GATE CSE 2014 | Set 1 | Question: 21

Question

Consider the relation scheme $R = (E, F, G, H, I, J, K, L, M, N)$ and the set of functional dependencies $$\left\{ \{E, F \} \to \{G\}, \{F\} \to \{I, J\}, \{E, H\} \to \{K, L\}, \\ \{K\} \to \{M\}, \{L\} \to \{N\}\right\}$$ on $R$. What is the key for $R$?

• A. $\{E, F\}$
• B. $\{E, F, H\}$
• C. $\{E, F, H, K, L\}$
• D. $\{E\}$

Comments

check by putting the options

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Option B is correct

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A simple analogy to understand why Option C is incorrect and B is correct,

Both of them are able to derive all the attributes i agree on this completely but the question specifically asked for key which mean candidate key.You can think of it like candidate key is the minimal and optimal way of doing things.

The option C is correct but its more of a brute force method , where as key is like a optimal approach for it.

Answer 1

Since $E,F,H$ cannot be derived from anything else $E,F,H$ should be there in key.

Using Find $\{EFH\}^+,$ it contains all the attributes of the relation.

Hence, it is key.

Correct Answer: $B$

Comments

Really awesome way of finding keys. Must watch. Check this

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Sir EFH must be in the key

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$\{E,F,H\}$ can be determine every attributes of the relation, but $\{E,F,H\}$ cannot be determined by any other attributes,

hence, $\{E,F,H\}$ is the primary key

Answer 2

A) {EF}+ = {EFGIJ} ≠ R(The given relation)

B) {EFH}+ = {EFGHIJKLMN} = R (Correct since each member of the
                                    given relation is determined)

C) {EFHKL}+ = {EFGHIJKLMN} = R (Not correct although each member
                                of the given relation can be determined
                                but it is not minimal, since by the definition
                                of Candidate key it should be minimal Super Key)

 D) {E}+ = {E} ≠ R

Answer 3

Any question related to functional dependencies can be solved by a simple method:

just look at the right side of all functional dependencies and note which attributes are not present at the right-hand side. Then the candidate key is definitely going to contain them because they can't be derived from the other.Now find out the closure and check the options.

If They are talking about key, they mean to say CANDIDATE KEY

for this question:- EFH is not present at right side so definitely the candidate key is going to contain them.

closure  (EFH) =  EFGHIJKLMN

So The correct option is B.

Answer 4

this is my answer.

Comments

m confused between commas in LHS part .{E,F,H} why we are taking as whole {EFH} ?

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CAN WE APPLY UNION RULE FOR COMBINING {E,F,H}

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my solution is based on given option that's why I take EFH as the combination. take a closer set of EFH you get all the attribute.