Three concurrent processes $X$, $Y$, and $Z$ execute three different code segments that access and update certain shared variables. Process $X$ executes the $P$ operation (i.e., wait) on semaphores $a$, $b$ and $c$; process $Y$ executes the $P$ operation on semaphores $b$, $c$ and $d$; process $Z$ executes the $P$ operation on semaphores $c$, $d$, and $a$ before entering the respective code segments. After completing the execution of its code segment, each process invokes the $V$ operation (i.e., signal) on its three semaphores. All semaphores are binary semaphores initialized to one. Which one of the following represents a deadlock-free order of invoking the $P$ operations by the processes?

To prevent deadlock among concurrent processes requesting multiple shared resources, a standard technique is resource ordering (Dijkstra's resource hierarchy). If all processes request resources in accordance with a strict total order, a circular wait condition is impossible.

Let us check the ordering in Option B:

• Process $X$: $P(b) \to P(a) \to P(c)$

• Process $Y$: $P(b) \to P(c) \to P(d)$

• Process $Z$: $P(a) \to P(c) \to P(d)$

Define the total ordering of semaphores as:

Let us verify each process against this total ordering:

• Process $X$ requests $b, a, c$, which follows $b < a < c$.

• Process $Y$ requests $b, c, d$, which follows $b < c < d$.

• Process $Z$ requests $a, c, d$, which follows $a < c < d$.

Since all processes request resources in strictly monotonically increasing order according to $b < a < c < d$, circular wait cannot occur, and hence execution is guaranteed to be deadlock-free.

In all other options, there is a cycle in the request graph, which can lead to a deadlock under interleaved execution.