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.// code of T1 wait(s1); x = x+1; print(x); wait(s2); signal(s1);// code of T2 wait(s1); x = x+1; print(x); signal(s2); signal(s1);Which of the following outcomes is/are possible when threads T1 and T2 execute concurrently?

Let us trace the possible execution orders of the two threads:

1. Case 1: Thread T2 runs first

   • T2 executes wait(s1): s1 becomes $0$.

   • T2 executes x = x + 1: x becomes $1$.

   • T2 executes print(x): prints 1.

   • T2 executes signal(s2): s2 becomes $1$.

   • T2 executes signal(s1): s1 becomes $1$.

   • Now, thread T1 can execute:

     • T1 executes wait(s1): s1 becomes $0$.

     • T1 executes x = x + 1: x becomes $2$.

     • T1 executes print(x): prints 2.

     • T1 executes wait(s2): s2 was $1$, decrements to $0$ without blocking.

     • T1 executes signal(s1): s1 becomes $1$.

   • Outcome:T2 runs first and prints 1, T1 runs next and prints 2. Thus, (B) is possible.

2. Case 2: Thread T1 runs first

   • T1 executes wait(s1): s1 becomes $0$.

   • T1 executes x = x + 1: x becomes $1$.

   • T1 executes print(x): prints 1.

   • T1 executes wait(s2): since s2 is initialized to $0$, T1 blocks waiting on s2.

   • Now, T2 tries to run and executes wait(s1): since s1 is $0$ (held by T1), T2 blocks waiting on s1.

   • T1 is waiting for s2 (which can only be signaled by T2), and T2 is waiting for s1 (which can only be signaled by T1).

   • Outcome: Both threads are blocked, resulting in a deadlock where T1 has printed 1 and T2 does not print anything. Thus, (C) is possible.

Therefore, the correct options are B and C.