# Recent questions tagged superkeys

1 vote
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Consider the relation $R\left ( A,B,C,D,E \right )$ $A\rightarrow BC$ $C\rightarrow E$ $B\rightarrow D$ $E\rightarrow A$ Total number of superkeys present in relation will be ____________________
1 vote
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Consider the following relation R(A1, A2,...A15) with (A1,A2, ... A6) of relation R are simple candidate key. The number of possible superkey in relation R is_
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Consider the following relation: R (A1, A2, ….An) and every (n-2) attributes of R forms a candidate key. How many super keys are there in R?
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Maximum number of superkeys for the relation schema $R(X,Y,Z,W)$ with $X$ as the key is $6$ $8$ $9$ $12$
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Consider the relation R(A,B,C,D,E) with the functional dependencies : A → B, B → C, C → A, D → E, and E → D The Maximum possible super-keys of R is ______________?
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How many Super keys Possible for R(A,B,C,D,E) with 1. {A,BC,DE} as the keys ? 2. {A,BC,CDE} as the keys ?
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Relation R(A1, A2, A3 ..... An) n attributes. On this relation m simple candidate key (m ≤ n). How many super keys possible in this relation? A. 2m – 1 B. (2n – 1) ⋅ 2n-m C. 2n – 1 D. (2m – 1) 2n-m One way is by taking instances and solve it, but plz without it what's the logic behind the answer?
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Consider the following relational schema R(ABCDE). The number of super keys in relation R if every two attributes of relation R is candidate keys are __________. Please explain a general method to solve
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How to calculate super key of any relation? Pls explain with example.
1 vote
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R (A1, A2, ….An) and every (n-2) attributes of R forms a candidate key. How many super keys are there in R? a) $_{n-2}^{n}\textrm{C}$ b) $_{n-2}^{n}\textrm{C}$ * 4 c) $_{n-2}^{n}\textrm{C}$ * n d) $_{n-2}^{n}\textrm{C}$ + n +1
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a relation R(A,B,C,D,E,F,G,H) and set of functional dependencies are {CH->G,A->BC,B->CFH,E->A,F->EG } then how many possible super keys are present ? https://gateoverflow.in/42238/to-calculate-how-many-no-of-super-keys Is my approach right? Since candidate keys are (AD) ... ,ADF,BDE,BDF, EDF,AED,ADEF,ABDE,AFBE,AFBD,ABDEF}= 15 ways total = 15* 8 = 120 ways ....Can someone plz confirm it.... thank u
Consider a relation $R$ with $2n$ attributes. Assume any $2n/2$ of the attributes constitutes key. Total number of super keys possible are: $^{2n}C_{2^{n-1}}\times \left(2\right)^{2^{n-1}}$ $^{2n}C_{2^{n-1}} + \left(2\right)^{2^{n-1}}$ $^{2n}C_{2^{n-1}} - \left(2\right)^{2^{n-1}}$ $^{2n}C_{2^{n-1}} \div \left(2\right)^{2^{n-1}}$