Stephan Holzer

ETH Zurich – Distributed Computing Group Stephan Holzer SODA 2012

1

Stephan HolzerSilvio Frischknecht Roger Wattenhofer

Networks Cannot Compute Their Diameter in Sublinear Time

ETH Zurich – Distributed Computing Group


Distributed network Graph G of n nodes

2

4

51

3



2

4

51

3



2

4

51

31 Local infor-

mation only



2

4

51

32

13

Local infor-mation only

Slide inspired by Danupon Nanongkai



2

4

51

32

13


?



2

4

51

3

Limitedbandwidth

log n

log

nlog n log n

log n

log n

2

13


?



2

4

51

3

Limitedbandwidth

log n

log

nlog n log n

log n

log n

2

13


?Time complexity:

number of communication rounds

Synchronized

Internal computations

negligible


Distributed algorithms:a simple example


Count the nodes!


Count the nodes!

1. Compute BFS-Tree


0

Count the nodes!

1. Compute BFS-Tree


1

1

0

1

1

Count the nodes!

1. Compute BFS-Tree


Count the nodes!

1

1

0

21

2

12

2

2

Runtime: ?Diameter

1. Compute BFS-Tree

2. Count nodes in subtrees


Diameter of a network

• Distance between two nodes = Number of hops of shortest path



• Distance between two nodes = Number of hops of shortest path



• Distance between two nodes = Number of hops of shortest path• Diameter of network = Maximum distance, between any two nodes




Diameter of this network?


Fundamental graph-problems

• Spanning Tree – Broadcasting, Aggregation, etc• Minimum Spanning Tree – Efficient

broadcasting, etc. • Shortest path – Routing, etc.• Steiner tree – Multicasting, etc. • Many other graph problems.

Thanks for slide to Danupon Nanongkai





• Global problems: Ω( D )2

13


?



• Maximal Independent Set• Coloring• Matching



• Maximal Independent Set• Coloring• Matching

• Local problems: runtime independent of / smaller than D

e.g. O(log n)


Fundamental problems



• Diameter appears frequently in distributed computing




• Diameter appears frequently in distributed computing

Fundamental problems 1. Formal definition?

Complexity of computing D? .

Known: Ω (D)

Ω(n)

≈Ω(1)

This talk:Even if D = 3


Networks cannot compute their diameter in sublinear time!



Diameter of this network?














Base graph has diameter 3



has diameter 3 2?



Pair of nodes not connected on both sides?

has diameter 3 2?




has diameter 3 2?



Now: slightly more formal




Label potential edges





1 1





1 12 2





1 12 23 3





1 12 23 3

4 4




1 12 23 3

4 4




24

2 31 1

34


Given graph



Θ(n) edges

24

2 31 1

344

2 34

A B




Θ(n) edges

24

2 31 1

34

2 34

Same as “A and B not disjoint?”


A

B

4

23

4



Θ(n) edges

24

2 31 1

34

2 34

Same as “A and B not disjoint?”


A [n⊆ 2]

B [n⊆ 2]

4

23

4


A [n⊆ 2]

B [n⊆ 2]

“A and B not disjoint?”

4

23

4




Θ(n) edges

A [n⊆ 2]

B [n⊆ 2]

“A and B not disjoint?”Communication Complexity randomized: Ω(n 2) bits

4

23

4Ω(n) time


Same as


Diameter Approximation


3/2-ε approximating the diameter takes W(n1/2)

Extend2 vs. 3

D = 9 base graph



Extend2 vs. 3

D = 9 base graph



Extend2 vs. 3

9

D = 9 base graph



Extend2 vs. 3

D = 9 base graph



Extend2 vs. 3

D = 9 base graph

5



Extend2 vs. 3

D = 9 base graphD = 7 good graph-------------Factor: 3/27

5



Extend2 vs. 3

D = 9 base graphD = 7 good graph-------------Factor: 3/2 -ε7


Technique is general


G



G



G

cut



cut



cut



cut

f’(Ga,Gb)


Alice Bob

cut

f’(Ga,Gb)


Alice Bob

cut

f’(Ga,Gb)


Alice Bob

cut

f’(Ga,Gb)


Alice Bob

cut

Time(g) Time(f) ≥ -----------

|cut|

f’(Ga,Gb)


Summary

Diameter Ω(n) 73/2-eps approximation takes Ω(n1/2)

cutgeneral technique


Thanks!

Documents

Stephan Holzer