CS 6400 EXAM 3 QUESTIONS AND ANSWERS
GRADED A+
Spurious tuples - ANSWER -Tuples generated by joining two relations
on attributes that are not keys or foreign keys on these relations.
Waste of storage space - ANSWER -due to NULLs and the difficulty of
performing selections, aggregation operations, and joins due to NULL
values
A functional dependency (A->B) - ANSWER -occurs when the
attribute A uniquely determines B. A = left-hand side, B=right-hand-
side.
-is a property of the semantics or meaning of the attributes.
FD or f.d. - ANSWER -abbreviation for functional dependency
Candidate Key - ANSWER -An attribute, or combination of attributes,
that uniquely identifies a row in a relation.
legal relation states (or legal extensions) of R - ANSWER -Relation
extensions r(R) that satisfy the functional dependency constraints
FD cannot be inferred automatically - ANSWER -must be defined
explicitly by someone who knows the semantics of the attributes of R.
diagrammatic notation - ANSWER -isplaying FDs: Each FD is
displayed as a horizontal
, Normalization of Data - ANSWER -a process of analyzing the given
relation schemas based on their FDs and primary keys to achieve the
desirable properties of (1) minimizing redundancy and (2) minimizing
the insertion, deletion, and update anomalies
Boyce-Codd Normal Form (BCNF) - ANSWER -A special type of
third normal form (3NF) in which every determinant is a candidate key.
A table in BCNF must be in 3NF.
normal form test - ANSWER -is decomposed into smaller relation
schemas that contain a subset of the attributes and meet the test that was
otherwise not met by the original relation.
nonadditive join - ANSWER -or lossless join property, guarantees that
the spurious tuple generation problem does not occur
dependency preservation property - ANSWER -ensures that each
functional dependency is represented in some individual relation
resulting after decomposition
superkey - ANSWER -any key that uniquely identifies each row
key - ANSWER -of a relation schema R = {A1, A2, ... , An} is a set of
attributes S ⊆ R with the property that no two tuples t1 and t2 in any
legal relation state r of R will have t1[S] = t2[S].
Candidate Key - ANSWER -An attribute, or combination of attributes,
that uniquely identifies a row in a relation.
, primary key - ANSWER -A field (or group of fields) that uniquely
identifies a given entity in a table
prime attribute - ANSWER -A key attribute; that is, an attribute that is
part of a key or is the whole key.
nonprime attribute - ANSWER -An attribute that is not part of a key.
First Normal Form (1NF) - ANSWER -A relation that has a primary
key and in which there are no repeating groups. 1NF disallows relations
within relations or relations as attribute values within tuples. all domain
values are atomic.
Second Normal Form (2NF) - ANSWER -is based on the concept of
full functional dependency. a relation is in 1NF and there are no partial
dependencies. every nonkey attribute is fully dependent on the key.
full functional dependency - ANSWER -removal of any attribute A
from X means that the dependency does not hold anymore
partial dependency - ANSWER -if some attribute A ε X can be
removed from X and the dependency still holds
Third Normal Form (3NF) - ANSWER -is based on the concept of
transitive dependency. R is 2NF and every nonkey attribute is non-
transitively dependent on the key.
transitive dependency - ANSWER -if X → Y and Y → Z, then X → Z
, general alternative definition of 3NF - ANSWER -A relation schema R
is in 3NF if every nonprime attribute of R meets both of the following
conditions: It is fully functionally dependent on every key of R.
all-key relation - ANSWER -obvious redundancy in the relation
X → → Y - ANSWER -X multi-determines Y
Fourth Normal Form (4NF) - ANSWER -A table is in 4NF if it is in
3NF and contains no multiple independent sets of multivalued
dependencies.
join dependency (JD) - ANSWER -specifies a constraint on the states r
of R.
Fifth Normal Form (5NF) - ANSWER -A normal form necessary to
eliminate an anomaly where a table can be split apart but not correctly
joined back together.
Also known as Project-Join Normal Form (PJ/NF).
relational design by synthesis - ANSWER -bottom-up approach to
design that presupposes that the known functional dependencies among
sets of attributes in the Universe of Discourse (UoD) have been given as
input.
properties of decompositions - ANSWER -the dependency preservation
property and the nonadditive (or lossless) join property
universal relation - ANSWER -hypothetical relation containing all the
attributes
GRADED A+
Spurious tuples - ANSWER -Tuples generated by joining two relations
on attributes that are not keys or foreign keys on these relations.
Waste of storage space - ANSWER -due to NULLs and the difficulty of
performing selections, aggregation operations, and joins due to NULL
values
A functional dependency (A->B) - ANSWER -occurs when the
attribute A uniquely determines B. A = left-hand side, B=right-hand-
side.
-is a property of the semantics or meaning of the attributes.
FD or f.d. - ANSWER -abbreviation for functional dependency
Candidate Key - ANSWER -An attribute, or combination of attributes,
that uniquely identifies a row in a relation.
legal relation states (or legal extensions) of R - ANSWER -Relation
extensions r(R) that satisfy the functional dependency constraints
FD cannot be inferred automatically - ANSWER -must be defined
explicitly by someone who knows the semantics of the attributes of R.
diagrammatic notation - ANSWER -isplaying FDs: Each FD is
displayed as a horizontal
, Normalization of Data - ANSWER -a process of analyzing the given
relation schemas based on their FDs and primary keys to achieve the
desirable properties of (1) minimizing redundancy and (2) minimizing
the insertion, deletion, and update anomalies
Boyce-Codd Normal Form (BCNF) - ANSWER -A special type of
third normal form (3NF) in which every determinant is a candidate key.
A table in BCNF must be in 3NF.
normal form test - ANSWER -is decomposed into smaller relation
schemas that contain a subset of the attributes and meet the test that was
otherwise not met by the original relation.
nonadditive join - ANSWER -or lossless join property, guarantees that
the spurious tuple generation problem does not occur
dependency preservation property - ANSWER -ensures that each
functional dependency is represented in some individual relation
resulting after decomposition
superkey - ANSWER -any key that uniquely identifies each row
key - ANSWER -of a relation schema R = {A1, A2, ... , An} is a set of
attributes S ⊆ R with the property that no two tuples t1 and t2 in any
legal relation state r of R will have t1[S] = t2[S].
Candidate Key - ANSWER -An attribute, or combination of attributes,
that uniquely identifies a row in a relation.
, primary key - ANSWER -A field (or group of fields) that uniquely
identifies a given entity in a table
prime attribute - ANSWER -A key attribute; that is, an attribute that is
part of a key or is the whole key.
nonprime attribute - ANSWER -An attribute that is not part of a key.
First Normal Form (1NF) - ANSWER -A relation that has a primary
key and in which there are no repeating groups. 1NF disallows relations
within relations or relations as attribute values within tuples. all domain
values are atomic.
Second Normal Form (2NF) - ANSWER -is based on the concept of
full functional dependency. a relation is in 1NF and there are no partial
dependencies. every nonkey attribute is fully dependent on the key.
full functional dependency - ANSWER -removal of any attribute A
from X means that the dependency does not hold anymore
partial dependency - ANSWER -if some attribute A ε X can be
removed from X and the dependency still holds
Third Normal Form (3NF) - ANSWER -is based on the concept of
transitive dependency. R is 2NF and every nonkey attribute is non-
transitively dependent on the key.
transitive dependency - ANSWER -if X → Y and Y → Z, then X → Z
, general alternative definition of 3NF - ANSWER -A relation schema R
is in 3NF if every nonprime attribute of R meets both of the following
conditions: It is fully functionally dependent on every key of R.
all-key relation - ANSWER -obvious redundancy in the relation
X → → Y - ANSWER -X multi-determines Y
Fourth Normal Form (4NF) - ANSWER -A table is in 4NF if it is in
3NF and contains no multiple independent sets of multivalued
dependencies.
join dependency (JD) - ANSWER -specifies a constraint on the states r
of R.
Fifth Normal Form (5NF) - ANSWER -A normal form necessary to
eliminate an anomaly where a table can be split apart but not correctly
joined back together.
Also known as Project-Join Normal Form (PJ/NF).
relational design by synthesis - ANSWER -bottom-up approach to
design that presupposes that the known functional dependencies among
sets of attributes in the Universe of Discourse (UoD) have been given as
input.
properties of decompositions - ANSWER -the dependency preservation
property and the nonadditive (or lossless) join property
universal relation - ANSWER -hypothetical relation containing all the
attributes
