Normalisation of something i dont know and will never know
- 1. 2nd Normal Form A relation schema is in 2NF if it satisfies 1NF and every non key attribute is fully functionally dependent on the primary key of Relation.
- 2. Example of 2NF EMP_PROJ is 1NF with Key SSN, PNUMBER but… SSN ENAME - ENAME, a Non-Prime Attribute, Depends Partially on SSN, PNUMBER, i.e., Depends on Only SSN and not Both PNUMBER {PNAME, PLOCATION} - PNAME, PLOCATION, two Non-Prime Attributes, Depends Partially on SSN, Pno i.e., Depends on Only PNo ¬ Both
- 3. Another Example of 2NF
- 4. Third Normal Form (3NF) It is based on the concept of transitive dependency. A functional dependency XY in a relation schema R is a transitive dependency if there is a set of attributes Z that is neither a candidate key nor a subset of any key of R, and both XZ and ZY hold.( Book Def)
- 5. EMP_DEPT ENAME SSN BDATE ADDRESS DNUMBER DNAME DMGRSS N
- 6. Here, SSNDNAME Since SSNDNUMBER and DNUMBERDNAME holds.
- 7. A relation schema R is said to be in 3NF if it satisfies 2NF and no nonprime attribute of R is transitively dependent on the Primary key. To normalize from 2NF to 3NF, decompose and set up a relation that includes the non-prime attribute that functionally determine other non-prime attribute and remove this attribute from the original relation. Relation Schema EMP_DEPT is not in 3NF but it is in 2NF as no partial dependency on a key exists.
- 8. To normalize EMP_DEPT into 3NF we decompose it into two 3NF relation schemas as shown:
- 9. ENAME SSN BDATE ADDRESS DNUMBER DNAME DMGRSS N ENAME SSN BDAT E ADDRESS DNUMBE R DNUMBE R DNAME DMGRSS N ED1 ED2
- 10. Consider a LOTS relation schema that describes plots for sale in various states of countries . It has two candidate keys PROPERTY_ID# and {COUNTRY_NAME,LOT#}. Also we assume that tax rate is determined by country name. Also price of a lot is determined by it’s area.
- 11. PROPERTY _ID# COUNTRY_N AME LOT# AREA PRICE TAX_RATE Determine all FDs. And Normalise the relation schema to 3rd Normal Form.
- 12. PROPERTY _ID# COUNTRY_N AME LOT# AREA PRICE TAX_RATE FD1 FD2 FD3 FD4
- 13. Decomposition into 2NF LOTS1 PROPERTY _ID# COUNTRY_N AME LOT# AREA PRICE LOTS2 COUNTRY_NAM E TAX_RATE
- 14. Normalizing LOTS1 into 3NF (lots 2 will remain as it is) PROPERTY _ID# COUNTRY_NAME LOT# AREA LOTS1A LOTS1B AREA PRICE
- 15. Boyce-Codd Normal Form A relation is in BCNF if and only if every nontrivial, left-irreducible FD has a candidate key as it’s determinant. Hence, for every nontrivial, left-irreducible FD XA, X should be a candidate key. The difference between 3NF and BCNF is that 3NF allows A to be a prime attribute. Every relation that in BCNF is also in 3NF, however a relation in 3NF is not necessarily in BCNF.For example,
- 16. In simple terms
- 18. Superkey
- 22. 4 NF Fourth normal form (4NF) is a level of database normalization where there are no non-trivial multivalued dependencies other than a candidate key. It builds on the first three normal forms (1NF, 2NF and 3NF) and the Boyce-Codd Normal Form (BCNF
- 23. Properties : A relation R is in 4NF if and only if the following conditions are satisfied: 1.It should be in the Boyce-Codd Normal Form (BCNF). 2.the table should not have any Multi- valued Dependency.
- 24. A table with a multivalued dependency violates the normalization standard of Fourth Normal Form (4NK) because it creates unnecessary redundancies and can contribute to inconsistent data. To bring this up to 4NF, it is necessary to break this information into two tables.
- 25. example SID SNAME S1 A S2 B Table – R1(SID, SNAME) Table – R2(CID, CNAME)
- 26. CID CNAME C1 C C2 D
- 27. SID SNAME CID CNAME S1 A C1 C S1 A C2 D S2 B C1 C S2 B C2 D Multivalued dependencies (MVD) are: SID->->CID; SID->->CNAME; SNAME->->CNAME
- 28. Normal Form (5NF): A relation R is in 5NF if and only if every join dependency in R is implied by the candidate keys of R. A relation decomposed into two relations must have loss-less join Property, which ensures that no spurious or extra tuples are generated, when relations are reunited through a natural join.
- 29. Properties – A relation R is in 5NF if and only if it satisfies following conditions: 1.R should be already in 4NF. 2.It cannot be further non loss decomposed (join dependency)
- 30. Agent Company Product A1 PQR Nut A1 PQR Bolt A1 XYZ Nut Table – ACP
- 31. The relation ACP is again decompose into 3 relations. Now, the natural Join of all the three relations will be shown as: Agent Company A1 PQR A1 XYZ A2 PQR Table – R1
- 32. Agent Product A1 Nut A1 Bolt A2 Nut able – R2
- 33. Table R3 Company Product PQR Nut PQR Bolt XYZ Nut XYZ Bolt
- 34. Result of Natural Join of R1 and R3 over ‘Company’ and then Natural Join of R13 and R2 over ‘Agent’and ‘Product’ will be table ACP. Hence, in this example, all the redundancies are eliminated, and the decomposition of ACP is a lossless join decomposition. Therefore, the relation is in 5NF as it does not violate the property of lossless join.