Synchronization Checkpoint

Semaphores and deadlock detection.

Get every question correct to pass this checkpoint. You can review feedback, revise your answers, and submit again.

1. Start with 0 permits. Run P(11), P(12), V(9), P(13), V(11). Which acquisition order is returned?

Select one answer.

[11,13,12] [11,12,13] [9,11] [11,12]

2. A signal wakes a queued waiter and also increments the available count. Which later event exposes the defect?

Select one answer.

The awakened waiter appears in the acquisition output The queue becomes shorter A new P succeeds even though the signal permit was already transferred The number of signals increases by one

3. Graph edges are 0→1, 0→2, 2→1. DFS finishes vertex 1 before following 0→2. What does edge 2→1 establish?

Select one answer.

A cycle because vertex 1 has been visited A self-loop at vertex 2 Deadlock because two processes wait for the same process No cycle; vertex 1 is finished, not on the current recursion path

4. A graph includes a cycle, but one request can be satisfied by either of two resource holders, one of whom is still runnable. Why is the simple deadlock conclusion unjustified?

Select one answer.

A directed cycle can never describe deadlock Edges may describe alternatives rather than mandatory waits FIFO wakeup always breaks cycles Any graph with more than two vertices is inconclusive

5. Initial permits 2; four signals have occurred; five acquisitions have completed. How many permits are available?

6. After all nodes in one component are finished, must a cycle detector also visit a disconnected unvisited component? Answer yes or no.

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