Consider the following schedules:
S1: r1(A); r1(B); w1(C); w2(D); w1(A); w2(A); r1(D);
S2: r1(A); r1(B); r1(C); r1(D); w1(A); w2(B); w3(C); w4(D)
Which of the following option is correct?
A) S1 has blind write S2 does not
B) S2 has blind write S1 does not
C) Both schedules are conflict serializable
D) None of the above
Correct me if I am wrong but I think S1 and S2 both have blind writes?