A decomposition is in 3NF if every decomposed relation is in 3NF. A decomposition is in BCNF if every decomposed relation is in BCNF. Let R(A,B,C,D,E,F,G) be a relation with the FDs: F = {CD → G, DG → C, AD→ G, G → A, AB → F}. Consider the decomposition of R into R1(C, D, G), R2(A, D, G), R3(A, B, F) and R4(C, E). Which of the following statements is TRUE about this decomposition? Select one: O Ris in BCNF but Not in 3NF. O R is in 3NF but Not in BCNF.

A decomposition is in 3NF if every decomposed relation is in 3NF. A decomposition is in BCNF if every decomposed relation is in BCNF. Let R(A,B,C,D,E,F,G) be a relation with the FDs: F = {CD → G, DG → C, AD→ G, G → A, AB → F}. Consider the decomposition of R into R1(C, D, G), R2(A, D, G), R3(A, B, F) and R4(C, E). Which of the following statements is TRUE about this decomposition? Select one: O Ris in BCNF but Not in 3NF. O R is in 3NF but Not in BCNF.

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Normalization Here is a set of FDs for a relation R {A, B, C, D, E, F, G}; AB BC>DE AEF>G AC>DE a. Compute the closure of {A, C}, i.e. (A, C}* b. What are the candidate keys for R? Justify your answer. c. Is R in BCNF? Justify your answer. d. Give a 3NF decomposition of this relation.
Question 1: Find the candidate key(s) for the following relations. a) R(A,B,C,D) F={D> A, D →B, ABC → D, A → B, B→ CD} b) R(A,B,C,D) F={A→ BCD, AB→ CD, ABC→> D, BD→ AB, C→ D} c) R(A,B,C,D,E) F={AB>CD, D→ A, BC→DE}
Solve the following:
Consider the relation R = {A, B, C, D, E, F, G, H, I,J} and the set of functional dependencies F = { {A, B} → {C}, {A} → {D, E}, {B} → {F}, {F} →{G, H}, {D} → {I, J} }. (b) Whether (A,B,F} is a super key? Whether {A,B,F} is a candidate key? Why? (c) Whether {A,D} is a super key? Whether {A,D} is a candidate key? Why?
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Task 4: Consider the relation schema R(A, B, C, D, E) with the following FDs FD1: {A,B,C} → {D,E} FD2: {B,C,D} → {A,E} FD3: {C} → {D} a) Show that R is not in BCNF. Ans) X = {C} then X+ = {C, D} which is not R therefore X is not a superkey, so R is not in BCNF. b) Decompose R into a set of BCNF relations (describe the process step by step). Ans) FD3 violates BCNF. Decomposing R using FD3. R1(C,D) with FDs: FD3, CK: {C} R2(A,B,C,E) with new FD: {A,B,C}→ E (Decomposed from FD1), CK: {A,B,C} {BC}+ = {BC} - Not a candidate key {ABC}+ = {ABCDE} - A candidate key {BCD}+ = {ABCDE} - A candidate key Candidate Keys = {ABC}, {BCD} The result of the decomposition consists of R1 and R2 %3D %3D
Solve the following:
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Suppose you are given a relation R-(A,B,C,D,E) with the following functional dependencies: {BC→ADE,D→B}. a. Find all candidate keys. b. Identify the best normal form that R satisfies (1NF , 2NF , 3NF , or BCNF). c. If the relation is not in BCNF, Decompose it until it becomes BCNF, At each step, identify a new relation, decompose and re-compute the keys the normal forms they satisfy.
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pen paper answer
The division operator of relational algebra, “÷”, is defined as follows. Let r(R)and s(S) be relations, and let S ⊆ R; that is, every attribute of schema S isalso in schema R. Given a tuple t, let t[S] denote the projection of tuple t onthe attributes in S. Then r ÷ s is a relation on schema R − S (that is, on theschema containing all attributes of schema R that are not in schema S). A tuplet is in r ÷ s if and only if both of two conditions hold:• t is in ΠR−S(r)• For every tuple ts in s, there is a tuple tr in r satisfying both of the following:a. tr[S] = ts[S]b. tr[R − S] = tGiven the above definition:a. Write a relational algebra expression using the division operator to findthe IDs of all students who have taken all Comp. Sci. courses. (Hint:project takes to just ID and course id, and generate the set of all Comp.Sci. course ids using a select expression, before doing the division.)b. Show how to write the above query in relational algebra, without usingdivision. (By doing so, you…