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 35E
Explanation of Solution
Ensuring that the resulting decomposition is lossless:
- Instead of using BCNF, it is possible to have a lossless decomposition into 3NF itself.
- For this, it is required that one schema at least should have a candidate key for the schema that is being composed.
- The set of functional dependencies F on a schema R is considered.
- The decomposition of R which is dependency-preserving, sigma={R1,R2,...,Rn} is assumed.
- Suppose X is a candidate key for R and a legal instance of R is r.
- Suppose j=ProjectX(r) join ProjectR1(r) join ProjectR2(r) ... Join ProjectRn(r).
- To prove that r=j, it is claimed that, t1 and t2 are two tuples, where t1=t2 such that t1[X]=t2[X].
- The inductive argument is used for proving this claim.
- Let F'=F1 union F2 union ... union Fn, where each Fi is the restriction of F to the schema Ri in sigma. Closure of X is computed under F'.
- Induction is applied the number of times that the for loop in the
algorithm executes.
Basis:
In the first step of the algorithm, result is assigned to X, and hence given that t1[X]=t2[X], it is known that t1[result]=t2[result] is true.
Induction Step:
- Let t1[result]=t2[result] be true at the end of the kth execution of the for loop...
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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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