Consider a multi-threaded program with two threads $T1$ and $T2$. The threads share two semaphores: $s1$ (initialized to 1) and $s2$ (initialized to 0). The threads also share a global variable $x$ (initialized to 0). The threads execute the code shown below. \begin{verbatim} // code of T1 // code of T2 wait(s1); wait(s1); x = x+1; x = x+1; print(x); print(x); wait(s2); signal(s2); signal(s1); signal(s1); \end{verbatim} Which of the following outcomes is/are possible when threads $T1$ and $T2$ execute concurrently?

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Outcome (A): This is not possible because $T2$ cannot proceed after $T1$ without acquiring $s1$, but $s1$ is released only when $T1$ completes. Thus, (A) is incorrect. Outcome (B): $T2$ can run first, acquire $s1$, increment $x$, and print $1$. Then $T1$ acquires $s1$, increments $x$, and prints $2$. This outcome is possible. Outcome (C): $T1$ acquires $s1$, increments $x$, and prints $1$. However, it waits on $s2$, which is never signaled by $T2$, causing a deadlock. This outcome is possible. Outcome (D): This is not possible because $T2$ signals $s2$ after printing, allowing $T1$ to proceed. Thus, (D) is incorrect. Final Answer: $$ \boxed{\text{(B), (C)}} $$

Answer

$T1$ runs first and prints 1, $T2$ runs next and prints 2

Answer

$T2$ runs first and prints 1, $T1$ runs next and prints 2

Answer

$T1$ runs first and prints 1, $T2$ does not print anything (deadlock)

Answer

$T2$ runs first and prints 1, $T1$ does not print anything (deadlock)

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The Correct Option is B

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