GATE CSE 2005 | Question: 78

Question

Consider a relation scheme $R = (A, B, C, D, E, H)$ on which the following functional dependencies hold: {$A \rightarrow B$, $BC \rightarrow D$, $E \rightarrow C$, $D \rightarrow A$}. What are the candidate keys R?

• A. $\text{AE, BE}$
• B. $\text{AE, BE, DE}$
• C. $\text{AEH, BEH, BCH}$
• D. $\text{AEH, BEH, DEH}$

Comments

Detailed Video Solution: https://youtu.be/l8OLJIw8Dq0?t=252&feature=shared

Normalization in DBMS Complete Playlist: https://youtube.com/playlist?list=PLIPZ2_p3RNHhJjQGnZB2jORa0JxLdMp4X&feature=shared

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H is not present in any of the FDs so, it must be included in CK , ----> option A, B eliminated
b/w  option C , D;   AEH & BEH are common so just check for last CK in both

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hey aspirants. here is the fastest trick. whatever attributes which are not in the rhs of all the fds, they must be part of all the candidate keys because they can't be determined by anyone else except themselves. in this question only EH are part of none of the fd's rhs. so you can eliminated all the options except D.

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Check in FD's right side which is not present like here

 EH is not present any one FD

So EH will present in every candidate keys (directly we cay say without solving)

Hence Option D is Correct

Answer 1

E, H → compulsory
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Choose A / B / D → completes the relation