CS 6400 Exam 2 Questions and Correct Answers Latest Update 2024

CS 6400 Exam 2 Questions and Correct Answers Latest Update 2024 Entity type ET with single value property - Answers relation ET has B as an attribute Entity type ET with identifying property A - Answers relation ET there will be an attribute A that is a primary key Entity ET with composite C, made up D and E - Answers Relation only includes D and E, C is lost Relation model is inherently: - Answers flat! Entity ET with identifying property A and multi-value property F - Answers 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 - Answers 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 - Answers A in ET2 (where the relationship is mandatory) is a foreign key to A in ET1 1:N relationship mapping - Answers 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 - Answers 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 - Answers 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 - Answers 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 - Answers 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 - Answers A in ET1 and ET2 are foreign keys to A in ET non-mandatory/disjoint - Answers A in ET1 and ET2 are foreign keys to A in ET union type - Answers 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 - Answers 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 - Answers union intersection set difference Cartesian product projection selection natural join outer join theta-join divideby rename projection - Answers eliminates columns from a result selection - Answers eliminates rows from a result rename - Answers allows you to manipulate names of attributes of relations so join and set operators work in the way they're intended simple expresion - Answers take attribute name and compare to a constant or compare attribute name to another attribute name composite expression - Answers use AND (conjunction) use OR (dis junction) (expression) - change order of evaluation NOT = negation Relations are: - Answers sets! Union - Answers The two relations in a union, intersection, or set difference must be type compatible, i.e. same number of attributes and pairwise same types intersection - Answers conjunction, think the center of a venn diagram Set difference - Answers given two sets, the set difference is the set of elements of the first set