Consider the following two threads T1 and T2 that update two shared variables $a$ and $b$. Assume that initially $a = b = 1$. Though context switching between threads can happen at any time, each statement of T1 or T2 is executed atomically without interruption. T1 a = a + 1;b = b + 1; T2 b = 2 * b;a = 2 * a; Which one of the following options lists all the possible combinations of values of $a$ and $b$ after both T1 and T2 finish execution?

Let the individual atomic statements be:

• $S_1$: a = a + 1; (from T1)

• $S_2$: b = b + 1; (from T1)

• $S_3$: b = 2 * b; (from T2)

• $S_4$: a = 2 * a; (from T2)

Within thread T1, $S_1$ must execute before $S_2$. Within thread T2, $S_3$ must execute before $S_4$. The total number of valid interleavings is $\binom{4}{2} = 6$:

1. $S_1 \rightarrow S_2 \rightarrow S_3 \rightarrow S_4$:

   • $S_1$: $a = 1 + 1 = 2$

   • $S_2$: $b = 1 + 1 = 2$

   • $S_3$: $b = 2 \times 2 = 4$

   • $S_4$: $a = 2 \times 2 = 4$

   • Result:$(a = 4, b = 4)$

2. $S_1 \rightarrow S_3 \rightarrow S_2 \rightarrow S_4$:

   • $S_1$: $a = 2$

   • $S_3$: $b = 2 \times 1 = 2$

   • $S_2$: $b = 2 + 1 = 3$

   • $S_4$: $a = 2 \times 2 = 4$

   • Result:$(a = 4, b = 3)$

3. $S_1 \rightarrow S_3 \rightarrow S_4 \rightarrow S_2$:

   • $S_1$: $a = 2$

   • $S_3$: $b = 2$

   • $S_4$: $a = 4$

   • $S_2$: $b = 3$

   • Result:$(a = 4, b = 3)$

4. $S_3 \rightarrow S_1 \rightarrow S_2 \rightarrow S_4$:

   • $S_3$: $b = 2$

   • $S_1$: $a = 2$

   • $S_2$: $b = 3$

   • $S_4$: $a = 4$

   • Result:$(a = 4, b = 3)$

5. $S_3 \rightarrow S_1 \rightarrow S_4 \rightarrow S_2$:

   • $S_3$: $b = 2$

   • $S_1$: $a = 2$

   • $S_4$: $a = 4$

   • $S_2$: $b = 3$

   • Result:$(a = 4, b = 3)$

6. $S_3 \rightarrow S_4 \rightarrow S_1 \rightarrow S_2$:

   • $S_3$: $b = 2 \times 1 = 2$

   • $S_4$: $a = 2 \times 1 = 2$

   • $S_1$: $a = 2 + 1 = 3$

   • $S_2$: $b = 2 + 1 = 3$

   • Result:$(a = 3, b = 3)$

Thus, the set of all distinct final values for $(a, b)$ is:

This matches Option (A).