GO Classes DPP | GATE CS | Operating System | Peterson's Algorithm

Answer 1

For process $P_i$, the entry code is:

$\texttt{flag[i] = TRUE}$

$\texttt{turn = i}$

and it waits while:

$\texttt{while(flag[j] \&\& turn == j);}$

Suppose $P_j$ enters the critical section before $P_i$.

After leaving the critical section, $P_j$ can again execute:

$\texttt{flag[j] = TRUE}$

$\texttt{turn = j}$

Now $P_i$ again finds $\texttt{flag[j] = TRUE}$ and $\texttt{turn == j}$, so it continues waiting.

This can happen repeatedly. Hence, $P_i$ may wait indefinitely.

Therefore, bounded waiting is violated.

Mutual exclusion and progress are still satisfied.

$\boxed{\text{Answer: C}}$

Comments

@GOClasses

If u can observe closely, two processes can enter C.S simultaneously violating Mutual Exclusion as well.

Let $P_0$, $P_1$

Execute in the following order:
 

$P_0$: flag[0] = TRUE;

$P_1$: flag[1] = TRUE;

$P_1$: turn = 1;

$P_0$: turn = 0;

 

Here, $P_1$ will be spinning and $P_0$ will enter C.S.

$P_0$: flag[0] = False;

Now $P_1$ enters C.S

Now execute $P_0$ from beginning again, it can easily enter C.S
 

And Hence, Answer is A as well.

 

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yeah , great observation . 
Mutual Exclusion also violating 
kindly check sir @GO Classes