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Exact bounds for distributed graph colouring
Joel Rybicki
SIROCCO 2015 July15, 2015
Helsinki Institute for Information Technology & Aalto University Max Planck Institute for Informatics
Exact bounds for distributed graph colouring Jukka Suomela Joel - - PowerPoint PPT Presentation
Exact bounds for distributed graph colouring Jukka Suomela Joel - - PowerPoint PPT Presentation
Exact bounds for distributed graph colouring Jukka Suomela Joel Rybicki Max Planck Institute for Helsinki Institute for Information Informatics Technology & Aalto University SIROCCO 2015 July15, 2015 Graph colouring Input: A cycle with
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Graph colouring
Input: A cycle with a consistent orientation Given a colouring f : V → {1, . . . , n} {u, v} 2 E ) f(u) 6= f(v) G = (V, E)
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Task: Colour reduction
3-colouring n-colouring Input: Output:
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Model of computing
- 1. sends messages
- 2. receives messages
- 3. updates local state
Synchronous rounds. Each node
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v
Local views
0 rounds
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v
Local views
1 round
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v
Local views
2 rounds
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Local views
r rounds v
A( ) ∈ { , , }
An algorithm is a map
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Time complexity
is the exact number of rounds it takes to 3-colour any n-coloured directed cycle C(n, 3)
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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 :
i
z }| { log · · · log n ≤ 1}
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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 :
i
z }| { log · · · log n ≤ 1}
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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 :
i
z }| { log · · · log n ≤ 1}
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Prior work
Complexity of 3-colouring C(n, 3) = 1 2 log∗ n + O(1) In “practice”, the additive term dominates: log∗ 1019728 = 5
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Our result
For infinitely many values of n, 3-colouring requires exactly 1 2 log∗ n rounds.
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The approach
Lower bound: Tighten Linial’s bound using new computational techniques Upper bound: A careful analysis of Naor– Stockmeyer (1995) colour reduction
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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.
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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”
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Two-sided ≈ one-sided
Two-sided view
v
One-sided view
v0
C(n, 3) = dT(n, 3)/2e r rounds 2r rounds C(n, 3) T(n, 3)
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The speed-up lemma
v a
- colouring in r rounds
c
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The speed-up lemma
v a v a
- colouring in r rounds
- colouring in r − 1 rounds
c
⇒
(2c − 2)
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New technique: Successor Graphs
Fix any (e.g. optimal) algorithm
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New technique: Successor Graphs
Fix any (e.g. optimal) algorithm and apply the speed-up lemma to get . . . At #colours #rounds t 23 − 2 3 t − 1 . . . . . . ≥ n A1 A0
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New technique: Successor Graphs
Fix any (e.g. optimal) algorithm and apply the speed-up lemma to get . . . At #colours #rounds t 23 − 2 3 t − 1 . . . . . . ≥ n A1 A0
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New technique: Successor Graphs
Fix any (e.g. optimal) algorithm and apply the speed-up lemma to get . . . At #colours #rounds t 23 − 2 3 t − 1 . . . . . . ≥ n A1 A0
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Colour is a successor of colour
Successor relation
Consider that outputs colours from
{ } . . . Ck =
. if outputs Ak u v Ak
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Successor graph
Nodes:
{ } . . . Ck =
Edges: the successor relation
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Starting from any algorithm we get
Algorithm: Successor graph:
A0 A1 . . . S0 S1 . . . St At
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Colourability lemma
Sk is c-colourable there is a c-colouring algorithm running in t-k rounds
⇒
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A finite super graph
For all k, there is a finite graph that contains the successor graph of any algorithm as a subgraph.
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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!
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For any t-time 3-colouring algorithm, the successor graph is 16-colourable
The key result
S2
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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 2 2 3 3 1 3 2 1 1 2 3
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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
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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
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Two-sided ≈ one-sided
Two-sided view
v
One-sided view
v0
C(n, 3) = dT(n, 3)/2e r rounds 2r rounds C(n, 3) T(n, 3)
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Conclusions
For infinitely many values C(n, 3) = 1 2 log∗ n. Use successor graphs and computers for lower bound proofs!
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