Consider the following database schedule with two transactions, $T_1$ and $T_2$. $$S = r_2(X); r_1(X); r_2(Y); w_1(X); r_1(Y); w_2(X); a_1; a_2$$ where $r_i(Z)$ denotes a *read* operation by transaction $T_i$ on a variable $Z$, $w_i(Z)$ denotes a *write* operation by $T_i$ on a variable $Z$ and $a_i$ denotes an *abort* by transaction $T_i$. Which one of the following statements about the above schedule is **TRUE**?

Let us analyze the sequence of operations in schedule $S$:

1. $r_2(X)$

2. $r_1(X)$

3. $r_2(Y)$

4. $w_1(X)$

5. $r_1(Y)$

6. $w_2(X)$

7. $a_1$

8. $a_2$

• Dirty Read Analysis: A dirty read occurs if a transaction reads data written by an uncommitted active transaction ($w_j(Z)$ followed by $r_i(Z)$ where $j \neq i$).

  • $w_1(X)$ is followed by $r_1(Y)$ (same transaction reads $Y$) and $w_2(X)$ (blind write by $T_2$). No transaction reads the value of $X$ written by $T_1$.

  • $w_2(X)$ is followed only by aborts $a_1$ and $a_2$. No transaction reads the value of $X$ written by $T_2$.

  • Therefore, there are no dirty reads in $S$.

• Cascading Abort: Since no transaction reads uncommitted data written by another transaction, the abort of $T_1$ ($a_1$) does not force $T_2$ to abort. Hence, $S$ does not have a cascading abort (it is cascadeless / avoids cascading aborts).

• Strictness: A schedule is strict if no transaction reads or writes data item $Z$ until the transaction that previously wrote $Z$ has committed or aborted. Here, $w_1(X)$ is overwritten by $w_2(X)$ before $T_1$ commits or aborts. Thus, $S$ is not strict.

Therefore, statement (C) is the only TRUE statement.