Database System Concepts
7th Edition
ISBN: 9789332901384
Author: Henry F. Korth & S. Sudarshan Abraham Silberschatz
Publisher: MCGRAW-HILL HIGHER EDUCATION
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Expert Solution & Answer
Chapter 7, Problem 14PE
Explanation of Solution
- Let F be a set of functional dependencies that hold on a schema R.
- Let σ= {R1, R2, . . . , Rn} be a dependency-preserving 3NF decomposition of R. Let X be a candidate key for R.
- Consider a legal instance r of R. Let j = ΠX(r) ⋈ΠR1 (r) ⋈ΠR2 (r) ⋈ΠRn (r). We want to prove that r = j.
- The user can claim that if t1 and t2 are two tuples in j such that t1[X] = t2[X], then t1 = t2. To prove this claim, we use the following inductive argument-
- Let F ′ = F1 ∪ F2 ∪ . . . ∪ Fn, where each Fi is the restriction of F to the schema Ri in s.
- Consider the use of the
algorithm given in Figure 8.8 to compute the closure of X under F ′ . We use induction on the number of times that the f or loop in this algorithm is executed.
- Basis : In the first step of the algorithm, result is assigned to X, and hence given that t1[X] = t2[X], we know that t1[result] = t2[result] is true.
- Induction Step : Let t1[result] = t2[result] be true at the end of the k th execution of the f or 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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