Computer Science Engineering (CSE) Exam > Computer Science Engineering (CSE) Questions > Let R (A, B, C, D, E, P, G) be a relational s...
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Let R (A, B, C, D, E, P, G) be a relational schema in which the following functional dependencies are known to hold: AB → CD, DE → P, C → E, P → C and B → G. The relational schema R is
- a)in BCNF
- b)in 3NF, but not in BCNF
- c)in 2NF, but not in 3NF
- d)not in 2NF
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
Let R (A, B, C, D, E, P, G) be a relational schema in which the follow...
Candidate key = AB
B->G is partial dependency
So, not in 2NF
B->G is partial dependency
So, not in 2NF
Most Upvoted Answer
Let R (A, B, C, D, E, P, G) be a relational schema in which the follow...
Given relational schema and functional dependencies
The relational schema R (A, B, C, D, E, P, G) and the following functional dependencies hold:
- AB → CD
- DE → P
- C → E
- P → C
- B → G
Explanation
To determine the normal form of the given schema, we need to check for violations of normal forms in the schema.
1. 1NF violation
The schema R (A, B, C, D, E, P, G) is already in 1NF as all the attributes are atomic and there are no repeating groups.
2. 2NF violation
- In the given schema, AB → CD violates 2NF as the determinant AB is not a candidate key and CD is only dependent on AB and not on any other attribute. To remove this violation, we can decompose R into two relations:
R1 (A, B, C, D)
R2 (B, G)
- DE → P also violates 2NF as the determinant DE is not a candidate key and P is only dependent on DE and not on any other attribute. To remove this violation, we can decompose R into two relations:
R1 (D, E, P)
R2 (A, B, C, G)
After decomposition, we get the following relations:
R1 (A, B, C, D)
R2 (B, G)
R3 (D, E, P)
R4 (A, B, C, G)
3. 3NF violation
- In the relation R1 (A, B, C, D), C → E violates 3NF as C is not a candidate key and E is only dependent on C and not on any other attribute. To remove this violation, we can decompose R1 into two relations:
R1 (A, B, C)
R2 (C, D, E)
- In the relation R3 (D, E, P), P → C violates 3NF as P is not a candidate key and C is only dependent on P and not on any other attribute. To remove this violation, we can decompose R3 into two relations:
R3 (D, E)
R4 (P, C)
After decomposition, we get the following relations:
R1 (A, B, C)
R2 (C, D, E)
R3 (D, E)
R4 (P, C)
R5 (B, G)
The given schema has been decomposed into 5 relations and all the relations are in 3NF. However, the relation R (A, B, C, D, E, P, G) is not in 2NF as it has partial dependencies.
The relational schema R (A, B, C, D, E, P, G) and the following functional dependencies hold:
- AB → CD
- DE → P
- C → E
- P → C
- B → G
Explanation
To determine the normal form of the given schema, we need to check for violations of normal forms in the schema.
1. 1NF violation
The schema R (A, B, C, D, E, P, G) is already in 1NF as all the attributes are atomic and there are no repeating groups.
2. 2NF violation
- In the given schema, AB → CD violates 2NF as the determinant AB is not a candidate key and CD is only dependent on AB and not on any other attribute. To remove this violation, we can decompose R into two relations:
R1 (A, B, C, D)
R2 (B, G)
- DE → P also violates 2NF as the determinant DE is not a candidate key and P is only dependent on DE and not on any other attribute. To remove this violation, we can decompose R into two relations:
R1 (D, E, P)
R2 (A, B, C, G)
After decomposition, we get the following relations:
R1 (A, B, C, D)
R2 (B, G)
R3 (D, E, P)
R4 (A, B, C, G)
3. 3NF violation
- In the relation R1 (A, B, C, D), C → E violates 3NF as C is not a candidate key and E is only dependent on C and not on any other attribute. To remove this violation, we can decompose R1 into two relations:
R1 (A, B, C)
R2 (C, D, E)
- In the relation R3 (D, E, P), P → C violates 3NF as P is not a candidate key and C is only dependent on P and not on any other attribute. To remove this violation, we can decompose R3 into two relations:
R3 (D, E)
R4 (P, C)
After decomposition, we get the following relations:
R1 (A, B, C)
R2 (C, D, E)
R3 (D, E)
R4 (P, C)
R5 (B, G)
The given schema has been decomposed into 5 relations and all the relations are in 3NF. However, the relation R (A, B, C, D, E, P, G) is not in 2NF as it has partial dependencies.
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Let R (A, B, C, D, E, P, G) be a relational schema in which the following functional dependencies are known to hold: AB → CD, DE → P, C → E, P → C and B → G. The relational schema R isa)in BCNFb)in 3NF, but not in BCNFc)in 2NF, but not in 3NFd)not in 2NFCorrect answer is option 'D'. Can you explain this answer? for Computer Science Engineering (CSE) 2025 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about Let R (A, B, C, D, E, P, G) be a relational schema in which the following functional dependencies are known to hold: AB → CD, DE → P, C → E, P → C and B → G. The relational schema R isa)in BCNFb)in 3NF, but not in BCNFc)in 2NF, but not in 3NFd)not in 2NFCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let R (A, B, C, D, E, P, G) be a relational schema in which the following functional dependencies are known to hold: AB → CD, DE → P, C → E, P → C and B → G. The relational schema R isa)in BCNFb)in 3NF, but not in BCNFc)in 2NF, but not in 3NFd)not in 2NFCorrect answer is option 'D'. Can you explain this answer?.
Let R (A, B, C, D, E, P, G) be a relational schema in which the following functional dependencies are known to hold: AB → CD, DE → P, C → E, P → C and B → G. The relational schema R isa)in BCNFb)in 3NF, but not in BCNFc)in 2NF, but not in 3NFd)not in 2NFCorrect answer is option 'D'. Can you explain this answer? for Computer Science Engineering (CSE) 2025 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about Let R (A, B, C, D, E, P, G) be a relational schema in which the following functional dependencies are known to hold: AB → CD, DE → P, C → E, P → C and B → G. The relational schema R isa)in BCNFb)in 3NF, but not in BCNFc)in 2NF, but not in 3NFd)not in 2NFCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Computer Science Engineering (CSE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let R (A, B, C, D, E, P, G) be a relational schema in which the following functional dependencies are known to hold: AB → CD, DE → P, C → E, P → C and B → G. The relational schema R isa)in BCNFb)in 3NF, but not in BCNFc)in 2NF, but not in 3NFd)not in 2NFCorrect answer is option 'D'. Can you explain this answer?.
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