Question 1 : Find all the possible candidate keys

Relation (A, B, C, D, E)

- Below are the functional dependencies
- A –> C
- C –> D
- D –> B
- E –> F

- Calculating Closure
- Items which are not present in right side of Functional dependencies i.e A and E we will start with that, if A and E if not enough to find all the elements of the relation then we will add new attributes to find candidate keys, if A and E by them self can find all the attributes of the relation then we can say that we have only 1 candidate key
- (AE)+ = AECDBF
- So AE is the only candidate key for this relation

Question 2 : Find all the possible candidate keys

Relation (A, B , C , D, E, F, G, H)

- Functional Dependencies
- CH –> G
- A –> BC
- B –> CFH
- E –> A
- F –> EG

Calculating Closure

- Items which are not present in right side of Functional dependencies is D, we will find D close and see if it is a candidate key, if not then we will keep adding one element at a time from the left side of functional dependencies to create candidate keys
- (D)+ = D –> D Closure Not a candidate key
- (DA)+ = DABCFHEG –> DA Closure is a candidate key
- (DB)+ = DBCFHGEA –> DB Closure is a candidate key
- (DC)+ = DC –> DC Closure not a candidate key
- (DE)+ = DEA –> Since DA is already a candidate key hence DE closure is a candidate key
- (DF)+ = DFEG –> Since DE is already a candidate key hence DF closure is a candidate key

- So all the candidate keys are
- DA, DB, DE, DF

- Calculating prime and not prime attributes
- Prime Attributes : If a attribute is a part of any of the candidate key it is a prime attribute in above example : ABDEF are the prime attributes and CGH are non prime attributes
- So you want to put the relation in 2NF it would be like
- R1(A, B, D, E, F)
- R2(B ,C ,H ,G) SInce B –>CFH and CH–> G

- By Definition you are part of any of the candidates key then you are a prime-attribute, other attributes which are not part of prime-attribute in a relation you call it a non prime attribute

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