1 votes 1 votes The given schedule: R1 (A) R2 (B) W2 (B) W1(A) W2 (A) Commit2 R1 (C) is revoreable, cascadeless or strict? Databases databases transaction-and-concurrency concurrency recoverable dirty + – budhu asked Jan 29, 2018 • reshown Jan 29, 2018 by budhu budhu 1.5k views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments Manu Thakur commented Jan 29, 2018 reply Follow Share yes, no dirty read hence cascadeless and recoverable. not strict because of W1(A) W2 (A) . 2 votes 2 votes royal shubham commented Feb 9, 2018 reply Follow Share I think sedule is non recoverable , why u all saying it is recoverable , T2 is writing someting written by T1 and consumer fisrt commit which is error in some case known as non recoverable . –1 votes –1 votes ganeshveluru08 commented Sep 8, 2020 reply Follow Share All cascade less. Schedules are recoverable 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes the schedule is recoverable and cascadeless but not strict. As there is no dirty read it is recoverable and cascadeless before write(A) operation in T2 there is no commit or abort in T1 so it is not strict scheduling $ruthi answered Jan 31, 2018 $ruthi comment Share Follow See all 0 reply Please log in or register to add a comment.
