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COMP 346 – WINTER 2015Tutorial # 5

Semaphores for Barrier SyncA barrier is a type of

synchronization method. A barrier for a group of threads or processes in the source code means any thread/process must stop at this point and cannot proceed until all other threads/processes reach this barrier.

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Semaphores for Barrier SyncTake a look at the typical problem:

all processes must finish their phase I before any of them starts phase II

processes must proceed to their phase II in a specific order, for example: 1, 2, 3…

This is called barrier synchronization.

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semaphore s1 = 0, s2 = 0; process P1 process P2 <phase I> <phase I> V (s1) V (s2) P (s2) P (s1) <phase II> <phase II>

Barrier Sync for Three ProcessesA possible copy-cat attempt:

Why it doesn’t work?

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semaphore s1 = 0, s2 = 0, s3 = 0; process P1 process P2 process P3 <phase I> <phase I> <phase I> V (s1) V (s2) V (s3) P (s3) P (s1) P (s2) <phase II> <phase II> <phase II>

Barrier Sync for Three ProcessesA possible copy-cat attempt:

Why it doesn’t work?Scenario: 1-6 , process 2 even hasn’t started!None of the requirements are met

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semaphore s1 = 0, s2 = 0, s3 = 0; process P1 process P2 process P33 <phase I> <phase I> 1 <phase I>4 V (s1) V (s2) 2 V (s3)5 P (s3) P (s1) P (s2)6 <phase II> <phase II> <phase II>

Barrier Sync for Three Processes (2)Another attempt:

What’s wrong now?

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semaphore s1 = 0, s2 = 0, s3 = 0; process P1 process P2 process P3 <phase I> <phase I> <phase I> P (s1) V (s1) V (s1) P (s1) P (s2) P (s3) V (s2) V (s3) <phase II> <phase II> <phase II>

Barrier Sync for Three Processes (2)Another attempt:

What’s wrong now?Scenario: 1-10, so far so good, but after…The second requirement isn’t met

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semaphore s1 = 0, s2 = 0, s3 = 0; process P1 process P2 process P37 <phase I> 4 <phase I> 1 <phase I>8 P (s1) 5 V (s1) 2 V (s1)9 P (s1) 6 P (s2) 3 P (s3)10 V (s2) V (s3) <phase II> <phase II> <phase II>

Barrier Sync for Three Processes (3)Last attempt:

A bit “optimized”:

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semaphore s1 = 0, s2 = 0, s3 = 0; process P1 process P2 process P3 <phase I> <phase I> <phase I> P (s1) V (s1) V (s1) P (s1) P (s2) P (s3) <phase II> <phase II> <phase II> V (s2) V (s3)

semaphore s1 = -1, s2 = 0, s3 = 0; process P1 process P2 process P3 <phase I> <phase I> <phase I> V (s1) V (s1) P (s1) P (s2) P (s3) <phase II> <phase II> <phase II> V (s2) V (s3)

Barrier Sync: Need for the General Solution

Problem with the proposed solution: # of semaphores == # of processes.

Semaphores as any other resource are limited and take space => overhead

Imagine you need to sync 10 processes in the same manner? 100? 1000?Complete mess and a high possibility of a

deadlock!

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Barrier Sync: Need for the General Solution (2)

Attempt for the first requirement:

The second requirement is left as an exercise to the curious student :-)

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semaphore s1 = -n + 2, s2 = 0; process P1 process P2 ... process Pn <phase I> <phase I> ... <phase I> P (s1) V (s1) ... V (s1) V (s2) P (s2) ... P (s2) V (s2) ... V (s2) <phase II> <phase II> ... <phase II>

Introducing the Semaphore ClassNOTE: Operations Signal and Wait are guaranteed to be atomic!

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class Semaphore{

private int value;

public Semaphore(int value){

this.value = value;}

public Semaphore(){

this(0);}

...

Introducing the Semaphore Class (2)

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...public synchronized void Wait(){

while(this.value <= 0){

try{

wait();}catch(InterruptedException e){

System.out.println(

"Semaphore::Wait() …” + e.getMessage()

);

e.printStackTrace();}

}

this.value--;}

...

Introducing the Semaphore Class (3)

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...public synchronized void Signal(){

++this.value;notify();

}

public synchronized void P(){

this.Wait();}

public synchronized void V(){

this.Signal();}

}

Referenceshttp://users.encs.concordia.ca/~mokhov/c

omp346/http://programmingexamples.wikidot.com/

java-barrier

http://docs.oracle.com/javase/tutorial/essential/concurrency/sync.html

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