# Database Design(Normal Forms)

Question 1 |

Relation R has eight attributes ABCDEFGH. Fields of R contain only atomic values. F = {CH -> G, A -> BC, B -> CFH, E -> A, F -> EG} is a set of functional dependencies (FDs) so that F+ is exactly the set of FDs that hold for R.
How many candidate keys does the relation R have?

3 | |

4 | |

5 | |

6 |

**Database Design(Normal Forms)**

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Question 1 Explanation:

A+ is ABCEFGH which is all attributes except D.
B+ is also ABCEFGH which is all attributes except D.
E+ is also ABCEFGH which is all attributes except D.
F+ is also ABCEFGH which is all attributes except D.
So there are total 4 candidate keys AD, BD, ED and FD

Question 2 |

Relation R has eight attributes ABCDEFGH. Fields of R contain only atomic values. F = {CH -> G, A -> BC, B -> CFH, E -> A, F -> EG} is a set of functional dependencies (FDs) so that F+ is exactly the set of FDs that hold for R.
How many candidate keys does the relation R have?

3 | |

4 | |

5 | |

6 |

**Database Design(Normal Forms)**

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Question 2 Explanation:

A+ is ABCEFGH which is all attributes except D.
B+ is also ABCEFGH which is all attributes except D.
E+ is also ABCEFGH which is all attributes except D.
F+ is also ABCEFGH which is all attributes except D.
So there are total 4 candidate keys AD, BD, ED and FD

Question 3 |

Consider the FDs given in above question. The relation R is

in 1NF, but not in 2NF. | |

in 2NF, but not in 3NF. | |

in 3NF, but not in BCNF. | |

in BCNF |

**GATE CS 2013**

**Database Design(Normal Forms)**

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Question 3 Explanation:

The table is not in 2nd Normal Form as the non-prime attributes are dependent on subsets of candidate keys.
The candidate keys are AD, BD, ED and FD. In all of the following FDs, the non-prime attributes are dependent on a partial candidate key.
A -> BC
B -> CFH
F -> EG

Question 4 |

Which of the following is

**TRUE**?Every relation in 3NF is also in BCNF | |

A relation R is in 3NF if every non-prime attribute of R is fully functionally dependent on every key of R | |

Every relation in BCNF is also in 3NF | |

No relation can be in both BCNF and 3NF |

**GATE CS 2012**

**Database Design(Normal Forms)**

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Question 5 |

Consider a relational table with a single record for each registered student with the following attributes.

1.Which one of the following option isRegistration_Num: Unique registration number of each registered student 2.UID: Unique identity number, unique at the national level for each citizen 3.BankAccount_Num: Unique account number at the bank. A student can have multiple accounts or join accounts. This attribute stores the primary account number. 4.Name: Name of the student 5.Hostel_Room: Room number of the hostel

**INCORRECT**?BankAccount_Num is candidate key | |

Registration_Num can be a primary key | |

UID is candidate key if all students are from the same country | |

If S is a superkey such that S∩UID is NULL then S∪UID is also a superkey |

**GATE CS 2011**

**Database Design(Normal Forms)**

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Question 5 Explanation:

A Candidate Key value must uniquely identify the corresponding row in table. BankAccount_Number is not a candidate key. As per the question “A student can have multiple accounts or joint accounts. This attributes stores the primary account number”. If two students have a joint account and if the joint account is their primary account, then BankAccount_Number value cannot uniquely identify a row.

Question 6 |

Consider the following relational schema:

Assume that, in the suppliers relation above, each supplier and each street within a city has a unique name, and (sname, city) forms a candidate key. No other functional dependencies are implied other than those implied by primary and candidate keys. Which one of the following is TRUE about the above schema?Suppliers(sid:integer, sname:string, city:string, street:string) Parts(pid:integer, pname:string, color:string) Catalog(sid:integer, pid:integer, cost:real)

The schema is in BCNF | |

The schema is in 3NF but not in BCNF | |

The schema is in 2NF but not in 3NF | |

The schema is not in 2NF |

**GATE-CS-2009**

**Database Design(Normal Forms)**

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Question 6 Explanation:

A relation is in BCNF if for every one of its dependencies X → Y, at least one of the following conditions hold:

X → Y is a trivial functional dependency (Y ⊆ X) X is a superkey for schema RSince (sname, city) forms a candidate key, there is no non-tirvial dependency X → Y where X is not a superkey

Question 7 |

Consider the following relational schemes for a library database:
Book (Title, Author, Catalog_no, Publisher, Year, Price)
Collection (Title, Author, Catalog_no)
with in the following functional dependencies:

I. Title Author --> Catalog_no II. Catalog_no --> Title, Author, Publisher, Year III. Publisher Title Year --> PriceAssume {Author, Title} is the key for both schemes. Which of the following statements is true?

Both Book and Collection are in BCNF | |

Both Book and Collection are in 3NF only | |

Book is in 2NF and Collection is in 3NF | |

Both Book and Collection are in 2NF only |

**Database Design(Normal Forms)**

**GATE CS 2008**

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Question 7 Explanation:

Book (Title, Author, Catalog_no, Publisher, Year, Price) Collection (Title, Author, Catalog_no)with in the following functional dependencies:

I. Title, Author --> Catalog_no II. Catalog_no --> Title, Author, Publisher, Year III. Publisher, Title, Year --> Price Assume {Author, Title} is the key for both schemes

- The table "Collection" is in BCNF as there is only one functional dependency “Title Author –> Catalog_no” and {Author, Title} is key for collection.
- Book is not in BCNF because Catalog_no is not a key and there is a functional dependency “Catalog_no –> Title Author Publisher Year”.
- Book is not in 3NF because non-prime attributes (Publisher Year) are transitively dependent on key [Title, Author].
- Book is in 2NF because every non-prime attribute of the table is either dependent on the whole of a candidate key [Title, Author], or on another non prime attribute. In table book, candidate keys are {Title, Author} and {Catalog_no}. In table Book, non-prime attributes (attributes that do not occur in any candidate key) are Publisher, Year and Prince

Question 8 |

Consider the relation scheme R = {E, F, G, H, I, J, K, L, M, M} and the set of functional dependencies {{E, F} -> {G}, {F} -> {I, J}, {E, H} -> {K, L}, K -> {M}, L -> {N} on R. What is the key for R?

{E, F} | |

{E, F, H} | |

{E, F, H, K, L} | |

{E} |

**Database Design(Normal Forms)**

**GATE-CS-2014-(Set-1)**

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Question 8 Explanation:

All attributes can be derived from {E, F, H}
To solve these kind of questions that are frequently asked in GATE paper, try to solve it by using shortcuts so that enough amount of time can be saved.

**Fist Method:**Using the given options try to obtain closure of each options. The solution is the one that contains R and also minimal Super Key, i.e Candidate Key.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

**Second Method:**Since, {EFGHIJKLMN}+ = {EFGHIJKLMN} {EFGHIJKLM}+ = {EFGHIJKLMN} ( Since L -> {N}, hence can replace N by L) In a similar way K -> {M} hence replace M by K {EFGHIJKL}+ = {EFGHIJKLMN} Again {EFGHIJ}+ = {EFGHIJKLMN} (Since {E, H} -> {K, L}, hence replace KL by EH) {EFGH}+ = {EFGHIJKLMN} (Since {F} -> {I, J} ) {EFH}+ = {EFGHIJKLMN} (Since {E, F} -> {G} )This explanation is contributed by

**Manish Rai.**Learn more here: Finding Attribute Closure and Candidate Keys using Functional DependenciesQuestion 9 |

Given the following two statements:

S1: Every table with two single-valued attributes is in 1NF, 2NF, 3NF and BCNF. S2: AB->C, D->E, E->C is a minimal cover for the set of functional dependencies AB->C, D->E, AB->E, E->C.Which one of the following is CORRECT?

S1 is TRUE and S2 is FALSE. | |

Both S1 and S2 are TRUE. | |

S1 is FALSE and S2 is TRUE. | |

Both S1 and S2 are FALSE. |

**Database Design(Normal Forms)**

**GATE-CS-2014-(Set-1)**

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Question 9 Explanation:

S1: Every table with two single-valued attributes is in 1NF, 2NF, 3NF and BCNF.A relational schema R is in BCNF iff in Every non-trivial Functional Dependency X->Y, X is Super Key. If we can prove the relation is in BCNF then by default it would be in 1NF, 2NF, 3NF also. Let R(AB) be a two attribute relation, then

- If {A->B} exists then BCNF since {A}+ = AB = R
- If {B->A} exists then BCNF since {B}+ = AB = R
- If {A->B,B->A} exists then BCNF since A and B both are Super Key now.
- If {No non trivial Functional Dependency} then default BCNF.

**Hence S1 is true.**

S2: AB->C, D->E, E->C is a minimal cover for the set of functional dependencies AB->C, D->E, AB->E, E->C.As we know Minimal Cover is the process of eliminating redundant Functional Dependencies and Extraneous attributes in Functional Dependency Set. So each dependency of F = {AB->C, D->E, AB->E, E->C} should be implied in minimal cover. As we can see AB->E is not covered in minimal cover since {AB}+ = ABC in the given cover {AB->C, D->E, E->C}

**Hence, S2 is false.**This explanation has been contributed by

**Manish Rai.**

**Learn more about Normal forms here:**Database Normalization | Introduction Database Normalization | Normal Forms

Question 10 |

The maximum number of superkeys for the relation schema R(E,F,G,H) with E as the key is

5 | |

6 | |

7 | |

8 |

**Database Design(Normal Forms)**

**GATE-CS-2014-(Set-2)**

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Question 10 Explanation:

Maximum no. of possible superkeys for a table with n attributes = 2^(n-1)
Here, n = 4.
So, the possible superkeys = 2

^{4-1}= 8 The possible superkeys are : E, EH, EG, EF, EGH, EFH, EFG, EFGHThere are 85 questions to complete.