Optimal Termination Detection for Rings

1
Optimal Termination Detection for Rings

• Murat Demirbas
• OSU

2
Termination Detection in D.S.

• Message passing, Asynchronous execution
• A process is either active or passive
• Active processes
• send and receive messages,
• can become passive spontaneously
• Passive processes
• can only receive messages,
• can become active only by receiving a message.

3
Termination detection problem

• The system is terminated iff
• all processes are passive
• no messages in transit
• The problem is to detect termination as and when
  the system terminated.

4
Related work

• Chandy Lamport snapshot alg.
• O(N2) time
• Dijkstra Safra token-based alg.
• O(N) time, 2N -- 3N
• Chandy alg.
• 2N integers per process, and message
• 2N2 integers at the detector
• Sivilotti improved the space complexity

5
Our algorithm

• Based on DijkstraSafra alg.
• Detection in 0 -- N time
• each process maintains 1 int. N bits
• token stores N int. N bits

6
Outline

• Dijkstra Safra alg
• Optimal alg
• Enhancement 1
• Enhancement 2
• Proof

7
Dijkstra Safra algorithm

• c.j mesgs sent - mesgs received
• The initiator obtains a snapshot
• sends a token to the ring
• gathers the sums of c.js
• If the snapshot is consistent and no mesg in
  transit, termination is detected.

8
Dijkstra Safra alg (cont.)
1
N
2
3

• Snapshot is inconsistent if the receipt of m is
  recorded but send of m is not.
• A process is blackened upon receiving a mesg
• A black node blackens the token

9
c.1 color.1
1
q color
c.2 color.2
c.N color.N
2
N
3
c.3 color.3
10
An optimal algorithm for Termination Detection on
Rings

• First enhancement (Enumeration bits)
• Second enhancement (Multiple initiators)
• Optimal algorithm 1st 2nd enh. combined

11
1st enhancement (enumeration bits)

• Dijkstra Safra blackens every process that
  receives a message.
• However, a message reception violates the
  snapshot iff the receive of the message is
  included in the snapshot whereas the send is not.

12
1
2
5
3
4
13
1st enhancement (enumeration bits)

• The messages that violate the consistency are
  those sent by a process in the visited region to
  another process in the unvisited region.
• A process sending a mesg piggybacks its
  enumeration bit its process id.
• j upon receiving m blackens itself iff
• enum.j ? enum.m
• j gt sender_id.m

14
An enumeration bit is sufficient ...
1
1
mlt1gt
1
2
5
0
1
0
3
4
15
1st enhancement N -- 2N
1
1
2
5
0
mlt1gt
1
0
3
4
16
2nd Enhancement (Multiple initiators)

• DS alg has a fixed initiator N
• A vector q1,q2,,qN maintains the sum, q,
  w.r.t. multiple initiators.
• N -- 2N time to detect termination

17
1
0,0,0,0,0
0,q2,q3,q4,q5
B,q2,q3,q4,0
q1,0,0,0,0
2
5
B,q2,q3,0,B
B,0,B,B,B
3
4
B,q2,0,B,B
18
The Optimal Algorithm

• 2nd enh. blackens ?jq.j and requires 2N.
• 0--N if we do not blacken any q at 2, and q2, q3
  at 4.

1
m2
2
4
m1
q1,q2,q3,q4
3
19
The Optimal Algorithm(cont.)

• We merge 1st 2nd enh.
• enum.j, enum.m, sender_id.m (1st enh.)
• q1,q2,,qN, tok_color.1N (2nd enh.)
• color.j.k
• color.j.k js color w.r.t. initiator k
• propagate and retransmit actions are merged into
  one action

20
The Optimal Algorithm(cont.)

• j receives m from l
• enum.j ? enum.m and j gt sender_id.m
• m violates the consistency of the snapshot ?kj
  ? k ? N ? 1 ? k ? l
• ?kj?k?N ? 1?k?l color.j.k black

21
Proof

• W detects X
• W ?X
• X leads-to W
• Proof
• X termination predicate
• W witness predicate
• I invariant
• (I ?W) ?X
• (I ?X) leads-to W

22
Proof (cont)

• X ( (?j idle.j)?(mesg_sent - mesg_rcvd 0)
  )
• W (?j (tok_at_j) ? (idle.j) ? (color.j.jwhite)
  ? (c.jq.j0) ? (tok_color.jwhite) )
• I ( (?jc.j) mesg_sent - mesg_rcvd
  (I1) ?(?i Q.i ? R.i ? S.i ? T.i) )
• Q.i ( (?jj?VSTD.i idle.j) ? q.i
  (?jj?VSTD.ic.j) )
• R.i ( q.i(?jj?VSTD.ic.j) gt 0 )
• S.i (?j j?VSTD.i color.j.iblack )
• T.i (tok_color.iblack )

23
Proof (I?W) ?X

• Token is at j
• W ? (1)tok_at_j ? (2)idle.j ? (3)color.j.jwhite ?
  (4)c.jq.j0 ? (5)tok_color.jwhite
• (1 ? 3) ? ?S.j
• (1 ? 4) ? ?R.j
• 5 ? ?T.j
• (I ? ?S.j ? ?R.j ? ?T.j) ? Q.j
• (1 ? 2 ? Q.j ? 4 ? I1) ? X

24
Proof (I?X) leads-toW in 0--N

• (I ? X) ? (?j idle.j) ? (?jc.j) 0
• The only enabled action is Propagate Token
• Let tok_at_j then q.j0, color.j.jwhite,
  tok_color.jwhite
• Claim (?k color.k.j white)
• Then tok_color.jwhite is stable.
• When tok_at_j again, (c.jq.j (?jc.j) 0 )
• Therefore, within 1 cycle of token W is
  satisfied.
• 0--N

25
Proof (?k color.k.j white)

• Assume (?k color.k.j black). 3 cases to
  consider
• kltj
• color.k.j black before token visits k leads-to
  color.k.j white
• color.k.j cannot be blackened by 1 ? i ? k, since
  enum.ienum.k
• color.k.j cannot be blackened by k ? i.
• kj color.j.jwhite
• kgtj
• sender_id ? j then color.k.j is not blackened
• sender_id gt j then color.k.j is not blackened,
  since enum.sender_id enum.k

26
Conclusion

• An optimal termination detection algorithm on
  rings 0--N

27
New Results T.D. in Trees Chandys model

• 2h--3h detection in trees
• h detection in trees
• Efficient T.D. in Chandys model
• 1--2 rounds to detect termination
• requires just 1 integer 1 bit in each process
  including DET.
• Chandy 2N integers in each process, 2N2 integers
  in DET.