5 votes 5 votes How many conflict equivalent schedules are possible for the given schedule ? $R_1(A), R_2(A), R_3(A), R_4(A), W_1(B), W_2(B), W_3(B), W_4(B)$ Databases databases conflict-serializable transaction-and-concurrency + – Hardik Maheshwari asked Oct 10, 2018 reopened Nov 2, 2018 by Arjun Hardik Maheshwari 975 views answer comment Share Follow See all 7 Comments See all 7 7 Comments reply Show 4 previous comments kumar.dilip commented Dec 1, 2018 reply Follow Share Rishi yadav How 24 please explain ?? 0 votes 0 votes Rishi yadav commented Dec 2, 2018 reply Follow Share I think i am wrong kumar.dilip 0 votes 0 votes Jaideep Bankoti commented Dec 3, 2018 reply Follow Share i am getting 1560, here's how i tried to solve it, the sequence of reads doesn't matter, whereas writes does matter, so number of possible ways to arrange 4 reads 24, and number of ways to permute 4 writes among 5 places between reads(counting the ends) in order is 65. so total was 65*24.(total number of schedules is 2520 so it is within bounds too).please verify! 0 votes 0 votes Please log in or register to add a comment.
Best answer 7 votes 7 votes I am getting 105 here's what I did : fixed r(a) of T1, now r2(A) has three places - before r1(A), after r1(A) and after w1(B) because of two items r1(a) and w1(b) now for r3(a), we have 5 places because of 4 items r1(a), r2(a) and w1(b) and w2(b) similarly, for r4(a) we have 7 places because of 6 items now total possible places would be - 3*5*7 = 105 Method 2 :- Fo with Topological sort method, Some more references to solve like this problems :- https://gateoverflow.in/253496/number-of-topological-order https://gateoverflow.in/245897/hashing-with-linear-probing Anuj Mishra answered Jan 23, 2019 edited Mar 1, 2019 by Shaik Masthan Anuj Mishra comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments balchandar reddy san commented Jan 23, 2019 reply Follow Share So you are saying that reads can be shuffled but not writes in order to maintain data consistency. 0 votes 0 votes Anuj Mishra commented Jan 23, 2019 reply Follow Share In question it is asked - " how many conflict equivalent schedules are possible " , so in whatever schedules we form , for them to be be conflict equivalent to given schedule , the conflicts need to be resolved in same fashion. So T2>T1 >T3 >T4 would not be be conflict equivalent . 0 votes 0 votes KUSHAGRA गुप्ता commented Oct 19, 2019 reply Follow Share @Anuj Mishra Can we start with $R_{3}$ after fixing $R_{1}$ and then $R_{2 }$ & then $R_{4}$ ?? 0 votes 0 votes Please log in or register to add a comment.