1 concurrent counting is harder than queuing costas busch rensselaer polytechnic intitute srikanta...
DESCRIPTION
3 Distributed Counting Some processors request a counter value countTRANSCRIPT
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Concurrent Counting is harder than Queuing
Costas BuschRensselaer Polytechnic Intitute
Srikanta TirthapuraIowa State University
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Arbitrary graph
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Distributed Counting
Some processors request a counter value
count
count
count
count
4
Distributed Counting
Final state
2
1
34
5
Distributed Queuing
Some processors perform enqueue operations
Enq(A) Enq(B)
Enq(D)
Enq(C)A B
C
D
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Distributed Queuing
A B
C
D
headTail
A BC D
Previous=nil
Previous=B
Previous=D
Previous=A
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Counting: - parallel scientific applications - load balancing (counting networks)
Queuing:- distributed directories for mobile
objects- distributed mutual exclusion
Applications
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Multicast with the condition:all messages received at all nodes in the same order
Either Queuing or Counting will do
Which is more efficient?
Ordered Multicast
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Queuing vs Counting?
Total ordersQueuing = finding predecessorNeeds local knowledge
Counting = finding rankNeeds global knowledge
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ProblemIs there a formal sense in which Counting
is harder problem than Queuing?
Reductions don’t seem to help
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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
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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 zeroNo more operations added later
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Cost Model : delay till v gets back queuing result
Cost of algorithm A on request set R is
Queuing Complexity =
Define Counting Complexity Similarly
)(vCQ
Rv
QQ vCRAC )(),(
)},({maxmin RACQVRA
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Lower Bounds on Counting
Counting Cost = )log( * nn
For arbitrary graphs:
For graphs with diameter :D)( 2DCounting Cost =
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Theorem:
Proof:
Consider some arbitrary algorithmfor counting
For graphs with diameter :D)( 2DCounting Cost =
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Take shortest path of length
Graph
D
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Graph
make these nodes to count
12 3 4567 8
18
k
Node of count decides after at least time steps
k
21k
Needs to be aware of other processors k-1
21k
21k
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Counting Cost: )(21 2
1DkD
k
End of Proof
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Theorem:
Counting Cost = )log( * nnFor arbitrary graphs:
Proof:
Consider some arbitrary algorithmfor counting
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Prove it for a complete graph with nodes:any algorithm on any graph with nodes can be simulated on the complete graph
nn
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The initial state affects the outcomeRed: countBlue: don’t count
v
Initial State
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Final state
Red: count
v
1
23
4
5
Blue: don’t count
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Initial state
Red: count
v
Blue: don’t count
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Final state
Red: count
v
1
23
45
6 78
Blue: don’t count
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v
Let be the set of nodes whose input may affect the decision of
)(vAv
)(vA
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Suppose that there is an initial statefor which decides v k
kvA |)(|Then:
v
)(vA
k
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v
)(vA
k
v
)(vA
k
These two initial states give same result forv
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v
)(vA
k
If , then would decide less than
kvA |)(|k
Thus, kvA |)(|
v
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Suppose that decides at timetv
We show:
22|)(| vA
22 t times
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Suppose that decides at timetv
kvA |)(|
22|)(| vA
22 t
timeskt *log
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kt *log
)log(log *
1
* nnkn
k
Cost of node : v
If nodes wish to count:n
Counting Cost =
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v
),( tvA
Nodes that affectup to time
vt
v
),( tvB
Nodes that affectsup to timev
t
|),(|max)( txAta x |),(|max)( txBtb x
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v
)1,( tvA
v
)1,( tvB
After , the sets grow1t
1)( ta 1)( tb
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),( tvA
v
)1,( tvA
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),( tvAv
)1,( tvA
),( tzA
Eligible to send messageat time
z
1t
There is an initial state such thatthat sends a message to z v
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),( tvAv
),( tzAz
)1,( tvA
),( tsAs
),(),( tzAtsA
Eligible to send messageat time 1t
Suppose thatThen, there is an initial state such that both send message tov
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),( tvAv
),( tzAz
)1,( tvA
),( tsAs
Eligible to send messageat time 1t
However, can receive one message at a time v
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),( tvAv
),( tzAz
)1,( tvA
),( tsAs
Eligible to send messageat time 1t
),(),( tzAtsATherefore:
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),( tzAz
),( tsAs
Number of nodes like : s )()( tbta
|),(|max txAx |),(|max txBx
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),( tvAv
),( tzAz
)1,( tvA
),( tsAs
Therefore: )()()(),()1,( tbtatatvAtvA
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)()(1)()1( tbtatata
We can also show: )(21)()1( tatbtb
Thus:
Which give:22)( a
22
times
End of Proof
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Upper Bound on Queuing
)log( nnOQueuing Cost =
For graphs with spanning trees of constant degree:
)(nOQueuing Cost =
For graphs whose spanning trees are lists or perfect binary trees:
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An arbitrary graph
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Spanning tree
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Spanning tree
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A
A
Distributed Queue
HeadTailTail
Previous = Nil
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A
B enq(B)
Tail AHeadTail
Previous = ?
Previous = Nil
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A
B
enq(B)
Tail AHeadTail
Previous = ?
Previous = Nil
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A
B
enq(B)
Tail AHeadTail
Previous = ?
Previous = Nil
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A
B
enq(B)
Tail AHeadTail
Previous = ?
Previous = Nil
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A
B
enq(B)
Tail AHeadTail
Previous = ?
Previous = Nil
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A
A
B
Tail
Previous = A
BA informs BHeadTail
Previous = Nil
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A
Concurrent Enqueue Requests
Tail
BC
AHeadTail
enq(B)enq(C)Previous = ? Previous = ?
Previous = Nil
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A
Tail
BC
AHeadTail
enq(B)enq(C)Previous = ? Previous = ?
Previous = Nil
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A
Tail
BC
AHeadTail
enq(B)
enq(C)
Previous = ? Previous = ?
Previous = Nil
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A
Tail
BC
enq(B)
ACHeadTail
Previous = A Previous = ?
Previous = Nil
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A
Tail
BCenq(B)
ACHeadTail
Previous = A Previous = ?
Previous = Nil
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A
Tail
BC
ACHeadTail
C informs B
Previous = CPrevious = A
B
Previous = Nil
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A
Tail
BC
ACHeadTail
Previous = CPrevious = A
B
Previous = Nil
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A
Tail
BC
ACHeadTail
Previous = CPrevious = A
B
Previous = Nil
Paths of enqueue requests
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Nearest-Neighbor TSP tour on Spanning tree
A
C B
D
Origin(first element in queue)
E F
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A
C B
Visit closest unused node in Tree
ED F
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A
C B
Visit closest unused node in Tree
ED F
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A
C B
ED F
Nearest-Neighbor TSP tour
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Queuing Cost 2 x Nearest-Neighbor TSP length
[Herlihy, Tirthapura, Wattenhofer PODC’01]
For spanning tree of constant degree:
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Nearest-NeighborTSP length Optimal
TSP length
[Rosenkratz, Stearns, Lewis SICOMP1977]
If a weighted graph satisfies triangular inequality:
nlogx
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A
C
D
B
E F
A
C
D
B
F2
4
41
3
1 2
31
weighted graph of distances
E1
23 4
21
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Satisfies triangular inequality
A
C
D21
11e
2e3e )()()( 321 ewewew
A
C
D
B
F2
4
41
3
1 2
31
E1
23 4
21
70
A
C
E
B
FD
Nearest Neighbor TSP tour
A
C
D
B
F2
4
41
3
1 2
31
E1
23 4
21
Length=8
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A
C
E
B
FD
A
C
D
B
F2
4
41
3
1 2
31
E1
23 4
21
Optimal TSP tour
Length=6
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It can be shown that:
Optimal TSP length n2
A
C
E
B
FD
(Nodes in graph)
Since every edge is visited twice
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Therefore, for constant degree spanning tree:
Queuing Cost = O(Nearest-Neighbor TSP)
)log( nnO= O(Optimal TSP x )nlog
=
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For special cases we can do better:
Spanning Tree is
Queuing Cost = )(nO
List balanced binary tree
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Graphs with Hamiltonian path,have spanning trees which are lists
Complete graph
Queuing Cost = )(nO
Mesh Hypercube
Counting Cost = )log( * nn
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If the spanning tree is a list, then
Theorem:
Queuing Cost = )(nO
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Nearest Neighbor TSP
Queuing Cost = O(Nearest-Neighbor TSP)
Proof:
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x yyx
ab
ab 2
79
Length doubles
Total length n2
n nodes
Even sides
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Length doubles
n nodes
Total length n2
Odd sides
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n nodes
Total length nnn 422
Even+Odd sides
End of proof