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Recent questions tagged databases

0 votes
0 answers 10 views
Consider the following schedule. $T1$ $T2$ $T3$ $R(Y)$ $R(X)$ $R(X)$ $W(Y)$ $R(Z)$ $W(Z)$ $R(X)$ denotes the read operation on data item $X$ by transaction $T_{i}$. W(X) denotes the write operation on data item $X$ by transaction $T_{i}$.how many conflict serializable schedule are possible for the above schedule? 6 5 3
asked 18 hours ago in Databases karanvashisht1026 7 points 10 views
0 votes
2 answers 9 views
Consider the relation employee(name,sex,manager) with name as the key, manager gives the name of the supervisor of the employee under consideration. What does the following Tuple Relational Calculus query produce? A. Names of employees who have a female subordinate B. Names of ... one female manager How to interpret this TRC expression? The y.name = e.manager =>x.sex=female part is troublesome.
asked 2 days ago in Databases Sambhrant Maurya 204 points 9 views
0 votes
0 answers 4 views
In translating from an entity-relationship (E-R) diagram to a relational schema, one piece of E-R logic that cannot be captured by primary keys, uniques, and foreign keys is (A) the weak entity. (B) any ternary relationship. (C) mandatory participation for one-time occurrence (that is, with the arrow). (D) mandatory participation for many-time occurrence (that is, without the arrow).
asked 2 days ago in Databases Sambhrant Maurya 204 points 4 views
0 votes
0 answers 4 views
Consider the following entity-relationship (ER) model For the following group of statements, which one does not coincide (Or not true) with the model ? (A) Each person has at most one e-mail address. (B) Each e-mail address belongs to exactly one person. ... 1 passport number? eg I can assign 3 passport numbers to the same person Pinkman. Passport Number Name 101 Pinkman 102 Pinkman 103 Pinkman
asked 2 days ago in Databases Sambhrant Maurya 204 points 4 views
0 votes
1 answer 7 views
Consider the following table Mystery A B 1 null 2 4 3 5 null 6 select count() from Mystery where A not in( select B from Mystery) What is the output of this query? (A) 1 (B) 0 (C) 4 (D) 3
asked 2 days ago in Databases Sambhrant Maurya 204 points 7 views
0 votes
1 answer 5 views
im getting 5, but answer is not 5
asked 3 days ago in Databases ummokkate 45 points 5 views
0 votes
0 answers 4 views
Can someone explain me group by and having clause in a clear way.
asked 4 days ago in Databases Priyansh Singh 12 points 4 views
1 vote
1 answer 44 views
What is the minimum number of tables needed in the relational model? 2 3 4 5
asked Sep 14 in Databases Sambhrant Maurya 204 points 44 views
0 votes
1 answer 19 views
Consider a relation R(A,B) where A is the primary key and B is a foreign key which references the same relation. Consider the following relation instance of R. A 2 3 5 7 4 3 B 4 4 2 2 5 4 Which of the following tuple to be deleted first to preserve the referential integrity constraint when the tuple (2,4) is deleted? (2,4) (5,2) (7,2) (4,5) p.s The obvious answer (5,2) is given wrong.
asked Sep 14 in Databases Sambhrant Maurya 204 points 19 views
0 votes
0 answers 2 views
'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
asked Sep 14 in Databases Sambhrant Maurya 204 points 2 views
0 votes
0 answers 4 views
Is there any difference between lossless decomposition and join dependency or both are same. both definition has same criteria. https://en.wikipedia.org/wiki/Join_dependency https://en.wikipedia.org/wiki/Lossless-Join_Decomposition Thanks
asked Sep 13 in Databases piyushhurpade 6 points 4 views
0 votes
0 answers 6 views
ENo SSN 1 15 2 20 3 15 4 null 5 null Relation R SELECT SSN,count(*) FROM R GROUP BY SSN Will the output be: SSN count 15 2 20 1 null 2 SSN count 15 2 20 1 null 1 null 1 SSN count 15 2 20 1
asked Sep 12 in Databases Sambhrant Maurya 204 points 6 views
0 votes
1 answer 9 views
Consider the following ER diagram. Assume that there are no multivalued attributes present in any of E1,E2,R and each entity set has a key. What is the minimum number of tables required to convert this ER diagram into relations when: (i) E1 has total participation in R (ii) E2 has total participation in R (iii) Both E1 and E2 have total participation in R
asked Sep 12 in Databases Sambhrant Maurya 204 points 9 views
0 votes
0 answers 12 views
What do you mean by Seek? For Natural Loop Join, we calculate No. of Seeks are (nr + br). Assume that, Relation r is called the Outer Relation and Relation s is called the Inner Relation. (ni,bj) – ni represents total no. of tuples and bj represents no. of blocks in that relation. Reference (Korth) – For Block Nested-Loop Join –
asked Sep 10 in Databases ayushsomani 13 points 12 views
0 votes
0 answers 10 views
From this link How The last query is safe?
asked Sep 10 in Databases Lakshman Patel RJIT 84 points 10 views
0 votes
0 answers 6 views
what does this statement means? “In TO the schedule is equivalent to the particular serial order corresponding to the order of the transaction timestamps”
asked Sep 7 in Databases aditi19 29 points 6 views
0 votes
1 answer 10 views
I In the above question whether my solution is correct ? If not can anyone please explain with an example
asked Sep 5 in Databases Manasa.M 19 points 10 views
0 votes
0 answers 15 views
0 votes
0 answers 20 views
Consider the following FDs : Sid → Sname Sname → Age The above set of FD is not - non trivial. I knw da definition of non - trivial in FD is that there should be no common element between X and Y attribute sets in the FD X → Y. The above set of FD is not non ... X and Y have something in common that is Sname. That is why the set of FDs is not non-trivial. Is my perception towards the FDs correct ?
asked Aug 19 in Databases user2525 1.5k points 20 views
0 votes
0 answers 5 views
Hi all. I have a slight doubt regarding checking equivalence of two given FD sets. The traditional method that I used to follow is that taking each FD from one set, and finding its closure on the other to check if one covers the another. This is the method I am ... if closure of each attribute matches then it's equivalent. I am not sure if this method is correct or not. Please help. Thank you
asked Aug 3 in Databases DukeThunders 132 points 5 views
0 votes
1 answer 9 views
Do cascadeless schedule allow committed blind writes ? Cascadeless schedule: A cascadeless schedule is one where, for each pair of transactions Ti and Tj such that Tj reads a data item previously written by Ti , the commit operation of Ti appears before the read operation ... W(x) W(x) Commit Abort If so, then this won't be recoverable.But cascadeless schedules come under recoverable schedules .
asked Jul 1 in Databases vk_9_1_9 6 points 9 views
0 votes
1 answer 11 views
Given an attribute X, another attribute Y is dependent on it, if for a given X (A) There are many Y values (B) There is only one value of Y (C) There is one or more Y values (D) There is none or one Y values
asked Jun 24 in Databases Prajna 15 points 11 views
0 votes
0 answers 12 views
Hi everyone, I had a little doubt in DBMS regarding BCNF. Can we say that in a relation if all the attributes are candidate keys, the relation will be always in BCNF? Taking a relation R(A,B,C,D) with FDs like A->B,B->C,C->D,D->A. Here all the atributes are CKs. The relation is in BCNF. Can it be said for all such relations? Thanks.
asked Jun 22 in Databases DukeThunders 132 points 12 views
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