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 %3D

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
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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
Transcribed Image Text: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
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