# Questions on Super Keys and Candidate Keys

## Questions on Super Keys and Candidate Keys using Closure

Identify Super Keys and Candidate keys :

```Question 1 :
Let R(ABCDE) is a relational schema, where
(AB)+ = ABCDE
(A)+ = ABCDE
Is AB: Candidate Key or Not??```
```Solution :
AB : Not a Candidate Key, AB is only : Super Key```
```Question 2 :
Let R(ABCDE) is a relational Schema having FDs
{AB→C, C→D, B→E}
Find out the Candidate Key ?```
```Solution :
(AB+) : {ABCDE}   ⇒ super key
(A+) : {A}   ×
(B+) : {EB}   ×
∴ AB : minimal superkey ⇒ Candidate Key. No subset of its attributes is a key.```
```Question 3 :
Let R(ABCDE) is a relational schema having FDs
{AB→C , C→D, B→EA}
Find Out the Candidate Key ?```
```Solution :
(AB+) : {ABCDE}   ⇒Superkey
(A+)  : {A}
(B+)  :  {EABCD}  ⇒ Superkey
⇒ B is Candidate Key.```
```Question 4 :
Let R(ABCDE) is a relational schema having FDs
{A→B, B→C, C→D}
Find out the Candidate Key ?```
```Solution :
(AE+) :{ABCDE} ⇒ SuperKey
(A+) : {ABCD}
(E+) : {E}
AE : Candidate Key. No subset of its attributes is a key.```
```Question 5 :
Let R(ABCDEF) is a relational schema having FDs
Find out the Candidate Key ?```
```Solution:
(A)+: ABCDEF  ⇒ (SuperKey)
(C)+ : {C}
{A, B} ⇐ Candidate Key. No subset of its attributes is a key.```
```Question 6:
Given the following set F of functional dependencies for relation schema R = {A, B, C, D, E}.
{A -> BC, CD -> E, B -> D, E -> A}
List the candidate keys for R.```
```Solution :
(A)+: ABCDE  ⇒ (SuperKey)
(E)+: EABCD  ⇒ (SuperKey)
(CD)+: CDEAB  ⇒ (SuperKey)
(CB)+: CBDEA  ⇒ (SuperKey)
Any combination of attributes that includes those is a superkey.
From above , the minimal super keys are  ⇒  A, E, CD and BC.
Hence, the candidate keys are A, E, CD, BC.```
```Question 7:
Consider a relation R(A,B,C,D,E) with the following dependencies:
{AB-> C, CD -> E, DE -> B}
Is AB a candidate key of this relation? If not, is ABD? Explain your answer.```
```No. The closure of AB does not give you all of the attributes of the relation.
For ABD,
(ABD)+ = ABDCE ⇒ Super Key
(A) = {A}
(B) = {B}
(D) = {D}
⇒ ABD is a candidate key. No subset of its attributes is a key.```
```Question 8 :
Consider a relation with schema R(A,B,C,D) and FDs {AB -> C, C -> D, D -> A}. What are all candidate keys of R?```
```(AB)+ : ABCD  ⇒ (SuperKey)
(A)+  : A     ⇒ (Not able to determine all the attributes)
(B)+  : B     ⇒ (not able to determine all the attributes)
(DB)+ : ABCD  ⇒ (SuperKey)
(CB)+ : ABCD  ⇒ (SuperKey)
(D)+  : DA    ⇒ (Not able to determine all the attributes)
(C)+  : CDA   ⇒ (Not able to determine all the attributes)

⇒ By calculating an attribute closure we can see the candidate keys are: AB, BC, and BD.```

### Incoming search terms:

• candidate key numaericals in dbms
• questions on keys in dbms
• candidate key 2013 gate questions
• super key and candidate keys proble
• super key problems