CS

Transcription

1 CS Relational Algebra

2 Relational algebra Relational algebra is a formal system for manipulating relations Operands are relations Operations are: Set operations Union, intersection, difference, and Cartesian product Specific relational operations Selection, projection, rename, join, division... Five primitive relational algebra operators: Selection, projection, product, union, and difference 2

3 Selection σ CONDITION (RELATION) Selects all tuples from RELATION that meet CONDITION Conditions are comparisons of attributes of RELATION using =,, <,, >, Simple conditions can be combined using,, and 3

4 Selection example Enrollment Stud_ID Course CS Math Math Art 707 σ Stud_ID = (Enrollment) Stud_ID Course CS Math 336 σ Stud_ID > Stud_ID= (Enrollment) σ Stud_ID > Course Math 422 (Enrollment) Stud_ID Course Art 707 Stud_ID Course CS Math Math Art 707 4

5 Projection π ATTRIBUTE_LIST (RELATION) Projects a relation down to a subset of its attributes Discards all attributes not listed in ATTRIBUTE_LIST Note that ATTRIBUTE_LIST must be a subset of the attributes of RELATION 5

6 Projection examples Students π Name (Students) Name ID Major GPA Name Alice CS 3.45 Alice Bob Math 3.23 Bob Charlie CS 2.75 Charlie Denise Art 4.0 Denise π Major,GPA (Students) π Name,ID,GPA (Students) Major GPA CS 3.45 Math 3.23 CS 2.75 Art 4.0 Name ID GPA Alice Bob Charlie Denise

7 What are the results of these operations? Relations! Which means they can be used as operands to further operations π Name,GPA (σ Major="CS" (Students)) σ Stud_ID > (σ Course "Math 422" (Enrollment)) 7

8 Properties of σ and π σ cond1 (σ cond2 (r(r))) = σ cond2 (σ cond1 (r(r))) σ cond1 (σ cond2 (r(r))) = σ cond2 cond1 (r(r)) = σ cond1 cond2 (r(r)) π list1 (π list2 (r(r))) = π list1 (r(r)) Need to be careful with π π Name,GPA (σ Major="CS" (Students)) == σ Major="CS" (π Name,GPA (Students))? Can be sequenced CSM σ Major="CS" (Students) RESULT π Name,GPA (CSM) 8

9 Rename Rename result relation: RESULT(Name, QPA) π Name,GPA (σ Major="CS" (Students)) Rename all attributes of a relation ρ (a1, a2, a3, a4, a5) (RELATION) (a1, a2, a3, a4, a5) are new attribute names RELATION is the relation having its attributes renamed Renaming individual attributes ρ NEW/OLD (RELATION) E.g., ρ StudentName/Name (Students) 9

10 Set operation particulars Union, intersection, and difference operands must be union-compatible Must have the same arity Matching attributes must have same domain Consider r(r) s(s) for relations r(r) and s(s) with arities of n and m Result will have an arity of n + m Name conflicts can be resolved using relation name prefixes (e.g., R.a1, R.a2, S.a1, S.a2) Result is "flattened" compared to set product operation r(r) s(s) = {(r 1, r 2,, r n, s 1, s 2,, s m ) (r 1, r 2,, r n ) r(r), (s 1, s 2,, s m ) s(s)} NOT {((r 1, r 2,, r n ), (s 1, s 2,, s m )) (r 1, r 2,, r n ) r(r), (s 1, s 2,, s m ) s(s)} 10

11 Combining relations Natural join: r(r) * s(s) Example: Courses * Sections Courses Name CNum CS CS CS CS Sections CNum Location Instructor Lee Farnan Farnan Misurda Farnan Garrison 11

12 Result Name CNum Location Instructor CS Lee CS Farnan CS Farnan CS Misurda CS Farnan CS Garrison 12

13 Why do we need join? We have product! Consider the 5 primitive relational algebra operations: Selection, projection, product, union, and difference Why isn't intersection listed? Can be expressed as a difference a(a) b(b) = a(a) - (a(a) - b(b)) Why isn't join listed? Can be expressed in terms of other operators right? Consider relations r(r) and s(s) with arities of n and m Let r 1, r 2,, r i be the attributes unique to r(r) Let s 1, s 2,, s j be the attributes unique to s(s) Let c 1, c 2,, c k be the attributes common to r(r) and s(s) TMP ρ x1/c1 (ρ x2/c2 ( ρ xk/ck (s(s)) )) RESULT σ c1=x1 c2=x2 ck=xk (r(r) TMP) REAL_RES π(r 1, r 2,, r i, c 1, c 2,, c k, s 1, s 2,, s j )(RESULT) 13

14 What about joining Students and Enrollment? Enrollment Stud_ID Course CS Math Math Art 707 Students Name ID Major GPA Alice CS 3.45 Bob Math 3.23 Charlie CS 2.75 Denise Art 4.0 Need a way to specify how to join these tables Θ-join: r(r) CONDITION s(s) CONDITION uses Θ to compare attributes of R and S Θ can be <,, >,, =, E.g., Enrollment Stud_ID = ID Students When Θ is =, this is called an equijoin 14

15 Result of Enrollment Stud_ID = ID Students Stud_ID Course CS Math Math Art 707 Name ID Major GPA Alice CS 3.45 Alice CS 3.45 Bob Math 3.23 Denise Art

16 Expressing Θ-join in terms of primitive operations 16

17 Semijoins Result tuples from only left or right relation Depts Dept Chair CS Znati Physics Leibovich Math Yotov English Bialostosky Depts Faculty Dept Chair CS Znati Math Yotov Faculty Fname Lname Dept Adal Raja History Ivan Yotov Math Henry Block Statistics Daniel Mosse CS Depts Faculty Fname Lname Dept Ivan Yotov Math Daniel Mosse CS 17

18 Antijoin Tuples that don't have a match Depts Dept Chair CS Znati Physics Leibovich Math Yotov English Bialostosky Faculty Fname Lname Dept Adal Raja History Ivan Yotov Math Henry Block Statistics Daniel Mosse CS Depts Faculty Dept Chair Physics Leibovich English Bialostosky 18

19 Division Assume r(r) and s(s) such that S R, Partial tuples from r(r) that appear with every tuple from s(s) r(r) A B C D s(s) C D r(r) s(s) A B

20 Who follows everyone that Susan follows? Followers Follower Target Susan Alex Mark Alex Kirk Mary Tom Alex Susan Mary Tom Lory Kirk Chad Kirk Alex 20

21 What do we do if everything doesn't line up? Courses Name CNum CS CS CS CS CS Sections CNum Location Instructor Lee Farnan Farnan Misurda Farnan Garrison NULL 5313 NULL Courses CNum = CNum Sections? Courses * Sections? 21

22 Extended relational operations Practical extensions to classical relational algebra Outer join Outer union Generalized projection Aggregation Grouping 22

23 What if we want to view all courses? Left outer join Courses CNum = CNum Sections C.Name C.CNum S.CNum S.Location S.Instructor CS Lee CS Farnan CS Farnan CS Misurda CS Farnan CS Garrison CS NULL NULL NULL 23

24 What if we want to view all sections? Right outer join Courses CNum = CNum Sections C.Name C.CNum S.CNum S.Location S.Instructor CS Lee CS Farnan CS Farnan CS Misurda CS Farnan CS Garrison NULL NULL NULL 5313 NULL 24

25 Full outer join Courses Sections CNum = CNum C.Name C.CNum S.CNum S.Location S.Instructor CS Lee CS Farnan CS Farnan CS Misurda CS Farnan CS Garrison CS NULL NULL NULL NULL NULL NULL 5313 NULL 25

26 Outer union Equivalent to full outer join on all union compatible attributes Name CNum Location Instructor CS Lee CS Farnan CS Farnan CS Misurda CS Farnan CS Garrison CS NULL NULL NULL NULL 5313 NULL 26

27 Generalized projection Consider the the relation: Employee = {Ssn, Salary, Deduction, Years} How do you easily generate a report for all employees listing: The net salary as Salary-Deduction A total bonus of 2000$ for every year an employee has been with the company The 25% tax that must be collected? Include math in projection's attribute list: π Ssn, Salary-Deduction, 2000*Years, 0.25*Salary (Employee) Use with rename to avoid confusion: ρ (Ssn, Net_Salary, Bonus, Tax) (π Ssn, Salary-Deduction, 2000*Years, (Employee)) 27

28 Aggregation Applies a function to attributes from groups of tuples COUNT SUM AVERAGE MAXIMUM MINIMUM Again consider Employee = {Ssn, Salary, Deduction, Years} What is the longest someone has been with the company? F MAXIMUM Years (Employee) How much is the company paying in employee salaries? F (Employee) SUM Salary 28

29 Grouping Specifies a group of tuples to apply an aggregate function to Years F COUNT Ssn, AVERAGE Salary (Employee) You'll also want to combine this with rename: ρ (Years, Num_Emp, Mean_Salary) ( Years F COUNT Ssn, AVERAGE Salary (Employee)) 29

30 Limitations of relational algebra Consider Employee = {Fname, Lname, Ssn, Supervisor_ssn} How to get a list of all employees (directly or indirectly) supervised by John Smith? JSMITH σ Fname = "John" Lname = "Smith" (Employee) DIRECT Employee Supervisor_ssn = Ssn JSMITH INDIRECT1 Employee Supervisor_ssn = Ssn DIRECT RESULT DIRECT INDIRECT1 30

31 Queries Consider again: Students = {Name, ID, Major, GPA} Enrollment = {Stud_ID, Course} Courses = {Name, CNum} Sections = {CNum, Location, Instructor} What instructors are teaching courses taken by students with the highest GPA? MAXES Students GPA = MAXIMUM_GPA (F MAXIMUM GPA (Students)) CNAMES π Course (Enrollment Stud_ID = ID MAXES) CNUMS Courses Name = Course CNAMES RESULT π Instructor (Sections * CNUMS) Try writing that without assigning partial results... 31

32 Query trees π Instructor * Name = Course Sections Courses π Course Stud_ID = ID Enrollment GPA = MAXIMUM_GPA Students F MAXIMUM GPA Students 32

33 Query trees π Instructor Name = Course π Course Stud_ID = ID * Enrollment GPA = MAXIMUM_GPA Students F MAXIMUM GPA Sections Courses Students 33

34 Practical concerns Relational algebra is procedural But we said the DBMS is going to take care of figuring out the best way to answer a user query! In order to realize our desire to offload this work to the DBMS from the user, the DBMS query language must be declarative SQL 34