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Normalization Sridhar Narayan narayans@uncw.edu

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SSNPNUMBERHOURSENAMEPNAMEPLOC E1P120JoeCIS RoofUNCW E1P220JoeRestaurantMayfaire E2P140JoeCIS RoofUNCW EMP_PROJ Something feels wrong about this design Try adding a row – Insertion anomaly Try deleting a row – Deletion anomaly Try updating a row – Update anomaly Need a formal way to reason about what is wrong with it and how to fix it

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Functional Dependency Constraints between attribute sets in a relation If X and Y are sets of attributes of a relation R, and whenever two tuples in R have the same X-values they also have the same Y-values, we say that X functionally determines Y.

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Functional Dependency Written as X -> Y – X functionally determines Y – Y is functionally determined by X – X is the determinant, Y is the dependent Examples – SSN -> SSN (trivial dependency) – PNUMBER -> PNAME – SSN -> ENAME – SSN, PNUMBER -> HOURS

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Functional Dependency Between sets of attributes, not just single attributes Holds for all time, not just for a particular instance (snapshot) of a relation Formally states constraints that exist for the relation – These constraints are in addition to those imposed by primary keys and foreign keys

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Functional dependencies and keys If X functionally determines all attributes of R, then X is a super key If X is irreducible, i.e. every member of X is essential for the functional dependencies to hold, then X is a candidate key. Attributes that are a part of a candidate key are key attributes

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Examples Super key: – SSN, PNUMBER, PNAME -> SSN, PNUMBER, HOURS, ENAME, PNAME, PLOC Candidate key: – SSN, PNUMBER -> SSN, PNUMBER, HOURS, ENAME, PNAME, PLOC SSNPNUMBERHOURSENAMEPNAMEPLOC E1P120JoeCIS RoofUNCW E1P220JoeRestaurantMayfaire E2P140JoeCIS RoofUNCW

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Redundancy If in a relation R, A -> B and A is not a candidate key for R, then R will involve some redundancy. SSNPNUMBERHOURSENAMEPNAMEPLOC Intuitively, all functional dependencies in a relation should involve candidate keys to eliminate redundancy

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Normalization A process that utilizes functional dependencies to identify relation schemas that have an undesirable form (redundancy) and decomposes them into smaller schema in which the redundancy has been eliminated.

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Decomposition Decomposition should be – Lossless join Allow exact recovery of the original schema (without spurious tuples) – Dependency preserving Allow dependencies to be checked without requiring a join

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Lossy decomposition SSNPNUMBERHOURSENAME E1P120Joe E1P220Joe E2P140Joe ENAMEPNAMEPLOC JoeCIS RoofUNCW JoeRestaurantMayfaire JoeCIS RoofUNCW

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Natural join to recover original SSNPNUMBERHOURSENAMEPNAMEPLOC E1P120JoeCIS RoofUNCW E1P220JoeRestaurantMayfaire E2P140JoeCIS RoofUNCW E2P140JoeRestaurantMayfaire

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Heath’s Theorem If relation R = {A,B,C} where A,B,C are attribute sets and A -> B then R 1 = {A, B} and R 2 = {A, C} represents a lossless decomposition

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Levels of normalization First normal form – 1NF Second normal form – 2NF Third normal form – 3NF Boyce-Codd Normal Form - BCNF Increasingly stringent requirements

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Normal Forms 1NF 2NF 3NF BCNF

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First normal form Relation is in 1NF if all attribute values are atomic (By definition, all relations are in 1NF) D_NAMED_NUMMGR_SSND_LOCATIONS RESEARCH5334619276{Lumberton, Red Springs, Raeford} Assume that a department can have multiple locations, like {Lumberton, Red Springs, Raeford} Relation not in 1NF

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Resolution? D_NAMED_NUMMGR_SSND_LOCATIONS RESEARCH5334619276Lumberton RESEARCH5334619276Red Springs RESEARCH5334619276Raeford

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Decomposition D_NAMED_NUMMGR_SSND_LOCATIONS D_NAMED_NUMMGR_SSND_NUMD_LOCATIONS

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Second Normal Form: 2NF A relation is in 2NF if – It is in 1NF, and – If the non-key attributes are fully (irreducibly) dependent on the primary key

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Example: EMP_PROJ SSNPNUMBERHOURSENAMEPNAMEPLOC Functional Dependencies? SSN -> ENAME PNUMBER -> PNAME, PLOC {SSN, PNUMBER} -> HOURS Relation not in 2NF Non-key attributes ENAME, and PLOC and PNAME, are not fully dependent on the primary key

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Solution? Decompose SSNPNUMBERENAMEPNAMEPLOC SSNPNUMBERHOURS

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Decompose further… SSNPNUMBERPNAMEPLOC SSNENAME

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And a little more… SSNPNUMBER 3b is a part of 1a, so drop it. PNUMBERPNAMEPLOC

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2NF Normalization SSNPNUMBERHOURS SSNENAMEPNUMBERPNAMEPLOC

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More than one way to get here SSNPNUMBERHOURSENAMEPNAMEPLOC PNUMBERPNAMEPLOC SSNPNUMBERHOURSENAME

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Decompose further… SSNPNUMBERHOURSSSNPNUMBERENAME

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And a little bit more SSNPNUMBER SSNENAME

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3NF Normalization A relation is in 3NF if – It is in 2NF, and – If the non-key attributes are mutually independent. That is, no functional dependencies exist between non-key attributes.

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Example: EMP_DEPT Functional Dependencies? SSN -> {ENAME, DOB, ADDRESS, DNUM} DNUM -> {DNAME, DMGRSSN} Redundancy? Relation in 1NF ? 2NF ? 3NF ? SSNENAMEDOBADDRESSDNUMDNAMEDMGRSSN

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3NF Normalization DNUMDNAMEDMGRSSN SSNENAMEDOBADDRESSDNUM

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BCNF Normalization S# and SNAME – Supplier# and Supplier Name are unique FDs – S# -> SNAME – SNAME -> S# – S#,P# -> QTY – SNAME, P# -> QTY Candidate keys – S#, P# and SNAME, P# S#SNAMEP#QTY S1Acme SupplyP1100 S2Gem MfgP1200 S1Acme SupplyP2400

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BCNF Normalization Redundancy? 1NF? 2NF? 3NF? S#SNAMEP#QTY S1Acme SupplyP1100 S2Gem MfgP1200 S1Acme SupplyP2400

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BCNF Relation is in BCNF if and only if the only determinants are candidate keys FDs – S# -> SNAME – SNAME -> S# – S#,P# -> QTY – SNAME, P# -> QTY

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BCNF Normalization S#P#QTY S1P1100 S2P1200 S1P2400 S#SNAME S1Acme Supply S2Gem Mfg S1Acme Supply Two candidate keys: S# SNAME

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1 Normalization Chapter 19. 2 What it’s all about Given a relation, R, and a set of functional dependencies, F, on R. Assume that R is not in a desirable.

1 Normalization Chapter 19. 2 What it’s all about Given a relation, R, and a set of functional dependencies, F, on R. Assume that R is not in a desirable.

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