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+15 votes
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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 $R$?

  1. $\{E, F\}$
  2. $\{E, F, H\}$
  3. $\{E, F, H, K, L\}$
  4. $\{E\}$
asked in Databases by Veteran (103k points) | 1k views

3 Answers

+23 votes
Best answer
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.
answered by Boss (11.5k points)
edited by
0
Question is about KEY. Its not mentioned that it should be a candidate key or a super key. Then how to determine what is the Answer, because option C is also a KEY coz it contains EFH too.
+8
if it is mentioned as key then its only candidate keys / primary key

a super key is any set of attribute that uniquely determines a tuple

whereas a key or candidate key is minimal super key

so option C is superkey. but set of attributes is not minimal

hence not a key
0
Thanks a lot... :)
+2
With H even E & F  cannot be derived from anything else so E,F,H need to be there in key!!
+1
Really awesome way of finding keys. Must watch. Check this
0
Sir EFH must be in the key
+3 votes

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

answered by Loyal (8.5k points)
+1 vote

this is my answer.

answered by Boss (13k points)
0
m confused between commas in LHS part .{E,F,H} why we are taking as whole {EFH} ?
0
CAN WE APPLY UNION RULE FOR COMBINING {E,F,H}
0
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.
Answer:

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