GATE CS 2026 Set 2 — Question 51

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MSQ+2 / -0HardSemaphoresProcess SynchronizationOperating System

Operating System → Process Synchronization → Semaphores

Last updated 5 September 2026

Question

Consider three processes P1, P2, and P3 running identical code, as shown in the pseudocode below. A and B are two binary semaphores initialized to 1 and 0, respectively. X is a shared variable initialized to 0. Each line in the pseudocode is executed atomically. Pseudocode of P1, P2, and P3 Wait(A); Print(*); X = X+1; If (X == 2) { Print($); Signal(B); } Signal(A); Wait(B); Print(#); Signal(B); Assume that any of the three processes can start to execute first and context switching can happen between these processes at any arbitrary time and in any arbitrary order. Which of the following patterns is/are possible to be generated as an outcome of the execution of these three processes?

Correct answer (A) $; (B) $; (C) $ Solution The code has two sections. The first section is protected by semaphore A (mutex). The second section is controlled by semaphore B. 1. First Section (Wait(A) ... Signal(A)): Processes enter sequentially. 1st process: Prints *, sets X=1. 2nd process: Prints *, sets X=2, enters if, prints $, signals B. 3rd process: Prints *, sets X=3. Because of Wait(A), the 3rd process cannot print * until the 2nd process releases A. The 2nd process prints $ before releasing A. Thus, the output must contain * (from 1st) and * $ (from 2nd) before the 3rd * can appear? No, the 3rd * appears after 2nd releases A. But $ is printed inside the critical section of the 2nd process. So * and $ from the 2nd process are atomic relative to other * s. The sequence of * and $ must be * (1st), * $ (2nd), * (3rd) in terms of when they are enabled. Specifically, $ must appear after the second * and before the 2nd process releases A. This implies the prefix **$ is mandatory. Option (D) ***$ is impossible because the 3rd * cannot occur before the 2nd process releases A, and the 2nd process prints $ before releasing A. 2. Second Section (Wait(B) ... Signal(B)): This is enabled by Signal(B) in the 2nd process. Once B becomes 1, processes can print # sequentially. The printing of # can be interleaved with the 3rd process's execution of the first section (printing the 3rd *), as these are independent once A is released by the 2nd process. Analysis of Options: (A) **$*### : 2nd proc finishes, 3rd proc enters and prints *, then all print # . Possible. (B) **$#*## : 2nd proc finishes, one proc prints #, then 3rd proc prints *, then others print # . Possible. (C) **$##*# : 2nd proc finishes, two procs print #, then 3rd proc prints *, then last # . Possible. (D) ***$### : Impossible as explained above. Thus, A, B, and C are possible. Turn this into a strength. Explore AI-powered practice and doubt support with Success Tracker.

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Correct answer (A) $; (B) $; (C) $ Solution The code has two sections. The first section is protected by semaphore A (mutex). The second section is controlled by semaphore B. 1. First Section (Wait(A) ... Signal(A)): Processes enter sequentially. 1st process: Prints *, sets X=1. 2nd process: Prints *, sets X=2, enters if, prints $, signals B. 3rd process: Prints *, sets X=3. Because of Wait(A), the 3rd process cannot print * until the 2nd process releases A. The 2nd process prints $ before releasing A. Thus, the output must contain * (from 1st) and * $ (from 2nd) before the 3rd * can appear? No, the 3rd * appears after 2nd releases A. But $ is printed inside the critical section of the 2nd process. So * and $ from the 2nd process are atomic relative to other * s. The sequence of * and $ must be * (1st), * $ (2nd), * (3rd) in terms of when they are enabled. Specifically, $ must appear after the second * and before the 2nd process releases A. This implies the prefix **$ is mandatory. Option (D) ***$ is impossible because the 3rd * cannot occur before the 2nd process releases A, and the 2nd process prints $ before releasing A. 2. Second Section (Wait(B) ... Signal(B)): This is enabled by Signal(B) in the 2nd process. Once B becomes 1, processes can print # sequentially. The printing of # can be interleaved with the 3rd process's execution of the first section (printing the 3rd *), as these are independent once A is released by the 2nd process. Analysis of Options: (A) **$*### : 2nd proc finishes, 3rd proc enters and prints *, then all print # . Possible. (B) **$#*## : 2nd proc finishes, one proc prints #, then 3rd proc prints *, then others print # . Possible. (C) **$##*# : 2nd proc finishes, two procs print #, then 3rd proc prints *, then last # . Possible. (D) ***$### : Impossible as explained above. Thus, A, B, and C are possible.

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GATE CS 2026 Set 2 — Question 51

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