Suppose a database schedule S involves transactions T1, …,Tn. Construct the precedence graph of S with vertices representing the transactions and edges representing the conflicts. If S is serializable, which one of the following orderings of the vertices of the precedence graph is guaranteed to yield a serial schedule?a)Topological orderb)Depth-first orderc)Breadth-first orderd)Ascending order of transaction indicesCorrect answer is option 'A'. Can you explain this answer?

Question

Suppose a database schedule S involves transactions T1, &hellip;,Tn. Construct the precedence graph of S with vertices representing the transactions and edges representing the conflicts. If S is serializable, which one of the following orderings of the vertices of the precedence graph is guaranteed to yield a serial schedule?a)Topological orderb)Depth-first orderc)Breadth-first orderd)Ascending order of transaction indicesCorrect answer is option 'A'. Can you explain this answer?

Answer

T2, and T3. Let's assume that T1 and T2 are executed concurrently, and T3 is executed after T2.

A serial schedule for the given transactions would be T1, T2, T3. In this schedule, T1 is executed first, followed by T2, and finally T3. Each transaction is executed in its entirety before the next transaction begins. This ensures that the transactions are executed one after the other, without any concurrency.

On the other hand, a concurrent schedule for the given transactions could be T1, T2, T3. In this schedule, T1 and T2 are executed concurrently, meaning that their operations can be interleaved. For example, T1 may perform some operations, then T2 may perform some operations, and so on. Finally, T3 is executed after T2 completes.

It is important to note that there can be multiple possible concurrent schedules for a given set of transactions, depending on how their operations are interleaved. Concurrent schedules can provide benefits such as improved performance and resource utilization but may also introduce concurrency-related issues such as conflicts and consistency problems.

Answer

Serial schedule is possible only when precedence graph doesn’t contain cycle. If precedence graph contains a cycle, it means schedule is not conflict serializable.
Breadth first search and Depth first search of a graph are possible even if graph contains cycle.
Topological sort in a graph will not work if graph contains a cycle.

Consider a directed acyclic graph:

Here Two orders possible: V₂, V₃, V₁, V₄, V₅, V₆ OR V₃, V₂, V₁, V₄, V₅, V₆.
In case of ascending order of transaction indices, two non-conflicting schedules can occur simultaneously.

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