GATE Overflow Test Series | Operating Systems | Test 1 | Question: 21

Question

The following program consists of $3$ concurrent processes and $3$ binary semaphores. The semaphores are initialized as $S_0=0, S_1=1$ and $S_2=1.$
$$\begin{array}{|l|l|l|}\hline \text{Process $P_0$} & \text{Process $P_1$} & \text{Process $P_2$}
\\\hline \text{while (true) \{} & \text{wait ($S_1$);} & \text{wait ($S_2$);}
\\ \quad\text{wait ($S_0$);} & \text{release ($S_0$);} & \text{release ($S_0$);}
\\ \quad\text{print ‘1';} & &
\\ \quad\text{release ($S_1$);} & &
\\ \quad\text{release ($S_2$);} & &
\\ \} & &
\\\hline \end{array}$$
How many times will process $P_0$ print '$1$'?

• A. At least twice
• B. Exactly once
• C. Exactly twice
• D. None of the above

Comments

why not 1 option @gatecse   @Lakshman Patel RJIT

https://gateoverflow.in/2347/gate2010-45 --please refer this 

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@val_pro20 please check the initial conditions of s0, s1 and s2.

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that release() is not signal()

Answer 1

Initially $S_0$ is locked and it can be unlocked by either $P_1$ or $P_2.$ Say if $P_1$ proceeds first and then $P_2$ proceeds. In this case $S_0$ will be unlocked and process $P_0$ will print $’1’$ once. If process $P_0$ does wait(S0) before process $P_2$ does release $S_0$ it would have printed $1$ twice. So, correct answer here is at least $’1’$. Option D.

Comments

At least 1 is correct and so is at most 2.

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@Arjun     According to your solution IF Process Po is printing two times, so the ans should be exactly twice. But you are saying at least one so it means it could be printed one time or 3 time, If it can be printed 3 times so at least 2 should be the answer, can you please tell is there any way to print one time One

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P0 is in loop(can be run any number of times)

At max P0 can print 1 twice.

On executing 3rd time it will stuck.