GATE CSE 2014 | Set 3 | Question: 29

Question

Consider the transactions $T1, T2, \:\text{and} \:T3$ and the schedules $S1 \:\text{and} \:S2$ given below. 

• $T1: r1(X); r1(Z); w1(X); w1(Z) $
• $T2: r2(Y); r2(Z); w2(Z) $
• $T3: r3(Y); r3(X); w3(Y) $
• $S1: r1(X); r3(Y); r3(X); r2(Y); r2(Z); w3(Y); w2(Z); r1(Z); w1(X); w1(Z) $
• $S2: r1(X); r3(Y); r2(Y); r3(X); r1(Z); r2(Z); w3(Y); w1(X); w2(Z); w1(Z) $

Which one of the following statements about the schedules is TRUE?

• A. Only $S1$ is conflict-serializable.
• B. Only $S2$ is conflict-serializable.
• C. Both $S1$ and $S2$ are conflict-serializable.
• D. Neither $S1$ nor $S2$ is conflict-serializable.

Comments

s1 is conflist serialezable but s2 is not confliect serizable

becz s2 precedence graph has cycle like 't1 ro t2 and t2 to t1'

---

Hope this will help !!

Answer 1

$S_1$ has no cycle hence, Conflict-Serializable

$S_2$ has cycle hence NOT Conflict-Serializable

Answer is option A.

Comments

In schedule S1: r1(z), w1(x), w1(z) are not taken into consideration for conflicts.

Is it because they are present at the last and all other operations have happened before them?

---

@Gupta731

Is it because they are present at the last and all other operations have happened before them?

yes,

---

Precedence graph is always between conflicting pairs. right?

like R(x)W(x), W(x)R(x) & W(x)W(x).

Answer 2

In s1 there is no cycle ..serial order of execution for s1 is t2 t3 and t1 in s2 there is cycle in precedence graph so not conflict serializable so ans is a

Answer 3

BEST ANSWER