CS 6400 Exam 2 UPDATED ACTUAL
Questions and CORRECT Answers
Entity type ET with single value property - CORRECT ANSWER - relation ET has B as
an attribute
Entity type ET with identifying property A - CORRECT ANSWER - relation ET there will
be an attribute A that is a primary key
Entity ET with composite C, made up D and E - CORRECT ANSWER - Relation only
includes D and E, C is lost
Relation model is inherently: - CORRECT ANSWER - flat!
Entity ET with identifying property A and multi-value property F - CORRECT
ANSWER - For each F a brand new relation is created, called ET-F. insert identifying
property so combination of A and F is a composite key in ET-F
1:1 relational mapping - CORRECT ANSWER - B in ET1 is a foreign key to B in ET2. or
A in ET2 is a foreign key to A in ET1. both are acceptable.
1:1 mandatory relationship - CORRECT ANSWER - A in ET2 (where the relationship is
mandatory) is a foreign key to A in ET1
1:N relationship mapping - CORRECT ANSWER - A in ET2 (the many side) is a foreign
key to A in ET1 (the one side). A in ET1 is unique for every single value of B
N:M Relation - CORRECT ANSWER - There needs to be a relation,R, for the relationship
that has a foreign key for each attribute. A in R is a foreign key to A in ET1 and B in R is a
foreign key to B in ET2
, Identifying relationship R to weak Entity ET2 - CORRECT ANSWER - ET2 requires a
reference to ET1, A in ET2 is a foreign key to A in ET1. A and B in combination are required to
define ET2
Mandatory/disjoint relation for super/sup type - CORRECT ANSWER - ET1 and ET2 are
disjoint so every ET distance has an A and B and every instance is hooked to ET1 or ET2. ET1
will inherit A and B and have C, ET2 will also and have D. only create a relation for ET1 and
ET2, not ET. ET1 is A, B, C; ET2 is A, B, D.
Mandatory Overlap relation - CORRECT ANSWER - Every instance of ET is an instance
of ET1, ET2, or both. A in ET1 and ET2 are foreign keys to A in ET. ET is A,B. ET1 is A, C.
ET2 is A, D
non-mandatory overlap relation - CORRECT ANSWER - A in ET1 and ET2 are foreign
keys to A in ET
non-mandatory/disjoint - CORRECT ANSWER - A in ET1 and ET2 are foreign keys to A
in ET
union type - CORRECT ANSWER - ET is a subset of the union of ET1 and ET2. ET-ID
in ET1 and ET2 are foreign keys to ET-ID in ET
closed algebra - CORRECT ANSWER - each time there is an operation, you get a rational
number back. this allows you to formulate high level models. You can use previous thoughts to
form new thoughts for high level models
Relational algebra operators - CORRECT ANSWER - union
intersection
set difference
Cartesian product
projection
selection
Questions and CORRECT Answers
Entity type ET with single value property - CORRECT ANSWER - relation ET has B as
an attribute
Entity type ET with identifying property A - CORRECT ANSWER - relation ET there will
be an attribute A that is a primary key
Entity ET with composite C, made up D and E - CORRECT ANSWER - Relation only
includes D and E, C is lost
Relation model is inherently: - CORRECT ANSWER - flat!
Entity ET with identifying property A and multi-value property F - CORRECT
ANSWER - For each F a brand new relation is created, called ET-F. insert identifying
property so combination of A and F is a composite key in ET-F
1:1 relational mapping - CORRECT ANSWER - B in ET1 is a foreign key to B in ET2. or
A in ET2 is a foreign key to A in ET1. both are acceptable.
1:1 mandatory relationship - CORRECT ANSWER - A in ET2 (where the relationship is
mandatory) is a foreign key to A in ET1
1:N relationship mapping - CORRECT ANSWER - A in ET2 (the many side) is a foreign
key to A in ET1 (the one side). A in ET1 is unique for every single value of B
N:M Relation - CORRECT ANSWER - There needs to be a relation,R, for the relationship
that has a foreign key for each attribute. A in R is a foreign key to A in ET1 and B in R is a
foreign key to B in ET2
, Identifying relationship R to weak Entity ET2 - CORRECT ANSWER - ET2 requires a
reference to ET1, A in ET2 is a foreign key to A in ET1. A and B in combination are required to
define ET2
Mandatory/disjoint relation for super/sup type - CORRECT ANSWER - ET1 and ET2 are
disjoint so every ET distance has an A and B and every instance is hooked to ET1 or ET2. ET1
will inherit A and B and have C, ET2 will also and have D. only create a relation for ET1 and
ET2, not ET. ET1 is A, B, C; ET2 is A, B, D.
Mandatory Overlap relation - CORRECT ANSWER - Every instance of ET is an instance
of ET1, ET2, or both. A in ET1 and ET2 are foreign keys to A in ET. ET is A,B. ET1 is A, C.
ET2 is A, D
non-mandatory overlap relation - CORRECT ANSWER - A in ET1 and ET2 are foreign
keys to A in ET
non-mandatory/disjoint - CORRECT ANSWER - A in ET1 and ET2 are foreign keys to A
in ET
union type - CORRECT ANSWER - ET is a subset of the union of ET1 and ET2. ET-ID
in ET1 and ET2 are foreign keys to ET-ID in ET
closed algebra - CORRECT ANSWER - each time there is an operation, you get a rational
number back. this allows you to formulate high level models. You can use previous thoughts to
form new thoughts for high level models
Relational algebra operators - CORRECT ANSWER - union
intersection
set difference
Cartesian product
projection
selection
