CS6400 - Exam Study Guide Questions
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Entity type ET with single value property - ✔✔relation ET has B as an attribute
Entity type ET with identifying property A - ✔✔relation ET there will be an attribute A that
is a primary key
Entity ET with composite C, made up D and E - ✔✔Relation only includes D and E, C is lost
Relation model is inherently: - ✔✔flat!
Entity ET with identifying property A and multi-value property F - ✔✔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 - ✔✔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 - ✔✔A in ET2 (where the relationship is mandatory) is a
foreign key to A in ET1
1:N relationship mapping - ✔✔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 - ✔✔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 - ✔✔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 - ✔✔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 - ✔✔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 - ✔✔A in ET1 and ET2 are foreign keys to A in ET
non-mandatory/disjoint - ✔✔A in ET1 and ET2 are foreign keys to A in ET
union type - ✔✔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 - ✔✔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
Solved 100% Correct | Verified Answers
Entity type ET with single value property - ✔✔relation ET has B as an attribute
Entity type ET with identifying property A - ✔✔relation ET there will be an attribute A that
is a primary key
Entity ET with composite C, made up D and E - ✔✔Relation only includes D and E, C is lost
Relation model is inherently: - ✔✔flat!
Entity ET with identifying property A and multi-value property F - ✔✔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 - ✔✔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 - ✔✔A in ET2 (where the relationship is mandatory) is a
foreign key to A in ET1
1:N relationship mapping - ✔✔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 - ✔✔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 - ✔✔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 - ✔✔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 - ✔✔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 - ✔✔A in ET1 and ET2 are foreign keys to A in ET
non-mandatory/disjoint - ✔✔A in ET1 and ET2 are foreign keys to A in ET
union type - ✔✔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 - ✔✔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
