Conflict Serializability

Find out conflict serializability for the given transactions

T1	T2	T3
R(X)
		R(Y)
		R(X)  t1 <- t3
	R(Y) t3 <- t2
	R(Z) t1 <- t2
		W(Y)
	W(Z) t1 <- t2 and  t1 <- t2
R(Z)
W(X)
W(Z)

Step 1: 

Conditions for read(R) and write(W)

R-W

W-R

W-W ->W-R

        ->W-W

Step 2:

Check out the conflict pairs in other transactions and then draw edges using Precedence Graph.

Precedence Graph:

Therefore the indegree of T1=2

Therefore the indegree of T2=0

Therefore the indegree of T3=1

Step 3:

Remove the indegree of T2 whose degree is 0 then the Precedence graph

Therefore the transaction to the given set is 

T2 --- T3 --- T1

Since there is no loop or cycle so it is conflict serializable which is serializable consistent.

T1 --- T2 --- T3

T1 --- T3 --- T2

T2 --- T1 --- T3

T2 --- T3 --- T1

T3 --- T1 --- T2

T3 --- T2 --- T1

Question:

check out whether the given schedule is conflict serializable or not  

T1	T2	T3
R(A) t2<-t1 t3<-t1
	W(A) t1<-t2
W(A) t3<-t1
		W(A)

Step 1: 

Conditions for read(R) and write(W)

R-W

W-R

W-W ->W-R

        ->W-W

Step 2:

Check out the conflict pairs in other transactions and then draw edges using Precedence Graph.

Precedence Graph:

Loop occurred i.e T1 --T2 -- T2 --T1 means it is not a conflict serializable