Consider the following threads, $T_1$, $T_2$, and $T_3$ executing on a single processor, synchronized using three binary semaphore variables, $S_1$, $S_2$, and $S_3$, operated upon using standard wait() and signal(). The threads can be context switched in any order and at any time. Which initialization of the semaphores would print the sequence BCABCABCA....?

To print the sequence BCABCABCA..., the execution must strictly follow the cyclic order $T_2 \rightarrow T_1 \rightarrow T_3 \rightarrow T_2 \dots$

1. First character 'B':

   • Thread $T_2$ prints 'B' after executing wait(S1).

   • For $T_2$ to execute first without allowing other threads to interleave, we must initialize $S_1 = 1$.

   • If either $S_2 = 1$ or $S_3 = 1$ initially, thread $T_3$ (prints 'A') or $T_1$ (prints 'C') could run first or concurrently, violating the required sequence. Thus, we must have $S_2 = 0$ and $S_3 = 0$.

2. Tracing execution with $S_1 = 1, S_2 = 0, S_3 = 0$:

   • $T_2$ runs:wait(S1) sets $S_1 = 0$, prints "B", and signal(S3) sets $S_3 = 1$.

   • $T_1$ runs:wait(S3) sets $S_3 = 0$, prints "C", and signal(S2) sets $S_2 = 1$.

   • $T_3$ runs:wait(S2) sets $S_2 = 0$, prints "A", and signal(S1) sets $S_1 = 1$.

   • The cycle repeats indefinitely, generating the output BCABCABCA... strictly.

Thus, the correct initialization is $S_1 = 1; S_2 = 0; S_3 = 0$, which corresponds to Option (C).