Wednesday, July 12, 2017

Calculate the key for relation R in DBMS

Find the candidate keys of a relation, How to find the candidate keys, Which is the key for the given table, concept of candidate key in dbms, candidate key examples

Question:
Consider the relation R = {A, B, C, D, E, F, G, H, I, J} and the set of functional dependencies F as follows;
F = { AB C, BD EF, AD GH, A I, H J}.
Find the key of relation R.

Solution:
We can find keys (candidate keys) for a relation by finding the closure of an/set of attributes. Checking each attribute or all subsets of the given set of attributes for a key is time consuming job. Hence, we may employ some of the following heuristics/assumptions in identifying the keys;

  • We may start checking all the left hand side attributes of any/all of the given set of functional dependencies. We can start with single LHS attributes.
  • If we find the closure of an attribute and that attribute is the candidate key then any superset cannot be the candidate key. For example, if A is a candidate key, then AB is not a candidate key but a super key.

LHS
Result
Description
A+
= AI from the functional dependency A → I (By reflexivity rule)
No more functional dependencies that has either of the attributes AI in LHS. Hence, A+ = AI.
A+ ≠ R.
So, A is not a candidate key.
H+
= HJ from H → J (By reflexivity rule)
No more functional dependencies that has either of the attributes HJ in LHS. Hence, H+ = HJ.
Note: We need not find H+ since H is available on RHS in AD → GH.
H+ ≠ R.
So, H cannot be a candidate key.
(AB)+
= ABC from AB C
= ABCI from A → I
No more functional dependencies that has either of the attributes ABCI in LHS. Hence, (AB)+ = ABCI.
(AB)+ ≠ R.
So, (AB) is not a candidate key.
(AD)+
= ADGH from AD → GH (By reflexivity rule)
= ADGHI from A → I (By union rule. AD → GH and A → I, so AD → GHI)
= ADGHIJ from H → J
No more functional dependencies that has either of the attributes ADGHIJ in LHS. Hence, (AD)+ = ADGHIJ.
(AD)+ ≠ R.
So, (AD) is not a candidate key.
(BD)+
= BDEF from BD → EF
No more functional dependencies that has either of the attributes BDEF in LHS. Hence, (BD)+ = BDEF.
(BD)+ ≠ R.
So, (BD) is not a candidate key.
(ABD)+
= ABDGH (By reflexivity and augmentation. That is, AD → GH and if we augment B on both sides we get ABD → BGH. Hence, ABDGH.)
= ABDGHIJ from (AD)+ - refer above.
= ABCDGHIJ from AB → C
= ABCDEFGHIJ from BD → EF.
(ABD)+ = R.
Hence, (ABD) is the candidate key.

Hint: The left hand side only attribute will definitely be the key or part of the key.

***************


Go to - 1NF,    2NF,    3NF,    BCNF




Find the candidate keys of a database relation

How to find the candidate keys of a relation if functional dependencies given

 

No comments:

Post a Comment

Featured Content

Multiple choice questions in Natural Language Processing Home

MCQ in Natural Language Processing, Quiz questions with answers in NLP, Top interview questions in NLP with answers Multiple Choice Que...

All time most popular contents

data recovery