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Normalization

Normalization. Ms. Hatoon Al-Sagri CCIS – IS Department. Informal Design Guidelines for Relational Databases. R elational database design: The grouping of attributes to form "good" relation schemas Two levels of relation schemas: The logical "user view" level

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Normalization

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  1. Normalization Ms. Hatoon Al-Sagri CCIS – IS Department

  2. Informal Design Guidelines for Relational Databases • Relational database design: The grouping of attributes to form "good" relation schemas • Two levels of relation schemas: • The logical "user view" level • The storage "base relation" level • Design is concerned mainly with base relations

  3. Informal Design Guidelines for Relational Databases Four informal measures of quality for relation schema design: • Semantics of the Relation Attributes • Reducing the redundant information in tuples • Reducing Null values in tuples • Disallowing the possibility of one generating spurious tuples.

  4. 1- Semantics of the Relation Attributes Each tuple in a relation should represent one entity or relationship instance Guideline #1: Design a schema that can be explained easily relation by relation. The semantics of attributes should be easy to interpret.

  5. 2- Redundant Information in Tuples and Update Anomalies • Mixing attributes of multiple entities may cause problems: • Information is stored redundantly wasting storage • Problems with update anomalies: • Insertion anomalies • Deletion anomalies • Modification anomalies Guideline #2:Design a schema that does not suffer from the insertion, deletion and update anomalies. If there are any present, then note them so that applications can be made to take them into account

  6. Base RelationsEMP_PROJ with redundant information

  7. 3- Null Values in Tuples Reasons for nulls: a. attribute not applicable or invalid b. attribute value unknown (may exist) c. value known to exist, but unavailable Guideline #3: Relations should be designed such that their tuples will have as few NULL values as possible Attributes that are NULL frequently could be placed in separate relations (with the primary key)

  8. 4- Spurious Tuples • Bad designs for a relational database may result in erroneous results for certain JOIN operations  Guideline #4: The relations should be designed to satisfy the lossless join condition. No spurious tuples should be generated by doing join of any relations.

  9. Functional Dependencies • Functional dependencies (FDs) are used to specify formal measures of the "goodness" of relational designs • FDs and keys are used to define normal forms for relations • FDs are constraints that are derived from the meaning and interrelationships of the data attributes

  10. Examples of FD constraints • Social security number determines employee name SSN -> ENAME • Project number determines project name and location PNUMBER -> {PNAME, PLOCATION} • Employee ssn and project number determines the hours per week that the employee works on the project {SSN, PNUMBER} -> HOURS

  11. Functional Dependencies Describes the relationship between attributes in a relation. If A and B are attributes of relation R, B is functionally dependent on A, denoted by A B, if each value of A is associated with exactly one value of B. B may have several values of A. Determinant Dependent B is functionally dependent on A A B

  12. Functional Dependencies Example 1:1 or M:1 relationship between attributes in a relation StaffNo position Position is functionally dependent on Staffno SL21 Manager 1:M relationship between attributes in a relation position StaffNo is NOT functionally dependent on position StaffNo Manager SL21 SG5

  13. Trivial Functional Dependencies A B is trivial if B  A StaffNo, Sname SName StaffNo, SName StaffNo We are not interested in trivial functional dependencies as it provides no genuine integrity constraints on the value held by these attributes.

  14. Question Find FDs of the relation shown below that lists dentist/patient appointment data; known that: • A patient is given an appointment at a specific time and date with a dentist located at a particular surgery. • On each day of patient appointments, a dentist is allocated to a specific surgery for that day. Dentist-patient (staffNo, aDate, aTime, dentistName, patNo, patName, surgeryNo)

  15. Question Dentist-patient (staffNo, aDate, aTime, dentistName, patNo, patName, surgeryNo) FDs list FD1: staffNo, aDate, aTime patNo, patName FD2: staffNo dentistName FD3: patNo patName, surgeryNo FD4: staffNo, aDate surgeryNo FD5: aDate, aTime, patNo dentistName, staffNo

  16. Introduction to Normalization • Normalization: Process of decomposing unsatisfactory "bad" relations by breaking up their attributes into smaller relations • Normal form: Condition using keys and FDs of a relation to certify whether a relation schema is in a particular normal form

  17. Normalization into 1NF

  18. ExamplesFirst Normal Form • EMP_PROJ (Ssn, Ename, {Phone#}) { } Mulitvalue attribute EMP_PROJ1 (Ssn, Ename) EMP_PROJ2 (Ssn, Phone#) • EMP_PROJ (Ssn, Ename (Fname, Lname))( ) composite attribute EMP_PROJ (Ssn, Fname,Lname) • EMP_PROJ(Ssn, Ename, {PROJS (Pnamber, Hours)}) EMP_PROJ1 (Ssn, Ename) EMP_PROJ2 (Ssn, Pnamber, Hours)

  19. Second Normal Form • Uses the concepts of FDs, primary key • Definitions: • Prime attribute - attribute that is member of the primary key K • Full functional dependency - a FD Y Z where removal of any attribute from Y means the FD does not hold any more

  20. Full Functional Dependency If A and B are attributes of a relation. B is fully functionally dependent on A if B is functionally dependent on A, but not on any proper subset of A. B is partial functional dependent on A if some attributes can be removed from A & the dependency still holds. StaffNo, Sname BranchNo Partial dependency ClientNo, PropertyNo RentDate Full dependency

  21. 1NF 2NF 1. Start with 1NF relation. 2. Find the FDs of a relation. 3. Test the FDs whose determinant attribute is part of the PK.

  22. ExamplesSecond Normal Form • {SSN, PNUMBER} HOURS is a full FD since neither SSN HOURS nor PNUMBER HOURS hold • {SSN, PNUMBER} ENAME is not a full FD (it is called a partial dependency ) since SSN ENAME also holds • A relation schema R is in second normal form (2NF) if every non-prime attribute A in R is fully functionally dependent on the primary key  • R can be decomposed into 2NF relations via the process of 2NF normalization

  23. ExamplesSecond Normal Form

  24. Second Normal Form Note: The test for 2NF involves testing for functional dependencies whose left-hand side attributes are part of the primary key. If the primary key contains a single attribute, the test need not be applied at all.

  25. Third Normal Form • Definition • Transitive functional dependency – a FD X  Y in R is a transitive dependency if there is a set of attributes Z that are neither a primary or candidate key and both X Z and Z  Y holds. • Examples: • SSN DMGRSSN is a transitive FD since SSN DNUMBER and DNUMBER DMGRSSN hold • SSN ENAME is non-transitive since there is no set of attributes X where SSN X and X ENAME

  26. 3rd Normal Form A relation schema R is in third normal form (3NF) if it is in 2NF and no non-prime attribute A in R is transitively dependent on the primary key

  27. ExamplesThird Normal Form

  28. SUMMARY OF NORMAL FORMS based on Primary Keys

  29. BCNF (Boyce-Codd Normal Form) • A relation schema R is in Boyce-Codd Normal Form (BCNF) if whenever an FD X A holds in R, then X is a superkey of R • Each normal form is strictly stronger than the previous one: • Every 2NF relation is in 1NF • Every 3NF relation is in 2NF • Every BCNF relation is in 3NF • There exist relations that are in 3NF but not in BCNF • The goal is to have each relation in BCNF (or 3NF)

  30. BCNF R1(A,C) R2(C,B)

  31. BCNF FDs: • {Student,course}  Instructor • Instructor  Course It is in 3NF not in BCNF • Decomposing into 2 schemas {Student, Instructor} {Instructor, Course}

  32. ExamplesBCNF • R ( Client#,Problem, Consultant _name) R1 (Client#, Consultant _name) • R2 (Consultant _name, Problem) • R (Stud#,Class#, Instructor, Grade) R1 (Stud#, Instructor, Grade) • R2 (Instructor, Class#)

  33. Example Consider the following relation for published books: BOOK (Book_title, Author_name, Book_type, Listprice, Author_affil, Publisher) - Author_affil referes to the affiliation of the author. Suppose thefollowing dependencies exist: Book_title -> Publisher, Book_type Book_type -> Listprice Author_name -> Author-affil (a) What normal form is the relation in? Explain your answer. (b) Apply normalization until you cannot decompose the relations further. State the reasons behind each decomposition.

  34. Answer BOOK (Book_title, Authorname, Book_type, Listprice, Author_affil, Publisher) (a) The key for this relation is (Book_title, Authorname). This relation is in 1NF and not in 2NF as no attributes are Full FD on the key. It is also not in 3NF. (b) 2NF decomposition: Book0(Book_title, Authorname) Book1(Book_title, Publisher, Book_type, Listprice) Book2(Authorname, Author_affil) This decomposition eliminates the partial dependencies. 3NF decomposition: Book0(Book_title, Authorname) Book1-1(Book_title, Publisher, Book_type) Book1-2(Book_type, Listprice) Book2(Authorname, Author_affil) This decomposition eliminates the transitive dependency of Listprice

  35. Example Given the relation schema Car_Sale (Car#, Salesman#, Date_sold, Commission%, Discount_amt) with the functional dependencies Date_sold -> Discount_amt Salesman# -> Commission% Car# -> Date_sold This relation satisfies 1NF but not 2NF (Car# -> Date_sold and Salesman# -> Commission%) so these two attributes are not Full FD on the primary key and not 3NF

  36. Answer To normalize, 2NF: Car_Sale1 (Car#, Salesman#) Car_Sale2 (Car#, Date_sold, Discount_amt) Car_Sale3 (Salesman#,Commission%) 3NF: Car_Sale1(Car#, Salesman#) Car_Sale2-1(Car#, Date_sold) Car_Sale2-2(Date_sold, Discount_amt) Car_Sale3(Salesman#,Commission%)

  37. Question Given the following Dentist-patient database schema: Dentist-patient (staffNo, aDate, aTime, dentistName, patNo, patName, surgeryNo) Normalize the above relation, showing appropriate dependency diagrams to justify decomposition. FDs List: • FD1: staffNo, aDate, aTime patNo, patName • FD2: staffNo dentistName • FD3: patNopatName, surgeryNo • FD4: staffNo, aDate surgeryNo • FD5: aDate, aTime, patNo dentistName, staffNo

  38. Answer 1NF Dentist-patient (staffNo, aDate, aTime, dentistName, patNo, patName, surgeryNo) 2NF (fd2 and fd4 violates 2NF) Dentist-patient (staffNo, aDate, aTime, patNo, patName) Surgery (staffNo, aDate, surgeryNo) Dentist (staffNo, dentistName) 3NF (Fd3’ violates 3NF) Dentist-patient (staffNo, aDate, aTime, patNo) Surgery (staffNo, aDate, surgeryNo) Dentist (staffNo, dentistName) Patient (patNo, patName)

  39. Answer BCNF (No violation) Dentist-patient (staffNo, aDate, aTime, patNo) Surgery (staffNo, aDate, surgeryNo) Dentist (staffNo, dentistName) Patient (patNo, patName)

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