###### NVS PGT CS 2019 Part-B

October 5, 2023###### NIC-NIELIT Scientist-B 2020

October 5, 2023# Database-Management-System

Question 18 |

The following functional dependencies hold true for the relational schema {V, W, X, Y, Z} :

V → W

VW → X

Y → VX

Y → Z

Which of the following is irreducible equivalent for this set of functional dependencies?

V → W

VW → X

Y → VX

Y → Z

Which of the following is irreducible equivalent for this set of functional dependencies?

V→W V→X Y→V Y→Z | |

V→W W→X Y→V Y→Z | |

V→W V→X Y→V Y→X Y→Z | |

V→W W→X Y→V Y→X Y→Z |

Question 18 Explanation:

Step 1:

V → W, VW → X, Y → V, Y → X, Y→ Z

Step 2:

V → W, VW → X, Y → V, Y → X, Y→ Z

(V)

(VW)

(Y)

(Y)

(Y)

Without Y → X, the closure of Y is deriving ‘X’ from the remaining attributes.

So, we can remove Y → X as its redundant.

Step 3:

V → W, VW → X, Y → V, Y → Z

(V)

So, W is redundant. We can remove it.

So, the final canonical form is

V→W, V→X, Y→V, Y→Z

⇾ So, option (A) is correct.

V → W, VW → X, Y → V, Y → X, Y→ Z

Step 2:

V → W, VW → X, Y → V, Y → X, Y→ Z

(V)

^{+}= V ×(VW)

^{+}= VW ×(Y)

^{+}= YXZ(Y)

^{+}= YVW ×(Y)

^{+}= YVWXWithout Y → X, the closure of Y is deriving ‘X’ from the remaining attributes.

So, we can remove Y → X as its redundant.

Step 3:

V → W, VW → X, Y → V, Y → Z

(V)

^{+}= VW, the closure of V is deriving W from the remaining FD’s.So, W is redundant. We can remove it.

So, the final canonical form is

V→W, V→X, Y→V, Y→Z

⇾ So, option (A) is correct.

Correct Answer: A

Question 18 Explanation:

Step 1:

V → W, VW → X, Y → V, Y → X, Y→ Z

Step 2:

V → W, VW → X, Y → V, Y → X, Y→ Z

(V)

(VW)

(Y)

(Y)

(Y)

Without Y → X, the closure of Y is deriving ‘X’ from the remaining attributes.

So, we can remove Y → X as its redundant.

Step 3:

V → W, VW → X, Y → V, Y → Z

(V)

So, W is redundant. We can remove it.

So, the final canonical form is

V→W, V→X, Y→V, Y→Z

⇾ So, option (A) is correct.

V → W, VW → X, Y → V, Y → X, Y→ Z

Step 2:

V → W, VW → X, Y → V, Y → X, Y→ Z

(V)

^{+}= V ×(VW)

^{+}= VW ×(Y)

^{+}= YXZ(Y)

^{+}= YVW ×(Y)

^{+}= YVWXWithout Y → X, the closure of Y is deriving ‘X’ from the remaining attributes.

So, we can remove Y → X as its redundant.

Step 3:

V → W, VW → X, Y → V, Y → Z

(V)

^{+}= VW, the closure of V is deriving W from the remaining FD’s.So, W is redundant. We can remove it.

So, the final canonical form is

V→W, V→X, Y→V, Y→Z

⇾ So, option (A) is correct.

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