Question

The following program consists of 3 concurrent processes and 3 binary semaphores. The semaphores are initialized as S0 = 1, S1 = 0, S2 = 0.

Process P0
while(true)
{
wait(S0);
print '0';
release(S1);
release(S2);
}

Process P1
wait(S1);
release(S0);

Process P2
wait(S2);
release(S0);
How many times will P0 print '0' ?
a) At least twice
b) Exactly twice
c) Exactly thrice
d) Exactly once

Answer 1

Process P0 will print '0' exactly once because after its first execution, there is a race condition with processes P1 and P2, which can release semaphore S0 before P0 has the chance to wait on it again.

Step-by-step explanation:

The question asks about the number of times process P0 will print '0' given a program with 3 concurrent processes and 3 binary semaphores. The semaphores are initialized as S0 = 1, S1 = 0, S2 = 0. After analyzing the given program, process P0 can only print '0' exactly once before it releases the semaphores S1 and S2. The first time P0 runs, it can enter its critical section because S0 is initialized to 1. However, after P0 prints '0' and releases semaphores S1 and S2, either P1 or P2 will proceed (as they are waiting on S1 and S2, respectively) and immediately release S0. This creates a race condition where either P1 or P2 can execute and release S0 again before P0 gets a chance to execute its wait(S0) commands. This means P0 might not immediately re-enter its critical section as it may lose out in the race condition, making it possible that P0 prints '0' only once before either P1 or P2 executes and releases S0.