Conflict-equvalent serial schedule

Question

Consider a schedule
S: r₁(A); W₁(B); W₁(C); r₂(A); W₂(B); W₂(C); abort₂; r₃(A); W₃(B); W₃(C); C₁; C₃

How many conflict equivalent serial schedules are possible for the given schedule ?

1.  2
2.  4
3.  3
4.  5

Comments

here only one shcedule is possible right?

that is T1, T2 then T3 as T3 is dependent on both T1 and T2 to complete and and T2 depends on T1 to complete according to transaction graph...

so only one schelude is posible then why not its in options??

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oh sorry i forgot to consider the significance of abort in T2.
answer will be 3 right?
T1-> T2->T3
T2-> T1->T3
T1-> T3->T2

Answer 1

T1	T2	T3
R(A)
W(B)
W(C)
	R(A)
	W(B)
	W(C)
	abort 2
		R(A)
		W(B)
		W(C)
C1;
		C3;

Here Transaction T2 is aborted in middle so no need of that transaction.it is independent.

No of conflict serializable schedules:

Construct precedence graph no of topological sortings equall to no of schedules.

No of conflict schedules=3

T1->T2->T3

T1->T3->T2

T2->T1->T3