In a schedule S having 4 transactions executed concurrently. The order of their operation is given below. Find out whether the given schedule is serializable or not?R1(A), R2(B), W2(A), W3(C), R4(C), R3(A), W3(B), R4(A), W2(B), W4(B), W3(A)Conflict SerializableView SerializableBoth (a) and (b)Not Serializable

To determine whether a schedule is serializable or not, we need to construct a precedence graph. A schedule is serializable if its precedence graph is acyclic.

Here are the steps to construct the precedence graph:

1. Create a node for each transaction in the schedule. In this case, we have 4 transactions, so we create 4 nodes: T1, T2, T3, and T4.

2. Draw an edge from Ti to Tj if an operation in Ti conflicts with an operation in Tj and Ti precedes Tj. Two operations are said to conflict if they belong to different transactions, they access the same data item, and at least one of them is a write operation.

   • R1(A) and W2(A) conflict, so draw an edge from T1 to T2.
   • R2(B) and W2(B) conflict, so draw an edge from T2 to T2. But this is a self-loop and can be ignored.
   • W2(A) and R3(A) conflict, so draw an edge from T2 to T3.
   • W2(B) and W4(B) conflict, so draw an edge from T2 to T4.
   • W3(C) and R4(C) conflict, so draw an edge from T3 to T4.
   • W3(B) and W4(B) conflict, so draw an edge from T3 to T4. But this edge already exists, so it can be ignored.
   • W3(A) and R4(A) conflict, so draw an edge from T3 to T4. But this edge already exists, so it can be ignored.

3. Check if the graph contains any cycles. If it does, the schedule is not serializable. If it doesn't, the schedule is serializable.

In this case, the precedence graph does not contain any cycles, so the schedule is serializable. Therefore, the answer is both (a) Conflict Serializable and (b) View Serializable.