The Synchronization Problem

1
The Synchronization Problem

• Synchronization problems occur because
• multiple processes or threads want to share data
• the executions of these processes interleave in
  arbitrary fashions.

2
A Simple Example

• Two processes trying to increment a common
  variable V (suppose V 0 initially)

load V,Rload V,Radd 1,Radd 1,Rstore R,Vstore
R,V
load V,Radd 1,Rstore R,Vload V,Radd 1,Rstore
R,V
Process1load V,Radd 1,Rstore R,V
Process2load V,Radd 1,Rstore R,V
V 1
V 2
V
3
Critical Section

• It can be easily seen that the load of one
  process must follow the store of the other
  process.
• The three instructions must be executed
  atomically, thus forming a critical section (CS).
• Ie., when one process is in its critical section,
  no other process can be in its critical section
  at the same time.

4
2-Process CS

• Algorithms 1 2 do not satisfy the progress
  requirement.
• To prove an algorithm not satisfying a
  requirement, we only need to find one case (ie.,
  there exists).
• To prove an algorithm satisfying a requirement,
  we have to prove that it does so for all cases
  (ie., for all).

5
One Failure Case of Algo. 1
while turn ? 0 do no-op CSturn 1 exit //
remainder sectionwhile turn ? 1 do
no-op CSturn 0 remainder sectionwhile
turn ? 1 do no-op
time
No more progress!
6
One Failure Case of Algo. 2
flag0 trueflag1 truewhile flag0 do
no-op
flag0 trueflag1 truewhile flag1 do
no-op
No progress either case!
7
Algorithm 3 Mutual Exclusion

• turn, a shared variable, is either 0 or 1, and
  hence only one process can pass through the while
  statement at a time.
• This is a direct proof.

8
Proving For All

• Instead of a direct proof, to prove for all, we
  could use proof by contradiction.
• Assume there exists a case not satisfying a
  requirement show that such a case wont exist.

9
Algorithm 3 - Progress

• Proof by contradiction
• Assume there exists a time when both processes
  are executing the while statement and cannot get
  into the CS.
• Then both turn 0 and turn 1 must be true
  an impossible case! The assumption cannot be
  true.

10
Algorithm 3 Bounded Waiting

• Suppose P0 is executing in the CS, and P1 is
  waiting (executing the while).
• flag0 and turn 0 in P1s while statement
  are true.
• When P0 exits the CS, it sets flag0 to false.
• Two possibilities at this point either P0
  continues to execute or P1 enters the CS.
• Contd

11
Bounded Waiting (contd)

• Suppose P0 continues to execute and arrives at
  its while statement.
• It will be stuck at the while statement because
  both flag1 and turn 1 would be true.
• Hence, in either case, P1 is the next one to
  enter the CS.

12
The Bakery Algorithm Solving the n-process CS

• The counter can serve only one customer at a
  time. The counter is the CS.
• When a customer enters the bakery, it picks a
  number which is 1 the largest number being held
  by a existing customer.
• Contd

13
Bakery Algorithm (contd)

• To get to the counter, the customer compares its
  number with every one of the existing customers.
• It holds a bit when coming to a customer, i, who
  is in the middle of getting its number (the
  choosingi variable is used).
• If its number is smaller than all existing
  numbers, the customer gets the counter.