Final answer:
Creating a digraph with three tasks such that every possible priority list results in a unique schedule is impossible, since the linear dependency of tasks in a digraph limits the number of unique schedules that can be obtained.
Step-by-step explanation:
It is possible to create a digraph with three tasks such that every possible priority list creates a different schedule. First, we should understand that a digraph or directed graph is a set of vertices connected by edges, where the edges have a direction associated with them. To create a digraph with three tasks that ensures each priority list results in a unique schedule, we can arrange them in a linear sequence where each task is dependent on the completion of the task before it.
For example, consider three tasks A, B, and C. We can organize them in the digraph as follows:
- Task A must be completed before Task B (A → B)
- Task B must be completed before Task C (B → C)
In this configuration, there is only one possible priority list (A, B, C), which correlates to one unique schedule. Since the question asks for every possible priority list to create a different schedule, the key is that with three tasks, there are only 3! = 6 possible priority lists or permutations. However, given the linear dependency represented in the digraph, these permutations will not result in unique schedules. Therefore, it is actually impossible to create a digraph with three tasks where every possible priority list creates a different schedule.