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1 vote
1 vote

a

in Databases retagged by
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yes, sandeep  PQ -> R must be there,    but It can be merged with PQ -> T

so, it becomes PQ -> RT
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yes, sorry , i read wrong fds , i solved PQ-->P , :P
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Then what's the ans 4 or 5?
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4 Answers

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Best answer

Answer will be 5 here .... 

here you will got very nice explaination to find minimal cover... go through this steps and answer will be 5 only.

https://gateoverflow.in/11276/find-minimal-cover-from-given-functional-dependencies

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Nonsense question

Minimal covers are not unique and minimal cover never means that no of FDs will be least.It just means that the obtained FD sets are irreducible
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no this is wrong answer =>only thing we can do here is apply union rule last

PQ->R &  PQ->T    ==== PQ->RT

answer is 4 but nothing is extraneous here
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The given FD’s are itself the minimal cover, as we don't have any extraneous attributes. so can’t remove any FD from it. 

Just apply the union rule :

To compute a canonical cover for F: repeat Use the union rule to replace any dependencies in F α1 → β1 and α1 → β2 with α1 → β1 β2

 If we apply union rule we get answer 4. otherwise answer will be 5.