1. 1 Relational Query Languages Relational Algebra (procedural) Relational Calculus (non-procedural)

2. Relational Languages Relational Algebra (procedural) –defines operations on tables Relational Calculus (declarative) –based on first-order predicate calculus Every relational algebra query can be translated to relational calculus Every safe relational calculus query can be translated to relational algebra. Any language that is atleast as expressive as relational algebra is said to be relationally complete.

3. 3 Relational Algebra Operators Relational Algebra Operators Select : given a relation R and a predicate P, select tuples from R that satisfy P. Project : given a relation R and a subset of its attributes X, return a relation which is the same as R except that all columns not in X are left out. Rename : given a relation R and a name N, return a relation that is exactly the same as R except that it has a name N. Cartesian Product : Given 2 relations R1and R2,.return a relation R3 whose tuples are the concatenation of tuples in R1 and R2 Union : Given relations R1 and R2, return a relation R3 which contains all tuples in R1 and R2 Set Difference : Given relations R1 and R2, return a relation R3 containing all tuples in R1 that are not in R2

4. 4 selection cond(R) or select[selection cond]R Example: Employee(name, dept, sal) Employeeselect [sal > 20,000] Employee namedeptsal janepharmacy30,000 jack hardware 30,000 jillpharmacy75,000 select [(dept = toy) or (sal < 20,000)] Employee namedeptsal joe toy20,000 billtoy12,000 Selection Operation name deptsal jane pharmacy30,000 jack hardware30,000 jill pharmacy75,000 joetoy20,000 billtoy12,000 

5. 5Projection Proj [list of attr of R] (R ) or  list of attr of R (R) R A B C S A C Jane Toy 10,000 Jane Toy Jim Toy 20,000 John Complaint June Complaint 20,000 Proj[A]R Proj[CB]R ACB Jane 10,000 Toy Jim 20,000Complaint June 20,000Toy

6. 6 Cartesian Product Denoted by R x S R: A BCS: AD joetoy10Kjoejill jackcom20Kjackjill RxS: R.ABCS.AD joetoy10Kjoejill joe toy10Kjackjill jackcom20Kjoejill jackcom20Kjackjill Notice attribute naming strategy to disambiguate attribute names attributes get the name, R.A, where A is attrib name, and R is the relation name from which attrib originates. If there is no possible ambiguity, relation name is dropped!

7. 7 Set Difference Denoted by R - S ( Illegal if R & S have different numbers of attributes or if respective domains mismatch!) R A B S A C Jane Toy Jane Toy Jim Toy John Complaint June Complaint R - S =AB Jim Toy June Complaint Note attributes in resulting relation take name from the first relation

8. 8 Rename Operator Strategy used to disambiguate attribute names: –For union and set difference operators, the resulting relation takes the attribute names of the first relation –For cartesian product, attributes are named as Relation- name.attribute-name, where Relation name refers to the relation from which the attribute originally came. –Strategy will not disambiguate when the same relation appears multiple times in the relational query. Let R(A,B) be a relation. Consider R x R---- what to name the attributes of the resulting relation? Define a rename operator: denoted by rename[N]R or by  N (R) –returns a relation which is exactly same as R except it has the name N

9. 9 Rename Operator Consider relation Employee(name, dept, sal) List all the employees who work in the same department as Jill We first find the department(s) for which Jill works –Proj[dept](select[name = Jill] Employee) ---list of departments for which Jill works To find out about all Employees working for this department, we need to reference the Employee table again: –select[P] ( Employee x Proj[dept](select[name = Jill] Employee) ) –where P is a selection predicate which requires dept values to be equal. If we use Employee.dept in P it is ambiguous which instance of Employee relation in the query the attribute refers to. To disambiguate, use rename operator: Proj[Employee.name](select[Employee.dept = Employee2.dept] Employee x (Proj[dept] (select[name = Jill]( rename[Employee2](Employee))))

10. 10 Formal Definition of Relational Algebra Basic Expressions: –Relation in a database –Constant Relations General Expressions: constructed from basic ones. Let E1 and E2 be relational algebra expressons. Then the following are also expressions: –E1 U E2, if arity of E1 = E2 and corresponding attributes are of the same type –E1 - E2, if arity of E1 = E2 and corresponding attributes are of the same type –E1 x E2, if attributes in E1 and E2 have different names –Select[P](E1), where P is a predicate on attributes in E1 –Proj[S](E1), where S is a list of some attributes in E1 –rename[X](E1), where X is a new name for relation E1

11. 11 Additional Operators Basic Relational Algebra operators are like assembly language. Define more powerful operators that make the task of writing relational algebra queries easier Each of these operators can be expressed in relational algebra and do not increase the expressibility of the language Example: Intersection R  S = R - (R - S) = { t | t  R & t  S }

12. 12 R join condition S = select[ join condition] (R x S) join condition is of the form: AND AND where each condiition is of the form Ai  Bj, where –Ai is attribute of R –Bj is attribute of S –  is a comparison operator {=,, =, <>} Joins Example: E(emp, dept) M(dept, mgr) List all employees and their managers. Proj[emp, mgr](select[E.dept = M.dept] (ExM)) can be represented as: Proj[emp,mgr] ( E E.dept = M.dept M )

13. 13 Types of Joins Theta-Join: if a join condition uses some comparison operator other than equality. –E.g., list names of all managers who manage departments other than Jill’s – Proj[mgr]( select[emp = Jill](E ) (E.dept  M.dept) M) Equi-Join: if join conditions use only equality operator. –E.g., list the manager’s name of Jill’s department –Proj[mgr]( select[emp = Jill](E ) (E.dept  M.dept) M) Natural Join: special type of equi-join.. –Let R and S be relations. Let attributes of R and S be denoted by R and S respectively. –Denote by R U S the union of the list of attributes of R and S –Let list of attributes common to both R and S be {A1, A2, …, An} –Natural join of R and S (denoted R S) is: – Proj[R U S ] (R R.A1 = S.A1 and R.A2 = S.A2 and … and R.An = S.An S) –E.g., Proj[mgr]( select[emp = Jill](E ) M)

14. 14 Assignment Operator Lots of time convenient to write relational algebra expressions in parts using assignment to temporary relational variables. For this purpose use assignment operator, denoted by := E.g., Who makes more than their manager? E(emp, dept, sal) M(mgr, dept) ESM(emp, sal, mgr) := Proj[emp, sal, mgr] (E  M) (Proj[ESM.emp](ESM [mgr = E.emp & ESM.sal >E.sal] E) ) With the assignment operator, a query can be written as a sequential program consisting of a series of assignments followed by an expression whose value is the result of the query.

15. Examples A query is a composition of basic relational algebra operators Consider relations: –customer(ssno, name, street, city) –account(acctno, custid, balance) list account balance of Sharad list names of all customers who have a higher balance than all customers in Champaign

16. 16 Diag Pat Dis Winslett Strep Liu Mono Harandi Meningitis Harandi Hepatitis Liu Hepatitis Outcome Pat Test Outcome Winslett a T Winslett b F Liu b T Harandi f T Winslett e F Harandi e F Harandi g F Winslett e T ToDiag Dis Test Strep A Mono B Meningitis C Hepatitis D Encehhalitis E Meningitis F Meningitis G

17. 17 1. Who has what disease? Diag 2. Who has a disease they have been tested for? –Proj[pat](Diag |  ToDiag |  Outcome) 3. Who has a disease they tested positively for? Proj[pat](Diag |  ToDiag |  (select[outcome = ‘T’])Outcome)) 4. Who has a disease that they tested both positively & negatively for? Temp1(pat, dis) := Proj[pat,dis](Diag |  ToDiag select[outcome = ‘T’])Outcome) Temp2(pat, dis) : = Proj[pat,dis](Diag |  ToDiag select[outcome = ‘F’])Outcome) Proj[pat](Temp1  Temp2) Use better names!!

18. 18 Example of Queries in Relational Algebra 5. Who tested both positively and negatively for a disease, whether or not they have it? Testpos(pat, dis) = Proj[pat,dis](ToDiag |  select[outcome = ‘T’]) Outcome) Testneg(pat, dis) = Proj[pat,dis](ToDiag |  select[outcome = ‘F’]) Outcome) (Testpos  Testneg)[pat] 6. Who tested both positively & negatively for the same test? (Winslett) Proj[pat](Outcome |  condition (rename[Outcome2](Outcome)) where condition is: [Outcome.pat = Outcome2.pat & Outcome.test = Outcome2.test & Outcome.outcome = Outcome2. outcome]

19. 19 7. What testable disease does no one have? (encephalitis) Proj[dis]ToDiag  Proj[dis]Diag Note technique: compute opposite of what you want, take a difference. Use in hard queries with negation (‘no one’) 8. What disease does more than one person have? Proj[dis](Diag condition rename[Diag2](Diag)) where, condition is [Diag.pat  Diag2.pat & Diag.dis = Diag2dis] 9. What disease does everyone have? clue that query is very hard Disease(dis) := diag[dis] Patients(pat) := diag[pat] DiseasesNotEveryoneHas(dis) := Proj[dis]((Patients x Disease) - Diag) Disease - Diseases Not Everyone Has Note technique used! A very hard query might require taking the difference several times.

20. Outer Joins E emp dept sal M mgr dept Jones Missiles 10K Mendez Tanks Chu Tanks 20K Frank Explosive Swami Tanks 50K Jones Missiles Barth Revolver 100K Right outer join of E with M emp dept sal mgr Jones Missiles 10K Jones Chu Tanks 20K Mendez null Explosives null Frank left outer join of E with M emp dept sal mgr Jones Missiles 10K Jones Chu Tanks 20K Mendez Swami Tanks 50K Mendez Barth Revolver 100K null Full outer join of E with M emp dept sal mgr Jones Missiles 10K Jones Chu Tanks 20K Mendez Swami Tanks 50K Mendez Barth Revolver 100K null null Explosives null Frank

21. 21 Aggretate Functions Mathematical aggregate functions (e.g., sum, count, average, min, max) on a collection of values from the database. [grouping attributes] F [function list] ( ) where grouping attributes specify how to group the tuples function list specify the aggregate functions to be applied to the individual groups relation expression is the expression that results in a relation. Example: Consider a relation Employee(dno, ssno, sal). An expression [dno] F [COUNT ssno, AVERAGE salary] (Employee) results in a relation: dno #employee average_salary 11010,000 2 238000 Queries involving aggregate functions cannot be expressed in pure relational algebra!

22. 22 Recursive Closure Queries Consider parent relation parent Child Mom Dad sam Anda Chuck Chuck Donna Harvey Harvey Betty Reggie Reggie Cristie John It may be interesting to query the relation for the following: retrieve all the female anscestors of sam. retrieve all male ansestors of chuck retrieve harvey’s family tree retrieve all the descendants of cristie Such queries (in general) require an unbounded application of joins of the parent relation to itself and CANNOT be represented in relational algebra.