Distributed database
- 2. Relational Data Model • Two major strengths • Three components • Relation database based on Relational DM
- 3. The relational model • Data is represented in the form of tables, and the model has 3 components • Data structure – data are organised in the form of tables with rows and columns • Data manipulation – powerful operations (using the SQL language) are used to manipulate data stored in the relations • Data integrity – facilities are included to specify business rules that maintain the integrity of data when they are manipulated
- 4. Relation
- 5. • EMP(ENO, ENAME, TITLE, SAL, PNO, RESP, DUR) • PROJ(PNO,PNAME, BUDGET) • Each attribute in both relations has a domain • Domains need not to be distinct
- 6. Database Table Keys Definition: A key of a relation is a subset of attributes with the following attributes: • Unique identification • Non-redundancy
- 8. Normalization of Database Tables
- 9. 9 In this chapter, you will learn: • What normalization is and what role it plays in the database design process • About the normal forms 1NF, 2NF, 3NF, BCNF, and 4NF • How normal forms can be transformed from lower normal forms to higher normal forms • How normalization and ER modeling are used concurrently to produce a good database design • How some situations require denormalization to generate information efficiently
- 10. 10 Database Tables and Normalization • Normalization – Process for evaluating and correcting table structures to minimize data redundancies • Reduces data anomalies – Works through a series of stages called normal forms: • First normal form (1NF) • Second normal form (2NF) • Third normal form (3NF)
- 11. 11 Database Tables and Normalization (continued) • Normalization (continued) – 2NF is better than 1NF; 3NF is better than 2NF – For most business database design purposes, 3NF is as high as we need to go in normalization process – Highest level of normalization is not always most desirable
- 12. 12 The Need for Normalization • Example: Company that manages building projects – Charges its clients by billing hours spent on each contract – Hourly billing rate is dependent on employee’s position – Periodically, report is generated that contains information displayed in Table 5.1
- 13. 13 The Need for Normalization (continued)
- 14. 14 The Need for Normalization (continued)
- 15. 15 The Need for Normalization (continued) • Structure of data set in Figure 5.1 does not handle data very well • The table structure appears to work; report generated with ease • Unfortunately, report may yield different results depending on what data anomaly has occurred
- 16. 16 The Normalization Process • Each table represents a single subject • No data item will be unnecessarily stored in more than one table • All attributes in a table are dependent on the primary key
- 18. 18 Conversion to First Normal Form • Repeating group – Derives its name from the fact that a group of multiple entries of same type can exist for any single key attribute occurrence • Relational table must not contain repeating groups • Normalizing table structure will reduce data redundancies • Normalization is three-step procedure
- 19. 19 Conversion to First Normal Form (continued) • Step 1: Eliminate the Repeating Groups – Present data in tabular format, where each cell has single value and there are no repeating groups – Eliminate repeating groups, eliminate nulls by making sure that each repeating group attribute contains an appropriate data value
- 20. 20 Conversion to First Normal Form (continued)
- 21. 21 Conversion to First Normal Form (continued) • Step 2: Identify the Primary Key – Primary key must uniquely identify attribute value – New key must be composed
- 22. 22 Conversion to First Normal Form (continued) • Step 3: Identify All Dependencies – Dependencies can be depicted with help of a diagram – Dependency diagram: • Depicts all dependencies found within given table structure • Helpful in getting bird’s-eye view of all relationships among table’s attributes • Makes it less likely that will overlook an important dependency
- 23. 23 Conversion to First Normal Form (continued)
- 24. 24 Conversion to First Normal Form (continued) • First normal form describes tabular format in which: – All key attributes are defined – There are no repeating groups in the table – All attributes are dependent on primary key • All relational tables satisfy 1NF requirements • Some tables contain partial dependencies – Dependencies based on only part of the primary key – Sometimes used for performance reasons, but should be used with caution – Still subject to data redundancies
- 25. 25 Conversion to Second Normal Form • Relational database design can be improved by converting the database into second normal form (2NF) • Two steps
- 26. 26 Conversion to Second Normal Form (continued) • Step 1: Write Each Key Component on a Separate Line – Write each key component on separate line, then write original (composite) key on last line – Each component will become key in new table
- 27. 27 Conversion to Second Normal Form (continued) • Step 2: Assign Corresponding Dependent Attributes – Determine those attributes that are dependent on other attributes – At this point, most anomalies have been eliminated
- 28. 28 Conversion to Second Normal Form (continued)
- 29. 29 Conversion to Second Normal Form (continued) • Table is in second normal form (2NF) when: – It is in 1NF and – It includes no partial dependencies: • No attribute is dependent on only portion of primary key
- 30. 30 Conversion to Third Normal Form • Data anomalies created are easily eliminated by completing three steps • Step 1: Identify Each New Determinant – For every transitive dependency, write its determinant as PK for new table • Determinant – Any attribute whose value determines other values within a row
- 31. 31 Conversion to Third Normal Form (continued) • Step 2: Identify the Dependent Attributes – Identify attributes dependent on each determinant identified in Step 1 and identify dependency – Name table to reflect its contents and function
- 32. 32 Conversion to Third Normal Form (continued) • Step 3: Remove the Dependent Attributes from Transitive Dependencies – Eliminate all dependent attributes in transitive relationship(s) from each of the tables that have such a transitive relationship – Draw new dependency diagram to show all tables defined in Steps 1–3 – Check new tables as well as tables modified in Step 3 to make sure that each table has determinant and that no table contains inappropriate dependencies
- 33. 33 Conversion to Third Normal Form (continued)
- 34. 34 Conversion to Third Normal Form (continued) • A table is in third normal form (3NF) when both of the following are true: – It is in 2NF – It contains no transitive dependencies
- 35. Formal Relational Query Languages • Two mathematical Query Languages form the basis for “real” languages (e.g. SQL), and for implementation: – Relational Algebra: More operational, very useful for representing execution plans. – Relational Calculus: Lets users describe what they want, rather than how to compute it. (Non-operational, declarative.)
- 36. Example Instances sid sname rating age 22 dustin 7 45.0 31 lubber 8 55.5 58 rusty 10 35.0 sid sname rating age 28 yuppy 9 35.0 31 lubber 8 55.5 44 guppy 5 35.0 58 rusty 10 35.0 sid bid day 22 101 10/10/96 58 103 11/12/96 R1 S1 S2 • “Sailors” and “Reserves” relations for our examples. • We’ll use positional or named field notation, assume that names of fields in query results are `inherited’ from names of fields in query input relations.
- 37. 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: – Intersection, join, division, renaming: Not essential, but (very!) useful. • Since each operation returns a relation, operations can be composed! (Algebra is “closed”.)
- 38. Projection sname rating yuppy 9 lubber 8 guppy 5 rusty 10 sname rating S , ( )2 age 35.0 55.5 age S( )2 • Deletes attributes that are not in projection list. • Schema of result contains exactly the fields in the projection list, with the same names that they had in the (only) input relation. • Projection operator has to eliminate duplicates! (Why??) – Note: real systems typically don’t do duplicate elimination unless the user explicitly asks for it. (Why not?)
- 39. Selection rating S 8 2( ) sid sname rating age 28 yuppy 9 35.0 58 rusty 10 35.0 sname rating yuppy 9 rusty 10 sname rating rating S , ( ( )) 8 2 • Selects rows that satisfy selection condition. • No duplicates in result! (Why?) • Schema of result identical to schema of (only) input relation. • Result relation can be the input for another relational algebra operation! (Operator composition.)
- 40. Union, Intersection, Set-Difference • All of these operations take two input relations, which must be union-compatible: – Same number of fields. – `Corresponding’ fields have the same type. • What is the schema of result? sid sname rating age 22 dustin 7 45.0 31 lubber 8 55.5 58 rusty 10 35.0 44 guppy 5 35.0 28 yuppy 9 35.0 sid sname rating age 31 lubber 8 55.5 58 rusty 10 35.0 S S1 2 S S1 2 sid sname rating age 22 dustin 7 45.0 S S1 2
- 41. Cross-Product • Each row of S1 is paired with each row of R1. • Result schema has one field per field of S1 and R1, with field names `inherited’ if possible. – Conflict: Both S1 and R1 have a field called sid. ( ( , ), )C sid sid S R1 1 5 2 1 1 (sid) sname rating age (sid) bid day 22 dustin 7 45.0 22 101 10/10/96 22 dustin 7 45.0 58 103 11/12/96 31 lubber 8 55.5 22 101 10/10/96 31 lubber 8 55.5 58 103 11/12/96 58 rusty 10 35.0 22 101 10/10/96 58 rusty 10 35.0 58 103 11/12/96 Renaming operator:
- 42. Joins • Combine information from two or more tables • Example: students enrolled in courses: S1 S1.sid=E.studidE Sid name gpa 50000 Dave 3.3 53666 Jones 3.4 53688 Smith 3.2 53650 Smith 3.8 53831 Madayan 1.8 53832 Guldu 2.0 cid grade studid Carnatic101 C 53831 Reggae203 B 53832 Topology112 A 53650 History 105 B 53666 S1 E
- 43. Joins sid name gpa cid grade studid 53666 Jones 3.4 History105 B 53666 53650 Smith 3.8 Topology112 A 53650 53831 Madayan 1.8 Carnatic101 C 53831 53832 Guldu 2.0 Reggae203 B 53832 Sid name gpa 50000 Dave 3.3 53666 Jones 3.4 53688 Smith 3.2 53650 Smith 3.8 53831 Madayan 1.8 53832 Guldu 2.0 cid grade studid Carnatic101 C 53831 Reggae203 B 53832 Topology112 A 53650 History 105 B 53666 S1 E
- 44. Joins • Condition Join: • Result schema same as that of cross-product. • Fewer tuples than cross-product, might be able to compute more efficiently • Sometimes called a theta-join. R c S c R S ( ) (sid) sname rating age (sid) bid day 22 dustin 7 45.0 58 103 11/12/96 31 lubber 8 55.5 58 103 11/12/96 S R S sid R sid 1 1 1 1 . .
- 45. Joins • Equi-Join: A special case of condition join where the condition c contains only equalities. • Result schema similar to cross-product, but only one copy of fields for which equality is specified. • Natural Join: Equijoin on all common fields. sid sname rating age bid day 22 dustin 7 45.0 101 10/10/96 58 rusty 10 35.0 103 11/12/96 S R sid 1 1
- 46. Division • Not supported as a primitive operator, but useful for expressing queries like: Find sailors who have reserved all boats. • Let A have 2 fields, x and y; B have only field y: – A/B = – i.e., A/B contains all x tuples (sailors) such that for every y tuple (boat) in B, there is an xy tuple in A. – Or: If the set of y values (boats) associated with an x value (sailor) in A contains all y values in B, the x value is in A/B. • In general, x and y can be any lists of fields; y is the list of fields in B, and x y is the list of fields of A. x x y A y B| ,
- 47. Examples of Division A/B sno pno s1 p1 s1 p2 s1 p3 s1 p4 s2 p1 s2 p2 s3 p2 s4 p2 s4 p4 pno p2 pno p2 p4 pno p1 p2 p4 sno s1 s2 s3 s4 sno s1 s4 sno s1 A B1 B2 B3 A/B1 A/B2 A/B3
- 48. Expressing A/B using Basic Operators • For A/B, compute all x values that are not `disqualified’ by some y value in B. – x value is disqualified if by attaching y value from B, we obtain an xy tuple that is not in A.
- 49. Find names of sailors who’ve reserved boat #103 • Solution 1: sname bid serves Sailors(( Re ) ) 103 Solution 2: ( , Re )Temp serves bid 1 103 ( , )Temp Temp Sailors2 1 sname Temp( )2 Solution 3: sname bid serves Sailors( (Re )) 103
- 50. Find names of sailors who’ve reserved a red boat • Information about boat color only available in Boats; so need an extra join: sname color red Boats serves Sailors(( ' ' ) Re ) A more efficient solution: sname sid bid color red Boats s Sailors( (( ' ' ) Re ) ) A query optimizer can find this, given the first solution!
- 51. Find sailors who’ve reserved a red or a green boat • Can identify all red or green boats, then find sailors who’ve reserved one of these boats: ( ,( ' ' ' ' ))Tempboats color red color green Boats sname Tempboats serves Sailors( Re ) Can also define Tempboats using union! (How?) What happens if is replaced by in this query?
- 52. Find the names of sailors who’ve reserved all boats • Uses division; schemas of the input relations to / must be carefully chosen: ( , ( , Re ) / ( ))Tempsids sid bid serves bid Boats sname Tempsids Sailors( ) To find sailors who’ve reserved all ‘Interlake’ boats: / ( ' ' ) bid bname Interlake Boats .....
- 53. Other Relational Languages • Relational calculus -tuple or domain (QBE) • SQL -you already know!
- 54. Summary • The relational model has rigorously defined query languages that are simple and powerful. • Relational algebra is more operational; useful as internal representation for query evaluation plans. • Several ways of expressing a given query; a query optimizer should choose the most efficient version.