Consider the following transactions with data items P and Q initialized to zero:

T1 : read (P) ; 
     read (Q) ; 
     if P = 0 then Q : = Q + 1 ; 
     write (Q). 

T2 : read (Q) ; 
     read (P) 
     if Q = 0 then P : = P + 1 ; 
     write (P).

Any non-serial interleaving of T₁ and T₂ for concurrent execution leads to

Consider the following schedule for transactions T1, T2 and T3: Which one of the schedules below is the correct serialization of the above?

Consider two transactions T₁ and T₂ and four schedules S₁, S₂, S₃, S₄ of T₁ and T₂ as given below:

T₁: R₁[ x ] W₁[ x ] W₁[ y ]
T₂: R₂[ x ] R₂[ y ] W₂[ y ]
S₁: R₁[ x ] R₂[ x ] R₂[ y ] W₁[ x ] W₁[ y ] W₂[ y ]
S₂: R₁[ x ] R₂[ x ] R₂[ y ] W₁[ x ] W₂[ y ] W₁[ y ]
S₃: R₁[ x ] W₁[ x ] R₂[ x ] W₁[ y ] R₂[ y ] W₂[ y ]
S₄: R₂[ x ] R₂[ y ] R₁[ x ] W₁[ x ] W₁[ y ] W₂[ y ]

Which of the above schedules are conflict-serializable?

Let R and S be relational schemes such that R = { a, b, c } and S = { c }. Now consider the following queries on the database:

I. $$\pi_{R-S}(r) - \pi_{R-S} \left (\pi_{R-S} (r) \times s - \pi_{R-S,S}(r)\right )$$

II. $$\left\{t \mid t \in \pi_{R-S} (r) \wedge \forall u \in s \left(\exists v \in r \left(u = v[S] \wedge t = v\left[R-S\right]\right )\right )\right\}$$

III.$$\left\{t \mid t \in \pi_{R-S} (r) \wedge \forall v \in r \left(\exists u \in s \left(u = v[S] \wedge t = v\left[R-S\right]\right )\right ) \right\}$$

IV. Select R.a, R.b
    From R, S
    Where R.c = S.c

Which of the above queries are equivalent?