00:01
Hello students, so the answer of the first question is precedence graph for the schedule s.
00:06
So here t1 to t3 read a read a, t2 to t3 read b read b, t3 to t1 write a read b and t3 to t2 write b read b.
00:16
So these are the precedence graph.
00:18
Now this is the precedence graph checking whether the schedule s is conflict serializable or not.
00:24
A schedule is conflict serializable if it can be transformed into a serial schedule by swapping non -conflicting operation.
00:35
Two operations are non -conflicting if they operate on the separate data items on the same data items, but at least one of them is a read operation.
00:46
In the schedule s we have the following conflicting operation.
00:50
So here t1 write a, t3 read a, after that t3 write b and t2 read b.
01:02
So these conflicting operations cannot be swapped without violating the order of the operation within each transaction.
01:10
Therefore the schedule is not conflict serializable.
01:13
No.
01:14
3 is constructing a serializable schedule s, serial schedule s which is conflict equivalent to the schedule s.
01:23
A serial schedule is a schedule in which all the transactions are executed in one specific order.
01:29
A conflict equivalent schedule is a schedule that process the same output as the original schedule even though the order of operation may be different.
01:41
So to construct a serial schedule s which is conflict equivalent to the schedule s, we can swap the non -conflicting operation in s as follows...