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Dr. Alexandra I. Cristea CS 319: Theory of Databases: C4.

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Presentation on theme: "Dr. Alexandra I. Cristea CS 319: Theory of Databases: C4."— Presentation transcript:

1 Dr. Alexandra I. Cristea CS 319: Theory of Databases: C4

2 (provisionary) Content 1.Generalities DB 2.Integrity constraints (FD revisited) 3.Relational Algebra (revisited) 4.Query optimisation 5.Tuple calculus 6.Domain calculus 7.Query equivalence 8.LLJ, DP and applications 9.Temporal Data 10.The Askew Wall

3 … previous FD

4 Relational model E.F. Codd: check wikipedia! Papers: 03/Databases/codd.pdfhttp://www.cl.cam.ac.uk/Teaching/20 03/Databases/codd.pdf –Reprint at: A relational Model of Data for Large Shared Data BanksA relational Model of Data for Large Shared Data Banks Relational Completeness of Data Base SublanguagesRelational Completeness of Data Base Sublanguages

5 Relational Model* Structure of Relational Databases Fundamental Relational-Algebra-Operations Additional Relational-Algebra-Operations *based on Silberschatz, Korth, Sudarshan Database System Concepts 5 th Edition, Chapter 2

6 Example of a Relation Instance

7 Basic Structure Formally, given sets D 1, D 2, …. D n a relation r is a subset of D 1 x D 2 x … x D n Thus, a relation is a set of n-tuples (a 1, a 2, …, a n ) where each a i D i

8 Example If –customer_name = {Jones, Smith, Curry, Lindsay, …} /* Set of all customer names */ –customer_street = {Main, North, Park, …} /* set of all street names*/ –customer_city = {Harrison, Rye, Pittsfield, …} /* set of all city names */ Then r = { (Jones, Main, Harrison), (Smith, North, Rye), (Curry, North, Rye), (Lindsay, Park, Pittsfield) } is a relation over customer_name x customer_street x customer_city

9 Attribute Types Each attribute of a relation has a name The set of allowed values for each attribute is called the domain of the attribute Attribute values are (normally) required to be atomic; that is, indivisible –E.g. the value of an attribute can be an account number, but cannot be a set of account numbers Domain is said to be atomic if all its members are atomic The special value null is a member of every domain The null value causes complications in the definition of many operations –We shall ignore the effect of null values in our main presentation and consider their effect later

10 Relation Schema A 1, A 2, …, A n are attributes R = (A 1, A 2, …, A n ) is a relation schema Example: Customer_schema = (customer_name, customer_street, customer_city) r(R) denotes a relation r on the relation schema R Example: customer (Customer_schema)

11 Relation Instance The current values (relation instance) of a relation are specified by a table An element t of r is a tuple, represented by a row in a table Jones Smith Curry Lindsay customer_name Main North Park customer_street Harrison Rye Pittsfield customer_city customer attributes (or columns) tuples (or rows)

12 Relations are Unordered Order of tuples is irrelevant (tuples may be stored in an arbitrary order) Example: account relation with unordered tuples

13 Database A database consists of multiple relations Information about an enterprise is broken up into parts, with each relation storing one part of the information Storing all information as a single relation such as bank(account_number, balance, customer_name,..) results in –repetition of information e.g.,if two customers own an account (What gets repeated?) –the need for null values e.g., to represent a customer without an account Normalization theory deals with how to design relational schemas

14 The customer Relation

15 The depositor Relation

16 Keys Let K R K is a superkey of R if values for K are sufficient to identify a unique tuple of each possible relation r(R) Example: {customer_name, customer_street} and {customer_name} are both superkeys of Customer, if no two customers can possibly have the same name

17 Keys (Cont.) K is a candidate key if K is minimal Example: {customer_name} : superkey + no subset of it is a superkey. Primary key: a candidate key chosen as the principal means of identifying tuples within a relation

18 Foreign Keys A relation schema may have an attribute that corresponds to the primary key of another relation. The attribute is called a foreign key.

19 Query Languages Language in which user requests information from the database. Categories of languages –Procedural –Non-procedural, or declarative Pure languages: –Relational algebra –Tuple relational calculus –Domain relational calculus Pure languages form underlying basis of query languages that people use.

20 Relational Algebra Procedural language Six basic operators –select: –project: –union: –set difference: – –Cartesian product: x –rename: The operators take one or two relations as inputs and produce a new relation as a result.

21 Select Operation – Example Relation r ABCD A=B ^ D > 5 (r) ABCD

22 Select Operation Notation: p (r) p is called the selection predicate Defined as: p (r) = {t | t r and p(t)} Where p is a formula in propositional calculus consisting of terms connected by : (and), (or), (not) Each term is one of: op or where op is one of: =,, >,. <. Example of selection: branch_name=Perryridge (account)

23 Project Operation – Example Relation r: ABC AC = AC A,C (r)

24 Project Operation Notation: where A 1, A 2 are attribute names and r is a relation name. The result is defined as the relation of k columns obtained by erasing the columns that are not listed Duplicate rows removed from result, since relations are sets Example: To eliminate the branch_name attribute of account account_number, balance (account)

25 Union Operation – Example Relations r, s: r s: AB AB 2323 r s AB

26 Union Operation Notation: r s Defined as: r s = {t | t r or t s} For r s to be valid. 1. r, s must have the same arity (same number of attributes) 2. The attribute domains must be compatible (example: 2 nd column of r deals with the same type of values as does the 2 nd column of s) Example: to find all customers with either an account or a loan customer_name (depositor) customer_name (borrower)

27 Set Difference Operation – Example Relations r, s: r – s: AB AB 2323 r s AB 1111

28 Set Difference Operation Notation r – s Defined as: r – s = {t | t r and t s} Set differences must be taken between compatible relations. –r and s must have the same arity –attribute domains of r and s must be compatible

29 Cartesian-Product Operation – Example Relations r, s: r x s: AB 1212 AB CD E aabbaabbaabbaabb CD E aabbaabb r s

30 Cartesian-Product Operation Notation r x s Defined as: r x s = {t q | t r and q s} Assume that attributes of r(R) and s(S) are disjoint. (That is, R S = ). If attributes of r(R) and s(S) are not disjoint, then renaming must be used.

31 Composition of Operations Can build expressions using multiple operations Example: A=C (r x s) r x s A=C (r x s) AB CD E aabbaabbaabbaabb ABCDE aabaab

32 Rename Operation Allows us to refer to results of RA expressions & to refer to a relation by more than one name. Example: x (E) returns the expression E under the name X expression E under name X, with attributes renamed to A 1, A 2, …., A n.

33 Revise basic operators!

34 Relational Algebra Operators thus … Procedural language with six basic operators –select: –project: –union: –set difference: – –Cartesian product: x –rename: The operators take one or two relations as inputs and produce a new relation as a result.

35 Banking Example branch (branch_name, branch_city, assets) customer (customer_name, customer_street, customer_city) account (account_number, branch_name, balance) loan (loan_number, branch_name, amount) depositor (customer_name, account_number) borrower (customer_name, loan_number)

36

37 Example Queries Find all loans of over $1200 Find the loan number for each loan of an amount greater than $1200 amount > 1200 (loan) loan_number ( amount > 1200 (loan)) Find the names of all customers who have a loan, an account, or both, from the bank customer_name (borrower) customer_name (depositor)

38 Example Queries Find the names of all customers who have a loan at the Perryridge branch. Find the names of all customers who have a loan at the Perryridge branch but do not have an account at any branch of the bank. customer_name ( branch_name = Perryridge ( borrower.loan_number = loan.loan_number (borrower x loan))) – customer_name (depositor) customer_name ( branch_name=Perryridge ( borrower.loan_number = loan.loan_number (borrower x loan)))

39 Example Queries Find the names of all customers who have a loan at the Perryridge branch. Query 2 customer_name ( loan.loan_number = borrower.loan_number ( ( branch_name = Perryridge (loan)) x borrower)) Query 1 customer_name ( branch_name = Perryridge ( borrower.loan_number = loan.loan_number (borrower x loan)))

40 Example Queries Find the largest account balance –Strategy: Find those balances that are not the largest –Rename account relation as d so that we can compare each account balance with all others Use set difference to find those account balances that were not found in the earlier step. –The query is: balance (account) - account.balance ( account.balance < d.balance (account x d (account)))

41 Formal Definition A basic expression in the relational algebra consists of either one of the following: –A relation in the database –A constant relation Let E 1 and E 2 be relational-algebra expressions; the following are all relational-algebra expressions: –E 1 E 2 –E 1 – E 2 –E 1 x E 2 – p (E 1 ), P is a predicate on attributes in E 1 – s (E 1 ), S is a list consisting of some of the attributes in E 1 – x (E 1 ), x is the new name for the result of E 1

42 Additional Operations We define additional operations that do not add any power to the relational algebra, but that simplify common queries. Set intersection Natural join Division Assignment

43 Set-Intersection Operation Notation: r s Defined as: r s = { t | t r and t s } Assume: –r, s have the same arity –attributes of r and s are compatible Note: r s = r – (r – s)

44 Set-Intersection Operation – Example Relation r, s: r s A B A B 2323 r s A B 2

45 Notation: r s Natural-Join Operation Let r and s be relations on schemas R and S respectively. Then, r s is a relation on schema R S obtained as follows: –Consider each pair of tuples t r from r and t s from s. –If t r and t s have the same value on each of the attributes in R S, add a tuple t to the result, where t has the same value as t r on r t has the same value as t s on s

46 Natural joint example Example: R = (A, B, C, D) S = (E, B, D) –Result schema = (A, B, C, D, E) –r s is defined as: r.A, r.B, r.C, r.D, s.E ( r.B = s.B r.D = s.D (r x s))

47 Natural Join Operation – Example Relations r, s: AB CD aababaabab B D aaabbaaabb E r AB CD aaaabaaaab E s r s

48 Division Operation Notation: Suited to queries that include the phrase for all*. Let r and s be relations on schemas R and S respectively where –R = (A 1, …, A m, B 1, …, B n ) –S = (B 1, …, B n ) The result of r s is a relation on schema R – S = (A 1, …, A m ) r s = { t | t R-S (r) u s ( tu r ) } Where tu means the concatenation of tuples t and u to produce a single tuple r s

49 Division Operation – Example Relations r, s: r s: A B 1212 AB r s

50 Another Division Example AB aaaaaaaaaaaaaaaa CD aabababbaabababb E Relations r, s: r s: D abab E 1111 AB aaaa C r s

51 Division Operation (Cont.) Property –Let q = r s –Then q is the largest relation satisfying q x s r Definition in terms of the basic algebra operation Let r(R) and s(S) be relations, and let S R r s = R-S (r ) – R-S ( ( R-S (r ) x s ) – R-S,S (r )) To see why – R-S,S (r) simply reorders attributes of r – R-S ( R-S (r ) x s ) – R-S,S (r) ) gives those tuples t in R-S (r ) such that for some tuple u s, tu r.

52 Assignment Operation The assignment operation ( ) provides a convenient way to express complex queries. – Write query as a sequential program consisting of a series of assignments followed by an expression whose value is displayed as a result of the query. –Assignment must always be made to a temporary relation variable. Example: Write r s as temp1 R-S (r ) temp2 R-S ((temp1 x s ) – R-S,S (r )) result = temp1 – temp2 –The result to the right of the is assigned to the relation variable on the left of the. –May use variable in subsequent expressions.

53 Bank Example Queries Find the names of all customers who have a loan and an account at the bank. customer_name (borrower) customer_name (depositor) Find the name of all customers who have a loan at the bank and the loan amount customer_name, loan_number, amount (borrower loan)

54 Query 1 customer_name ( branch_name = Downtown (depositor account )) customer_name ( branch_name = Uptown (depositor account)) Query 2 customer_name, branch_name (depositor account) temp(branch_name ) ({(Downtown ), (Uptown )}) Note that Query 2 uses a constant relation. Bank Example Queries Find all customers who have an account from at least the Downtown and the Uptown branches.

55 Find all customers who have an account at all branches located in Brooklyn city. Bank Example Queries customer_name, branch_name (depositor account) branch_name ( branch_city = Brooklyn (branch))

56 Library Case author book reservation copy borrow title publisher year date name Initials name department cancelled department name from to present cpYear

57 Library database book ( ISBN, title, publisher, year ) author (ISBN, initials, name ) copy (barcode, ISBN, department, cpYear, present ) reservation (name, department, ISBN, date, cancelled ) borrow (name, department, barcode, from, to )

58 Library Questions (RA) 1List all the authors of books of which the library has a copy that has never been borrowed. 2List all the authors of books that have never been borrowed. 3List all the authors of which no book has ever been borrowed.

59 (simple) Employee database employee(person_name, street, city) works(person_name, company_name, salary) company(company_name, city) manages(person_name, manager_name)

60 Exercise 2.1.c Find the names of all employees who earn more than every employee of SBC. –The more than every... clause cannot be expressed directly in the relational algebra. –For all other employees there is an employee of SBC earning more. This can be described using a selection on a cartesian product (of works with works). –We then use the difference to find the correct result.

61 Solution 2.1.c person-name (W) – - person-name( ( w.salary<=w2.salary ^ w2.company-name = SBC (W x W2 (W) ))

62 Exercise 2.5.e Find all companies located in every city in which SBC is located.

63 Case 1: 1 city per company C.company-name ( ( C.city=SBC.city ( C (company ) x ( SBC.company-name=SBC ( SBC (company ) ) ) Trivial!

64 Case 2: multiple cities per company Find all companies located in every city in which SBC is located. –The every city... clause cannot be expressed directly in the relational algebra (except for division). –Lets see …

65 Case 2a: multiple cities: division All cities where SBC is located: SBCCity = city ( company-name=SBC (company ) ) Is SBCCity constant? Yes! So we can use it on the right hand side of the division. companies located in every city in which SBC is located: company SBCCity Still quite neat!!

66 Case 2b: multiple cities: no division Find all companies located in every city in which SBC is located. –The every city... clause cannot be expressed directly in the relational algebra (except for division). –For the other companies there is a city in which SBC is located and the other company is not. –The cities where a company is not located are all the cities except the ones where the company is located. –We thus need 2 times a difference in this query. –Can also be formulated using a difference operator

67 Case 2b: multiple cities: no division For the other companies there is a city in which SBC is located and the other company is not = IntRes. E = company-name (company ) x SBCCity IntRes = E – company We want not (IntRes): company-name (company ) - company-name (IntRes)

68 What is the meaning of: SBCCity

69 Recognizing Types of Queries Identify the type of the following queries, and afterwards also translate them to the algebra (PSJ, union, intersection, difference, division): 1.Give the name of customers that have a loan with a branch where they also have an account. 2.Give the name of customers who have a loan at a branch where they do not have an account. 3.Give the name of customers who have a loan at every branch where they have an account. 4.Give the name of customers who have loans only at branches where they have an account.

70 Reading Queries customer-name ( balance>amount ( account depositor borrower loan)) branch-name (branch) branch-name ( branch-city customer-city ( branch account depositor customer)) X.customer-name ( X.account-number = Y.account-number X.customer-name Y.customer-name ( X (depositor) Y (depositor)))

71 Beer Database visits(drinker, bar) serves(bar, beer) likes(drinker, beer).

72 Beer questions with a difference 1.Give all drinkers that visit bars that dont serve any beer they like 2.Give all drinkers that only visit bars that serve a beer they like 3.Give all drinkers that only visit bars that serve no beer they like 4.Give all drinkers that only visit bars that serve all beers they like (and maybe other beers as well) 5.Give all drinkers that only visit bars that only serve beers they like (and thus serve nothing else)

73 Summary We have learned RA We have learned to perform simple and more complex queries in RA

74 … to follow Query optimisation


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