Test for Conflict Serializability
Let S be he schedule. Get the precedence graph G = (V,E).
V, Vertices are Transactions and E are edges Ti ? Tj for
- Ti Write(Q) before Tj Read(Q)
- Ti Read(Q) before Tj Write(Q)
- Ti Write(Q) before Tj Write(Q)
If there are no cycles in G then it is conflict-serializable.
Serial schedule is obtained by topological sorting. See examples.