GATE CSE 2004 | Question: 48

Question

Consider two processes $P_1$ and $P_2$ accessing the shared variables $X$ and $Y$ protected by two binary semaphores $S_X$ and $S_Y$ respectively, both initialized to 1. $P$ and $V$ denote the usual semaphore operators, where $P$ decrements the semaphore value, and $V$ increments the semaphore value. The pseudo-code of $P_1$ and $P_2$ is as follows:
$$\begin{array}{|l|l|}
\hline 
P_1: & P_2: \\
\hline  
\text{While true do \{} & \text{While true do \{} \\ 
L_1:\dots\dots & L_3:\dots\dots \\  
L_2:\dots\dots & L_4:\dots\dots \\ 
\text{X = X + 1;} & \text{Y = Y + 1;} \\ 
Y = Y - 1; & X = Y - 1; \\ 
V(S_X); & V(S_Y); \\  
V(S_Y); & V(S_X); \\
\} & \} \\
\hline  
\end{array}$$
In order to avoid deadlock, the correct operators at $L_1$, $L_2$, $L_3$ and $L_4$ are respectively.

• A. $P(S_Y), P(S_X); P(S_X), P(S_Y)$

• B. $P(S_X), P(S_Y); P(S_Y), P(S_X)$

• C. $P(S_X), P(S_X); P(S_Y), P(S_Y)$

• D. $P(S_X), P(S_Y); P(S_X), P(S_Y)$

Comments

Look at the first locks of both processes (L1 and L3). If both hold different locks, that is, L1 holds Sx and L3 holds Sy (or vice versa), there will always be deadlock.

Answer 1

• A. deadlock $p1$ : line$1$|$p2$:line$3$| $p1$: line$2$(block) |$p2$ :line$4$(block)
  So, here $p1$ want $s(x)$ which is held by $p2$ and $p2$ want $s(y$) which is held by $p1$.
  So, its circular wait (hold and wait condition). So. there is deadlock.
   
• B. deadlock $p1$ : line $1$| $p2$ line $3$|$p1$: line $2$(block) |$p2$ : line $4$(block)
  Som here $p1$ wants sy which is held by $p2$ and $p2$ wants $sx$ which is held by $p1$. So its circular wait (hold and wait ) so, deadlock.
   
• C. $p1$ :line $1$|$p2$ : line $3| p2$ line $4$ (block) $|p1$ line $2$ (block) here, $p1$ wants $sx$ and $p2$ wants $sy$, but both will not be release by its process $p1$ and $p2$ because there is no way to release them. So, stuck in deadlock.
   
• D. $p1$ :line $1$ $|p2$ : line $3$ (block because need sx ) $|p1$ line $2 |p2$ : still block $|p1$: execute cs then up the value of sx |p2 :line 3 line $4$(block need $sy$)$| p1$ up the$ sy$ $|p2$ :lin$4$ $4$ and easily get $cs$.
  We can start from $p2$ also, as I answered according only $p1$, but we get same answer.
  So, option (D) is correct 

Comments

E) Sy,Sx,Sy,Sx is also another valid option, although not in options

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one more option if given P(Sy) P(Sx) P(Sy) P(Sx) is correct or not

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@moh_haris Yes it would also be correct

Answer 2

Option (A):

$\Rightarrow$ Deadlock possible

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Option (B) . Same as Option (A). Deadlock possible.

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Option (C):

No need to think of another process to get a deadlock situation:

$\Rightarrow$ Both processes will be blocked once they start.

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Option (D) is ok ! as far as deadlock is concerned.

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Comments

@debashidh

in option A there may b not deadlock .

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QS asks to avoid options having non zero probability of deadlock. Qs does not ask which option may have deadlock.

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well

Answer 3

let execute P1 till L1 and then execute P2 till L3..
now check options.
Option A,B,C results deadlock.
even Option C always deadlocked..

Option D works fine ..

Answer 4

Deadlock is possible when there is no order defined how resources will be consumed by processes. Therefore graph based protocol is free from deadlock as every process try checking from root and if first resource is occupied then the process is blocked and the process which has locked it first completes its execution. After checking the option you can see both processes are consuming resources in an order in option d. Sx followed by Sy which prevents deadlock.

Answer 5

Option D

P1                                          P2
down Sx                              down Sx     (here Sx becomes  -1)
down Sy                              down Sy     (here Sy  becomes  -1)
after that P1 can continue execute  and up both Sx  &  Sy

Answer 6

Video Explanation by Vishhwadeep Gothi sir ---https://youtu.be/2yIqMvQtYzY?list=PLG9aCp4uE-s3klreqEhbzOBQDg5Ha0U38&t=1373