MadeEasy Test Series: Databases - Functional Dependencies

Question

Consider the following relation R(A, B, C, D, E, F, G) and set of functional dependencies.

F={BCD → A, BC → E, A → F, F → G, A→G, C → D}

Which of the following is minimal cover of F?

• A. {BC → A, BC → E, A → F, C → D, A → G}
• B. {BC → A, B → E, A → F, F → G, C → D}
• C. {BC → A, BC → E, A → F, F → G, C → D}
• D. {BC → A, BC → E, A → F, C → D}

Comments

got it @adarsh_1997 I thought that only minimal cover FDs implies Given FDs is needed,

minimal cover should cover given FDs and Given FDs should also cover minimal cover.

M i Right? 

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yes @Shivam Kasat thats right

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C is correct.

Answer 1

Given that $R(A,B,C,D,E,F,G)$ and

set of Functional dependencies$:F = \left\{BCD → A, BC→ E, A→ F, F→ G, C→ D, A→ G\right\}$

 

Comments

@Lakshman Patel RJIT

I am talking about Closure of (BC) when we remove BC gives E. (BCD)+ is correct.

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@Lakshman Patel RJIT

Can you find a quick method to solve this question - https://gateoverflow.in/116026/number-of-different-minimal-cover?

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@ayushsomani

Yes, you are right. But I write only which is required for answers.

see this

• https://stackoverflow.com/questions/10284004/minimal-cover-and-functional-dependencies
• https://www.inf.usi.ch/faculty/soule/teaching/2014-spring/cover.pdf
• https://dba.stackexchange.com/questions/214669/how-to-solve-multiple-minimal-covers-for-functional-dependencies
• http://www.mathcs.emory.edu/~cheung/Courses/377/Syllabus/9-NormalForms/FD-equi.html
• https://gateoverflow.in/6346/is-canonical-cover-and-minimal-cover-the-same-thing

Answer 2

first delete  D AND find  the closure of remaining, ie- (BC+) =  if you cand the D again then writing BCD WILL BE MORE COSLTY SO, SIMPLY WRITE IT LIKE- BC----->A

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Comments

Thanks  sir krmanish043

Answer 3

answer can't be B, it doesn't covers F

as in (B) it is given  B$\rightarrow$ E, but to derive E we need both B and C

$\therefore$ BC $\rightarrow$ E

Comments

Closure(BC)=BECD

Thus BC-->E in option B

M i wrong?

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ya, so the correct answer is C

{BC → A, BC → E, A → F, F → G, C → D}