Consider the following pseudocode, where $S$ is a semaphore initialized to 5 in line#2 and counter is a shared variable initialized to 0 in line#1. Assume that the increment operation in line#7 is not atomic.1. int counter = 0; 2. Semaphore S = init(5); 3. void parop(void) 4. { 5. wait(S); 6. wait(S); 7. counter++; 8. signal(S); 9. signal(S); 10. }If five threads execute the function parop concurrently, which of the following program behavior(s) is/are possible?

Let us analyze the possible executions when five threads run parop concurrently:

1. Option D (Deadlock):

   • The initial semaphore value is $S = 5$.

   • If all $5$ threads execute the first wait(S) at line 5 concurrently, each decrements $S$ by $1$. The value of $S$ becomes $5 - 5 = 0$.

   • When all $5$ threads subsequently attempt to execute the second wait(S) at line 6, all of them block because $S = 0$.

   • No thread can proceed to execute signal(S), leading to a deadlock involving all five threads. Hence, (D) is possible.

2. Option A (Counter = 5):

   • If all five threads execute sequentially one after another without preemption during critical operations, each thread successfully increments counter by $1$.

   • After all 5 threads finish, the final value of counter is $5$. Hence, (A) is possible.

3. Option B (Counter = 1):

   • The operation counter++ is non-atomic and consists of three machine instructions: LOAD R, counter, INCR R, and STORE counter, R.

   • Since $S = 5$, two threads (say $T_1$ and $T_2$) can acquire two wait(S) each simultaneously ($2 + 2 = 4 \le 5$).

   • $T_1$ reads counter = 0, increments its local register to $1$, and is then preempted before storing.

   • The remaining threads ($T_2, T_3, T_4, T_5$) execute one by one and increment counter up to $4$.

   • Finally, $T_1$ resumes and stores $1$ into counter, overwriting the previous value.

   • Hence, a final value of $1$ is possible, so (B) is possible.

4. Option C (Counter = 0):

   • For all threads to successfully complete execution, each thread must store at least $R + 1 \ge 1$ into counter at its completion. It is impossible for counter to remain $0$ after successful completion of all threads. Thus, (C) is not possible.

Therefore, the correct options are A, B, and D.