P={P1,P2,P3,P4}, R={R1,R2,R3,R4} (single instances each).Assignment: R1→P1, R2→P2, R3→P3, R4→P4.Clai

Q: What is GATE-CS?

GATE-CS is a competitive examination conducted in India. This question is from the 2025 GATE-CS Operating Systems paper.

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To prepare for GATE-CS Operating Systems, practice previous year questions (PYQs), understand concepts thoroughly, and solve mock tests regularly. This 2025 question covers Deadlock (Detection, Prevention, Avoidance) topic.

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This GATE-CS 2025 Operating Systems question has a hard difficulty level and is marked as an important question.

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Question

P={P1,P2,P3,P4}, R={R1,R2,R3,R4} (single instances each).Assignment: R1→P1, R2→P2, R3→P3, R4→P4.Claim: P1→R2, P2→R3, P3→R1, P2→R4, P4→R2.Which statement(s) is/are CORRECT?

Accepted Answer

Correct Answer: (C) Aborting P2Deadlock cycles: P1→R2→P2→R3→P3→R1→P1 (cycle 1); P2→R4→P4→R2→P2 (cycle 2). P2 is in BOTH cycles.Abort P2: R2 and R3, R4 (eventually) freed. P1 gets R2 → completes → releases R1. P3 gets R1 → completes. P4 gets R2 (once P1 releases) → completes. System deadlock-free. ✓Abort P1 only (A): Cycle P2↔P4 (via R2↔R4) remains. ✗Abort P3 only (B): P3 releases R3; P2 gets R3 but still waits for R4 (P4 holds); P4 waits for R2 (P2 holds). Deadlock P2↔P4 remains. ✗Abort P1 and P4 (D): P3 gets R1 → completes. P2 (has R2, gets R3 from P3, R4 from P4) → completes. System deadlock-free. This IS correct per ACE solution — but official key shows only C. Per official key: C only.

P={P1,P2,P3,P4}, R={R1,R2,R3,R4} (single instances each).Assignment: R1→P1, R2→P2, R3→P3, R4→P4.Claim: P1→R2, P2→R3, P3→R1, P2→R4, P4→R2.Which statement(s) is/are CORRECT?

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P={P1,P2,P3,P4}, R={R1,R2,R3,R4} (single instances each).Assignment: R1→P1, R2→P2, R3→P3, R4→P4.Claim: P1→R2, P2→R3, P3→R1, P2→R4, P4→R2.Which statement(s) is/are CORRECT?

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