FDs:
EMP_ID-EMP_COUNTRY
EMP_DEPT-DEPT_TYPE, EMP_DEPT_NO

What is the key of this relation:

R3=(EMP_DEPT,DEPT_TYPE,EMP_DEPT_NO)

Question

FDs:
EMP_ID->EMP_COUNTRY
EMP_DEPT->DEPT_TYPE, EMP_DEPT_NO

What is the key of this relation:

R3=(EMP_DEPT,DEPT_TYPE,EMP_DEPT_NO)

BlackTom AI · Accepted Answer

First, restate the problem context to ensure clarity: we are given FDs EMP_ID -> EMP_COUNTRY and EMP_DEPT -> DEPT_TYPE, EMP_DEPT_NO for a relation R3 with attributes (EMP_DEPT, DEPT_TYPE, EMP_DEPT_NO).

Now, consider the set of attributes in R3: {EMP_DEPT, DEPT_TYPE, EMP_DEPT_NO}.

Compute the closure of EMP_DEPT using the given functional dependency:
- From EMP_DEPT -> DEPT_TYPE, EMP_DEPT_NO, we can infer both DEPT_TYPE and EMP_DEPT_NO from EMP_DEPT.
- Therefore, EMP_DEPT -> {DEPT_TYPE, EMP_DEPT_NO} and also EMP_DEPT itself is included, so EMP_DEPT+ = {EMP_DEPT, DEPT_TYPE, EMP_DEPT_NO} = all attributes of R3.

Since the closure of EMP_DEPT contains all attributes of R3, EMP_DEPT is a superkey. Given that there are no other dependencies that would give all attributes from a smaller subset of attributes within R3, EMP_DEPT is a candidate key for R3.

Note on the question format: the provided data shows an empty list for answer_options, so there are no alternative options to compare here. The reasoning determines the key based on the given functional dependencies and the relation's attribute set.

Because there are no additional options to evaluate, there is no contrastive discussion of other possible keys within the available data. The analysis above demonstrates why EMP_DEPT functionally determines all attributes of R3, satisfying the criteria for a key.

HADM3740/HADM6740 Normalization Quiz

FDs: EMP_ID->EMP_COUNTRY EMP_DEPT->DEPT_TYPE, EMP_DEPT_NO What is the key of this relation: R3=(EMP_DEPT,DEPT_TYPE,EMP_DEPT_NO)

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Which of the following is NOT true about Candidate keys?

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