QUESTION 23 Given the relation R(V, W, X, Y, Z) and functional dependencies: • VY -> Y • W-> XZ • VY -> W What is the highest normal form R is in? O Un-normalized form O First Normal Form O Second Normal Form O Third Normal Form QUESTION 24 Click the box next to the multivalued attribute(s) in this conceptua Borrower id (PK)

QUESTION 23 Given the relation R(V, W, X, Y, Z) and functional dependencies: • VY -> Y • W-> XZ • VY -> W What is the highest normal form R is in? O Un-normalized form O First Normal Form O Second Normal Form O Third Normal Form QUESTION 24 Click the box next to the multivalued attribute(s) in this conceptua Borrower id (PK)

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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We have a relation with the following attributes: R={A,B,C,D,E} and all the attributes are atomic. Primary key of this relation is AB. What is the normal form of this relation given the following functional dependencies: AC→D, AB→C, AB→D, AB→E Select one: a. First Normal Form b. This relation is not normal. c. Third Normal Form d. Second Normal Form
You are given a relation R with the following attributes R(A,B,C,D,E). Assume that you were given the following functional dependencies: • FD1= AB -> E • FD2: = D -> C (a) What is a valid key for this relation? justify your response. (b) Is this relation in 3NF (Third Normal Form)? Justify your response. If it is not in 3NF, normalize the relation and demonstrate all steps of normalization.
Consider a relation schema R = {M, N, P, Q, S} and its functional dependency set F = {N → Q,S → M,M →P,P → S}. Is MN a super key of R? Is MN a candidate key of R? (a) MN is not a super key for R; MN is not a candidate key for R.(b) MN is a super key for R; MN is not a candidate key for R.(c) MN is a super key for R; MN is a candidate key for R.(d) MN is not a super key for R; MN is a candidate key for R.
Consider a relation schema R = {A, B, C, D, E, G} and its functional dependency set F = {A → BCD,BC →DE,B → D,D → A}. Compute B+and choose the answer.(a) B+= AB(b) B+= ABC(c) B+= ABCD(d) B+= ABCDE(e) none of the above
Consider a relation named as BOOK that contains data about the books in a library. BOOK relation was initially created with the attributes BookID (an id that library assigns), ISBN, CopyNr (used to differentiate copies of the same book), Title, PublYear, Author, and AuthorBornYear. Primary key: BookID, CopyNr and Author Candidate key: ISBN, CopyNr and Author Below is three dependencies of attributes: Dependency 1 (D1): ISBN, Title, PublYear, AuthorBornYear were dependent on BookID, CopyNr and Author Dependency 2 (D2): Title and PublYear were dependent on BookID, CopyNr Dependency 3 (D3): AuthorBornYear was dependent on Author . First determine which normal form (1NF, 2NF, 3NF or BCNF) the above relation is, and why. Then, if necessary, convert the above relation to the highest normal form (BCNF). Write any assumptions that you make.
Consider a relation schema R = {A, B, C, D, E} and its functional dependency set F = {A → B,CD → E,B →D,E → A}. Compute A+ and choose the answer.(a) A+ = AB(b) A+ = ABC(c) A+ = ABCD(d) A+= ABCDE(e) none of the above
Given:1. The universal relation schema Q(CTHRSG), where C = course, T = teacher, H = hour, R = room, S= student, and G = grade.2. The minimal set of functional dependencies F are assumed: C ->T each course has one teacher HR -> C only one course can meet in a room at one time. HT -> R a teacher can be in only one room at one time. CS -> G each student has one grade in each course. HS -> R a student can be in only one room at one time. 3. The decomposition of the universal schema Q(CTHRSG) based on the algorithm (Lossless Join and Dependency-Preserving Decomposition into 3NF Relation Schema) is:D = {R1, R2, R3, R4, R5}, where R1 = {CT} R2 = {HRC} R3 = {HTR} R4 = {CSG} R5 = {HSR} 4. HS is the key of R, and is a part of R5. Please apply the Lossless Join Testing algorithm to see if the decomposition D is a lossless join decomposition. Please provide testing details step by step when you apply the set of functional dependency for testing.
Given a relation R (W,I,S,D,O,M) with thefollowing eight functional dependencies:F: {WI → S; D → OM; S → W; IO → S; IS → D; SM→ ID; WSD → I; SO → WM}.Perform the following(i) Remove Extraneous Attributes(ii) Remove redundant Functional dependencies(iii) Find the minimal Cover(iv) Find the closure for all key attributes(v) Find the Candidate Keys.
Question 3Discuss in detailed explanation that the concept that an attribute that is part of a composite primary key is also a candidatekey.
Consider a relation with schema R(A,B,C,D,E,F,G). Choose the key for the relation for the following sets of functional dependencies. BCD→A BC→E A→F F→ G C→ D A→ G ? A) B B) C C) BC D) All combinations of attributes are key
1. Consider a relation schema R1(A, B, C, D) that satisfies the set of functional dependencies F1 ={A→B, B→C, C→A}.a) Calculate A + and (AC)+b) Can AB→ C? Justify your answer.c) Find all the candidate keys of R1 using F1 and show your steps to find them. 2. Consider a relation schema R2(A, B, C, D) that satisfies the set of functional dependencies F2 = {A→B, C→D, D→AC}.a) Is (AD) is candidate key of R2? Justify your answer.b) Is R2 in BCNF? Explain the reason to your answer.c) If R2 is not in BCNF, give a lossless-join BCNF decomposition of R2. Is this decomposition dependency preserving? 3. Consider a relation schema R3(A, B, C, D, E) that satisfies the set of functional dependencies F3 = {A → BC, AC → BD}.a) Calculate (AC)+ and (AE)+b) Is AB → C in F3 +? Justify your answer.c) Find the canonical cover Fc of F3.d) Is R3 in 3NF? Explain the reason to your answer. If R3 is not in 3NF, give a lossless-join, dependency preserving 3NF decomposition of R3.
Functional Dependencies and Normalization Exercise 1: Consider the following relation: Stock(#prod, #dep, pname, quantity) Determine the functional dependencies on Stock. For each of them, explain what they mean in plain English. Is Stock 3NF? Is Stock BCNF? Exercise 2: Consider the following relation: Plane(#plane, type, constructor, capacity, owner) Determine the functional dependencies on Plane. For each of them, explain what they mean in plain English. Is Plane 3NF? Is Plane BCNF? Exercise 3: Compute the attribute closure of XZA with respect to the following set of dependencies: F = {XZ => ZYB, YA -> GC, C -> W, B -> G, XZ -> G} Is the dependency XZA -> YB implied by F? Hint: Use the Armstrong axioms. Exercise 4: What are the properties of the BCNF decomposition algorithm? Explain lossless and dependency preservation with your own words. Apply the decomposition algorithm on Stock.