Venn Diagrams: Database Principles
Venn Diagrams: Database Principles
Venn Diagrams: Database Principles
Database Principles
Venn Diagram
Venn Diagrams are used to represent relationships between sets. They can also be used to represent set operations like union, intersection and set difference. Since Relational Algebra has operations corresponding to these set operations, Venn Diagrams are a useful design tool for Relational Algebra Queries.
union
Table1 Table2
Table1
intersection
Table2
set difference
Table1 Table2
Database Principles
Exercise:
Draw a Venn Diagram showing the following three sets: The set of all suppliers The set of suppliers of red parts The set of suppliers of non-red parts Label each part of the diagram.
All Suppliers
Database Principles
Query Types:
Hard Queries: These are queries that have words such as only, all and no in the query condition.
Find the suppliers of all/no/only red parts
What makes a query easy is that is can be answered with at most join, select and project operators. What makes a query hard is that you must use set difference or quotient to answer the query.
Database Principles
Easy Queries:
Venn Diagrams do not play much part in solving easy queries. The problem is too easy to need a design phase. Easy queries are solved by joining all necessary tables, selecting the rows of interest and projecting the columns in the answer.
Database Principles
Hard Queries:
Remember Pr(A) = 1 Pr(A)
In probability, it is often easier to calculate the probability of the complement of an event (A) than the probability of the event itself (A). This same approach can be used to solve hard queries. Asked to find something hard: Start by stating and finding its complement Then use the set difference operator to throw away the newly found complement You are left with what you want
Database Principles
Only Query
OnlyRedPartSuppliers)
Database Principles
Exercise:
In the previous Venn Diagram, find out where the Suppliers who supply nothing at all are located. It turns out they are not in the answer set. What if we want them in the answer set? Replace
PartSuppliers = SNo(Supplies)
with
PartSuppliers = SNo(Supplier)
Database Principles
Negation Queries:
Entity Properties: Location of a Supplier is a property modeled as an attribute What Parts a Supplier supplies is also a property modeled as a relationship
supplies
Supplier PK Sno Sname Location PK Part Pno Pdesc Colour
(0,n) O_date
(1,n)
Database Principles
(0,n) O_date
(1,n)
Find all entity instances where the value of Location is not Boston
Query Mechanism: Look at all the entity instances and as you find one where the value of Location is something other than Boston, pick out and return that instance as part of the answer. SuppliersNotFromBoston = Location != Boston (Supplier)
Database Principles
(0,n) O_date
(1,n)
Find all entity instances that do not participate in the <supplies> relationship.
Query Mechanism: Pick a Supplier. Look at every row in the Supplies table and if you fail to see that Supplier even once then consider that Supplier as part of the answer set. Then move on to a new Supplier. So instead of selecting members of the answer set looking at rows one at a time (which is what happens when we use join) we need to consider all the rows in Supplies as a set and see if a particular Supplier is completely missing or not. NonSuppliers = SNo(Supplier) \ SNo(Supplies)
Database Principles
Negation queries that negate the value of an attribute are easy and can be resolved using join, select & project. Negation queries that negate the participation in a relationship are hard and need set difference.
Table1 Table2
Database Principles
Quotient Queries:
Queries that contain all or every in the query condition. Venn diagrams not very useful.
Q = R/S where: Q = the key to the thing you are looking for S = the key to the thing described in the all condition R = a table that contains the above two keys and is related to the query. S = the list of key values that satisfy the all condition Example: Find the books reserved by all cardholders from New Paltz Q = {isbn}, since we are looking for books S = {borrowerid}, since borrowerid is the key for cardholders S = borrowerid(b_addr = New Paltz (Cardholder)) R = {isbn,borrowerid}, must combine Q and S. R = isbn,borrowerid(Reserves) Q = R/S = {the books reserved by all cardholders from New paltz}
Database Principles