recategorized by
11,084 views
28 votes
28 votes

Consider the relation scheme $R(A, B, C)$ with the following functional dependencies:

  • $A, B \rightarrow C,$
  • $C \rightarrow A$
  1. Show that the scheme $R$ is in $3\text{NF}$ but not in $\text{BCNF}$.
  2. Determine the minimal keys of relation $R$. 
recategorized by

6 Answers

Best answer
42 votes
42 votes
The Candidate Keys are $AB$ and $BC$.

None of the given functional dependencies are partial. So, the scheme qualifies for $2\text{NF}$.

There is no transitive dependency. So, the scheme qualifies for $3\text{NF}$.

All determinants are not Candidate Keys. So, the scheme do not qualify for $\text{BCNF}$.
edited by
23 votes
23 votes
Candidate keys  AB, CB

AB->C  (no issues)
C->A  ( On RHS  A is a part of candidate key so you can not say it is partial or transitive dependency hence in 3NF)

For BCNF ,  C is not a super key hence not in BCNF
15 votes
15 votes
Candidate keys are AB , BC ... So all the attributes are prime attributes ...

If all the attributes of a relation are prime attributes, then the relation is guaranteed to be in 3 NF.

It is not in BCNF as C is not a Super key
4 votes
4 votes
BCNF :-

1.IF X----->Y (then X is either Super key or Candidate Key).

2.LHS is super key(For every dependency)

3. it should be in the 3NF.

 

3NF-

       X--------->Y

1.X is super key

OR

2.Y is prime Attribute.

now given question,

R(A,B,C)

AB----->C

C------>A          CANDIDATE KEY---(AB)

                        PRIME ATTRIBUTE --(A,B)

BCNF :-

AB----->C (then AB is either Super key or Candidate Key).   (CORRECT)

 C-------->A (then C is either Super key or Candidate Key).  AND  (A IS PRIME ATTRIBUTE ) ( NOT CORRECT)

Go for 3nf

 

AB----->C (then AB is either Super key or Candidate Key).   (CORRECT)

 C-------->A (then C is either Super key or Candidate Key).  OR  (A IS PRIME ATTRIBUTE ) (  CORRECT)

SO the given functional dependencies: is 3NF, NOT BCNF

Related questions

24 votes
24 votes
3 answers
1
Kathleen asked Oct 8, 2014
4,415 views
Consider the relation scheme.$$\begin{array}{ll} \text{AUTHOR} & \text{(ANAME, INSTITUTION, ACITY, AGE)} \\\hline \text{PUBLISHER} & \text{(PNAME, PCITY)} \\\hline \te...
7 votes
7 votes
1 answer
2
go_editor asked Feb 12, 2018
2,512 views
What is the equivalent minimal Boolean expression (in sum of products form) for the Karnaugh map given below?
25 votes
25 votes
3 answers
3
Kathleen asked Oct 8, 2014
3,638 views
What is the number of binary trees with $3$ nodes which when traversed in post-order give the sequence $A, B, C ?$ Draw all these binary trees.
27 votes
27 votes
5 answers
4
Kathleen asked Oct 8, 2014
6,465 views
Let $G_1$ and $G_2$ be subgroups of a group $G$.Show that $G_1 \cap G_2$ is also a subgroup of $G$.Is $G_1 \cup G_2$ always a subgroup of $G$?.