BoyceCodd Normal Form / BCNF Explained / What is BCNF? / BoyceCodd Normal Form with Example / A relation which is in 3NF but not in BCNF Example.
BoyceCodd Normal Form (BCNF)
BCNF was jointly proposed by Raymond F. Boyce and Edgar F. Codd in the
year 1974 to address certain types of anomaly which were present even after the
schema is normalized to 3NF.
For a relation schema which is in 3NF, the presence of modification anomalies which could not be treated well by
3NF is due to one of the following reasons;
Reason 1: A relation schema might contain more than one candidate
keys.
Reason 2: In case of more than one candidate keys presents, all of
them might be composite.
Reason 3: If the above two reasons exist, then there is a possibility
of overlap between the candidate keys.
A relation schema R is in BCNF if and only if,
 For all the Functional Dependencies (FDs) hold in the relation R, if the FD is nontrivial then the determinant (LHS of FD) of that FD should be a Super key.
Through this definition, BCNF insists that all the determinants of any
Functional Dependency must be a Candidate Key. Due to this reason, BCNF sometimes
referred as strict 3NF.
Note: A relation schema R which is in BCNF also in 3NF automatically.
But, a relation schema R which is in 3NF need not be in BCNF.
Explanation:
If we have set of FDs in R such that X → Y, then X must be a super key. In other words, if X
is not a key, then the relation R is not in BCNF.
A relation which is not in BCNF.
RegNo

SName

Gen

PR

Phone

R1

Sundar

M

BTech

9898786756

R2

Ram

M

MS

9897786776

R3

Karthik

M

MCA

8798987867

R4

John

M

BSc

7898886756

R1

Sundar

M

BTech

9898781158

R3

Karthik

M

MCA

8798987867

Table 1 – STUDENT
Let us find out the set F of FDs holding on Table 1 STUDENT.
F = { RegNo → SName, RegNo →
Gen, RegNo → PR }
But, Phone cannot be uniquely identified by RegNo. Hence, the key for
this relation is a composite key (RegNo, Phone). We can write the set F of FDs
as follows;
F = { (RegNo → SName Gen PR), (RegNo Phone →
SName Gen PR) }
Among these two FDs, the first one which is having RegNo as the LHS is violating the rule of
BCNF. The reason is, the determinant
RegNo of the FD is not a key for this relation. Hence, the above table is
not in BCNF.
How do we convert a table STUDENT into BCNF table?
We have to decompose the table into two or more tables as discussed in
2NF, and 3NF. For this, we can consider the set F of FDs already listed above.
Hence, the resultant schemas would be;
 STUDENT(RegNo, SName, Gen, PR)
 STU_PHONE(RegNo, Phone)
Now, the relation STUDENT with the FD {RegNo → SName Gen PR}
satisfies,
1NF (because no repeating group of values/no
multivalued attributes),
2NF (because no partial key FD as the key is simple
attribute),
3NF (because there
is no transitive FD since there is no FDs with nonkey attribute on the LHS), BCNF
(because there is only one key for the relation).
For the relation, STU_PHONE the key is trivial. That is, composite key
(RegNo, Phone). Hence, it is holding all the Normal Forms in it.
Note: The relation schema with the following properties need not be
checked for every normal form conditions;
1. If only two attributes are present
2. If more than one attribute present and the key is simple attribute.
A relation schema which is in 3NF but not in BCNF
Let us take the following table, DEALER for our discussion. This table
stores the information about various dealers for various products. And it
stores the dealer details along with the products we purchase from them and the
product count. According to the sample table given below, it is very evident
that same products may be purchased from different dealers.
Dealer_ID

DName

Product_ID

Quantity

D101

Sun Flower

P101

100

D101

Sun Flower

P103

150

D102

Market One

P102

200

D105

A One

P101

100

D104

Indian Curry

P110

250

D104

Indian Curry

P102

50

Table 2 – DEALER
For this table, let us identify the set of Functional Dependencies;
1. Attribute Dealer_ID
cannot be a determinant for this table. Because, it does not uniquely identify
any other attribute other than dealer name (DName).
Hence,
Dealer_ID → DName
2. Attribute DName determine
the value of Dealer_ID. That is, dealers have unique names. Hence,
DName →
Dealer_ID
3. Attribute Product_ID cannot uniquely identify anything. Because, same
products purchased from different dealers in different quantify.
If we go on analyzing this, we would get the following set of FDs;
Dealer_ID → DName
DName → Dealer_ID
Dealer_ID Product_ID → DName Quantity
DName Product_ID → Dealer_ID Quantity
From the above set of FDs, we can derive Two Candidate Keys. Those are;
(Dealer_ID,
Product_ID) and (DName, Product_ID)
From the derived FDs, it is clear that the relation DEALER is in 3NF,
because, no Transitive Functional Dependency is held.
According to the properties of BCNF, the relation DEALER must have FDs
with the LHS as candidate keys. But, for DEALER among the set of FDs, we have
the FDs Dealer_ID →
DName and DName →
Dealer_ID where neither Dealer_ID nor DName are the candidate keys. Thus, these
two FDs violate the rules of BCNF. Hence, DEALER is not in BCNF.
How do we convert DEALER into BCNF table?
We need to decompose DEALER using the FDs which are violating the BCNF
rules. Thus, we get following schemas as a result of decomposition.
 DEALER1(Dealer_ID, DName)
 DEALER_PROD(Dealer_ID, Product_ID, Quantity)
Relation DEALER1 satisfies all the Normal Forms.
Relation DEALER_PROD has the following set of FDs;
Dealer_ID Product_ID → Quantity
As there are no other FDs, the relation DEALER_PROD is in BCNF.
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