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

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

---

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

---

@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 .