Database System Concepts
Database System Concepts
7th Edition
ISBN: 9780078022159
Author: Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher: McGraw-Hill Education
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Chapter 7, Problem 33E

a.

Explanation of Solution

Candidate keys

  • A key is known as a candidate key when it has the ability to derive all the attributes from a relation...

b.

Explanation of Solution

Canonical cover for F

  • The steps for canonical conversion is to break the functional dependencies so that right hand side will have only one single attribute.
  • Then all the extra unnecessary functional dependencies are removed.
  • Then all the functional dependencies are combined.
  • Hence here the given functional dependencies include AB -> G, AB -> D ...

c.

Explanation of Solution

Remaining steps of algorithm

  • The given relation is in 1NF as there is no multivalued attribute or complex attribute.
  • The given relation is not in 2NF as there is partial functional dependency...

d.

Explanation of Solution

Final decomposition

  • Given functional dependencies are AB -> CD, ADE -> GDE, B -> GC, G -> DE.
  • Here ADE -> GDE have some trivial attributes.
  • So those attributes can be removed and hence the dependency is ADE -> G...

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Students have asked these similar questions
Consider the schema R = (A, B, C, D, E, G,H) and the set F of functional dependencies:AB → CDD → CDE → BDEH → ABAC → DCUse the 3NF decomposition algorithm to generate a 3NF decomposition of R,and show your work. This means:a. A list of all candidate keysb. A canonical cover for Fc. The steps of the algorithm, with explanationd. The final decomposition
Consider a schema R, a set F of functional dependencies on R, and two candidate keys (k1, k2) as follows:R = (X,Y,Z).F = {Y → Z,XZ → Y}.k1 = XZ.k2 =XY Is R in BCNF? Is R in 3NF?(a) R is in both BCNF and 3NF(b) R is in BCNF but not in 3NF(c) R is not in BCNF but in 3NF(d) R is in neither BCNF not 3NF
Consider a schema R and a set F of functional dependencies as follows: R= (A,B,C,D,E,G). F = {A → BCD, BC DE,B→D,D→A}. Prove that AG is a superkey for R.
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