Suppose a database schedule 5 involves transactions T 1, T 2, .............T nConsider 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 ?
Topological order
Understanding Database Schedule Serialization
A database schedule is a sequence of operations (like read, write, commit, abort) from a set of concurrent transactions. When multiple transactions run at the same time, their operations can interleave. A key property for ensuring correctness in concurrent execution is serializability.
A schedule is considered serializable if its result is equivalent to the result of some serial schedule. A serial schedule is one where transactions are executed one after another, with no interleaving of operations.
Precedence Graph and Serializable Schedules
To determine if a schedule is serializable, we use a precedence graph. This directed graph has vertices representing the transactions involved in the schedule ($T_1, T_2, \dots, T_n$). An edge is drawn from transaction $T_i$ to transaction $T_j$ ($T_i \rightarrow T_j$) if:
- $T_i$ and $T_j$ access the same data item.
- At least one of these accesses is a write operation (this is a conflict).
- The conflicting operation of $T_i$ occurs before the conflicting operation of $T_j$ in the schedule.
A fundamental theorem in concurrency control states that a schedule is serializable if and only if its precedence graph contains no cycles. If the precedence graph has no cycles, it is a Directed Acyclic Graph (DAG).
Topological Ordering for Serializable Schedules
The question asks what ordering of the vertices (transactions) in the precedence graph guarantees a serial schedule equivalent to the original serializable schedule S.
If a schedule S is serializable, its precedence graph is a DAG. For any DAG, a topological sort is possible. A topological sort of a DAG is a linear ordering of its vertices such that for every directed edge $u \rightarrow v$, vertex $u$ comes before vertex $v$ in the ordering.
Consider a topological ordering of the transactions from the precedence graph of the serializable schedule S, say $T_{p_1}, T_{p_2}, \dots, T_{p_n}$. If we construct a new serial schedule $S'$ by executing the transactions in this order ($T_{p_1}$ runs completely, then $T_{p_2}$ runs completely, and so on), this serial schedule $S'$ is guaranteed to be equivalent to the original schedule S.
This is because the topological order respects all the conflicts captured by the edges in the precedence graph. If there is a conflict dependency $T_i \rightarrow T_j$ in the original schedule S, it means $T_i$ performed a conflicting operation before $T_j$. In the topological sort, $T_i$ will necessarily appear before $T_j$. Therefore, in the serial schedule $S'$ constructed based on this topological order, $T_i$ will complete all its operations before $T_j$ starts, thus preserving the dependency and resulting in the same outcome as S.
Why Other Orderings Don't Guarantee a Serial Schedule
- Depth-first order and Breadth-first order: These are graph traversal algorithms. While they visit all nodes, the order in which they finish visiting nodes or add them to a queue does not necessarily respect the dependency edges in a way that guarantees $u$ appears before $v$ for every edge $u \rightarrow v$. They are not topological sorts.
- Ascending order of transaction indices: Ordering transactions simply by their numerical index ($T_1, T_2, \dots, T_n$) has no relation to the actual dependencies or conflicts defined by the schedule. This arbitrary order would likely violate dependencies ($T_i \rightarrow T_j$ where $i > j$) and thus would not generally produce an equivalent serial schedule, even if the original schedule was serializable.
Therefore, only a topological ordering of the precedence graph vertices is guaranteed to yield an equivalent serial schedule for a serializable schedule.
Revision Table: Database Schedule Concepts
| Concept | Description | Relevance to Serialization |
|---|---|---|
| Schedule | Sequence of operations from concurrent transactions. | The execution order being analyzed. |
| Serial Schedule | Transactions execute one after another without interleaving. | The target equivalent execution type for serializability. |
| Precedence Graph | Graph representing conflict dependencies between transactions. | Used to test for serializability (cycle detection). |
| Serializable Schedule | Equivalent to some serial schedule. | Precedence graph is a DAG (no cycles). |
| Topological Order | Linear ordering of vertices in a DAG respecting edge directions. | Provides an execution order for a serial schedule equivalent to a serializable schedule. |
Additional Information: Types of Conflicts and Serializability
Conflicts in schedules arise when two transactions access the same data item and at least one access is a write operation. There are three types of conflicts:
- Read-Write (RW) Conflict: $T_i$ reads item A, then $T_j$ writes item A. ($R_i(A); W_j(A)$) An edge $T_i \rightarrow T_j$ is added if $R_i(A)$ is before $W_j(A)$.
- Write-Read (WR) Conflict: $T_i$ writes item A, then $T_j$ reads item A. ($W_i(A); R_j(A)$) An edge $T_i \rightarrow T_j$ is added if $W_i(A)$ is before $R_j(A)$.
- Write-Write (WW) Conflict: $T_i$ writes item A, then $T_j$ writes item A. ($W_i(A); W_j(A)$) An edge $T_i \rightarrow T_j$ is added if $W_i(A)$ is before $W_j(A)$.
The existence of any of these conflicts $T_i \rightarrow T_j$ means $T_i$ must logically precede $T_j$ to maintain the schedule's outcome. If the precedence graph contains a cycle, it implies a circular dependency ($T_1 \rightarrow T_2 \rightarrow \dots \rightarrow T_k \rightarrow T_1$), which cannot be resolved in any serial order, making the schedule non-serializable.
Topological sorting provides a valid execution order because it ensures that for every conflict $T_i \rightarrow T_j$, $T_i$ is completely finished before $T_j$ begins, thus preserving the order of conflicting operations seen in the original serializable schedule.
Important Questions from Serializability of Schedules - Teaching
Consider the following statements:
Statement I: Conservative 2 PL is a deadlock-free protocol.
Statement II: Thomas's write rule enforces conflict serializability.
Statement III: Timestamp ordering protocol ensures serializability based on the order of transaction timestamps.
Which of the following is correct?
Consider the following schedules involving two transactions.
S 1: r 1(X) ; r 1(Y) ; r 2(X) ; r 2(Y) ; w 2(Y) ; w 1(X)
S2 : r 1(X) ; r 2(X) ; r 2(Y) ; w 2(Y) ; r 1(Y) ; w 1(X)
Which one of the following statements is correct with respect to above?Consider the following four schedules due to three transactions (indicated by the subscript) using read and write on a data item X, denoted by r(X) and w(X) respectively. Which one of them is conflict serializable ?
S 1 : r 1(X); r 2(X); w 1(X); r 3(X); w 2(X)
S2: r 2(X); r 1(X); w 2(X); r 3(X); w 1(X)
S 3 : r 3(X); r 2(X); r 1(X); w 2(X); w 1(X)
S4 : r 2(X); w 2(X); r 3(X); r 1(X); w 1(X)
