exact bounds for distributed graph...

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!