Database-Management-System
August 29, 2024Database-Management-System
August 29, 2024Functional-Dependency
| Question 28 |
Comprehension:
Answer question (96-100) based on the problem statement given below:
An organization needs to maintain database having five attributes A, B, C, D, E. These attributes are functionally dependent on each other for which functionally dependency set F is given as : F: {A→ BC, D → E, BC → D, A →D}. Consider a universal relation R(A, B, C, D, E) with functional dependency set F. Also all attributes are simple and take atomic values only.
:→ Minimal cover F’ of functional dependency set F is
| F’ = {A → B, A → C, BC → D, D →E} | |
| F’ = {A → BC, B → D, D → E} | |
| F’ = {A → B, A → C, A → D, D → E} | |
| F’ = {A → B, A → C, B → D, C → D.D → E} |
Question 28 Explanation:
Steps to find minimal cover:
Step1: Write all FDs in such a way that the RHS of each FD contain only one attribute.
A→ B
A→ C
D → E
BC → D
A →D
Step2: Then for each FD see whether that RHS attribute can be driven by the LHS attribute using remaining FDs, if yes then remove that FD otherwise keep it. So step 1 results in following FDs:
A→ B
A→ C
D → E
BC → D
Step3: Now see the FD which is having 2 or more attributes in its LHS.Then find the closure of LHS attributes and then eliminate the attributes from LHS which are common in clsure.
Above BC are two attributes in LHS.
B+ = {B}
C+ = {C}
Since nothing is common in closure so keep both attributes in LHS.
Hence minimal cover is
A→ B
A→ C
D → E
BC → D
Step1: Write all FDs in such a way that the RHS of each FD contain only one attribute.
A→ B
A→ C
D → E
BC → D
A →D
Step2: Then for each FD see whether that RHS attribute can be driven by the LHS attribute using remaining FDs, if yes then remove that FD otherwise keep it. So step 1 results in following FDs:
A→ B
A→ C
D → E
BC → D
Step3: Now see the FD which is having 2 or more attributes in its LHS.Then find the closure of LHS attributes and then eliminate the attributes from LHS which are common in clsure.
Above BC are two attributes in LHS.
B+ = {B}
C+ = {C}
Since nothing is common in closure so keep both attributes in LHS.
Hence minimal cover is
A→ B
A→ C
D → E
BC → D
Correct Answer: A
Question 28 Explanation:
Steps to find minimal cover:
Step1: Write all FDs in such a way that the RHS of each FD contain only one attribute.
A→ B
A→ C
D → E
BC → D
A →D
Step2: Then for each FD see whether that RHS attribute can be driven by the LHS attribute using remaining FDs, if yes then remove that FD otherwise keep it. So step 1 results in following FDs:
A→ B
A→ C
D → E
BC → D
Step3: Now see the FD which is having 2 or more attributes in its LHS.Then find the closure of LHS attributes and then eliminate the attributes from LHS which are common in clsure.
Above BC are two attributes in LHS.
B+ = {B}
C+ = {C}
Since nothing is common in closure so keep both attributes in LHS.
Hence minimal cover is
A→ B
A→ C
D → E
BC → D
Step1: Write all FDs in such a way that the RHS of each FD contain only one attribute.
A→ B
A→ C
D → E
BC → D
A →D
Step2: Then for each FD see whether that RHS attribute can be driven by the LHS attribute using remaining FDs, if yes then remove that FD otherwise keep it. So step 1 results in following FDs:
A→ B
A→ C
D → E
BC → D
Step3: Now see the FD which is having 2 or more attributes in its LHS.Then find the closure of LHS attributes and then eliminate the attributes from LHS which are common in clsure.
Above BC are two attributes in LHS.
B+ = {B}
C+ = {C}
Since nothing is common in closure so keep both attributes in LHS.
Hence minimal cover is
A→ B
A→ C
D → E
BC → D