+20 votes
2.2k views

Consider three data items $D1, D2,$ and $D3,$ and the following execution schedule of transactions $T1, T2,$ and $T3.$ In the diagram, $R(D)$ and $W(D)$ denote the actions reading and writing the data item $D$ respectively.
$${\textbf{time}} {\Huge{\downarrow}} \begin{array}{|l|l|l|}\hline \textbf{T1} & \textbf{T2} & \textbf{T3} \\\hline \text{} & \text{R(D3);} & \text{}\\ \text{} & \text{R(D2);} & \text{} \\ \text{} & \text{W(D2);} & \text{} \\ \text{} & \text{} & \text{R(D2);} \\ \text{} & \text{} & \text{R(D3);} \\ \text{R(D1);} & \text{} & \text{} \\ \text{W(D1);} & \text{} & \text{} \\ \text{} & \text{} & \text{W(D2);} \\ \text{} & \text{} & \text{W(D3);} \\ \text{} & \text{R(D1);} & \text{} \\ \text{R(D2);} & \text{} & \text{} \\ \text{W(D2);} & \text{} & \text{} \\ \text{} & \text{W(D1);} & \text{}\\\hline \end{array}$$Which of the following statements is correct?

1. The schedule is serializable as $T2; T3; T1$

2. The schedule is serializable as $T2; T1; T3$

3. The schedule is serializable as $T3; T2; T1$

4. The schedule is not serializable

asked
edited | 2.2k views
0
Can't we check the schedule by trying to convert it in a serial schedule???
0
I hava a doubt...if trans T1 is writing a irem which is final means bo else trans is writing after this but reading it. Schenario:

T1.   . T2

R(a)

W(a)

R(a)

So here can we take T1 as final write ?
0
I have a doubt that if this transaction is view serializable then can we say that it is serializable. Even if it is not conflict serializable.

Please someone explain
+2

Swami yes if a schedule is not conflict but view serializable then it is also Serializable.

## 4 Answers

+28 votes
Best answer

There is a cycle in precedence graph so schedule is not conflict serialisable.
Check View Serializability:
Checking View Serializability is NPC problem so proving by contradiction..

1. Initial Read
$T2$ read $D2$ value from initial database and $T1$ modify $D2$ so $T2$ should execute before $T1$.
i.e., $T2 \to T1$

2. Final write.
final write of $D1$ in given schedule done by $T2$ and $T1$ modify $D1$ i.e. $W(D1)$..
that means $T2$ should execute after $T1$.
i.e., $T1 \to T2$

So, schedule not even View Serializable.
Not Serializable.

Correct Answer: $D$

answered by Veteran (59.9k points)
edited
+25
for checking view serialization, there is a more simple explanation than checking NPC and contradictory proofing,

ie. just check if there is any blind write in the execution schedule of transactions, if yes then only we need to form view equivalence graph.

in this case there is no blind write under either of the three transactions, so we can say that its not view serializable.hence, not serializable.

option d is right one
+3
we can also draw the polygraph and check for view serializability
+8

Having blind writes does not imply that it is view serializable.A schedule which is not conflict serializable will be having two choices :

1)Not containing Blind writes : then surely it will not be view serializable and thus not Serializable.

2)Containing Blind writes : it may/may not be view serializable. Then we will go for the Polygraph.

+1
+1
1)if schedule(s) not conflict serializable schedule && no blind write in schedule(s),then then schedule also not view serializable schedule.

2)if schedule(s) not conflict-serializable schedule && schedule(s) is view serializable schedule , then atleast one blind write must exist.
0

@AakS I totally agree with yur statement ,

but in this particular example we can easily see no blind writes are there hence we can reach conclusion (as in my case proving with contradiction may takes longer time in real exam).

+8 votes

Conflict Serializability is  the sufficient condition for a schedule to be Serializable while View Serializability is the Neccassary and Sufficient condition for a Serializable schedule.

So first we will check conflict serializability using precendence graph. Here it containing the cycle and thus it fails for conflict serializability.Now we will go for the Blind Writes check.

if a schedule is not conflict serializable and not containing the blind writes then it FAILS for view serializability also.

As given schedule not containing any Blind write, it is not View serializable.Thus, the given schedule is not Serializable.

answered by Active (1.8k points)
+5 votes
The above schedule is not conflict serializable as there is a cycle in the precedence graph .Also no blind writ in any of the three transactions (i.e write without read ) which is a must condition for view serializable  . So it is not serializable.

For Serializable : A schedule is serializble if it is either conflict serializable or view serializable  So simple and easy
answered by Junior (971 points)
–5 votes
APPROPIATE OPTION IS D IS  CORRECT
answered by Junior (593 points)
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