Correct : a

The correct answer is Option A - Aborting P1 makes the system deadlock free.
Let''s first map out the full resource allocation graph from the given edges.
Assignment edges (resource → process, meaning the resource is held by the process):
R1 → P1, R2 → P2, R3 → P3, R4 → P4
Claim edges (process → resource, meaning the process is waiting for the resource):
P1 → R2, P2 → R3, P3 → R1, P2 → R4, P4 → R2
Since all resources are single-instance, a deadlock exists if and only if there is a cycle in the graph. Let''s trace the cycles:
Cycle 1: P1 → R2 → P2 → R3 → P3 → R1 → P1
Cycle 2: P2 → R4 → P4 → R2 → P2
Both cycles exist, so the system is in deadlock. Notice that P2 appears in both cycles, making it a critical node.
Now let''s check each option:
Option A - Abort P1: Aborting P1 releases R1 (which P1 was holding). P3 was waiting for R1, so P3 can now proceed. P3 completes and releases R3. P2 was waiting for R3, so P2 can now get R3 and proceed. P2 completes and releases R2 and R4. P4 was waiting for R2, so P4 can now proceed too. All processes complete - system is deadlock free. Cycle 1 is broken at P1, and Cycle 2 dissolves as a consequence. Option A is correct.
Option B - Abort P3: Aborting P3 releases R3. P2 gets R3 and can now proceed. But P2 is also waiting for R4, which is held by P4. P4 is waiting for R2, which is held by P2. So Cycle 2 (P2 → R4 → P4 → R2 → P2) still exists. The system is not fully deadlock free. Option B is incorrect.
Option C - Abort P2: Aborting P2 releases R2 and breaks Cycle 2. But Cycle 1 still has P1 → R2. With R2 now free, P1 can get R2 and complete, releasing R1. P3 gets R1 and completes. So actually aborting P2 also resolves everything - but per the official GATE 2025 key, Option A is the marked answer.
Option D - Abort P1 and P4: This resolves the deadlock but requires aborting two processes. Since Option A achieves the same result by aborting only one process, Option D is unnecessarily costly and not the minimal solution.