Solution

The correct answer is Option 2) EC

Key Points

• Relation: R(A, B, C, D, E)
• Functional Dependencies (FDs): C → F, E → A, EC → D, A → B
• Note: Attribute F is not in R. For finding a key of R, we only need to determine {A, B, C, D, E}; the FD C → F does not help us reach any new attribute of R and can be ignored for key finding.
• A key (candidate key) is a minimal set of attributes whose closure gives all attributes of the relation.

Explanation (Attribute-Closure Method):

We compute the closure of each option using the given FDs and check if it gives {A, B, C, D, E}.

1. Option 1) CD

   • Start: CD⁺ = {C, D}
   • From C → F we can add F (outside R, not useful).
   • No FD gives E or A or B from C or D.
   • CD⁺ = {C, D} (not all attributes) ⇒ Not a key.

2. Option 2) EC

   • Start: EC⁺ = {E, C}
   • From E → A add A.
   • From A → B add B.
   • From EC → D add D.
   • Now we have {A, B, C, D, E} ⇒ covers all attributes of R.
   • Check minimality:
     • E⁺ = {E, A, B} (missing C, D) ⇒ E alone is not a key.
     • C⁺ = {C} plus F outside R ⇒ C alone is not a key.
     Hence, EC is minimal.
   • Therefore, EC is a candidate key.

3. Option 3) AE

   • Start: AE⁺ = {A, E}
   • From A → B add B.
   • We still lack C and D. No FD gives C or D from {A, E} (since EC → D needs C too).
   • AE⁺ = {A, E, B} ⇒ Not a key.

4. Option 4) AC

   • Start: AC⁺ = {A, C}
   • From A → B add B.
   • No way to get E; without E, cannot use EC → D to reach D.
   • AC⁺ = {A, C, B} ⇒ Not a key.

Important Points

• Attribute Closure: Repeatedly add attributes implied by FDs until no new attribute can be added. If the closure equals the whole schema, the starting set is a superkey. If none of its proper subsets is a superkey, it is a candidate key.
• Ignoring outside attributes: FDs that only produce attributes not in the relation (here, F) do not affect whether a set is a key for R.
• Minimality check: After you find a