##
**Set of solved exercises in
Normalization / Normalization Solved Examples / How to find candidate keys, and
primary keys in database? / Sets of examples to find the keys of a tables /
Process of Key finding in a database – Examples**

__Question:__
Consider the relation scheme R
= (E, F, G, H, I, J, K, L, M, N) and the set of functional dependencies {{E, F}
→ {G}, {F}
→ {I, J}, {E,
H} → {K, L} → {M}, {K}
→ {M}, {L} → {N}} on
R. What is the key for R?

**Solution:****, we need to find the closure of attributes. If any attribute’s or set of attributes’ closure gives all the attributes of the relation, then we would say that attribute/set of attributes as the key for that relation.**

*To find the key of a relation*
To simplify this task or to avoid
finding closure of all attributes, let us do find the closure for left hand
side (LHS) attributes of the functional dependencies.

For the given question,
attributes F, (E, F), (E, H), K, and L are the LHS attributes.

- The closure of F, i.e., (F)
^{+}= FIJ ≠ EFGHIJKLMN. Hence F is not a candidate key. [Refer, How to find closure]

- (EF)
^{+}= EFGIJ ≠ EFGHIJKLMN. Hence EF is not a candidate key.

- (EH)
^{+}= EHKLMN ≠ EFGHIJKLMN. Hence EH is not a candidate key.

- (K)
^{+}= KM ≠ EFGHIJKLMN. Hence, K is not a candidate key.

We observed that the left side
attributes of any FDs alone cannot form a key. Let us try with the combination
of two FDs LHS.

- (EFH)
^{+}= EFGHIJKLMN = R.

Hence,

*.***EFH is the key for this relation**
[

__Derivation of EFH+__:

Step 1: result = EFH

Step 2: result = EFGH from {E, F}
→ {G}

Step 3: result = EFGHIJ from {F} → {I, J}

Step 4: result = EFGHIJKLM from {E,
H} → {K, L} → {M}

Step 5: result = EFGHIJKLMN from
{L} → {N}

We have reached that the result =
R. hence, EFH is the key.

]

Go back to Normalization – solved exercises page.

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