Database-Management-System
August 29, 2024Normalization
August 29, 2024Database-Management-System
| Question 252 |
Consider a relational schema S=(U,V,W,X,Y,Z) on which the following functional dependence hold:
{ U → V, VW → X, Y → W, X → U}
Which are the candidate keys among following options?
{ U → V, VW → X, Y → W, X → U}
Which are the candidate keys among following options?
| UY, VY
| |
| UY, VY, XY | |
| UYZ, VYZ, VWZ | |
| UYZ, VYZ, XYZ |
Question 252 Explanation:
YZ is not present in the RHS of any FD, so the key must include YZ as its attributes.
Now lets see whether YZ is the candidate key or not?
YZ ={ Y, Z, W}
So YZ is not the candidate key.
Now let’s check the candidate key of length 3 .
YZX ={U, V, W ,X, Y, Z}
YZU ={U, V, W ,X, Y, Z}
YZV ={U, V, W ,X, Y, Z}
Now lets see whether YZ is the candidate key or not?
YZ ={ Y, Z, W}
So YZ is not the candidate key.
Now let’s check the candidate key of length 3 .
YZX ={U, V, W ,X, Y, Z}
YZU ={U, V, W ,X, Y, Z}
YZV ={U, V, W ,X, Y, Z}
Correct Answer: D
Question 252 Explanation:
YZ is not present in the RHS of any FD, so the key must include YZ as its attributes.
Now lets see whether YZ is the candidate key or not?
YZ ={ Y, Z, W}
So YZ is not the candidate key.
Now let’s check the candidate key of length 3 .
YZX ={U, V, W ,X, Y, Z}
YZU ={U, V, W ,X, Y, Z}
YZV ={U, V, W ,X, Y, Z}
Now lets see whether YZ is the candidate key or not?
YZ ={ Y, Z, W}
So YZ is not the candidate key.
Now let’s check the candidate key of length 3 .
YZX ={U, V, W ,X, Y, Z}
YZU ={U, V, W ,X, Y, Z}
YZV ={U, V, W ,X, Y, Z}
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