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:
A, C only
A canonical cover Fc of a set F of functional dependencies is a minimal set equivalent to F: it implies every dependency of F, and no dependency in it is redundant. A dependency X → Y in F is redundant if Y is still in the closure X+ computed from the remaining dependencies.
Step 1 — From F, each attribute determines the others (stuid ↔ dept, stuid ↔ clsroll), so any valid cover must still let each of stuid, dept and clsroll derive the other two.
Step 2 — A (remove FD2, FD3): left with stuid → dept, dept → clsroll, clsroll → stuid. This cycle gives every attribute the closure {stuid, dept, clsroll}, and dropping any one of the three breaks the cycle, so it is a canonical cover. Valid.
Step 3 — B (remove FD1, FD5): left with dept → stuid, stuid → clsroll, clsroll → stuid. Now stuid+ = {stuid, clsroll}; dept can no longer be derived from stuid, so the set is not equivalent to F. Invalid.
Step 4 — C (remove FD5): left with FD1–FD4. dept → stuid → clsroll still yields dept → clsroll, and none of the four can be removed without losing one direction, so it is a canonical cover. Valid.
Step 5 — D (remove FD1): stuid+ = {stuid, clsroll} again, so stuid → dept is lost. Invalid.
Only A and C give a minimal equivalent set; any choice containing B fails because removing FD1 together with FD5 loses stuid → dept.
A, C only option is correct.
Similar Questions
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(a) \(R_{1} \cap R_{2} \twoheadrightarrow (R_{1} - R_{2})\)
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(d) \(R_{2} \rightarrow R_{1} \cap R_{2}\)
Codes :
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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
