Simple User Solutions to the C. S. Problem

Tom Kelliher, CS 318

Feb. 16, 1998

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Outline:

Assignment:

Software Solutions

Two Process Solutions

Assumptions:

1. Only two cooperating processes.

2. We have P0 and P1.

3. Replace i with appropriate integer.

Try 1

int turn = 0;        // Shared control variable.

// mutexbegin:
while (turn != i)
   ;                 // Busy wait.

// mutexend:
turn = 1 - i;

1. Guarantees mutual exclusion.

2. Does not guarantee progress --- enforces strict alternation of processes entering CS's.

3. Bounded waiting violated --- suppose one process terminates while its its turn?

Try 2

Remove strict alternation requirement.

int flag[2] = { FALSE, FALSE }   // flag[i] indicates that Pi is in its
                                 //  critical section.

// mutexbegin:
while (flag[1 - i])
   ;
flag[i] = TRUE;

// mutexend:
flag[i] = FALSE;

1. Mutual exclusion violated.

2. Progress ok.

3. Bounded waiting?

Try 3

Restore mutual exclusion.

int flag[2] = { FALSE, FALSE }   // flag[i] indicates that Pi wants to
                                 //  enter its critical section.

// mutexbegin:
flag[i] = TRUE;
while (flag[1 - i])
   ;

// mutexend:
flag[i] = FALSE;

1. Guarantees mutual exclusion.

2. Violates progress --- both processes could set flag and then deadlock on the while.

3. Bounded waiting?

Try 4

Attempt to remove the deadlock.

int flag[2] = { FALSE, FALSE }   // flag[i] indicates that Pi wants to
                                 //  enter its critical section.

// mutexbegin:
flag[i] = TRUE;
while (flag[1 - i])
{
   flag[i] = FALSE;
   delay;                  // Sleep for some time.
   flag[i] = TRUE;
}

// mutexend:
flag[i] = FALSE;

1. Mutual exclusion guaranteed.

2. Progress violated (processes can ``dance'').

3. Bounded waiting violated.

Peterson's Algorithm

int flag[2] = { FALSE, FALSE }   // flag[i] indicates that Pi wants to
                                 //  enter its critical section.
int turn = 0;                    // turn indicates which process has
                                 //  priority in entering its critical
                                 //  section.

// mutexbegin:
flag[i] = TRUE;
turn = 1 - i;
while (flag[1 - i] && turn == 1 - i)
   ;

// mutexend:
flag[i] = FALSE;

1. Satisfies all solution requirements. Why?

Multiple Process Solution

Lamport's Bakery algorithm.

Assumptions:

1. NPROCS is the number of processes.

2. max(int *array) returns the maximum value in array.

3. Each process has a unique ID, so ties on the number chosen are broken by comparing IDs.

4. Replace i with the appropriate process ID.

// Global initialization:
int choosing[NPROCS] = { FALSE };
int number[NPROCS] = { 0 };

// mutexbegin:
choosing[i] = TRUE;
number[i] = max(number) + 1;
choosing[i] = FALSE;
for (j = 0; j < NPROCS; ++j)
{
   while (choosing[j])
      ;

   while (number[j] != 0 && (number[j] < number[i] ||
                             number[j] == number[i] && j < i) )
      ;
}

// mutexend:
number[i] = 0;

1. Is it correct?

2. What can happen to number? Is that likely?