Exact bounds for distributed graph colouring
Joel Rybicki
SIROCCO 2015 July15, 2015
Helsinki Institute for Information Technology & Aalto University
Max Planck Institute for Informatics
Jukka Suomela
Graph colouring
G = (V,E)
Input: A cycle with a consistent orientation
Graph colouring
Input: A cycle with a consistent orientation
Given a colouringf : V ! {1, . . . , n}
{u, v} 2 E ) f(u) 6= f(v)G = (V,E)
Task: Colour reduction
3-colouringn-colouringInput: Output:
Model of computing
1. sends messages
2. receives messages
3. updates local state
Synchronous rounds. Each node
v
Local views
0 rounds
v
Local views
1 round
v
Local views
2 rounds
Local views
r roundsv
A( ) 2 { , , }An algorithm is a map
Time complexity
is the exact number of rounds it takes to 3-colour any n-coloured directed cycle
C(n, 3)
Prior work
Linial (1992)
1
2
log
⇤ n� 1 C(n, 3) 1
2
log
⇤ n+ 3
Complexity of 3-colouring
log
⇤ n = min{i :iz }| {
log · · · log n 1}
Prior work
1
2
log
⇤ n� 1 C(n, 3) 1
2
log
⇤ n+ 3
Cole & Vishkin (1987)
Complexity of 3-colouring
log
⇤ n = min{i :iz }| {
log · · · log n 1}
Prior work
1
2
log
⇤ n� 1 C(n, 3) 1
2
log
⇤ n+ 3
Complexity of 3-colouring
Cole & Vishkin (1987)Linial (1992)
log
⇤ n = min{i :iz }| {
log · · · log n 1}
Prior workComplexity of 3-colouring
C(n, 3) =1
2
log
⇤ n+O(1)
In “practice”, the additive term dominates:
log
⇤10
19728= 5
Our result
For infinitely many values of n, 3-colouring requires exactly
1
2
log
⇤ n rounds.
The approach
Lower bound: Tighten Linial’s bound using new computational techniques
Upper bound: A careful analysis of Naor–Stockmeyer (1995) colour reduction
The lower bound
Show that a fast 3-colouring algorithm implies a fast 16-colouring algorithm
Bound the complexity of finding a 16-colouring
Step 1.
Step 2.
The lower bound
Show that a fast 3-colouring algorithm implies a fast 16-colouring algorithm
Bound the complexity of finding a 16-colouring
Step 1.
Step 2.
“Dependence on n”
“The additive O(1) term”
Two-sided ≈ one-sidedTwo-sided view
v
One-sided viewv0
C(n, 3) = dT (n, 3)/2e
r rounds
2r rounds
C(n, 3)
T (n, 3)
The speed-up lemma
va
-colouring in r roundsc
The speed-up lemma
va
va
-colouring in r rounds
-colouring in r− 1 rounds
c
)
(2c � 2)
New technique: Successor GraphsFix any (e.g. optimal) algorithm
New technique: Successor GraphsFix any (e.g. optimal) algorithmand apply the speed-up lemma to get
. . . At
#colours
#rounds t
23 � 23
t� 1 0
. . .
. . .
� n
A1A0
New technique: Successor GraphsFix any (e.g. optimal) algorithmand apply the speed-up lemma to get
. . . At
#colours
#rounds t
23 � 23
t� 1 0
. . .
. . .
� n
A1A0
New technique: Successor GraphsFix any (e.g. optimal) algorithmand apply the speed-up lemma to get
. . . At
#colours
#rounds t
23 � 23
t� 1 0
. . .
. . .
� n
A1A0
Colour is a successor of colour
Successor relationConsider that outputs colours from
{ }. . .Ck = .
if outputsAku v
Ak
Successor graph
Nodes: { }. . .Ck =
Edges: the successor relation
Starting from any algorithm we get
Algorithm:
Successor graph:
A0 A1. . .
S0 S1. . . St
At
Colourability lemma
Sk is c-colourable
there is a c-colouring algorithm running in t-k rounds
)
A finite super graph
For all k, there is a finite graph that contains the successor graph of
any algorithm as a subgraph.
Proving lower bounds
Super graph + colorability lemma:Chromatic number an upper bound for all successor graphs!
Finite super graph:Easy to use a computer search for small enough super graphs!
For any t-time 3-colouring algorithm, the successor graph is 16-colourable
The key result
S2
Complement of S₂
1 3 12 13
3 12 13
2 3 12 13
1 12 13
12 13
1 2 12 13
1 2 3 12 13
2 12 13
1 2 3 12 23
3 12 23
12 23
2 3 12 23
1 3 12 23
1 12 23
1 2 12 13
2 12 23
1 3 13 23
1 13 23
2 3 13 23
2 13 23
1 2 13 23
13 23
3 13 23
1 2 3 13 23
1 2 3 13
1 2 13
2 3 13
2 13
1 2 23
1 3 23
1 23
1 2 3 23
3 12
1 2 3 12
1 3 12
2 3 12
3 13
1 13
1 3 13
13
2 12
1 2 12
12
1 12
23
2 23
2 3 23
3 23
1 22 3
3
1 3
21
1 2 3
For any t-time 3-colouring algorithm, the successor graph is 16-colourable
The key result
S2
By colourability lemma, there exists a 16-colouring algorithm running in t− 2 rounds
The lower bound
Iterated speed-up lemma:16-colouring takes rounds
Step 1.
Step 2.
log
⇤ n� 2
Successor graph bound:3-colouring takes rounds
log
⇤ n
Two-sided ≈ one-sidedTwo-sided view
v
One-sided viewv0
C(n, 3) = dT (n, 3)/2e
r rounds
2r rounds
C(n, 3)
T (n, 3)
Conclusions
For infinitely many values
C(n, 3) =1
2
log
⇤ n.
Use successor graphs and computers for lower bound proofs!
Conclusions
For infinitely many values
C(n, 3) =1
2
log
⇤ n.
Use successor graphs and computers for lower bound proofs!
Thanks!