0 votes 0 votes 'A' is set of all possible schedules 'C' is set of all possible schedules that are guaranteed to produce a correct final result 'S' is the set of all serializable schedules 'P' is the set of all schedules possible under 2-phase locking protocol Which is FALSE? (A) P⊆C (B) S⊂P (C) S⊆P (D) P⊂C Please provide reason for the answer. (Given answer is A) Databases databases transaction-and-concurrency test-series + – !KARAN asked Nov 26, 2018 !KARAN 516 views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply aambazinga commented Nov 26, 2018 reply Follow Share S includes view serialisable as well as conflict serialisable, then how can S is a subset of P correct, when P is containing all conflict serialisable schedule (as 2PL guarantees conflict serialisability)..., when we know that #view serialisable schedules are greater than equal to #conflict serialisable schedule. 0 votes 0 votes Magma commented Nov 26, 2018 reply Follow Share where you find difficulty here..it's simple set theory logic if you already know the concept of conflict serializability , 2 phase locking system + you didn't post the given answer with the question , until and unless someone ask you because it just spoil the spirit to do the question 0 votes 0 votes srestha commented Nov 26, 2018 reply Follow Share how serializable schedule is proper subset of 2PL is a correct option? 0 votes 0 votes Magma commented Nov 26, 2018 reply Follow Share yeah option B is also false 1 votes 1 votes !KARAN commented Nov 27, 2018 reply Follow Share @Magma could you please give your explanation for the answer. According to me both option $B$ and $C$ are wrong as $2PL$ schedules are small schedules under all serializable schedules. Similarly option $A$ and $D$ are true bcz either set of all $2PL$ schedules produce correct result or some of them. Then how option $A$ is answer? 0 votes 0 votes Please log in or register to add a comment.