Computer Science Engineering (CSE)

 An e-NFA is ___________ in representation.
  • a)
    Quintuple
  • b)
    Quadruple
  • c)
    Triple
  • d)
    None of the mentioned
Correct answer is option 'A'. Can you explain this answer?

NAVDEEP SINGH answered  •  3 days ago
Correct Answer :- A
Explanation : An e-NFA consist of 5 tuples: A=(Q, S, d, q0, F)
Note: e is never a member of SAn e-NFA consist of 5 tuples: A=(Q, S, d, q0, F)
Note: e is never a member of S.

If every non-key attribute is functionally dependent on the primary key, then the relation will be in
  • a)
    First normal form
  • b)
    Third normal form
  • c)
    Second normal form
  • d)
    Fourth normal form
Correct answer is option 'C'. Can you explain this answer?

DHIRENDRA GUPTA answered  •  3 days ago
Conditions for various normal forms:
  1. 1 NF – A relation R is in first normal form (1NF) if and only if all underlying domains contain atomic values only.
  2. 2 NF – A relation R is in second normal form (2NF) if and only if it is in 1NF and every non-key attribute is fully dependent on the primary key.
  3. 3 NF – A relation R is in third normal form (3NF) if and only if it is in 2NF and every non-key attribute is non-transitively dependent on the primary key.
  4. BCNF – A relation R is in Boyce-Codd normal form (BCNF) if and only if every determinant is a candidate key.
Example:
Relation R(XYZ) with functional dependencies {X -> Y, Y -> Z, X -> Z}.
Notice here Y -> Z, in question it is not mention that non prime attribute is only dependent on primary key so this FD is perfectly valid.
This relation is in 2NF but not in 3NF because of every non-key attribute is transitively dependent on the primary key. Here {X} will be candidate key.

Fetching relevant content for you