1. Global state collection
Chandy Lamport algorithm does a partial job. Each process collects a fragment of the global state, but these pieces have to be put together.
Applications
- computing network topology
- termination detection
- deadlock detection
etc

2. A simple exercise s(i) s(j) i j s(k) s(l) k l Collect global state
via all-to-all broadcast
At the end, each process
will compute a set V, where
V= {s(i): 0 ≤ i ≤ N-1 }
s(i)
s(j) i j
s(k)
s(l) k l

3. A simple exercise V.i W.i (i,k) V.k W.k
Program broadcast (for process i}
define V.i, W.i : set of values;
initially V.i={s(i)}, W.i = 
andevery channel is empty
do V.i ≠ W.i send (V.i \ W.i) to every outgoing channel; W.i := V.i
 ¬ empty (k, i) receive X from channel (k, i); V.i := V.i  X od
V.i
W.i
(i,k)
V.k
W.k

4. Proof empty (i. k)  W.i  V.k. V.i W.i (i,k) V.k W.k
(Upon termination) i: V.i = W.i,
and all channels are empty.
So, V.i  V.k.
On a cyclic path, V.i = V.k must be
true. Since s(i) V.i, s(i) V.k
V.i
W.i
(i,k)
V.k
W.k

5. Termination detection
In a network, processes may be active or passive.
Termination implies:
(a) every process is passive,
(b) all channels are empty, and
(c) the global state of the system satisfies the
desired postcondition

6. Termination detection
The progress of a computation. Notice how
one process engages another process. Eventually
All processes turn white - this signals termination.

7. Dijkstra-Scholten algorithm
An initiator initiates termination
detection by sending a signal
down the edge when it engages
another node.
At a “suitable time,” the recipient
sends an ack back.
When the initiator receives ack, it
detects termination.
Node j engages node k. j k j k j k

8. Dijkstra-Scholten algorithm
Deficit (e) = # of signals on e - # of ack on e
For any node,
C= deficit along incoming edges
D= deficit along outgoing edges edges
Invariant 1. (C ≥ 0)  (D ≥ 0)
Invariant 2. (C > 0)  (D = 0) 1 2 3 4 5

9. Dijkstra-Scholten algorithm
Invariant 1. (C ≥ 0)  (D ≥ 0)
Invariant 2. (C > 0)  (D = 0)
The invariants must hold when
an interim node sends an ack. So
(C-1 ≥ 0)  (C-1> 0 D=0)
{follows from INV1 and INV2}
= (C>1)  (C≥1  D=0)
= (C>1) (C=1  D=0) 1 2 3 4 5

10. Dijkstra-Scholten algorithm
program detect {for an internal node i}
initially C=0, D=0, parent = i
do m = signal  (C=0) 
C:=1; state:= active; parent := sender
m = ack  D:= D-1
(C=1 D=0)  state = passive 
send ack to parent; C:= 0; parent := i
m = signal  (C=1) 
send ack to the sender; od 1 2 3 4 5

11. Dijkstra-Scholten algorithm
The engagement edges form a spanning tree. The tree grows and shrinks.
Signals are acknowledged on very-first-in-very-last-out basis
Termination is detected when the initiator receives ack. 1 2 3 4 5