Database Design(Normal Forms)

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

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

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

BCNF is a stronger version 3NF. So every relation in BCNF will also be in 3NF.

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.

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 R 

Since (sname, city) forms a candidate key, there is no non-tirvial dependency X → Y where X is not a superkey

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

Please refer Database Normalization | Normal Forms for details of normal forms.

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 Dependencies

 
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

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

Hence it's proved that a Relation with two single - valued attributes is in BCNF hence its also in 1NF, 2NF, 3NF. 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

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, EFGH