DATABASE SYSTEM CONCEPTS LCPO
7th Edition
ISBN: 9781265586577
Author: SILBERSCHATZ
Publisher: MCG
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Expert Solution & Answer
Chapter 7, Problem 32E
a.
Explanation of Solution
Nontrivial functional dependency
- The functional dependency CD → AB can also be written as follows:
CD → A
CD → B
- Now, CD+ = {C D A B C E}
- Thus, CD+ contains B without using the functional dependency, CD → B.
- Thus, CD → B can be deleted...
b.
Explanation of Solution
BCNF decomposition
- Here A → BC violates third normal form.
- Hence R(ABCDE) can be decomposed i...
c.
Explanation of Solution
Loseless decomposition
- The common attribute is A in relations R1(ABC) and R2(ADE)...
d.
Explanation of Solution
Dependency preserving decomposition
- The functional dependencies of original relation R is BD → E and CD → AB...
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Students have asked these similar questions
Suppose that we decompose the schema R = (A, B, C, D, E) into (A, B, C) (A, D, E). Show that this decomposition is a lossless decomposition if the following set F of functional dependencies holds: A → BC CD → E B → D E → A
Consider the schema R = (A, B, C, D, E, G) and the set F of functional dependencies:AB → CDB → DDE → BDEG → ABAC → DER is not in BCNF for many reasons, one of which arises from the functionaldependency AB → CD. Explain why AB → CD shows that R is not in BCNFand then use the BCNF decomposition algorithm starting with AB → CD togenerate a BCNF decomposition of R. Once that is done, determine whetheryour result is or is not dependency preserving, and explain your reasoning.
Consider the relation schema R (A, B, C, D) with all possible functional dependencies. For each of the following situations, identify the highest normal form for this relation R.
A, C -> B, D
C -> D
A, C -> B, D
B -> D
A, C -> B, D
Chapter 7 Solutions
DATABASE SYSTEM CONCEPTS LCPO
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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Similar questions
- Consider the schema R = (A, B, C, D, E, G) and the set F of functional dependencies:AB → CDADE → GDEB → GCG → DEUse the 3NF decomposition algorithm to generate a 3NF decomposition of R,and show your work. This means: A canonical cover for F, along with an explanation of the steps you tookto generate it.arrow_forwardGiven the relation R (A, B, C, D, E) and the following set of functional dependencies: F = {ABC, A➜E, B➜D, C⇒B} a. Specify the key of R. b. Is R in the 3NF? If not give a decomposition of R that is in the 3NF. c. Explain why this decomposition is not in BCNF. d. Give a decomposition of R that is in BCNF.arrow_forwardConsider a relation schema R = {A, B, C, D, E} and its functional dependency set F = {A → BC,CD → E,B →D,E → A}. Let R be decomposed into R1 and R2, where R1 = (A,B,C) and R2 = (A,D,E). Is the decompositionlossless? Is the decomposition dependency preserving? (a) The decomposition is lossless and dependency preserving.(b) The decomposition is neither lossless nor dependency preserving.(c) The decomposition is not lossless but dependency preserving.(d) The decomposition is lossless but not dependency preserving.arrow_forward
- Consider the relation schema R(A, B, C, D, E, F) and the set S = {AB->C, BC->AD, D->E, C->B) of functional dependencies. Assuming that R is decomposed into the relation schemas R1(A,B), R2(B,C), R3(A,B,D,E) and R4(E,F). Use the Chase test to show if this is a lossless decomposition.arrow_forwardConsider the relation r(A,B ) with the following instance of r. Which of the functional dependencies do NOThold on this instance?(a) C → A(b) B → A(c) AB → C(d) A → C(e) B → C Instance of r is attachedarrow_forwardConsider the relation schema R=VWXYZ, with the set of functional dependencies F = {VW->XY, V>X, W->Z, YZ->Z, Z→V} A. Give three independent reasons why the following table cannot possibly be a legal instance of R. V W X Y v1 w1 х1 yl z1 v2 w1 x2 y2 z1 v2 w2 x1 y2 z2 B. Show the functional dependencies that hold on each of R1=VWX and R2=WYZ. You do not need to show trivial functional dependencies, or those implied by others in the same set, but you must show all others. C. Is the decomposition of R into R1=VWX and R2=WYZ lossless-join? Why or why not?arrow_forward
- Consider the schema R = (A, B, C, D, E, G) and the set F of functional dependencies:A → BCBD → ECD → AB For your decomposition, state whether it is dependency preserving and explain why ?arrow_forwardConsider the schema R = (A, B, C, D, E, G) and the set F of functional dependencies:AB → CDADE → GDEB → GCG → DEUse the 3NF decomposition algorithm to generate a 3NF decomposition of R,and show your work. This means: The final decomposition.arrow_forwardConsider the relation schema R(A, B, C, D, E, F) with the following set of functional dependencies FDs F = {ABC → E, DE → A, ACE → D, BC → A, BD → E}. Is the decomposition of R into R into R1(A, B, C, E), R2(B, C, D) and R3(B, D, F) a lossless decomposition? (Explain)arrow_forward
- Our discussion of lossless decomposition implicitly assumed that attributes on the left-hand side of a functional dependency cannot take on null values. What could go wrong on decomposition, if this property is violated?arrow_forwardConsider the schema R=ABCDEG and the set of functional dependenciesF={BC→AG, BG→CD, C→AE, D→AG}2a) Determine the keys of the schema2b) Decide whether the schema is 3NF and motivate your answer 2c) Find a decomposition of the schema such that:- every subschema is 3NF- preserves the dependencies- has a lossless join.arrow_forwardConsider a relation schema R = {A, B, C, D, E, F, G} with the following set of functional dependencies: F = {BG A, A BC, BC A, BD E, E →F, CF B} The decomposition of R into R1 =A, B, C, G), R2 = {B, D, E), R3 = {B, C, F} and R4 = {E, F} is O In 3NF but Not in BCNF. O In 3NF and in BCNF. O Not in 3NF and Not in BCNF O In BCNF but Not in 3NF.arrow_forward
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