The relation schemas \(R_{1}\) and \(R_{2}\) form a Lossless join decomposition of R if and only if : (a) \(R_{1} \cap R_{2} \twoheadrightarrow (R_{1} - R_{2})\) (b) \(R_{1} \rightarrow R_{2}\) (c) \(R_{1} \cap R_{2} \twoheadrightarrow (R_{2} - R_{1})\) (d) \(R_{2} \rightarrow R_{1} \cap R_{2}\) Codes :
(a) and (c) happens
A decomposition of R into \(R_{1}\) and \(R_{2}\) is lossless exactly when the common attributes multidetermine the remaining attributes of one side.
By the complementation rule, \(R_{1} \cap R_{2} \twoheadrightarrow (R_{1} - R_{2})\) holds if and only if \(R_{1} \cap R_{2} \twoheadrightarrow (R_{2} - R_{1})\) holds, so these two conditions always occur together.
Conditions such as \(R_{1} \rightarrow R_{2}\) or \(R_{2} \rightarrow R_{1} \cap R_{2}\) are not the lossless-join test; the second is trivially true and the first is not required.
So the multivalued conditions on both differences describe the lossless join property.
Hence, (a) and (c) happens.
Similar Questions
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Student(stuid, dept, clsroll)
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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
