Concurrent Counting is harder than Queuing

Transcription

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

2 rbitrary graph 2

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

4 Distributed Counting Final state 4

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

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

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

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? 8

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

10 Problem Is there a formal sense in which Counting is harder problem than Queuing? Reductions don t seem to help 10

11 Our Result Concurrent Counting is harder than Concurrent Queuing on a variety of graphs including: many common interconnection topologies complete graph, mesh hypercube perfect binary trees 11

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 12

13 C Q (v) Cost Model : delay till v gets back queuing result Cost of algorithm on request set R is C (, R) C ( v) Q v R Q Queuing Complexity = min {max R V C Q (, R)} Define Counting Complexity Similarly 13

14 Lower Bounds on Counting For arbitrary graphs: * n Counting Cost = ( n log ) For graphs with diameter D : Counting Cost = ( D 2 ) 14

15 Theorem: For graphs with diameter D : Counting Cost = ( D 2 ) Proof: Consider some arbitrary algorithm for counting 15

16 Graph Take shortest path of length D 16

17 Graph make these nodes to count 17

18 Node of count at least k 1 2 k decides after time steps k 1 2 k 1 2 k Needs to be aware of k-1 other processors 18

19 Counting Cost: ( D ) D k 1 k End of Proof 19

20 Theorem: For arbitrary graphs: * n Counting Cost = ( n log ) Proof: Consider some arbitrary algorithm for counting 20

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

22 The initial state affects the outcome Red: count Blue: don t count Initial State v 22

23 Red: count Blue: don t count 1 3 Final state v 23

24 Red: count Blue: don t count Initial state v 24

25 Red: count Blue: don t count Final state v

26 Let (v ) be the set of nodes whose input may affect the decision of v (v ) v 26

27 Suppose that there is an initial state for which v decides k Then: ( v ) k (v ) v k 27

28 (v ) v k These two initial states give same result for v (v ) v k 28

29 (v ) v k If ( v ) k, then v would decide less than k Thus, ( v ) k 29

30 Suppose that v decides at time t We show: ( v ) t times 30

31 Suppose that v decides at time t ( v ) t times t log * k ( v ) k 31

32 Cost of node v : t log * k If n nodes wish to count: Counting Cost = n k 1 log * k ( n log * n ) 32

33 ( v, t ) B ( v, t ) v v Nodes that affect v up to time t Nodes that v affects up to time t a( t ) max ( x, t ) b( t ) max B( x, t ) x x 33

34 ( v, t 1) B( v, t 1) v v a ( t ) 1 b( t ) 1 fter t 1, the sets grow 34

35 ( v, t 1) ( v, t ) v 35

36 ( z, t ) ( v, t 1) z ( v, t v ) Eligible to send message at time t 1 There is an initial state such that that z sends a message to v 36

37 ( s, t s ) ( z, t ) z ( v, t 1) ( v, t v ) Eligible to send message at timet 1 Suppose that ( s, t ) ( z, t ) Then, there is an initial state such that both send message to v 37

38 ( s, t s ) ( z, t ) z ( v, t 1) ( v, t v ) Eligible to send message at timet 1 However, v can receive one message at a time 38

39 ( s, t s ) ( z, t ) z ( v, t 1) ( v, t v ) Eligible to send message at timet 1 Therefore: ( s, t ) ( z, t ) 39

40 ( s, t s ) ( z, t z ) Number of nodes like s : a( t ) b( t ) max ( x, t ) x maxx B( x, t ) 40

41 ( v, t 1) ( s, t s ) ( z, t z ) ( v, t ) v Therefore: ( v, t 1) ( v, t ) a( t ) a ( t ) b( t ) 41

42 Thus: a( t 1) a( t ) 1 a( t ) b( t ) We can also show: a ( t ) b( t 1) b( t ) 1 2 Which give: a( ) times End of Proof 42

43 Upper Bound on Queuing For graphs with spanning trees of constant degree: Queuing Cost = O( n logn ) For graphs whose spanning trees are lists or perfect binary trees: Queuing Cost = O (n ) 43

44 n arbitrary graph 44

45 Spanning tree 45

46 Spanning tree 46

47 Distributed Queue Previous = Nil Tail Tail Head 47

48 B enq(b) Previous =? Previous = Nil Tail Tail Head 48

49 B Previous =? enq(b) Previous = Nil Tail Tail Head 49

50 enq(b) B Previous =? Previous = Nil Tail Tail Head 50

51 enq(b) B Previous =? Previous = Nil Tail Tail Head 51

52 B Previous =? enq(b) Previous = Nil Tail Tail Head 52

53 Tail B Previous = Previous = Nil informs B Tail B Head 53

54 Concurrent Enqueue Requests enq(c) C B enq(b) Previous =? Previous =? Previous = Nil Tail Tail Head 54

55 C B Previous =? Previous =? enq(c) enq(b) Previous = Nil Tail Tail Head 55

56 C B Previous =? Previous =? enq(b) enq(c) Previous = Nil Tail Tail Head 56

57 Tail C B Previous = Previous =? enq(b) Previous = Nil C Tail Head 57

58 Tail enq(b) C B Previous = Previous =? Previous = Nil C Tail Head 58

59 C informs B Tail C Previous = B Previous = C Previous = Nil B C Tail Head 59

60 Tail C Previous = B Previous = C Previous = Nil B C Tail Head 60

61 Paths of enqueue requests Tail C Previous = B Previous = C Previous = Nil B C Tail Head 61

62 Nearest-Neighbor TSP tour on Spanning tree C B D E F Origin (first element in queue) 62

63 Visit closest unused node in Tree C B D E F 63

64 Visit closest unused node in Tree C B D E F 64

65 Nearest-Neighbor TSP tour C B D E F 65

66 [Herlihy, Tirthapura, Wattenhofer PODC 01] For spanning tree of constant degree: Queuing Cost 2 x Nearest-Neighbor TSP length 66

67 [Rosenkratz, Stearns, Lewis SICOMP1977] If a weighted graph satisfies triangular inequality: Nearest-Neighbor TSP length Optimal TSP length x logn 67

68 C B D E F weighted graph of distances C D E 3 1 F 1 B

69 69 Satisfies triangular inequality C D e 1 e 2 e 3 ) ( ) ( ) ( e w e w e w C D B F E

70 C B D E F Length=8 Nearest Neighbor TSP tour C D E 3 1 F 1 B

71 C B D E F Length=6 Optimal TSP tour C D E 3 1 F 1 B

72 It can be shown that: Optimal TSP length 2n (Nodes in graph) C B D E F Since every edge is visited twice 72

73 Therefore, for constant degree spanning tree: Queuing Cost = O(Nearest-Neighbor TSP) = O(Optimal TSP x ) logn = O( n logn ) 73

74 For special cases we can do better: Spanning Tree is List balanced binary tree Queuing Cost = O (n ) 74

75 Graphs with Hamiltonian path, have spanning trees which are lists Complete graph Mesh Hypercube Queuing Cost = O (n ) * n Counting Cost = ( n log ) 75

76 Theorem: If the spanning tree is a list, then Queuing Cost = O (n ) 76

77 Proof: Nearest Neighbor TSP Queuing Cost = O(Nearest-Neighbor TSP) 77

78 x x y y a b 2a b 78

79 n nodes Even sides Length doubles Total length 2n 79

80 n nodes Odd sides Length doubles Total length 2n 80

81 n nodes Even+Odd sides Total length 2n 2n 4n End of proof 81