RelAlg: 1 Relational Algebra An Algebra is a pair: (set of values, set of operations) Note that an algebra is the same idea as an ADT Relational Algebra: (relations, relational operators) –set of values = relations –set of operations = relational operators Relational Operators: { , , , , , , , |  |, | , ,  }

RelAlg: 2  – Selection General Form:  Examples:  Cost > 75 r  ArrivalDate = 10 May  NrDays > 2 s RoomNr Name NrBeds Cost Kennedy Nixon Carter 2 80 GuestNr RoomNr ArrivalDate NrDays May 5

RelAlg: 3  – Projection General Form:  Examples:  City g  GuestNr,RoomNr s City Boston Hartford Providence GuestNr RoomNr

RelAlg: 4 Closed Set of Operators Results are relations Closed implies we can nest operations in expressions Example:  GuestNr,RoomNr  ArrivalDate = 10 May s GuestNr RoomNr

RelAlg: 5 Set Operators: , ,  Name Carter Green RoomNr RoomNr  Name r   Name g  RoomNr r   RoomNr s  RoomNr  Cost < 75 r   RoomNr  ArrivalDate = 10 May s Note: schemes must be “union compatible.”

RelAlg: 6  – Renaming General Form:   Examples:  RoomName  Name  RoomName  Cost < 75 r  Nr, Name  RoomNr  Nr r   Nr, Name  GuestNr  Nr g Note: The old attribute must be in the scheme and the new attribute must not be in the scheme. RoomName Blue Green Nr Name

RelAlg: 7  – Cross Product  NrBeds r   RoomNr s  NrBeds RoomNr NrBeds RoomNr = Note: The intersection of the schemes must be empty.

RelAlg: 8 |  | – Natural Join r |  | g =  RoomNr,Name,NrBeds,Cost,GuestNr,StreetNr,City  Name = Name (r   Name  Name g) RoomNr Name NrBeds Cost GuestNr StreetNr City Carter Main Hartford 5 Green Main Boston

RelAlg: 9 Natural Join – Examples A B B C A B C A B C D A B C D A B C B C D A B C D = = = |||| |||| ||||

RelAlg: 10 |  |  – Theta-Join  NrBeds r |  | NrBeds<RoomNr  RoomNr s |  | NrBeds<RoomNr NrBeds RoomNr NrBeds RoomNr = Note: Like cross-product, the intersection of the schemes must be empty.

RelAlg: 11 , ,  – Outerjoins  GuestNr,RoomNr  ArrivalDate = 10 May s   RoomNr,Name  NrBeds = 2 r GuestNr RoomNr RoomNr Name Kennedy 2 Nixon 3 Carter GuestNr RoomNr Name Kennedy Carter   2 Nixon =  Note: The outer join retains all “dangling tuples.”

RelAlg: 12 , ,  – Outerjoins  GuestNr,RoomNr  ArrivalDate = 10 May s   RoomNr,Name  NrBeds = 2 r GuestNr RoomNr RoomNr Name Kennedy 2 Nixon 3 Carter GuestNr RoomNr Name Kennedy Carter  =  Note: The left outer join retains “dangling tuples” on the left.

RelAlg: 13 , ,  – Outerjoins  GuestNr,RoomNr  ArrivalDate = 10 May s   RoomNr,Name  NrBeds = 2 r GuestNr RoomNr RoomNr Name Kennedy 2 Nixon 3 Carter GuestNr RoomNr Name Kennedy Carter  2 Nixon =  Note: The right outer join retains “dangling tuples” on the right.

RelAlg: 14 |× – Semijoin Where might the semijoin be useful? (optimization of distributed DB systems)  GuestNr,RoomNr  ArrivalDate = 10 May s ×  RoomNr,Name  NrBeds = 2 r GuestNr RoomNr RoomNr Name Kennedy 2 Nixon 3 Carter GuestNr RoomNr = ×

RelAlg: 15  – Division  GuestNr,RoomNr s   RoomNr  NrBeds = 2 r GuestNr RoomNr RoomNr GuestNr = Note: The scheme of the divisor must be a proper subset of the scheme of the dividend. The scheme of the quotient is the difference.   A point easy to overlook: The division into groups is by the attributes in the dividend that are not in the divisor  GuestNr in our example. As an example of incorrectly setting this up, suppose we don’t project on GuestNr,RoomNr, then the division would be by GuestNr,ArrivalDate,NrDays, which would make the answer empty.

RelAlg: 16 Query Examples List names and cities of guests arriving on 15 May.  Name,City  ArrivalDate = 15 May (g |  | s) List names of each guest who has a reservation for a room that has the same name as the guest’s name.  Name (g |  | s |  | r) List names of guests who have a reservation for rooms with two beds.  Name (g |  | s |  |  RoomNr  NrBeds = 2 r)

RelAlg: 17 More Query Examples List the names of guests from Hartford who are arriving after 10 May.  Name (  GuestNr,Name  City = Hartford g |  |  GuestNr  ArrivalDate > 10 May s) List names of rooms for which no guest is arriving on 10 May.  Name (r |  | (  RoomNr r   RoomNr  ArrivalDate = 10 May s))  RoomNr r   ArrivalDate = 10 May s List all rooms along with their reservations for May 10 th (if any).

RelAlg: 18 Even More Query Examples List the names of guests who have reservations for all presidential suites (i.e., those that have two beds).  Name (g |  | (  GuestNr s   GuestNr ((  GuestNr s |  |  RoomNr  NrBeds = 2 r)   GuestNr,RoomNr s))) all possible GuestNr- RoomNr combinations minus the reservations we have guests who don’t have reservations for all presidential suites guests who do  Name (g |  | (  GuestNr,RoomNr s   RoomNr  NrBeds = 2 r)) GuestNr’s whose RoomNr includes all RoomNr’s with 2 beds

RelAlg: 19 Are All Operators Necessary? Relational Operators: { , , , , , , , |  |, |  | , | ,  } A Complete Set of Operators: { , , , , , |  |} –Can do  with , , |  | –Can do  with |  |, |  |  with |  | and , and |  with |  | and  –Can do  with |  |, also with , i.e., (r  s = r  (r  s)) Why might this be of value? –Fewer operators to implement –Query rewriting for optimization