Each of a set of $n$ processes executes the following code using two semaphores a and b initialized to 1 and 0, respectively. Assume that count is a shared variable initialized to 0 and not used in CODE SECTION P.CODE SECTION P wait(a); count=count+1; if (count==n) signal(b); signal(a); wait(b); signal(b); CODE SECTION QWhat does the code achieve?

This synchronization pattern implements a barrier (rendezvous):

1. a is a mutex semaphore initialized to $1$, protecting the critical section where count is incremented.

2. When the first $n-1$ processes finish CODE SECTION P:

   • Each enters the mutex, increments count, checks count == n (which evaluates to false), releases mutex a, and then blocks on wait(b) since semaphore b is initialized to $0$.

3. When the $n$-th (last) process finishes CODE SECTION P:

   • It increments count to $n$, evaluates count == n as true, and executes signal(b).

   • It then executes signal(a) and reaches wait(b). It consumes the signal on b and immediately performs signal(b).

4. The signal(b) executed after wait(b) initiates a chain reaction (turnstile mechanism) that unblocks the next waiting process, which in turn unblocks the next, until all $n$ processes proceed to CODE SECTION Q.

Thus, no process can execute CODE SECTION Q until all $n$ processes have completed CODE SECTION P. Therefore, option (A) is correct.