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Termination Detection. Presented by: Yonni Edelist. The model. A finite graph of machines (nodes) and communication channels (edges), in which a distributed computation is taking place Machines can be either ‘ active ’ or ‘ passive ’ Only active machines may send messages

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termination detection

Termination Detection

Presented by:

Yonni Edelist

the model
The model
  • A finite graph of machines (nodes) and communication channels (edges), in which a distributed computation is taking place
  • Machines can be either ‘active’ or ‘passive’
  • Only active machines may send messages
  • Passive machine may become active only when it receives a message
  • Active machine may become passive at any time
the problem
The problem
  • If, in the computation, a finite number of messages are sent, the graph will reach a stable state in which no more messages are sent
  • One machine, say machine no. 0, wants to detect that the graph has reached that state
  • We will build an algorithm which enables machine 0 to detect termination
example
example
  • We denote messages sent by the termination detection algorithm ‘signals’

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the model1
The model
  • N machines, 0..N-1
  • Machine 0 initiates the detection algorithm
  • Does so by sending a signal (a token) to machine 1
  • Machine i can propagate the token to machine i+1
  • We denote the machine currently holding the token ‘t’
system invariant
System Invariant
  • We describe an invariant, which will be true for every configuration of the system (for every value of t)
  • Machine 0 will know that all other machines are passive based on:
    • t=0
    • The invariant
    • (possibly) further information available at machine 0
first invariant p0
First Invariant- P0
  • First step- we assume that there are no messages
  • Meaning: an active node may become passive, but not vice-versa
  • P0: (∀ i: 0≤i<t machine i is passive)
p0 example
P0- Example

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Red outline represents an active machine

rule 0
Rule 0
  • In order to preserve the invariant, we must follow the rule:
  • Rule 0: Machine i transmits the token only if it’s passive
  • (remember that a passive machine can’t become active again)
invariant p1
Invariant P1
  • If machine i<t receives a message, it becomes active, thus falsifying P0
  • Message must have been sent by machine j≥t
  • P1 is established whenever P0 is falsified in the above manner

P0: (∀ i: 0≤i<t machine i is passive)

invariant p1 cont
Invariant P1 (cont.)
  • A machine can be either black or white
  • P1: (∃ j: t ≤ j ≤ N: machine j is black)
  • Rule 1: When machine j sends a message to machine i<j, it becomes black
invariant p1 cont1
Invariant P1 (cont.)
  • Now P= P0 ∨ P1 is not falsified when a message is sent to i < j.
  • Machine 0 knows:
  • (t=0 ∧ machine 0 is white)  ¬P1
  • P0: (∀ i: 0≤i<t machine i is passive)
  • Therefore- machine 0 can still detect termination
slide18
But…

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invariant p2
Invariant P2
  • P1 may be falsified when a black machine hands the token over
  • We want to ensure that every token propagation that falsifies P1 establishes P2
  • We define that the token may be either black or white
invariant p2 cont
Invariant P2 (cont)
  • P2: the token is black
  • And the accompanying rule:
  • Rule 2: A black machine transmits a black token. A white machine doesn’t change the colour of the token
invariant p2 cont1
Invariant P2 (cont)
  • Machine 0 knows:
  • (t=0 ∧ machine 0 is white)  ¬P1

- Token is white  ¬P2

  • P0: (∀ i: 0≤i<t machine i is passive)
  • Therefore- machine 0 can still detect termination
rule 3
Rule 3
  • The algorithm may finish unsuccessfully
  • Rule 3: after an unsuccessful round, machine 0 initiates another round
slide24
But…

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rule 4
Rule 4
  • Since a round that begun with a black token is doomed, we adopt:
  • Rule 4: at the beginning of a new round, machine 0 makes itself white and transmits a white token
rule 5
Rule 5
  • Whitening a black machine i may falsify only P1, and only in the case i > t
  • Rule 5: when transmitting the token to machine i+1, machine i makes itself white

P1: (∃ j: t ≤ j ≤ N: machine j is black)

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