$P = \{P_1, P_2, P_3, P_4\}$ consists of all active processes in an operating system. $R = \{R_1, R_2, R_3, R_4\}$ consists of single instances of distinct types of resources in the system. The resource allocation graph has the following assignment and claim edges. Assignment edges: $R_1 \rightarrow P_1$, $R_2 \rightarrow P_2$, $R_3 \rightarrow P_3$, $R_4 \rightarrow P_4$ (the assignment edge $R_1 \rightarrow P_1$ means resource $R_1$ is assigned to process $P_1$ , and so on for others) Claim edges: $P_1 \rightarrow R_2$, $P_2 \rightarrow R_3$, $P_3 \rightarrow R_1$, $P_2 \rightarrow R_4$, $P_4 \rightarrow R_2$ (the claim edge $P_1 \rightarrow R_2$ means process $P_1$ is waiting for resource $R_2$ , and so on for others) Which of the following statement(s) is/are CORRECT?

To determine which processes are involved in deadlock, let us construct the resource allocation graph (RAG) and identify all directed cycles (since each resource has a single instance, a cycle is necessary and sufficient for deadlock):

1. Cycle 1:

   $$P_1 \rightarrow R_2 \rightarrow P_2 \rightarrow R_3 \rightarrow P_3 \rightarrow R_1 \rightarrow P_1$$

   Involving processes $\{P_1, P_2, P_3\}$ and resources $\{R_1, R_2, R_3\}$.

2. Cycle 2:

   $$P_2 \rightarrow R_4 \rightarrow P_4 \rightarrow R_2 \rightarrow P_2$$

   Involving processes $\{P_2, P_4\}$ and resources $\{R_2, R_4\}$.

Now let us evaluate each option:

• (A) Aborting $P_1$: Breaks Cycle 1, but Cycle 2 ($P_2 \rightarrow R_4 \rightarrow P_4 \rightarrow R_2 \rightarrow P_2$) still persists. Thus, the system is not deadlock-free. *(Incorrect)*

• (B) Aborting $P_3$: Breaks Cycle 1, but Cycle 2 still persists. Thus, the system is not deadlock-free. *(Incorrect)*

• (C) Aborting $P_2$: Since $P_2$ is part of both Cycle 1 and Cycle 2, aborting $P_2$ breaks both cycles. Resource $R_2$ is released. Then $P_1$ (waiting for $R_2$) and $P_4$ (waiting for $R_2$) can acquire $R_2$ and execute to completion sequentially, releasing all other resources so $P_3$ also finishes. Hence, the system becomes deadlock-free. *(Correct)*

• (D) Aborting $P_1$ and $P_4$: Aborting $P_1$ breaks Cycle 1 and releases $R_1$. Aborting $P_4$ breaks Cycle 2 and releases $R_4$. With $R_1$ released, $P_3$ (waiting for $R_1$) acquires $R_1$, finishes, and releases $R_3$. Then $P_2$ (waiting for $R_3$ and $R_4$) gets both resources, executes to completion, and releases all resources. Thus, the system becomes deadlock-free. *(Correct)*

Hence, (C) and (D) are the correct statements.