Question

Given the schema S=  <  { A,B,C,D,E,G,H }, F>, where F represents the following dependencies:

AB→D

A→D

E→B

E→C

G→C

E→A

EB→GH

H → A

1. Find a minimal cover for this schema.

2. Find a key for this schema.

3. Find a third normal form decomposition for this schema.

4. Find a BCN form decomposition for this schema.

Answer 1

b) Attribute E are not present in Rhs of any FO. Letus try computing their closure! b) te ]={E, ,A,B,C, D, G,4} Since tt yt can determine the rest of attribute of R, RE f is the key ABAD is trivial FD since we aheady have A>D. . This FD AB> D can be deleted In FD EBGH, E can be deleted since we have E→B. . The minimal cover is | AD LI I AD I E → AGCGHS HA . E → BCA GUC E->GH НЭА Ј. ABD violates QNF since it is a partial FD. PAB]*- {ABD} G-C violates Qne. {G}t= {G,c}. HA violates QnF .0437:4HA}. EB → GH is in 3NF since EB is super key. R CABCDEFG) RICABOS R2 (GC) R3 (AH) R4 (ABGHE) AB →D A+D > violates 3NF. R5(AB) RoCAD) The BNF relations are (AB),(AD),(GC), (HA) (AGGHE)

D) All the decomposed 3NF relation are also in BCNF since all the FD in all relation is of type superkey -> monkey.

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