Database-Management-System
August 29, 2024Computer-Networks
August 29, 2024Database-Management-System
Question 903 |
C->F,E->A,EC->D,A->B
Which of the following is a key for R?
CD
| |
EC | |
AE
| |
AC |
C -> F
E -> A
EC -> D
A -> B
We can use the closure of attributes to determine a key. A key is a set of attributes that uniquely determines all other attributes in the relation.
Let’s start with the attribute set {C, E}:
Closure({C, E}) = {C, E, F, A, B, D}
Since the closure of {C, E} contains all attributes (C, E, F, A, B, D), {C, E} is a superkey.
To check if {C, E} is a key, we need to see if it’s minimal. We can do this by removing each attribute one at a time and checking if the closure remains the same. If it does, that attribute can be removed without changing the closure.
Removing C from {C, E}:
Closure({E}) = {E, A}
The closure no longer contains all attributes (C, E, F, A, B, D), so {C, E} is minimal.
Therefore, {C, E} is a key for the given relational schema R.
C -> F
E -> A
EC -> D
A -> B
We can use the closure of attributes to determine a key. A key is a set of attributes that uniquely determines all other attributes in the relation.
Let’s start with the attribute set {C, E}:
Closure({C, E}) = {C, E, F, A, B, D}
Since the closure of {C, E} contains all attributes (C, E, F, A, B, D), {C, E} is a superkey.
To check if {C, E} is a key, we need to see if it’s minimal. We can do this by removing each attribute one at a time and checking if the closure remains the same. If it does, that attribute can be removed without changing the closure.
Removing C from {C, E}:
Closure({E}) = {E, A}
The closure no longer contains all attributes (C, E, F, A, B, D), so {C, E} is minimal.
Therefore, {C, E} is a key for the given relational schema R.