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Introduction to Database

Introduction to Database. CHAPTER 2 RELATIONAL MODEL. 2.1 Structure of Relational Databases 2.2 Fundamental Relational-Algebra Operations 2.3 Additional Relational-Algebra Operations 2.4 Extended Relational-Algebra Operations 2.5 Null Values 2.6 Modification of the Database.

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Introduction to Database

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  1. Introduction to Database CHAPTER 2 RELATIONAL MODEL 2.1 Structure of Relational Databases 2.2 Fundamental Relational-Algebra Operations 2.3 Additional Relational-Algebra Operations 2.4 Extended Relational-Algebra Operations 2.5 Null Values 2.6 Modification of the Database

  2. PART 1: Relational Databases • Relational Database: a shared repository of data that perceived by the users as a collection of tables. • To make database available to users: • Requests for data by • SQL (Chapter 3, 4) • QBE (Chapter 5) • Datalog (Chapter 5) • Data Integrity: protect data from damage by unintentional (Chapter 8) • Data Security: protect data from damage by intentional (Chapter 8) • Database Design (Chapter 7) • Design of database schema, tables • Normalization: Normal forms • Tradeoff: Possibility of inconsistency vs. efficiency 容器

  3. account 2.1 Structure of Relational Databases • Relational Database: a collection of tables • Table has a unique name • A row (tuple) in a table: a relationship of a set of values • Table: mathematical concept of relation • Relational Model: proposed by Codd, 1970, ref. p.1108: Bibliography [Codd 1970]E. F. Codd, "A Relational Model for Large Shared Data banks," CACM Vol. 13, No.6,(1970), pp. 377-387

  4. 2.1.1 Basic Structure • Relation: • Formally, given sets D1, D2, …. Dn • D1 x D2 x … x Dn = {(a1, a2, …, an) | where each ai Di} • a Relation r is a subset of D1 x D2 x … x Dn • Thus a relation is a set of n-tuples (a1, a2, …, an) where each ai Di • Example: if customer-name = {Jones, Smith, Curry, Lindsay}customer-street = {Main, North, Park}customer-city = {Harrison, Rye, Pittsfield}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

  5. account Example: The account Relation, Fig. 2.1 • D1 = { } • D2 = { } • D3 = { } • D1 x D2 x D3= • account =

  6. account Attribute Types • Each attribute of a relation has a name • The set of allowedvalues for each attribute is called the domain of the attribute • Attribute values are (normally) required to be atomic, that is, indivisible • E.g. multivalued attribute values are not atomic, (A-201, A-217) • E.g. composite attribute values are not atomic, BirthDate: (5, 17, 1950) • 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

  7. 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 2.1.2 Database Schema • Attributes:SupposeA1, A2, …, Anareattributes • Relation schema:R = (A1, A2, …, An) is a relation schema e.g. Customer-schema = (customer-name, customer-street, customer-city) • Relation:r(R) is a relation on the relation schema R e.g. customer(Customer-schema) customer

  8. attributes (or columns) customer-name customer-street customer-city Jones Smith Curry Lindsay Main North North Park Harrison Rye Rye Pittsfield tuples (or rows) Relation Instance • Relation Instance: The current values (relation instance) of a relation are specified by a table • Tuple: An element t of r is a tuple, represented by a row in a table customer

  9. account Relations are Unordered Order of tuples is irrelevant (tuples may be stored in an arbitrary order) e.g. The account relation with unordered tuples (unordered tuples) (ordered by account-number)

  10. bank … Database • Database: A database consists of multiple relations (ref. p. 1-??) • Information about an enterprise is broken up into parts, with each relation storing one part of the information E.g.: account: stores information about accountsdepositor: stores information about which customer owns which account customer: stores information about customers • Storing all information as a single relation such as bank(account-number, balance, customer-name, ..)results in • repetition of information (e.g. two customers own an account) • the need for null values (e.g. represent a customer without an account) • Normalization theory (Chapter 7) deals with how to design relational schemas

  11. Example: Banking Database • Banking Database: consists 6 relations: • branch (branch-name, branch-city, assets) • customer (customer-name, customer-street, customer-only) • account (account-number, branch-name, balance) • loan (loan-number, branch-name, amount) • depositor (customer-name, account-number) • borrower (customer-name, loan-number)

  12. 6. loan 3. depositor 5. account 存款帳 4. borrower 存款戶 貸款帳 貸款戶 Example: Banking Database 1. branch 2. customer 客戶(存款戶,貸款戶) 分公司

  13. 2.1.3 Keys • Let K  R = (A1, A2, …, An), set of attributes of relation r(R) • Superkey: K is a superkeyof 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. • Candidate key:K is a candidate key if K is minimal e.g: {customer-name} is a candidate key for Customer, since it is a superkey and no subset of it is a superkey. • Primary Key • Foreign key customer

  14. Schema Diagram, Fig. 2.8 Primary Key and foreign key can be depicted by schema diagram • Schema Diagram for the Banking Enterprise, Fig. 2.8

  15. 2.1.4 Query Languages • Query Language: user requests information from the database. • Categories of languages • procedural • non-procedural • SQL (ch. 3, ch. 4) • QBE (Section 5.3) • “Pure” languages: • Relational Algebra (Section 2.2, 2.3, 2.4) • Tuple Relational Calculus (Section 5.1) • Domain Relational Calculus (Section 5.2) • Pure languages form underlying basis of query languages that people use (e.g. SQL).

  16. Introduction to Database CHAPTER 3 RELATIONAL MODEL 2.1 Structure of Relational Databases 2.2 Fundamental Relational-Algebra Operations 2.3 Additional Relational-Algebra Operations 2.4 Extended Relational-Algebra Operations 2.5 Null Values 2.6 Modification of the Database

  17. Relational-Algebra Operations • Procedural language • The operators take one or more relations as inputs and give a new relation as a result. • Fundamental Relational-Algebra Operations • Select • Project • Union • Set difference • Cartesian product • Rename • Additional Relational-Algebra Operations • Intersection • Natural Join • Division • Assignment

  18. Select () Project () Product (x) a b c x y a a b b c c x y x y x y Difference () Union ( 2.2 Fundamental Relational-Algebra Operations • Select • Project • Union • Set difference • Cartesian product • Rename

  19. Relationr A B C D         1 5 12 23 7 7 3 10 • A=B ^ D > 5(r) A B C D     1 23 7 10 2.2.1 Select Operation: Example

  20. Select Operation • Notation: p(r) • p is called the selection predicate • Defined as: p(r) = {t | t  rand p(t)} Where p is a formula in propositional calculus consisting of terms connected by :  (and),  (or),  (not)Each term is one of: <attribute> op <attribute> or <constant> where op is one of: =, , >, . <.  • Example of selection:branch-name=“Perryridge”(account)

  21. Example Queries 1: Select • Find all loans of over $1200 amount > 1200 (loan) 5. loan

  22. A B C     10 20 30 40 1 1 1 2 2.2.2 Project Operation: Example • Relationr: A C A C A,C(r)     1 1 1 2    1 1 2 =

  23. Project Operation • Notation:A1, A2, …, Ak (r) where A1, A2 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 • e.g. To eliminate the branch-name attribute of accountaccount-number, balance (account)

  24. Example Queries 2: Project • List all all loan numbers and the amount of the loans loan-number, amount (loan) Fig. 2.10

  25. Example Queries 3: Project • Can use =, =, < > <=, >=, and, or, not … • Find all loans of over $1200 made by the Perryridge branch 梨崗山 amount > 1200 ^ branch-name=“Perryridge” (loan) loan Fig. 2.9

  26. A B C D E         1 1 1 1 2 2 2 2         10 10 20 10 10 10 20 10 a a b b a a b b A B r   1 2 C D E a a b b     10 10 20 10 A B C D E       10 20 20 a a b 1 2 2 2.2.3 Composition of Operations • We can build expressions by using multiple operations • Example: expression: A=C(r x s) op1:r x s op2: A=C(r x s) s

  27. Example Queries 4: Select/Project Find the loan number for each loan of an amount greater than $1200 loan-number (amount> 1200 (loan)) 5. loan

  28. 2. customer Example Queries 5: Composition • Find those customers who live in Harrison customer-name (customer-city=“Harrison” (customer))

  29. 2.2.4 Union Operation: Example • Relations r, s: A B A B    1 2 1   2 3 s r r s: A B     1 2 1 3

  30. 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, same heading) 2. The attribute domains must be compatible (e.g., 2nd column of r deals with the same type of values as does the 2nd column of s) • E.g. to find all customers with either an account or a loancustomer-name (depositor)  customer-name (borrower)

  31. Example Queries 6: Union/Intersection • Find the names of all customers who have a loan, an account, or both, from the bank customer-name (borrower)  customer-name (depositor) Find the names of all customers who have a loan and an account at bank. Fig. 2.11 customer-name (borrower)  customer-name (depositor)

  32. 2.2.5 Set-Difference Operation: Example • Relations r, s: A B A B    1 2 1   2 3 s r r – s: A B   1 1

  33. Set Difference Operation • Notation r – s • Defined as: r – s = {t | t rand 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

  34. 4. depositor 6. borrower Example Queries 7: Set Difference • Find the names of all customers who have an account, but not a loan customer-name (depositor) - customer-name (borrower) Fig. 2.12

  35. Product (x) a b c x y a a b b c c x y x y x y A B C D E rxs:         1 1 1 1 2 2 2 2         10 10 20 10 10 10 20 10 a a b b a a b b 2.2.6 Cartesian-Product Operation: Example A B C D E s r Relations r, s:   1 2     10 10 20 10 a a b b

  36. Product (x) a b c x y a a b b c c x y x y x y 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.

  37. 5. loan 6. borrower Example Queries 8: Cartesian-Product borrower x loan

  38. Queries 8:borrower loan (Fig. 2.13) 8 x 7 = 56 tuples

  39. Queries 8.1:Find the names of all customers who have a loan at the Perryridge branch.  branch-name = “Perryridge”(borrower  loan) 8 x 2 = 16 tuples (Fig. 2.14)

  40.  branch-name = “Perryridge”(borrower  loan),(Fig. 2.14) 8 x 2 = 16 tuples

  41. Queries 8.2:(borrower.loan-number = loan.loan-number (borrower x loan)) 8 x 1 = 8 tuples

  42. Queries 8.3:(branch-name=“Perryridge” (borrower.loan-number = loan.loan-number (borrower x loan))) 2 tuples

  43. 5. loan 6. borrower Example Queries 8.4: Cartesian-Product • Query: Find the names of all customers who have a loan at the Perryridge branch. customer-name (branch-name=“Perryridge” (borrower.loan-number = loan.loan-number (borrower x loan))) (Fig. 2.15)

  44. Example Queries 9 Query: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)

  45. Example Queries 10: Comparison • Query:“Find the names of all customers who have a loan at the Perryridge branch. • Query 1customer-name (branch-name = “Perryridge” ( borrower.loan-number = loan.loan-number (borrower x loan))) 8 x 7 = 56 tuples • Query 2 customer-name (loan.loan-number = borrower.loan-number ( (branch-name = “Perryridge” (loan)) x borrower)) 2 x 8 = 16 tuples Which one is better?

  46. 2.2.7 Rename Operation • Rename Operation: Allows us to name, and therefore to refer to, the results of relational-algebra expressions. • E.g.1: x (E) returns the expression E under the name X • E.g. 2. If a relational-algebra expression E has arity n, then x (A1, A2, …, An) (E) returns the result of expression E under the name X, and with the attributes renamed to A1, A2, …., An. • E.g. 3. x (r) ? • SQL: Rename r As x

  47. 3. Account = d 3. account Example Queries 11: Rename, p.53 Query:Find the largest account balance • Rename account relation as d • The query is: balance(account) - account.balance (account.balance < d.balance(account x rd (account)))

  48. 3. Account = d 3. account Example Queries 11: Rename (cont.) account.balance(account.balance < d.balance(account x rd (account))) (Fig. 2.17) (Fig. 2.16)

  49. 2. customer Fig. 2.18 Query: “Find the names of all customers who live on the same street and in the same city as Smith” Example Queries 12: Rename Algebra: p. 54

  50. 2.2.8 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 E1 and E2 be relational-algebra expressions; the following are all relational-algebra expressions: • E1 E2 • E1 - E2 • E1 x E2 • p (E1), P is a predicate on attributes in E1 • s(E1), S is a list consisting of some of the attributes in E1 •  x(E1), x is the new name for the result of E1

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