Identify the minimal key for relational scheme R(A, B, C, D, E) with functional dependencies \(F = \{A \rightarrow B, B \rightarrow C, AC \rightarrow D\}\)
AE
A key is a set of attributes whose closure contains all attributes of R; a minimal key has no proper subset with this property.
E never appears on the right side of any dependency, and A appears on no right side either, so both A and E must be in every key.
\(AE^+\): A → B gives B, B → C gives C, AC → D gives D, so \(AE^+ = ABCDE\). Its subsets fail: \(A^+ = ABCD\) lacks E and \(E^+ = E\).
The other sets do not work: \(BE^+ = BCE\) and \(CE^+ = CE\) cannot reach A or D.
Hence, the answer is AE.
Similar Questions
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F = {AD → E, BE → F, B → C, AF → G}
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(a) \(R_{1} \cap R_{2} \twoheadrightarrow (R_{1} - R_{2})\)
(b) \(R_{1} \rightarrow R_{2}\)
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(d) \(R_{2} \rightarrow R_{1} \cap R_{2}\)
Codes :
Let R ={A, B, C, D, E, F} be a relation schema with the following dependencies
\(C \rightarrow F, E \rightarrow A, EC \rightarrow D, A \rightarrow B\)
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{P, Q, S} → T
R → Y
{P, Q} → V
{R, X} → X
Q → U
V→{W, X}
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Consider a relation schema R = (U, V, W, X, Y, Z), on which the following functional dependencies hold :
{U→V, VW→X, Y→W; X→U}
The candidate keys of R are :
Consider a schema R(P,Q,R,S) and the following functional dependencies P→Q, Q→R, R→S, S→Q.
Then decomposition of R1 (P,Q), R2 (Q,R) and R3 (Q,S) is :
Consider the relation
Student(stuid, dept, clsroll)
The following functional dependencies are given:
FD1: stuid → dept
FD2: dept → stuid
FD3: stuid → clsroll
FD4: clsroll → stuid
FD5: dept → clsroll
Which functional dependencies should be removed to obtain the canonical cover of the set?
A. FD2 and FD3
B. FD1 and FD5
C. FD5
D. FD1
Choose the correct answer from the options given below:
Consider a schema R(MNPQ) and functional dependencies M → N, P → Q. Then the decomposition of R into R1(MN) and R2(PQ) is ________.
Which of the following statements is TRUE ?
D1 : The decomposition of the schema R(A, B, C) into R1(A, B) and R2 (A, C) is always lossless.
D2 : The decomposition of the schema R(A, B, C, D, E) having AD → B, C → DE, B → AE and AE → C, into R1 (A, B, D) and R2 (A, C, D, E) is lossless.
Important Questions from Functional Dependency
Relation R has eight attributes ABCDEFGH. Fields of R contain only atomic values.
F = {CH → G, A → BC, B → CFH, E → A, F → EG} is a set of functional dependencies (FDs) so that F+ is exactly the set of FDs that hold for R.
The relation R is
- Consider the following database table having A, B, C and D as its four attributes and four possible candidate keys (I, II, III and IV) for this table :
A B C D $a_1$ $b_1$ $c_1$ $d_1$ $a_2$ $b_3$ $c_3$ $d_1$ $a_1$ $b_2$ $c_1$ $d_2$
I : {B} II : {B, C} III : {A, D} IV : {C, D}
If different symbols stand for different values in the table (e.g., $d_1$ is definitely not equal to $d_2$), then which of the above could not be the candidate key for the database table ? - Let $P, Q, R$ and $S$ be the attributes of a relation in a relational schema. Let $X \rightarrow Y$ indicate functional dependency in the context of a relational database, where $X, Y \subseteq \{P, Q, R, S\}$.
Which of the following options is/are always true? - In the context of schema normalization in relational DBMS, consider a set F of functional dependencies. The set of all functional dependencies implied by F is called the closure of F. To compute the closure of F, Armstrong's Axioms can be applied. Consider $X$, $Y$, and $Z$ as sets of attributes over a relational schema. The three rules of Armstrong's Axioms are described as follows.
Reflexivity: If $Y \subseteq X$, then $X \rightarrow Y$
Augmentation: If $X \rightarrow Y$, then $XZ \rightarrow YZ$ for any $Z$
Transitivity: If $X \rightarrow Y$ and $Y \rightarrow Z$, then $X \rightarrow Z$
The additional rule of Union is defined as follows.
Union: If $X \rightarrow Y$ and $X \rightarrow Z$, then $X \rightarrow YZ$
It can be proved that the additional rule of Union is also implied by the three rules of Armstrong's Axioms. Listed below are four combinations of these three rules. Which one of these combinations is both necessary and sufficient for the proof ? Let R = ABCDE is a relational scheme with functional dependency set F = {A → B, B → C, AC → D}. The attribute closures of A and E are
