Relational Model. Prepared by: Ms. Pankti Dharwa ( Asst. Prof, SVBIT)

Transcription

1 2. Relational Model Prepared by: Ms. Pankti Dharwa ( Asst. Prof, SVBIT)

2 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

3 Relation Schema Formally, given domains 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 Schema of a relation consists of attribute definitions name type/domain integrity constraints

4 Database schema The database schema, it is a logical design of the database, and the database instance, which is a snapshot of the database at a given instant in time. Lower case is written for relation and upper case is written for relation schema. We use Account_schema to denote the relation schema for relation account. Account_schema=(account_number,branch_name,balanc e) We denote the face that account is a relation on Account_schema account(account_schema)

5 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 Order of tuples is irrelevant (tuples may be stored in an arbitrary order) attributes (or columns) customer_name customer_street customer_city Jones Smith Curry Lindsay Main North North Park Harrison Rye Rye Pittsfield tuples (or rows) customer

6 Example of the relation instance Consider the branch relation and the schema of that relation is: Branch_schema(Branch_name, Branch_city,assests) Branch_name Branch_city assests Nehrunagar Ahmedabad Satellite Ahmedabad Sector-12 G nagar Kalawad Rajkot Sector-25 G nagar

7 Why Split Information Across Relations? 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 (Chapter) deals with how to design relational schemas

8 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) by possible r we mean a relation r that could exist in the enterprise we are modeling. Example: {customer_name, customer_street} and {customer_name} are both superkeys of Customer, if no two customers can possibly have the same name In real life, an attribute such as customer_id would be used instead of customer_name to uniquely identify customers, but we omit it to keep our examples small, and instead assume customer names are unique.

9 Keys (Cont.) K is a candidate key if K is minimal Example: {customer_name} is a candidate key for Customer, since it is a superkey and no subset of it is a superkey. Primary key: a candidate key chosen as the principal means of identifying tuples within a relation Should choose an attribute whose value never, or very rarely, changes. E.g. address is unique, but may change

10 Schema diagram

11 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.

12 Select,project & rename are unary operations as they operate on 1 relation.

13 Relational Algebra Basic operations: Selection ( ) Selects a subset of rows from relation. Projection ( ) Deletes unwanted columns from relation. Cross-product ( ) Allows us to combine two relations. Set-difference ( ) Tuples in reln. 1, but not in reln. 2. Union ( ) Tuples in reln. 1 and in reln. 2. Additional operations: Set Intersection, natural join, division, assignment

14 Set Operations These are binary operations; that is, each is applied to two sets. Binary operations form mathematical set theory: UNION: R1 U R2, INTERSECTION: R1 n R2, SET DIFFERENCE: R1 - R2, CARTESIAN PRODUCT: R1 X R2. Rename: When these operations are adapted to relational databases, the two relations on which any of the above three operations are applied must have the same type of tuples; this condition is called Union Compatible if they have the same degree n. This means that the two relations have the same number of attributes and that each pair of corresponding attributes have the same domain.

15 UNION: The result of this operation, denoted by R S is a relation that includes all tuples that are either in R or in S or in both R and S. Duplicate tuples are eliminated. INTERSECTION: The result of this operation, denoted by R n S, is a relation that includes all tuples that are in both R and S. SET DIFFERENCE: The result of this operation denoted by R S, is a relation that includes all tuples that are in R but not in S. CARTESIAN PRODUCT: It is also known as CROSS PRODUCT or CROSS JOIN denoted by X, which is also a binary set operation, but the relations on which it is applied do not have to be union compatible. This operation is used to combine tuples from two relations in a combinatorial fashion.

16 Union Operation () Union of the two entity set id,name (borrower) id,name (depositor) (select id,name from borrower) union (select id,name from depositor); Option: union all Conditions: The relations r and s must be of the same arity The domains of the i th attribute of r and the i th attribute of s must be the same for all i

17 The union operation Consider a query to find the names of all bank customers who have either an account or loan or both. we know how to find out the names of all customers with the loan in the bank. Πcustomer_name (borrower). We know how to find out the names of all customers with an account in the bank. Πcustomer_name (depositor). Now union of these two sets that is we need all customer names that appear in either both of the two relations.we find this data by binary operation union,denoted as U,so this expression needed is Πcustomer_name (borrower) U Πcustomer_name (depositor)

18 List of all customers who have either a loan or an account Πcustomer_name (borrower) U Πcustomer_name (depositor) Cust_name Adams Curry Hayes Jackson Jones Smith Lindsay turner

19 Union Operation Example Relations r, s: A B 1 A B 2 r 2 1 s 3 r s: A B

20 Set Intersection Operation () Intersection of the two entity set id,name (borrower) id,name (depositor) (select id,name from borrower) intersection (select id,name from depositor); Option: intersection all Conditions: The relations r and s must be of the same arity The domains of the i th attribute of r and the i th attribute of s must be the same for all i

21 Intersect Operation Example Relations r, s: A B 1 A B 2 r 2 1 s 3 r n s: A B 2

22 Set Difference Operation (-) Give tuples which are in relation r but not in s id,name (borrower) id,name (depositor) (select id,name from borrower) except (select id,name from depositor); Option: except all Conditions: The relations r and s must be of the same arity The domains of the i th attribute of r and the i th attribute of s must be the same for all i

23 The set difference operation It is denoted by -, allows us to find tuples that are in one relation but not in another. We can find the customer who have an account but not a loan by: Πcustomer_name (depositor) - Πcustomer_name (borrower) Customer_name Johnson Lindsay turner

24 Set Difference Operation Example Relations r, s: A B 1 A B 2 r 2 1 s 3 r s: A B 1 1

25 unary operations The select,project,and rename are called unary operations because they operate on one relation. Loan_number Branch_name amount L-11 Round-hill 900 L-14 Downtown 1500 L-15 Perryridge 1300 L-16 Downtown 1000 L-17 Redwood 2000 L-93 Mianus 500 L-23 Perryridge 1300

26 select operation () Select operation select the tuples from the table with satisfy given condition or predicate condition (Relation) select <attributes> from <table name> where <condition>; select * from customer where id=1234;

27 The select operation The select operation selects tuples that satisfy given predicate.it is denoted by (σ). Thus, to select those tuples of the loan relation where the branch is Perryridge we write, σ branch_name= Perryridge (loan) Loan_number Branch_name amount L-23 Perryridge 1300 L-15 Perryridge 1300

28 We can find all tuples in which the amount lent is more than $1200 by writing σ amount>1200(loan) In general we allow comparisions using <,<=,>,>= in the selection predicate,further we can combine several predicates by using the connections and(^ ),or(v) and not(-) σ branch_name= perryridge ^amount>1200(loan)

29 project operation () Selects the particular attributes or columns from the table attributes (Relation) select <attributes> from <table name>; select id, name from customer;

30 Projection operation Suppose we want to list out all loan numbers and the amount of the loans,but donot care about the branch name. the project operations allows us to produce this relation. Since a relation is a set, any duplicate rows are eliminated. Projection is denoted by the upper case π

31 Write a query to list all loan numbers and the amount of loan as π loan_number,amount(loan) Loan_number L-11 L-14 L-15 L-16 L-17 L-93 L-23 amount

32 Composition of Relational Operations Composition provide the information of the table with specified columns with satisfy conditions or predictions id,name ( account>1000 (customer)) select <attributes> from <table name> where <predictions>; select id,name from customer where account>1000;

33 Composition of relational operations Find those customers who live in Harrisions Πcustomer_name (σ customer_city= Harrisions (customer))

34 Rename Operation () x (E) Returns the result of expression E under the name x x(a1,a 2,,A n ) (E) With the attributes rename to A 1, A 2,, A n e.g. find largest balance in account entity set balance (account) account,balance ( account.balance < d.balance (account x d (account))) (select balance from account) except (select account,balance from account,account as d where account.balance < d.balance)

35 Formal Definition of Relational Algebra For E 1 and E 2 Relational-Algebra Expressions: E 1 E 2 E 1 E 2 E 1 E 2 E 1 E 2 P (E 1 ) S (E 1 ) Union Operation Set Intersection Operation Set Difference Operation Cartesian Product Select: P is a predicate Project: S is set of attributes x (E 1 ) Rename: x is a new name for result of E 1

36 Cartesian-Product Operation () Allow us to combine information from any two relations Cartesian product of relations r1 & r2 is r1 r2 customer loan (customer.id, customer.name, customer.street, customer.city, loan.number, loan.amount) select id,name,number,amount from customer,loan If relations r1 has n rows and r2 has m rows than r1 x r2 is size of n x m.

37 Cartesian - Product Operation Example Relations r, s: r x s: A A B 1 2 r B C D C D s E a a b b a a b b E a a b b

38

39 Cartesian product operation It is denoted by a cross( ) allows us to combine information from any two relations. We write as r1 r2. For example relation schema for r=borrower loan is (borrower.customer_name, borrower.loan_num, loan. loan_number, loan.branch_name, loan.amount) We shall drop the relation-name prefix.this simplification doesnot lead to any ambiguity. We can write relation schema for r as (customer_name, borrower.loan_num,loan.loan_number,branch_name,a mount)

40 The borrower relation The loan relation

41 Borrower * loan

42 Suppose we want to find the names of all customers who have a loan at perryridge branch. σ branch_name= perryridge (borrower loan))

43 This is the table of Query: σ branch_name= Perryridge (borrower loan)

44 Since cartesian product operates every tuple of loan with ` every tuple of borrower σ borrower.loan_num=loan.loan_number (σ branch_name= perryridge (borrower loan)) Now if I want to find the names of customers,then we do projection: customer_name (σ borrower.loan_num=loan.loan_number (σ branch_name= perryridge (borrower loan))) Customer_name Adams Hayes

45 Emp( eno, fname, lname, gender, dob, address, salary, supno,dno) Dependent(eeno,depname,gender,bdate,relationship) Retrive names of each female employee s dependent

46 Emp( eno, fname, lname, gender, dob, address, salary, supno,dno) Dependent(eeno,depname,dob,relationship) Retrive names of each female employee s dependent Π fname,depname(σeno=eeno (σ gender = f (emp*dependent)))

47 Assignment operations Its denoted by <-.its r-s For eg:- temp1 <- R-S(r1) temp2<- R-S(r2) result-=temp1-temp2 Temp1 Πeno,fname (σ gender= f (emp)) Temp2 temp1 * dependent Temp3 Π fname,depname(σeno=eeno(temp2)) OR Π fname,depname(σeno=eeno (σ gender = f (emp*dependent)))

48 Natural Join Operation ( ) Takes the Cartesian product of an Entity Sets and apply the selection forcing equality on those attributes that appear in both relation schemas and finally removes duplicate attributes name,number,amount (borrower loan) select <attributes> from <table name>; select name,number,amount from borrower natural join loan; If r(r) and s(s) be relations without any attributes in common, That is R S =. Then r s = r x s

49 Natural-join Find the names of all customers who have a loan at the bank along with the loan number and the loan amount customer_name,loan.loan_num,amount (σ borrower.loan_num=loan.loan_number(borrower loan)) Natural join is a binary operation that allows us to combine certain selections and a cartesian product into one operation.it is denoted by symbol It removes the duplicate attributes. customer_name, loan_num, amount(borrower loan) Cust_name Loan_number Amount Adams L Curry L Hayes L Jackson L Jones L Smith L Smith L Williams L

50 Account Custom er depositor

51 Find the names of all branches with customers who have an account in bank and who lives in harrisions. branch_name(σ customer_city= harrison (customer account depositor) Natural join is associative, means (customer account) depositor customer (account depositor) branch_name Brighton perryridge

52 Schema diagram

53 FROM SCHEMA OF EMP perform foll. Queries Q>>Retrieve the name of manager of each department To get name we need to combine each dept tuple with emp tuple whose ENO matches the MGRNO value in dept tuple. Dname,lname,fname(Dept mgrno=eno Emp)

54 Exercise time Retrieve name & address of all emp who work for research dept. For every project located in chennai list project no, its dept number, dept manager name & dob.

55 Express foll. in relational algebra SAILORS (Salid, Salname, Rating, Age) RESERVES (Sailid, Boatid, Day) BOATS (Boatid, Boat-name, Color) i) Find the name of sailors who reserved green boat. ii) Find the colors of boats reserved by Ramesh iii) Find the names of sailors who have reserved a red or green boat. iv) Find the Sailid s of sailors with age over 20 who have not registered a red boat.

56 Extended Relational Algebra Operation

57 Generalisation projection customer_name,limit - credit_bal(credit_info) Table: credit info Cust_name Limit Credit_bal Curry Hayes Jones Smith customer_name(limit - credit_bal)as credit_available(credit_info)

58 Aggregate Functions Functions take a collection of values and return a single value as a result sum avg count min max Returns sum of the values Returns the average of the values Returns the number of elements in the collection Returns minimum values in the collection Returns maximum values in the collection Multisets: The collections on which aggregate functions operate can have multiple occurrences of a value; the order in which the valued appear is not relevant. Such collections are called multisets

59 Aggregate Functions Example e.g. Take a set valueset(num) = { 1, 1, 3, 5, 2, 5, 8 } sum 25 avg count 7 min 1 max 8 G sum(num) valueset G avg(num) valueset G count(num) valueset G min(num) valueset G max(num) valueset select sum(num) from valueset; select avg(num) from valueset; select count(num) from valueset; select min(num) from valueset; select max(num) from valueset;

60 Aggregate Operation Example Relation r: A B C G sum(c) (r) sum(c ) 27

61 Aggregate Operation Example Relation (pt_works) employee_name Branch_name balance Adams Brown Gopal Johnson Lorrena Peterson Rao sato Perryridge Perryridge Perryridge downtown Downtown downtown austin austin ranch_name sum(balance) G sum(balance) (pt_works) Austin 3100 Downtown 5300 * Branch_nameg sum(salary) (pt_works) perryridge 8100 Branch_nameG sum(salary) (pt_works)

62 Aggregate Functions (Cont.) Suppose we want to find the maximum salary,sum of salary for part-time emp at each branch Branch_name Sum_salary Max_salary Austin Downtown Perryridge branch_name G sum(balance), G max(salary)(pt_works)

63 Outer join Outer join operation is an extension of the join operation to deal with the missing information. Emp_name Street City Coyote Toon Hollywood Rabbit Tunnel Carrotville Smith Revovler Death valley Williams Seaview seattle Emp_name Branch_name salary Coyote Mesa 1500 Rabbit Mesa 1300 gates Redmond 5300 Williams Redmond 1500 Table:EMP Table:ft_works

64 The result of emp ft_works Emp_name Street City Coyote Toon Hollywood Rabbit Tunnel Carrotville Williams Seaview seattle Branch_name salary Mesa 1500 Mesa 1300 Redmond 1500

65 There is loss of information of street and city information about smith. Since smith is absent from the ft_works relation and similary we have lost the information about Gates, We can use outer join to avoid loss of information, three forms of operations: 1) left outer join 2) right outer join 3) full outer join

66 LEFT OUTER JOIN Emp_name Street City Branch_nam e salary Coyote Toon Hollywood Mesa 1500 Rabbit Tunnel Carrotville Mesa 1300 Williams Seaview seattle Redmond 1500 Smith Revovler Death valley NULL NULL

67 Right outer join Emp_name Street City Branch_nam e salary Coyote Toon Hollywood Mesa 1500 Rabbit Tunnel Carrotville Mesa 1300 Williams Seaview seattle Redmond 1500 Gates Null Null Redmond 5300

68 FULL outer join Emp_name Street City Branch_nam e salary Coyote Toon Hollywood Mesa 1500 Rabbit Tunnel Carrotville Mesa 1300 Williams Seaview seattle Redmond 1500 Gates NULL NULL Redmond 5300 Smith Revovler Death valley NULL NULL

69 Division Operation () Let r(r) and s(s) be relations, and let S R, Then relation r s is defined on R-S. A tuple t is in r s if two conditions are satisfied: 1. t is in R-S (r) 2. For every tuple t s in s, there is tuple t r in r satisfying both of the following: a. t r [S] = t s [S] b. t r [R S] = t customer custadd where customer = {(id,name,street,city)} custadd = {(name,street,city)}

70 Division Operation Notation: r s 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

71 Division Operation Example r s: Relations r, s: A A B r B 1 2 s

72 Another Division Example A B a a a a a a a a C D a a b a b a b b E Relations r, s: r s: D a b E 1 1 A B a a C r s

73 Bank Example Queries R1---- branch_name ( branch_city = Brooklyn (branch)) R2-- customer_name, branch_name (depositor account) Ans R2 R1 Cust_name Johnson

74 Bank Example Queries Find all customers who have an account at all branches located in Brooklyn city. R1---- branch_name ( branch_city = Brooklyn (branch)) R2-- customer_name, branch_name (depositor account) Ans R2 R1 customer_name, branch_name (depositor account) branch_name ( branch_city = Brooklyn (branch))

75 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.

76 Exercise time It is useful in special kind of queries that include the phrase for all. B3 A B1 B2 X Y X1 Y1 X1 Y3 X1 Y2 X4 Y3 X4 Y5 X2 Y3 X3 Y4 X4 Y1 X4 Y4 X5 Y5 X5 Y1 Y Y Y1 Y1 Y3 Y5 Y Y5 Y2 Y4 A B1 A B2 A B3 X X X

77 Modification of the Database The content of the database may be modified using the following operations: Deletion Insertion Updating All these operations are expressed using the assignment operator.

78 Deletion A delete request is expressed similarly to a query, except instead of displaying tuples to the user, the selected tuples are removed from the database. Can delete only whole tuples; cannot delete values on only particular attributes A deletion is expressed in relational algebra by: r r E where r is a relation and E is a relational algebra query.

79 Deletion Examples Delete all account records in the Perryridge branch. account account branch_name = Perryridge (account ) Delete all loan records with amount in the range of 0 to 50 loan loan amount 0and amount 50 (loan) Delete all accounts at branches located in Needham. r 1 branch_city = Needham (account branch ) r 2 account_number, branch_name, balance (r 1) r 3 customer_name, account_number (r 2 depositor) account account r 2 depositor depositor r 3

80 Deletion Delete all data of customer named smith from depositor depositor depositor - σ cust_name= smith (depositor) Delete all loans with amount range 0 to 50 Loan loan σ amount >= 0 ^ amount <=50(loan)

81 Insertion To insert data into a relation, we either: specify a tuple to be inserted write a query whose result is a set of tuples to be inserted in relational algebra, an insertion is expressed by: r r E where r is a relation and E is a relational algebra expression. The insertion of a single tuple is expressed by letting E be a constant relation containing one tuple.

82 Insertion Examples Insert information in the database specifying that Smith has $1200 in account A-973 at the Perryridge branch. account account {( A-973, Perryridge, 1200)} depositor depositor {( Smith, A-973 )} Provide as a gift for all loan customers in the Perryridge branch, a $200 savings account. Let the loan number serve as the account number for the new savings account. r 1 ( branch_name = Perryridge (borrowe loan)) account account loan_number, branch_name, 200 (r 1 ) depositor depositor customer_name, loan_number (r 1 )

83 Updating A mechanism to change a value in a tuple without charging all values in the tuple Use the generalized projection operator to do this task r F ( 1, 2 r F,, Fl, Each F i is either the I th attribute of r, if the I th attribute is not updated, or, ) if the attribute is to be updated F i is an expression, involving only constants and the attributes of r, which gives the new value for the attribute

84 Update Examples Make interest payments by increasing all balances by 5 percent. account account_number, branch_name, balance * 1.05 (account) Pay all accounts with balances over $10,000 6 percent interest and pay all others 5 percent account account_number, branch_name, balance * 1.06 ( BAL (account )) account_number, branch_name, balance * 1.05 ( BAL (account))

85 Null Values It is possible for tuples to have a null value, denoted by null, for some of their attributes null signifies an unknown value or that a value does not exist. The result of any arithmetic expression involving null is null. Aggregate functions simply ignore null values (as in SQL) For duplicate elimination and grouping, null is treated like any other value, and two nulls are assumed to be the same (as in SQL)

86 Null Values Comparisons with null values return the special truth value: unknown If false was used instead of unknown, then not (A < 5) would not be equivalent to A >= 5 Three-valued logic using the truth value unknown: OR: (unknown or true) = true, (unknown or false) = unknown (unknown or unknown) = unknown AND: (true and unknown) = unknown, (false and unknown) = false, (unknown and unknown) = unknown NOT: (not unknown) = unknown In SQL P is unknown evaluates to true if predicate P evaluates to unknown Result of select predicate is treated as false if it evaluates to unknown