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# Consistent cut - PowerPoint PPT Presentation

( a  consistent cut C)  (b happened before a)  b  C If this is not true, then the cut is inconsistent. Consistent cut. A cut is a set of events . b. g. c. a. d. P1. e. m. f. P2. P3. k. h. i. j. Cut 1. Cut 2. (Not consistent). (Consistent).

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If this is not true, then the cut is inconsistent

Consistent cut

A cut is a set of events.

b

g

c

a

d

P1

e

m

f

P2

P3

k

h

i

j

Cut 1

Cut 2

(Not consistent)

(Consistent)

Consistent snapshot

The set of states immediately following a consistent cut forms a consistent snapshot of a distributed system.

• A snapshot that is of practical interest is the most recent one. Let C1 and C2 be two consistent cuts and C1C2. Then C2 is more recent than C1.
• Analyze why certain cuts in the one-dollar bank are inconsistent.
Consistent snapshot

How to record a consistent snapshot? Note that

1. The recording must be non-invasive

2. Recording must be done on-the-fly.

You cannot stop the system.

Works on a

(1) strongly connected graph

(2) each channel is FIFO.

An initiator initiates the algorithm by sending out a marker ( )

Chandy-Lamport Algorithm
Initially every process is white.

When a process receives a marker,

it turns red if it has not already done so.

Every action by a process, and every

message sent by a process gets the

color of that process.

White and red processes
Step 1. In one atomic action, the initiator (a) Turns red (b) Records its own state (c) sends a marker along all outgoing channels

Step 2. Every other process, upon receiving a marker for the first time (and before doing anything else) (a) Turns red (b) Records its own state (c) sends markers along all outgoing channels

The algorithm terminates when (1) every process turns red, and (2) Every process has received a marker through each incoming channel.

Two steps
Theorem. The global state recorded by Chandy-Lamport algorithm is equivalent to the ideal snapshot state SSS.

Hint. A pair of actions (a, b) can be scheduled in any order, if there is no causal order between them, so (a; b) is equivalent to (b; a)

Why does it work?

All white

All red

SSS

Easy conceptualization of the snapshot state

Why does it work?

Let an observer observe the following actions:

w[i] w[k] r[k] w[j] r[i] w[l] r[j] r[l] …

w[i] w[k] w[j] r[k] r[i] w[l] r[j] r[l] … [Lemma 1]

w[i] w[k] w[j] r[k] w[l] r[i] r[j] r[l] … [Lemma 1]

w[i] w[k] w[j] w[l] r[k] r[i] r[j] r[l] … [done!]

Recorded state

Let us verify that Chandy-Lamport snapshot algorithm correctly counts

the tokens circulating in the system

Example 1. Count the tokens

token

C

token

no token

A

no token

token

no token

B

A

no token

token

no token

C

B

Are these consistent cuts?

How to account for the channel states?

Use sent and received variables for each process.

Something unusual

Let machine i start Chandy-lamport snapshot before it has sent M along ch1. Also, let machine j receive the marker after it sends out M’ along ch2. Observe that the snapshot state is

down  up M’

Doesn’t this appear strange? This state was never reached during the computation!

Understanding snapshot

The observed state is a feasible state that is reachable

from the initial configuration. It may not actually be visited

during a specific execution.

The final state of the original computation is always

reachable from the observed state.

Discussions

What good is a snapshot if that state has never been visited by the system?

- It is relevant for the detection of stable predicates.

- Useful for checkpointing.

Discussions

What if the channels are not FIFO?

Study how Lai-Yang algorithm works. It does not use any marker

LY1. The initiator records its own state. When it needs to send a message m to another process, it sends a message (m, red).

LY2. When a process receives a message (m, red), it records its state if it has not already done so, and then accepts the message m.

Question 1. Why will it work?

Question 1 Are there any limitations of this approach?

Questions

Distributed snapshot = distributed read.

Distributed reset = distributed write

How difficult is distributed reset?