Question

Consider the following read-write schedule \( S \) over three transactions \( T_1, T_2, \text{and } T_3 \), where the subscripts in the schedule indicate transaction IDs: \[ S: r_1(z); w_1(z); r_2(x); r_3(y); w_3(y); r_2(y); w_2(x); w_2(y); \]
Which of the following transaction schedules is/are conflict equivalent to \( S \)?

• A. \( T_1 T_2 T_3 \)
• B. \( T_1 T_3 T_2 \)
• C. \( T_3 T_2 T_1 \)
• D. \( T_3 T_1 T_2 \)

Hint:

Conflict equivalence ensures consistency in read/write operations across transactions. Always analyze dependencies and conflicting operations carefully.

Answer 1

The Correct Option is B

Step 1: Conflict equivalence definition.
Two schedules are conflict equivalent if they preserve the same order of conflicting operations (read/write or write/write) across all transactions. Step 2: Analyze the given schedule \( S \).
Transaction \( T_1 \) performs \( r_1(z); w_1(z) \). Transaction \( T_2 \) performs \( r_2(x); w_2(x); r_2(y); w_2(y) \). Transaction \( T_3 \) performs \( r_3(y); w_3(y) \). Conflicts occur between \( T_1 \) and \( T_2 \) on \( z \) and \( x \), and between \( T_3 \) and \( T_2 \) on \( y \). Step 3: Test each schedule.
For \( T_1 T_2 T_3 \), conflicts are not preserved, as \( T_3 \)'s operations on \( y \) may overlap incorrectly with \( T_2 \). For \( T_1 T_3 T_2 \), conflicts are preserved, maintaining the correct order of operations between \( T_1, T_3, \text{and } T_2 \). For \( T_3 T_2 T_1 \), conflicts are preserved, maintaining correct order of operations between \( T_3, T_2, \text{and } T_1 \). For \( T_3 T_1 T_2 \), conflicts are preserved, maintaining correct order of operations between \( T_3, T_1, \text{and } T_2 \). Final Answer: \[ \boxed{T_1 T_3 T_2, T_3 T_2 T_1, T_3 T_1 T_2} \]