Chapter 6 HW
_______ means putting a relation into a higher normal form.
Normalization
If a relation is 3NF but not BCNF, it must have a non-trivial functional dependency X → A such that
X is not a superkey
We can ensure that a binary decomposition is lossless if the set of common attributes in the two relations is
a superkey of one of the relations
A relation is first normal form if
all of these
In relational decomposition of a universal relation, the property of dependency preservation requires that
all the attributes on both sides of each dependency appear in the same relation
The normalization process that starts with a universal relation and uses decomposition to produce a set of normalized relations is called
analysis
A relation is in first normal form (1NF) if, and only if
every attribute is single-valued for each tuple.
In relational decomposition of a universal relation, the property of attribute preservation requires that
every attribute of the universal relation appears in at least one relation
A relation is Boyce-Codd Normal Form if
every determinant is a superkey
A relation is second normal form if it is 1NF and
every nonkey attribute is dependent on the entire key
If X and Y are sets of attributes of relation R, we say that Y is functionally dependent on X if
for each X value there is only one Y value
If a relational decomposition is not lossless, doing joins to recreate the original table may produce a table that
has more rows than the original table
The standard decomposition algorithm for BCNF requires
identifying and removing each violation of BCNF from a universal relation
In the relational model, if A functional determines B, it means that
if two rows have the same A value, they must have the same B value
Which of the following is not one of the major objectives of normalization?
improving efficiency
In the relational model, a spurious tuple is one that
is created by a lossy join
It is always possible to find a Boyce-Codd Normal Form decomposition that
is lossless, but may not preserve dependencies
For a relational decomposition, the property of being able to get back exactly the original relation by joins is
lossless decomposition
A functional dependency is actually a _______ relationship from attribute set A to attribute set B.
many-to-one
A relation having only one candidate key is third normal form if it is 2NF and
no nonkey attribute is dependent on another nonkey attribute
If set S{A,B,C} is a candidate key for the relation R(A,B,C,D,E), then which of the following must be true
none of these
In the relational model, every determinant is always
none of these
In the relational model, all of the following are types of dependencies that have been shown to lead to update anomalies EXCEPT
simple
Using an E-R diagram can be helpful in designing a normalized schema because
standard mapping of an E-R diagram to a relational model results in a schema that is close to normalized
The normalization process in which we find and group together functional dependencies with the same determinant is called
synthesis
Examining an instance of a relation can be sufficient to prove
that is it not in 3NF
In the relational model, a projection is lossless if
the join produces the same tuples as the original relation
If a relation is 2NF but not 3NF, it must have which type of functional dependency?
transitive
A functional dependency in which the attributes on the right hand side are included in the determinant is called
trivial
If set S{A,B,C} is a superkey for the relation R(A,B,C,D,E), then
{A,B,C,D} is also a superkey for R
It is always possible to find a dependency preserving lossless decomposition for _______.
3NF
The highest normal form that always allows us to preserve functional dependencies is
3NF
In the relation R( A, B, C, D), having the composite key {A,B}, which of the following FDs would demonstrate that the relation is not 2NF?
A → C
In a relation R( A,B,C,D), which of the following would prove that R is not 3NF?
B → C
In the relation R( A, B, C, D), having the composite key {A,B}, which of the following FDs would demonstrate that the relation is not 3NF?
C → D
In the relation R(A,B,C,D) if D is multivalued and the apparent key is A, which of the following is NOT an acceptable way to create a first normal form table for this schema?
Decompose R into R1(A,B,C,D) and R2(D)
All of the following are reasons for leaving a schema in a lower normal form except
Desire to reduce replication