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1 .Distributed Mutual Exclusion
2 .Distributed Mutual Exclusion
p0 CS
p1 CS
p2 CS
p3 CS
3 . Distributed Mutual Exclusion
Distributed mutual exclusion Classical mutual exclusion
4 . Why mutual exclusion?
Some applications are:
1. Resource sharing
2. Avoiding concurrent update on shared data
3. Implementing atomic operations
4. Medium Access Control in Ethernet
5. Collision avoidance in wireless broadcasts
5 . Mutual Exclusion Problem
Specifications
ME1. At most one process in the CS. (Safety property)
ME2. No deadlock. (Safety property)
ME3. Every process trying to enter its CS must eventually succeed.
This is called progress. (Liveness property)
Progress is quantified by the criterion of bounded waiting. It measures
a form of fairness by answering the question:
Between two consecutive CS trips by one process, how many times
other processes can enter the CS?
There are many solutions, both on the shared memory model and on the
message-passing model. We first focus on the message-passing model.
6 . Client-server based solution
Using message passing model
CLIENT
repeat
Send request and wait for reply;
Enter CS;
Send release and exit CS
busy
forever server
SERVER
repeat request ∧ ￢ busy send reply; busy:= true
[] request ∧ busy enqueue sender
client
[] release ∧ queue is empty busy:= false
[] release ∧ queue not empty queue not empty send reply to head of queue
forever
7 . Comments
- Centralized solution is simple.
- But the central server is a single point of failure. This is BAD.
- ME1-ME3 is satisfied, but FIFO fairness is not guaranteed. Why?
Can we do without a central server? Yes!
8 . Decentralized solution 1
{Lamport’s algorithm}
1. Broadcast a timestamped request to all.
Q0 Q1
2. Request received enqueue it in local Q. Not in CS
send ack, else postpone sending ack until exit from CS. 0 1
3. Enter CS, when
(i) You are at the “head” of your Q
(ii) You have received ack from all
4. To exit from the CS,
(i) Delete the request from your Q, and
(ii) Broadcast a timestamped release 2 3
5. When a process receives a release message, it removes Q2 Q3
the sender from its Q.
Completely connected topology
9 . Analysis of Lamport’s algorithm
Can you show that it satisfies all the properties
(i.e. ME1, ME2, ME3) of a correct solution? Q0 Q1
0 1
Observation. Processes taking a decision to enter CS
must have identical views of their local queues,
when all acks have been received.
Proof of ME1. At most one process can be in its CS at
any time.
Suppose not, and both j,k enter their CS. But 2 3
j in CS ⇒ Qj.ts(j) < Qk.ts(k) Qj.ts(j) < Qk.ts(k) Q2 Q3
k in CS ⇒ Qj.ts(j) < Qk.ts(k) Qk.ts(k) < Qj.ts(j)
Impossible.
10 . Analysis of Lamport’s algorithm
Proof of ME2. (No deadlock)
The waiting chain is acyclic. Q0 Q1
i waits for j 0 1
⇒ Qj.ts(j) < Qk.ts(k) i is behind j in all queues
(or j is in its CS)
⇒ Qj.ts(j) < Qk.ts(k) j does not wait for i
Proof of ME3. (progress)
2 3
New requests join the end of the Q2 Q3
queues, so new requests do not
pass the old ones
11 . Analysis of Lamport’s algorithm
Proof of FIFO fairness.
timestamp (j) < timestamp (k)
Req (30)
⇒ Qj.ts(j) < Qk.ts(k) j enters its CS before k does so
k j
Suppose not. So, k enters its CS before j. So k
did not receive j’s request. But k received the
ack from j for its own req.
This is impossible if the channels are FIFO Req ack
. (20)
Message complexity = 3(N-1) (per trip to CS)
(N-1 requests + N-1 ack + N-1 release)
12 . Decentralized algorithm 2
{Ricart & Agrawala’s algorithm}
What is new?
1. Broadcast a timestamped request to all.
2. Upon receiving a request, send ack if
-You do not want to enter your CS, or
-You are trying to enter your CS, but your timestamp is
larger than that of the sender.
(If you are already in CS or your request has a
smaller timestamp, then buffer the request)
3. Enter CS, when you receive ack from all.
4. Upon exit from CS, send ack to each buffered request
before making a new request.
(No release message is necessary)
13 . Ricart & Agrawala’s algorithm
{Ricart & Agrawala’s algorithm}
TS(j) < TS(k)
ME1. Prove that at most one process can be in CS.
Req(k) Ack(j)
ME2. Prove that deadlock is not possible.
ME3. Prove that FIFO fairness holds even if
k j
channels are not FIFO
Message complexity = 2(N-1) Req(j)
(N-1 requests + N-1 acks - no release message)
14 . Unbounded timestamps
Timestamps grow in an unbounded manner.
This makes real implementations impossible.
Can we somehow bounded timestamps?
Think about it.
15 . Decentralized algorithm 3
{Maekawa’s algorithm}
- First solution with a sublinear O(sqrt N) message
complexity.
- “Close to” Ricart-Agrawala’s solution, but each process
is required to obtain permission from only a subset
of peers
16 . Maekawa’s algorithm
• With each process i, associate a subset Si.
Divide the set of processes into subsets S0 S1
that satisfy the following two conditions:
0,1,2 1,3,5
i ∈ Si
∀i,j : 0≤i,j ≤ n-1 | Si ⋂ S Sj ≠ ∅
2,4,5
• Main idea. Each process i is required to
receive permission from Si only. S2
Correctness requires that multiple
processes will never receive permission
from all members of their respective
subsets.
17 . Maekawa’s algorithm
Example. Let there be seven processes 0, 1, 2, 3, 4, 5, 6
S0 = {0, 1, 2}
S1 = {1, 3, 5}
S2 = {2, 4, 5}
S3 = {0, 3, 4}
S4 = {1, 4, 6}
S5 = {0, 5, 6}
S6 = {2, 3, 6}
18 . Maekawa’s algorithm
Version 1 {Life of process I}
S0 = {0, 1, 2}
1. Send timestamped request to each process in Si. S1 = {1, 3, 5}
2. Request received send ack to process with the S2 = {2, 4, 5}
S3 = {0, 3, 4}
lowest timestamp. Thereafter, "lock" (i.e. commit)
S4 = {1, 4, 6}
yourself to that process, and keep others waiting.
S5 = {0, 5, 6}
3. Enter CS if you receive an ack from each member S6 = {2, 3, 6}
in Si.
4. To exit CS, send release to every process in Si.
5. Release received unlock yourself. Then send
ack to the next process with the lowest timestamp.
19 . Maekawa’s algorithm-version 1
ME1. At most one process can enter its critical S0 = {0, 1, 2}
section at any time. S1 = {1, 3, 5}
S2 = {2, 4, 5}
S3 = {0, 3, 4}
Let i and j attempt to enter their Critical Sections
S4 = {1, 4, 6}
Si ∩ Sj ≠ ∅ implies there is a process k ∊ Si ⋂ S Sj S5 = {0, 5, 6}
Process k will never send ack to both. S6 = {2, 3, 6}
So it will act as the arbitrator and establishes ME1
20 . Maekawa’s algorithm-version 1
ME2. No deadlock. Unfortunately deadlock is
possible! Assume 0, 1, 2 want to enter their
critical sections. S0 = {0, 1, 2}
S1 = {1, 3, 5}
From S0= {0,1,2}, 0,2 send ack to 0, but 1 sends ack to 1; S2 = {2, 4, 5}
From S1= {1,3,5}, 1,3 send ack to 1, but 5 sends ack to 2; S3 = {0, 3, 4}
S4 = {1, 4, 6}
From S2= {2,4,5}, 4,5 send ack to 2, but 2 sends ack to 0;
S5 = {0, 5, 6}
Now, 0 waits for 1 (to send a release), 1 waits for 2 (to send a
S6 = {2, 3, 6}
release), , and 2 waits for 0 (to send a release), . So deadlock
is possible!
21 . Maekawa’s algorithm-Version 2
Avoiding deadlock
S0 = {0, 1, 2}
If processes always receive messages in
S1 = {1, 3, 5}
increasing order of timestamp, then
deadlock “could be” avoided. But this is too S2 = {2, 4, 5}
strong an assumption. S3 = {0, 3, 4}
S4 = {1, 4, 6}
Version 2 uses three additional messages: S5 = {0, 5, 6}
S6 = {2, 3, 6}
- failed
- inquire
- relinquish
22 . Maekawa’s algorithm-Version 2
New features in version 2 S0 = {0, 1, 2}
S1 = {1, 3, 5}
- Send ack and set lock as usual. S2 = {2, 4, 5}
- If lock is set and a request with a larger S3 = {0, 3, 4}
timestamp arrives, send failed (you have no
S4 = {1, 4, 6}
chance). If the incoming request has a lower
timestamp, then send inquire (are you in S5 = {0, 5, 6}
CS?) to the locked process. S6 = {2, 3, 6}
- Receive inquire and at least one failed
message send relinquish. The recipient
resets the lock.
23 . Maekawa’s algorithm-Version 2
0
0
6
1
6
1
25
failed
req
ack inquire
18
18
5
req 2 5
2
12
12
req
4
3
4
3
24 . Comments
- Let K = |Si|. Let each process be a member of
D subsets. When N = 7, K = D = 3. When K=D,
N = K(K-1)+1. So K =O(√N)
- The message complexity of Version 1 is 3√N.
Maekawa’s analysis of Version 2 reveals a
complexity of 7√N
• Sanders identified a bug in version 2 …
25 . Token-passing Algorithms for
mutual exclusion
Suzuki-Kasami algorithm
The Main idea
Completely connected network of processes
There is one token in the network. The holder
of the token has the permission to enter CS.
Any other process trying to enter CS must
acquire that token. Thus the token will move
from one process to another based on
demand.
I want to enter CS
I want to enter CS
26 . Suzuki-Kasami Algorithm
req
last
Process i broadcasts (i, num) req
Sequence number
queue Q
Each process maintains of the request
-an array req: req[j] denotes the sequence
no of the latest request from process j
(Some requests will be stale soon)
req
req
Additionally, the holder of the token maintains
-an array last: last[j] denotes the
sequence number of the latest visit to CS req
for process j.
- a queue Q of waiting processes req: array[0..n-1] of integer
last: array [0..n-1] of integer
27 . Suzuki-Kasami Algorithm
When a process i receives a request (k, num) from
process k, it sets req[k] to max(req[k], num).
The holder of the token
--Completes its CS
--Sets last[i]:= its own num
--Updates Q by retaining each process k only if
1+ last[k] = req[k]
(This guarantees the freshness of the request)
--Sends the token to the head of Q, along with
the array last and the tail of Q
Req: array[0..n-1] of integer
In fact, token ≡ (Q, last) Last: Array [0..n-1] of integer
28 . Suzuki-Kasami’s algorithm
{Program of process j}
Initially, ∀i: req[i] = last[i] = 0
* Entry protocol *
req[j] := req[j] + 1;
Send (j, req[j]) to all;
Wait until token (Q, last) arrives;
Critical Section
* Exit protocol *
last[j] := req[j]
∀k ≠ j: k is in Q ⋀ req[k] = last[k] + 1 req[k] = last[k] + 1 append k to Q;
if Q is not empty send (tail-of-Q, last) to head-of-Q fi
* Upon receiving a request (k, num) *
req[k] := max(req[k], num)
29 . Example
req=[1,0,0,0,0]
req=[1,0,0,0,0]
last=[0,0,0,0,0] 1
0
2
req=[1,0,0,0,0]
4
req=[1,0,0,0,0]
3
req=[1,0,0,0,0]
initial state: process 0 has sent a
request to all, and grabbed the token
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