ci-1, if e has a neighbor e’
in Ci-1 which is added to Ci by rule R1, then add e to Ci
with the symbol $.
Rule - F
i-1
j
j
el
e
er
R3 : For each remaining cell e in Ci-1 , if e has a neighbor e' in
Ci-1 and e' is neighbor of e down to the level j, and
e(j , i) > (i - j)/, then add e to Ci with the label “F”.
All other columns become inactive
Rules are applied in the order R1, R2 and R3.
Final Picture
m’
F
F
F
F
F
F
C4
F
F
F
$
R
$
R
$
F
C3
F
$
R
R
R
$
F
F
F
C2
C1
$
R
R
$
F
$
R
R
$
R
R
R
R
R
R
R
R
R
Analysis of the Height

m ≤ m’.


For 1 ≤ i ≤ m’ and e є Ci ,
e(i) ≥ (m - e (i)) / 


For each i ≤ j, an interval colored j intersects Ci
Proof by induction on i
e (m') ≤ 2m'/

m’ crucial here. Without this there was a weaker
upper bound, and hence 10 competitiveness.
Proof Outline
i1
i2
i3
j1
j2
j3
j4
e
e (m') ≤ 2m'/
⇒ e(m') + e(m') ≥ m' – 2m'/
⇒ e(m') ≥ m'/(1 -2/)
⇒ e(m) ≥ m/(1 - 2/)
⇒ e(m) ≥ m/8, for =4
New Observations


C1 needs only to be cells corresponding to a
minimal clique cover of the underlying interval
graph.
Analysis works by changing the order of rules to
R1, R3, and R2. This yields a nicer proof of

For 1 ≤ i ≤ m’ and e є Ci , e(i) ≥ (m - e (i)) / 
Thank You
18
Online Vs Offline
Offline:
3
2
4
1
Online: (1,3,2,4)
[First fit]
4
2
1
3
Back
1-Wall: Examples
1-wall
Back
2-Wall: Examples
2-wall
Back
3-Wall: Examples
3-wall
Back