Consider the following proposed solution for the critical section problem. There are $n$ processes: $P_0 \dots P_{n-1}$. In the code, function pmax returns an integer not smaller than any of its arguments. For all $i$, t[i] is initialized to zero. Code for $P_i$:do { c[i]=1; t[i] = pmax(t[0],...,t[n-1])+1; c[i]=0; for every j ≠ i in {0,...,n-1} { while (c[j]); while (t[j] != 0 && t[j]<=t[i]); } Critical Section; t[i]=0; Remainder Section; } while (true);Which one of the following is TRUE about the above solution?

Let us analyze the given synchronization algorithm (a variant of Lamport's Bakery Algorithm):

1. Mutual Exclusion / At most one process in Critical Section:

   • To enter the Critical Section, process $P_i$ checks for every $j \neq i$ whether $t[j] \neq 0$ and $t[j] \le t[i]$.

   • If all active processes have distinct ticket values $t$, the process with the strictly smallest ticket will pass the while loop and enter the Critical Section, while all other processes wait. Hence, two processes with different ticket numbers cannot enter simultaneously.

   • If two processes $P_i$ and $P_j$ happen to receive the same ticket value ($t[i] = t[j]$), both will wait on each other indefinitely because the condition t[j] <= t[i] evaluates to true for both $P_i$ (checking $P_j$) and $P_j$ (checking $P_i$). This causes a deadlock, meaning neither enters the critical section.

   • Thus, under no circumstances can more than one process be in the critical section at the same time. Hence, at most one process can be in the critical section at any time.

2. Deadlock / Progress:

   • In standard Bakery algorithms, ties in ticket numbers are broken using process IDs (e.g., $(t[i], i) < (t[j], j)$). Here, there is no tie-breaking rule; the condition is simply t[j] <= t[i].

   • If $t[i] = t[j]$, both processes wait for each other, resulting in a deadlock. Therefore, it *can* cause deadlock, and the progress and bounded waiting conditions fail.

Thus, statement (A) is TRUE.