1 / 81

Concurrent Counting is harder than Queuing

Concurrent Counting is harder than Queuing. Costas Busch Rensselaer Polytechnic Intitute Srikanta Tirthapura Iowa State University. Arbitrary graph. Distributed Counting. count. count. count. count. Some processors request a counter value. Distributed Counting. Final state.

sonya-james
Télécharger la présentation

Concurrent Counting is harder than Queuing

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Concurrent Counting is harder than Queuing Costas Busch Rensselaer Polytechnic Intitute Srikanta Tirthapura Iowa State University

  2. Arbitrary graph

  3. Distributed Counting count count count count Some processors request a counter value

  4. Distributed Counting Final state

  5. Distributed Queuing Enq(B) Enq(A) A B Enq(C) C Enq(D) D Some processors perform enqueue operations

  6. Distributed Queuing Previous=D Previous=nil A B Previous=A C D Previous=B A D B C Tail head

  7. Applications Counting:- parallel scientific applications - load balancing (counting networks) Queuing:- distributed directories for mobile objects- distributed mutual exclusion

  8. Ordered Multicast Multicast with the condition:all messages received at all nodes in the same order Either Queuing or Counting will do Which is more efficient?

  9. Queuing vs Counting? • Total orders • Queuing = finding predecessor • Needs local knowledge • Counting = finding rank • Needs global knowledge

  10. Problem • Is there a formal sense in which Counting is harder problem than Queuing? • Reductions don’t seem to help

  11. Our Result • Concurrent Countingis harder thanConcurrent Queuing • on a variety of graphs including: • many common interconnection topologies • complete graph, • mesh • hypercube • perfect binary trees

  12. Model • Synchronous system G=(V,E) - edges of unit delay • Congestion: Each node can process only one message in a single time step • Concurrent one-shot scenario: • a set R subset V of nodes issue queuing (or counting) operations at time zero • No more operations added later

  13. Cost Model • : delay till v gets back queuing result • Cost of algorithm A on request set R is • Queuing Complexity = • Define Counting Complexity Similarly

  14. Lower Bounds on Counting For arbitrary graphs: Counting Cost = For graphs with diameter : Counting Cost =

  15. Theorem: For graphs with diameter : Counting Cost = Proof: Consider some arbitrary algorithm for counting

  16. Graph Take shortest path of length

  17. Graph 6 2 7 1 5 3 8 4 make these nodes to count

  18. Node of count decides after at least time steps k Needs to be aware of other processors k-1

  19. Counting Cost: End of Proof

  20. Theorem: For arbitrary graphs: Counting Cost = Proof: Consider some arbitrary algorithm for counting

  21. Prove it for a complete graph with nodes: any algorithm on any graph with nodes can be simulated on the complete graph

  22. The initial state affects the outcome Red: count Blue: don’t count Initial State

  23. Red: count Blue: don’t count 1 Final state 3 2 4 5

  24. Red: count Blue: don’t count Initial state

  25. Red: count Blue: don’t count 2 Final state 3 5 4 1 8 7 6

  26. Let be the set of nodes whose input may affect the decision of

  27. Suppose that there is an initial state for which decides Then:

  28. These two initial states give same result for

  29. If , then would decide less than Thus,

  30. Suppose that decides at time We show: times

  31. Suppose that decides at time times

  32. Cost of node : If nodes wish to count: Counting Cost =

  33. Nodes that affect up to time Nodes that affects up to time

  34. After , the sets grow

  35. Eligible to send message at time There is an initial state such that that sends a message to

  36. Eligible to send message at time Suppose that Then, there is an initial state such that both send message to

  37. Eligible to send message at time However, can receive one message at a time

  38. Eligible to send message at time Therefore:

  39. Number of nodes like :

  40. Therefore:

  41. Thus: We can also show: Which give: times End of Proof

  42. Upper Bound on Queuing For graphs with spanning trees of constant degree: Queuing Cost = For graphs whose spanning trees are lists or perfect binary trees: Queuing Cost =

  43. An arbitrary graph

  44. Spanning tree

  45. Spanning tree

  46. Distributed Queue A Previous = Nil Tail A Tail Head

  47. B enq(B) Previous = ? A Previous = Nil Tail A Tail Head

  48. B Previous = ? enq(B) A Previous = Nil Tail A Tail Head

  49. B Previous = ? enq(B) A Previous = Nil Tail A Tail Head

More Related