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Stephan Holzer

Networks Cannot Compute Their Diameter in Sublinear Time. Stephan Holzer. Silvio Frischknecht Roger Wattenhofer. ETH Zurich – Distributed Computing Group . Distributed network G raph G of n nodes. 2. 1. 5. 3. 4.

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Stephan Holzer

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  1. Networks Cannot Compute Their Diameter in Sublinear Time Stephan Holzer Silvio Frischknecht Roger Wattenhofer ETH Zurich – Distributed Computing Group

  2. Distributed network Graph G of n nodes 2 1 5 3 4

  3. Distributed network Graph G of n nodes 2 1 5 3 4

  4. Distributed network Graph G of n nodes 2 Localinfor- mationonly 1 1 5 3 4

  5. Distributed network Graph G of n nodes 2 Localinfor- mationonly 1 5 3 2 4 1 3 Slide inspiredbyDanuponNanongkai

  6. Distributed network Graph G of n nodes 2 Localinfor- mationonly 1 5 3 2 ? 4 1 3

  7. Distributed network Graph G of n nodes Limited bandwidth 2 log n Localinfor- mationonly log n log n 1 log n 5 log n 3 2 log n ? 4 1 3

  8. Distributed network Graph G of n nodes Limited bandwidth 2 Synchronized Internal computations negligible log n Localinfor- mationonly log n log n 1 log n 5 log n 3 2 log n 4 1 ? Time complexity: number of communication rounds 3

  9. Distributed algorithms:a simple example

  10. Count thenodes!

  11. Count thenodes! Compute BFS-Tree

  12. Count thenodes! Compute BFS-Tree 0

  13. Count thenodes! Compute BFS-Tree 1 1 1 0 1

  14. Count thenodes! Runtime: ? 2 Compute BFS-Tree 2. Count nodes in subtrees 1 2 1 2 2 1 0 2 1 Diameter

  15. Diameter of a network • Distancebetweentwonodes = Numberof hops ofshortestpath

  16. Diameter of a network • Distancebetweentwonodes = Numberof hops ofshortestpath

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

  18. Diameter of a network • Distancebetweentwonodes = Number of hops of shortestpath • Diameter of network = Maximum distance, betweenanytwonodes Diameter of thisnetwork?

  19. 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. ThanksforslidetoDanuponNanongkai

  20. 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. • Global problems:Ω( D ) Localinfor- mationonly 2 ? 1 3

  21. Fundamental graph-problems • Maximal Independent Set • Coloring • Matching

  22. Fundamental graph-problems • Maximal Independent Set • Coloring • Matching • Local problems: runtime independent of / smaller than D e.g. O(log n)

  23. Fundamental problems • Spanning Tree – Broadcasting, Aggregation, etc • Minimum Spanning Tree – Efficient broadcasting, etc. • Shortest path – Routing, etc. • Steiner tree – Multicasting, etc. • Many other graph problems. • Diameter appears frequently in distributed computing

  24. Fundamental problems 1. Formal definition? • Spanning Tree – Broadcasting, Aggregation, etc • Minimum Spanning Tree – Efficient broadcasting, etc. • Shortest path – Routing, etc. • Steiner tree – Multicasting, etc. • Many other graph problems. • Diameter appears frequently in distributed computing Complexity of computing D? . Known: Ω (D) ≈Ω(1) Even if D = 3 This talk: Ω(n)

  25. Networks cannot compute their diameter in sublinear time!

  26. Diameter of a network Diameter of thisnetwork? • Distance between two nodes = Number of hops of shortest path • Diameter of network = Maximum distance, between any two nodes

  27. Networks cannot compute their diameter in sublinear time!

  28. Networks cannot compute their diameter in sublinear time!

  29. Networks cannot compute their diameter in sublinear time!

  30. Networks cannot compute their diameter in sublinear time!

  31. Networks cannot compute their diameter in sublinear time!

  32. Networks cannot compute their diameter in sublinear time! Base graph has diameter 3

  33. Networks cannot compute their diameter in sublinear time! has diameter 3 2?

  34. Networks cannot compute their diameter in sublinear time! Pair of nodes not connected on both sides? has diameter 3 2?

  35. Networks cannot compute their diameter in sublinear time! Pair of nodes not connected on both sides? has diameter 3 2?

  36. Networks cannot compute their diameter in sublinear time! Pair of nodes not connected on both sides? Now: slightly more formal

  37. Networks cannot compute their diameter in sublinear time! Pair of nodes not connected on both sides? Label potential edges

  38. Networks cannot compute their diameter in sublinear time! Pair of nodes not connected on both sides? Label potential edges 1 1

  39. Networks cannot compute their diameter in sublinear time! Pair of nodes not connected on both sides? Label potential edges 1 1 2 2

  40. Networks cannot compute their diameter in sublinear time! Pair of nodes not connected on both sides? Label potential edges 1 1 3 3 2 2

  41. Networks cannot compute their diameter in sublinear time! Pair of nodes not connected on both sides? Label potential edges 1 1 3 3 2 2 4 4

  42. Networks cannot compute their diameter in sublinear time! Pair of nodes not connected on both sides? 1 1 3 3 2 2 4 4

  43. Networks cannot compute their diameter in sublinear time! Pair of nodes not connected on both sides? Given graph 1 1 3 3 2 2 4 4

  44. Networks cannot compute their diameter in sublinear time! Pair of nodes not connected on both sides? B A 1 1 3 3 3 2 2 2 4 4 4 4 Θ(n) edges

  45. Networks cannot compute their diameter in sublinear time! Pair of nodes not connected on both sides? Same as “A and B not disjoint?” B A 1 1 2 3 3 3 2 2 2 3 4 4 4 4 4 Θ(n) edges

  46. Networks cannot compute their diameter in sublinear time! Pair of nodes not connected on both sides? Same as “A and B not disjoint?” B ⊆ [n2] A ⊆ [n2] 1 1 2 3 3 3 2 2 2 3 4 4 4 4 4 Θ(n) edges

  47. Networks cannot compute their diameter in sublinear time! “A and B not disjoint?” B ⊆ [n2] A ⊆ [n2] 2 3 4 4

  48. Networks cannot compute their diameter in sublinear time! Pair of nodes not connected on both sides? Same as “A and B not disjoint?” Communication Complexity randomized: Ω(n 2) bits B ⊆ [n2] A ⊆ [n2] 2 3 Ω(n) time 4 4 Θ(n) edges

  49. Diameter Approximation

  50. 3/2-εapproximating the diameter takes W(n1/2) Extend 2 vs. 3 D = 9 basegraph

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