# Chapter 5 Relational Algebra and Relational Calculus Introduction

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Chapter 5 Relational Algebra and Relational Calculus

Introduction u u Relational algebra & relational calculus – formal languages associated with the relational model Relational algebra – (high-level) procedural language Relational calculus – non-procedural language. A language that produces a relation that can be derived using relational calculus is relationally complete 2

Relational Algebra u Relational algebra operations work on one or more relations to define another relation without changing original relations u Both operands and results are relations – Output from one operation can be input to another u Allows nested expressions – Closure property 3

Relational Algebra u Five basic operations – Selection, Projection, Cartesian product, Union, and Set Difference u Perform most required data retrieval operations u Additional operators – Join, Intersection, and Division – Can be expressed in terms of 5 basic operations 4

Selection (or Restriction) u predicate (R) – Works on single relation R – Defines relation that contains only tuples (rows) of R that satisfy specified condition (predicate) 5

Example - Selection (or Restriction) u List all staff with a salary greater than £ 10, 000. salary > 10000 (Staff) 6

Projection u col 1, . . . , coln(R) – Works on a single relation R – Defines relation that contains vertical subset of R – Extracts values of specified attributes – Eliminates duplicates 7

Example - Projection u Produce a list of salaries for all staff, showing only staff. No, f. Name, l. Name, and salary details. staff. No, f. Name, l. Name, salary(Staff) 8

Union u R S – Defines relation that contains all tuples of R, or S, or both R and S – Duplicate tuples eliminated – R and S must be union-compatible u For relations R and S with I and J tuples, respectively – Union is concatenation of R & S into one relation with maximum of (I + J) tuples 9

Example - Union u List all cities where there is either a branch office or a property for rent. city(Branch) city(Property. For. Rent) 10

Set Difference u. R –S – Defines relation consisting of tuples in relation R, but not in S – R and S must be union-compatible 11

Example - Set Difference u List all cities where there is a branch office but no properties for rent. city(Branch) – city(Property. For. Rent) 12

Intersection S – Defines relation consisting of set of all tuples in both R and S – R and S must be union-compatible u. R u Expressed using basic operations: R S = R – (R – S) 13

Example - Intersection u List all cities where there is both a branch office and at least one property for rent. city(Branch) city(Property. For. Rent) 14

Cartesian product u. R XS – Binary operation – Defines relation that is concatenation of every tuple of relation R with every tuple of relation S 15

Example - Cartesian product u List the names and comments of all clients who have viewed a property for rent. ( client. No, f. Name, l. Name(Client)) X ( client. No, property. No, comment (Viewing)) 16

Example - Cartesian product and Selection u Use selection operation to extract those tuples where Client. client. No = Viewing. client. No((Õclient. No, f. Name, l. Name(Client)) (Õclient. No, property. No, comment(Viewing))) Cartesian product and Selection can be reduced to single operation u. Join u 17

Relational Algebra Operations 18

Join Operations u Join - derivative of Cartesian product u Equivalent to performing Selection, using join predicate as selection formula, over Cartesian product of two operand relations u One of most difficult operations to implement efficiently in an RDBMS One reason RDBMSs have intrinsic performance problems u 19

Join Operations u Various forms of join operation – Theta join – Equijoin (a particular type of Theta join) – Natural join – Outer join – Semijoin 20

Theta join ( -join) u. R FS – Defines relation that contains tuples satisfying predicate F from Cartesian product of R and S – Predicate F is of the form R. ai S. bi where may be comparison operators (<, , >, , =, ) 21

Theta join ( -join) u Can rewrite Theta join using basic Selection and Cartesian product operations R FS = F(R S) u Degree of a Theta join u Sum of degrees of operand relations R and S u If predicate F contains only equality (=) u. Equijoin 22

Example - Equijoin u List the names and comments of all clients who have viewed a property for rent. ( client. No, f. Name, l. Name(Client)) Client. client. No = Viewing. client. No ( client. No, property. No, comment(Viewing)) 23

Natural join u. R S – Equijoin of relations R and S over all common attributes x – One occurrence of each common attribute eliminated from result 24

Example - Natural join u List the names and comments of all clients who have viewed a property for rent. ( client. No, f. Name, l. Name(Client)) ( client. No, property. No, comment(Viewing)) 25

Outer join u Used to display rows in result that do not have matching values in join column u. R S – (Left) outer join » Tuples from R that do not have matching values in common columns of S included in result relation 26

Example - Left Outer join u Produce a status report on property viewings. property. No, street, city(Property. For. Rent) Viewing 27

Semijoin u. R FS – Defines relation that contains tuples of R that participate in join of R with S u Can R rewrite Semijoin using Projection and Join: FS = A(R F S) 28

Example - Semijoin u List complete details of all staff who work at the branch in Glasgow. Staff. branch. No=Branch. branch. No( city=‘Glasgow’(Branch)) 29

Division u. R S – Defines relation over attributes C that consists of set of tuples from R that match combination of every tuple in S u Expressed using basic operations: T 1 C(R) T 2 C((S X T 1) – R) T T 1 – T 2 30

Example - Division u Identify all clients who have viewed all properties with three rooms. ( client. No, property. No(Viewing)) ( property. No( rooms = 3 (Property. For. Rent))) 31

Relational Algebra Operations 32

Aggregate Operations u AL(R) – Applies aggregate function list, AL, to R to define relation over aggregate list – AL contains one or more (<aggregate_function>, <attribute>) pairs u Main aggregate functions: – COUNT, SUM, AVG, MIN and MAX 33

Example – Aggregate Operations u How many properties cost more than £ 350 per month to rent? R(my. Count) COUNT property. No (σrent > 350 (Property. For. Rent)) 34

Grouping Operation u GA AL(R) – Groups tuples of R by grouping attributes, GA – Then applies aggregate function list, AL, to define new relation – AL contains one or more (<aggregate_function>, <attribute>) pairs – Resulting relation contains grouping attributes, GA, along with results of each aggregate function 35

Example – Grouping Operation u Find the number of staff working in each branch and the sum of their salaries. R(branch. No, my. Count, my. Sum) branch. No COUNT staff. No, SUM salary (Staff) 36

Order of Preference u Precedence of relational operators: – [σ, π, ρ] (highest) – [Χ, ⋈] –∩ – [∪, —] 37

Relational Calculus u Relational calculus query specifies what is to be retrieved rather than how to retrieve it – No description of how to evaluate a query u In first-order logic (or predicate calculus), predicate is a truth-valued function with arguments u Proposition – Substitution of values for arguments in predicate – Can be either true or false 38

Relational Calculus u If predicate contains a variable, must be range for x u Substitution of some values of range for x, proposition may be true; for other values, false u When applied to databases, relational calculus has forms: tuple and domain 39

Tuple Relational Calculus u Finds tuples for which predicate is true u Uses tuple variables u Tuple variable – Variable that ‘ranges over’ a named relation: i. e. , variable whose only permitted values are tuples of the relation u Specify range of a tuple variable S as the Staff relation as: Staff(S) u To find set of all tuples S such that F(S) is true: {S | F(S)} where F is a formula 40

Tuple Relational Calculus - Example u To find details of all staff earning more than £ 10, 000: {S | Staff(S) S. salary > 10000} u To find a particular attribute, such as salary, write: {S. salary | Staff(S) S. salary > 10000} 41

Tuple Relational Calculus u Quantifiers - tell how many instances the predicate applies to: – Existential quantifier \$ (‘there exists’) – Universal quantifier (‘for all’) variables qualified by or \$ are called bound variables, otherwise called free variables u Tuple 42

Tuple Relational Calculus u Existential quantifier used in formulae that must be true for at least one instance, such as: Staff(S) (\$B)(Branch(B) (B. branch. No = S. branch. No) B. city = ‘London’) u Means ‘There exists a Branch tuple with same branch. No as the branch. No of the current Staff tuple, S, and is located in London’ 43

Tuple Relational Calculus u Universal quantifier used in statements about every instance, such as: ( B) (B. city ‘Paris’) u Means ‘For all Branch tuples, the address is not in Paris’ u Can use ~(\$B) (B. city = ‘Paris’) which means ‘There are no branches with an address in Paris’ 44

Tuple Relational Calculus u u Formulae should be unambiguous and make sense A (well-formed) formula is made out of atoms: » R(Si), where Si is a tuple variable and R is a relation » Si. a 1 q Sj. a 2 » Si. a 1 q c u Can recursively build up formulae from atoms: » An atom is a formula » If F 1 and F 2 are formulae, so are their conjunction, F 1 F 2; disjunction, F 1 F 2; and negation, ~F 1 » If F is a formula with free variable X, then (\$X)(F) and ( X)(F) are also formulae 45

Example - Tuple Relational Calculus u List the names of all managers who earn more than £ 25, 000. {S. f. Name, S. l. Name | Staff(S) S. position = ‘Manager’ S. salary > 25000} u List the staff who manage properties for rent in Glasgow. {S | Staff(S) (\$P) (Property. For. Rent(P) (P. staff. No = S. staff. No) P. city = ‘Glasgow’)} 46

Example - Tuple Relational Calculus u List the names of staff who currently do not manage any properties. {S. f. Name, S. l. Name | Staff(S) (~(\$P) (Property. For. Rent(P) (S. staff. No = P. staff. No)))} Or {S. f. Name, S. l. Name | Staff(S) (( P) (~Property. For. Rent(P) ~(S. staff. No = P. staff. No)))} 47

Example - Tuple Relational Calculus u List the names of clients who have viewed a property for rent in Glasgow. {C. f. Name, C. l. Name | Client(C) ((\$V)(\$P) (Viewing(V) Property. For. Rent(P) (C. client. No = V. client. No) (V. property. No=P. property. No) P. city =‘Glasgow’))} 48

Tuple Relational Calculus u Expressions can generate infinite set – Example {S | ~Staff(S)} u Eliminate by: – Adding restriction that all values in result must be values in domain of expression E, dom(E) u Domain of E – Set of all values that appear explicitly in E or in relations whose names appear in E 49

Domain Relational Calculus u u Uses variables that take values from domains A general domain relational calculus expression: {d 1, d 2, . . . , dn | F(d 1, d 2, . . . , dn)} » R(di), where di is a domain variable and R is a relation » di q dj » di q c u Can recursively build up formulae from atoms: » An atom is a formula » If F 1 and F 2 are formulae, so are their conjunction, F 1 F 2; disjunction, F 1 F 2; and negation, ~F 1 » If F is a formula with free variable X, then (\$X)(F) and ( X)(F) are also formulae 50

Example - Domain Relational Calculus u Find the names of all managers who earn more than £ 25, 000. {f. N, l. N | (\$s. N, posn, sex, DOB, sal, b. N) (Staff (s. N, f. N, l. N, posn, sex, DOB, sal, b. N) posn = ‘Manager’ sal > 25000)} 51

Example - Domain Relational Calculus u List the staff who manage properties for rent in Glasgow. {s. N, f. N, l. N, posn, sex, DOB, sal, b. N | (\$s. N 1, cty)(Staff(s. N, f. N, l. N, posn, sex, DOB, sal, b. N) Property. For. Rent(p. N, st, cty, pc, typ, rms, rnt, o. N, s. N 1, b. N 1) (s. N=s. N 1) cty=‘Glasgow’)} 52

Example - Domain Relational Calculus u List the names of staff who currently do not manage any properties for rent. {f. N, l. N | (\$s. N) (Staff(s. N, f. N, l. N, posn, sex, DOB, sal, b. N) (~(\$s. N 1) (Property. For. Rent(p. N, st, cty, pc, typ, rms, rnt, o. N, s. N 1, b. N 1) (s. N=s. N 1))))} 53

Example - Domain Relational Calculus u List the names of clients who have viewed a property for rent in Glasgow. {f. N, l. N | (\$c. N, c. N 1, p. N 1, cty) (Client(c. N, f. N, l. N, tel, p. T, m. R) Viewing(c. N 1, p. N 1, dt, cmt) Property. For. Rent(p. N, st, cty, pc, typ, rms, rnt, o. N, s. N, b. N) (c. N = c. N 1) (p. N = p. N 1) cty = ‘Glasgow’)} 54

Other Languages u Transform-oriented languages – Non-procedural languages – Use relations to transform input data into required outputs (e. g. SQL) u Graphical languages – Provide user with picture of structure of relation – User fills in example of what is wanted and system returns required data in that format (e. g. QBE) 55

Other Languages u 4 GLs – Create complete customized application – Uses limited set of commands in a userfriendly, often menu-driven environment u Some systems accept a form of natural language, sometimes called a 5 GL, although this development is still at an early stage 56

In-class Questions Suppliers (sid: integer, sname: string, address: string) Parts (pid: integer, pname: string, color: string) Catalog (sid: integer, pid: integer, cost: real) u Write the following queries in relational algebra and tuple relational calculus: 1. Find the names of suppliers who supply some red part. Find the sids of suppliers who supply some red part or are at 221 Packer Ave. Find the sids of suppliers who supply some red part and some green part. Find the sids of suppliers who supply every red part. 2. 3. 4. 57

Relational Algebra Solutions 1. Find the names of suppliers who supply some red part. sname (Suppliers) sid = sid (Catalog) sid, pid pid = pid Sid, sname ( (Parts) color=‘red’ ) 58

2. Find the sids of suppliers who supply some red part or are at 221 Packer Ave. ( sid (Suppliers) ) address = ‘ 221 Packer Ave’ sid = sid (Catalog) sid, pid ( pid = pid (Parts) Color=‘red’ ) 59

3. Find the sids of suppliers who supply some red part and some green part sid=sid (Catalog) pid, sid ( pid (Catalog) pid, sid pid=pid (Parts) color=‘red’ ) ( pid=pid (Parts) color=‘green’ ) 60

3. Find the sids of suppliers who supply some red part and some green part (Alternate Solution). sid (Catalog) pid, sid ( pid (Catalog) pid, sid pid=pid (Parts) color=‘red’ sid ) ( pid=pid (Parts) color=‘green’ ) 61

4. Find the sids of suppliers who supply every red part ( ) (Catalog) pid, sid ( pid ) (Parts) color=‘red’ 62

Tuple Relational Calculus Solution 1. 2. 3. 4. {S. sname | Suppliers(S) ^ ( C)( P)(Catalog(C) ^ Parts(P) ^ (C. sid = S. sid) ^ (P. pid = C. pid) ^ (P. color = ‘red’))} {S. sid | Suppliers(S) ^ ((S. address = ‘ 221 Packer Ave. ’) v (( C)( P)(Catalog(C) ^ Parts(P) ^ (S. sid = C. sid) ^ (C. pid = P. pid) ^ (P. color = ‘red’))))} {C. sid | (Catalog(C) ^ ( P)(Parts(P) ^ (C. pid = P. pid) ^ (P. color = ‘red’))) ^ (Catalog(C 1) ^ ( P)(Parts(P) ^ (C 1. pid = P. pid) ^ (P. color = ‘green’))) ^ (C. sid = C 1. sid)} {S. sid | Suppliers(S) ^ (( P)(Parts(P) ^ (P. color=‘red’))) ^ (( C)(Catalog(C) ^ (C. pid = P. pid))) ^ (C. sid = S. sid)} 63

Chapters Covered u Chapter 5 u Assignment #3 64