Database System Concepts
7th Edition
ISBN: 9780078022159
Author: Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher: McGraw-Hill Education
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Expert Solution & Answer
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
- 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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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.
Chapter 7 Solutions
Database System Concepts
Ch. 7 - Prob. 1PECh. 7 - Prob. 2PECh. 7 - Explain how functional dependencies can be used to...Ch. 7 - Prob. 4PECh. 7 - Prob. 5PECh. 7 - Prob. 6PECh. 7 - Prob. 7PECh. 7 - Prob. 8PECh. 7 - Prob. 9PECh. 7 - Prob. 10PE
Ch. 7 - Prob. 11PECh. 7 - Prob. 12PECh. 7 - Prob. 13PECh. 7 - Prob. 14PECh. 7 - Prob. 15PECh. 7 - Prob. 16PECh. 7 - Prob. 17PECh. 7 - Prob. 18PECh. 7 - Prob. 19PECh. 7 - Prob. 20PECh. 7 - Prob. 21ECh. 7 - Prob. 22ECh. 7 -
Explain what is meant by repetition of...Ch. 7 -
Why are certain functional dependencies called...Ch. 7 - Prob. 25ECh. 7 - Prob. 26ECh. 7 - Prob. 27ECh. 7 - Prob. 28ECh. 7 - Prob. 29ECh. 7 - Prob. 30ECh. 7 - Prob. 32ECh. 7 - Prob. 33ECh. 7 - Prob. 35ECh. 7 - Prob. 36ECh. 7 - Prob. 37ECh. 7 - Prob. 38ECh. 7 - Prob. 39ECh. 7 - Prob. 40ECh. 7 - Prob. 41ECh. 7 - Prob. 42ECh. 7 - Prob. 43E
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