1. 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: { , , , , , , , |  |, | , ,  }

2. RelAlg: 2  – Selection General Form:  Examples:  Cost > 75 r  ArrivalDate = 10 May  NrDays > 2 s RoomNr Name NrBeds Cost --------------------------------------- 1 Kennedy 2 90 2 Nixon 2 80 3 Carter 2 80 GuestNr RoomNr ArrivalDate NrDays ---------------------------------------------------- 102 3 10 May 5

3. RelAlg: 3  – Projection General Form:  Examples:  City g  GuestNr,RoomNr s City ------ Boston Hartford Providence GuestNr RoomNr ------------------------ 101 1 101 2 101 3 102 3 103 1 104 4 105 1 106 2

4. 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 ------------------------ 101 1 102 3 104 4

5. RelAlg: 5 Set Operators: , ,  Name -------- Carter Green RoomNr ----------- 5 RoomNr ----------- 4 5 1 3  Name r   Name g  RoomNr r   RoomNr s  RoomNr  Cost < 75 r   RoomNr  ArrivalDate = 10 May s Note: schemes must be “union compatible.”

6. 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 --------------

7. RelAlg: 7  – Cross Product  NrBeds r   RoomNr s  NrBeds ---------- 2 1 RoomNr ----------- 1 2 3 4 NrBeds RoomNr ----------------------- 2 1 2 2 2 3 2 4 1 1 1 2 1 3 1 4 = Note: The intersection of the schemes must be empty.

8. 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 ------------------------------------------------------------------------------- 3 Carter 2 80 102 10 Main Hartford 5 Green 1 50 105 10 Main Boston

9. RelAlg: 9 Natural Join – Examples A B ------ 1 2 3 2 4 5 6 7 B C ------ 1 2 2 3 5 7 5 8 A B C ---------- 1 2 3 3 2 3 4 5 7 4 5 8 A B ------ 1 2 3 4 C D ------ 1 3 2 4 A B C D -------------- 1 2 1 3 1 2 2 4 3 4 1 3 3 4 2 4 A B C ---------- 1 2 3 2 2 3 4 5 6 B C D ---------- 2 3 4 5 6 7 2 6 0 A B C D -------------- 1 2 3 4 2 2 3 4 4 5 6 7 = = = |||| |||| ||||

10. RelAlg: 10 |  |  – Theta-Join  NrBeds r |  | NrBeds<RoomNr  RoomNr s |  | NrBeds<RoomNr NrBeds ---------- 2 1 RoomNr ----------- 1 2 3 4 NrBeds RoomNr ----------------------- 2 3 2 4 1 2 1 3 1 4 = Note: Like cross-product, the intersection of the schemes must be empty.

11. RelAlg: 11 , ,  – Outerjoins  GuestNr,RoomNr  ArrivalDate = 10 May s   RoomNr,Name  NrBeds = 2 r GuestNr RoomNr ------------------------ 101 1 102 3 104 4 RoomNr Name ------------------------ 1 Kennedy 2 Nixon 3 Carter GuestNr RoomNr Name ------------------------------------- 101 1 Kennedy 102 3 Carter 104 4   2 Nixon =  Note: The outer join retains all “dangling tuples.”

12. RelAlg: 12 , ,  – Outerjoins  GuestNr,RoomNr  ArrivalDate = 10 May s   RoomNr,Name  NrBeds = 2 r GuestNr RoomNr ------------------------ 101 1 102 3 104 4 RoomNr Name ------------------------ 1 Kennedy 2 Nixon 3 Carter GuestNr RoomNr Name ------------------------------------- 101 1 Kennedy 102 3 Carter 104 4  =  Note: The left outer join retains “dangling tuples” on the left.

13. RelAlg: 13 , ,  – Outerjoins  GuestNr,RoomNr  ArrivalDate = 10 May s   RoomNr,Name  NrBeds = 2 r GuestNr RoomNr ------------------------ 101 1 102 3 104 4 RoomNr Name ------------------------ 1 Kennedy 2 Nixon 3 Carter GuestNr RoomNr Name ------------------------------------- 101 1 Kennedy 102 3 Carter  2 Nixon =  Note: The right outer join retains “dangling tuples” on the right.

14. 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 ------------------------ 101 1 102 3 104 4 RoomNr Name ------------------------ 1 Kennedy 2 Nixon 3 Carter GuestNr RoomNr ------------------------- 101 1 102 3 = ×

15. RelAlg: 15  – Division  GuestNr,RoomNr s   RoomNr  NrBeds = 2 r GuestNr RoomNr ------------------------ 101 1 101 2 101 3 102 3 103 1 104 4 105 1 106 2 RoomNr ----------- 1 2 3 GuestNr ----------- 101 = 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.

16. 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)

17. 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).

18. 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

19. 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