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.
Only D2
A decomposition of R into R1 and R2 is lossless-join if and only if the common attributes functionally determine all attributes of R1 or all attributes of R2: \(R_1 \cap R_2 \to R_1\) or \(R_1 \cap R_2 \to R_2\).
Step 1 — D1: \(R_1 \cap R_2 = \{A\}\). No functional dependencies are given, so A need not determine B or C. For example, R = {(1, 1, 1), (1, 2, 2)} splits into {(1, 1), (1, 2)} and {(1, 1), (1, 2)}, whose join has four tuples and creates the spurious tuples (1, 1, 2) and (1, 2, 1). So D1 is not always lossless and is false.
Step 2 — D2: \(R_1 \cap R_2 = \{A, D\}\). The dependency AD → B gives \(\{A, D\}^+ \supseteq \{A, B, D\}\), so the common attributes determine every attribute of R1(A, B, D). The test is satisfied, so D2 is lossless and true.
Since D1 is false and D2 is true, choices claiming both, neither, or only D1 are all wrong.
Only D2 option is correct.
Similar Questions
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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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Codes :
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Student(stuid, dept, clsroll)
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FD1: stuid → dept
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FD3: stuid → clsroll
FD4: clsroll → stuid
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A. FD2 and FD3
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Choose the correct answer from the options given below:
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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
