reopened by
975 views
5 votes
5 votes
How many conflict equivalent schedules are possible for the given schedule ?

$R_1(A), R_2(A), R_3(A), R_4(A), W_1(B), W_2(B), W_3(B), W_4(B)$
reopened by

1 Answer

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

edited by

Related questions

0 votes
0 votes
1 answer
1
Harsh Saini_1 asked Dec 27, 2023
334 views
How many total $Conflict$ $Serializable$ $Schedules$ are possible that can be formed by $T1$ and $T2?$$T1:$ $r_1(A)$ $r_1(B)$ $w_1(B)$$T2:$ $r_2(B)$ $r_2(A)$ $w_2(B)$
2 votes
2 votes
4 answers
3