Database Design(Normal Forms)
Question 1 |
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Question 2 |
For the relation R(ABCDEFGH) with FD's= {CH->G, A->BC, B->CHF, E->A, F->EG such that F+ is exactly the set of FDs that hold for R.} 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 |
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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 3 |
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 |
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Question 4 |
1. Registration_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 hostelWhich one of the following option is 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 |
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Question 5 |
Suppliers(sid:integer, sname:string, city:string, street:string) Parts(pid:integer, pname:string, color:string) Catalog(sid:integer, pid:integer, cost:real)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?
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 |
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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 6 |
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 |
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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 7 |
Consider the relation scheme R = {E, F, G, H, I, J, K, L, M, N} 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} |
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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
Question 8 |
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. |
Discuss it
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.
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
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Question 10 |

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