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Unformatted text preview: Schema Refinement and Normal Forms 199 Answer 19.17 Answer omitted. Exercise 19.18 Consider a relation R with attributes ABCDE . Let the following FDs be given: A → BC , BC → E , and E → DA . Similarly, let S be a relation with attributes ABCDE and let the following FDs be given: A → BC , B → E , and E → DA . (Only the second dependency differs from those that hold over R .) You do not know whether or which other (join) dependencies hold. 1. Is R in BCNF? 2. Is R in 4NF? 3. Is R in 5NF? 4. Is S in BCNF? 5. Is S in 4NF? 6. Is S in 5NF? Answer 19.18 Answer omitted. Exercise 19.19 Let R be a relation schema with a set F of FDs. Prove that the decomposition of R into R 1 and R 2 is lossless-join if and only if F + contains R 1 ∩ R 2 → R 1 or R 1 ∩ R 2 → R 2 . Answer 19.19 For both directions (if and only-if) we use the notation C = R 1 ∩ R 2 , X = R 1 − C, Y = R 2 − C , so that R 1 = XC, R 2 = CY , and R = XCY . ( ⇐ ): For this direction, assume we are given the dependency C → X . (The other case C → Y is similar.) So let r be an instance of schema R and let ( x 1 , c, y 1 ) and ( x 2 , c, y 2 ) be two tuples in r . The FD, C → X implies that x 1 = x 2 . Thus, ( x 1 , c, y 2 ) is the same as ( x 2 , c, y 2 ) and ( x 2 , c, y 1 ) is the same as ( x 1 , c, y 1 ), so that both these tuples ( x 1 , c, y 2 ) and ( x 2 , c, y 1 ) are in r . Thus r satisfies the JD: R = R 1 ./ R 2 . Since r is an arbitrary instance, we have proved that the decomposition is lossless....
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This note was uploaded on 01/17/2012 for the course EGN 4302 taught by Professor Dr.vishak during the Fall '12 term at University of Central Florida.
- Fall '12