MADE EASY TEST SERIES

Question

Let F = {PQ → R, PR→ Q, Q → S, QR → P, PQ → T} where F is set of FD’s of a relation R(PQRST). The number of functional dependencies are in minimal cover determine single attribute of F is ________.

Comments

whats is your doubt?

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Language of this question is not clear to me. What are they asking us to find? If number of functional dependencies in minimal cover of F has been asked then according to me correct answer should be 4 and NOT 5 (as provided by MADE EASY). Please correct me if I am wrong.

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question asked find minimal cover then give number of Functional dependenct left . initialy 5.

How you get 4 [ which one you removed ]

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@Anirudh I think PQ-> R and PQ->T can be merged to form PQ-> RT...

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Prajwal  PQ-> RT are two depndency ryt PQ-> R and PQ->T  ?

A->BCDE not one dependency ryt ?

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@Anirudh I am not sure we need to count it as one FD or 2 .!

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now we can see PQ -> R is and PQ-> T . PQ indivisualy identify R and T not RT as both.

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PQ --> R and PQ --> T can be merged together as PQ --> RT. Please correct me if I am wrong.

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@Anirudh Check the page no.5 here http://www.inf.usi.ch/faculty/soule/teaching/2014-spring/cover.pdf

which says

 Simplifying an FD by the Union Rule Let X, Y , and Z be sets of attributes. If X → Y and X → Z, then X → Y Z

Answer 1

If the question is to find minimal cover of F = {PQ → R, PR→ Q, Q → S, QR → P, PQ → T} then it would be 

• No Trivial dependency
• Simplifying an FD by the Union Rule

PQ-> R and PQ->T can be merged to form PQ-> RT

• RHS Simplification and LHS Simplification is not possible here

Finally we are left with F = {PQ → RT, PR→ Q, Q → S, QR → P} 

But question says The number of functional dependencies are in minimal cover determine single attribute of F is 

In that case the given FD itself is in minimal form F = {PQ → R, PR→ Q, Q → S, QR → P, PQ → T}

Thus there will be 5 FD's in Minimal cover

Ref: http://www.inf.usi.ch/faculty/soule/teaching/2014-spring/cover.pdf

Comments

where written AB -->CD is one FD. they apply union rule .

Answer 2

Huluhulu test series

Huluhulu question.

Answer 3

A minimal cover, also known as an irreducible set of functional dependencies (FDs), does not have a unique solution. The number of FDs in a minimal cover can vary, so different valid solutions may contain 3, 4, or 5 FDs, depending on how the cover is derived. If a set of FDs cannot be further reduced and remains irreducible, it is considered a minimal cover.

Therefore, the number of FDs in a minimal cover cannot be determined.