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Normal forms - 1NF, 2NF, 3NF 1 Normal forms - 1NF, 2NF and 3NF (under the assumption of relations with just one candidate key)

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Normal forms - 1NF, 2NF, 3NF 2 Aim to familiarise with the concept of a normal form simplifying assumption (SA) every relation has one candidate key only, which is, therefore, the primary key

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Normal forms - 1NF, 2NF, 3NF 3 Generalities normal forms expressing FD constraints through choosing the structure of relations (structure the component attributes) 3NF 2NF 1NF transforming a relation into a higher normal form (vertical) decomposition - projection non-loss decomposition

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Normal forms - 1NF, 2NF, 3NF 4 1NF a relation is in 1NF if and only if all the domains of its attributes contain only scalar values

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Normal forms - 1NF, 2NF, 3NF 5 Relation in 1NF : Students-Info

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Normal forms - 1NF, 2NF, 3NF 6 1NF - update anomalies insert: exercise delete: exercise modify (update) suppose Databases changes to 3sem and 1.5cu the modification must be performed in more than one tuple; it implies a search; this structure (1NF) does not enforce consistent data; i.e. inconsistencies can be generated (e.g. by mistake) (implementation aspect) an update (involving a search) can be expressed by means of a single SQL statement, but –the search still has to be performed –an incorrect statement can generate inconsistencies

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Normal forms - 1NF, 2NF, 3NF 7 FD diagram for 1NF not all the determinants are the primary key

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Normal forms - 1NF, 2NF, 3NF 8 2NF A relation (with just one CK) is in 2NF if and only if it is in 1NF and every non-key attribute is irreducibly dependent on the primary key Students-Info is not 2NF; it can be decomposed into a set of relations in 2NF the equivalence must be preserved

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Normal forms - 1NF, 2NF, 3NF 9 Non-loss decomposition a decomposition of R into its projections R 1, …, R n is non-loss if and only if the natural join of R 1, …, R n results in the initial R

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Normal forms - 1NF, 2NF, 3NF 10 Non-loss decomposition

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Normal forms - 1NF, 2NF, 3NF 11 Activity consider different relations and experiment with different decompositions - identify which ones are lossy and which ones are non-loss

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Normal forms - 1NF, 2NF, 3NF 12 Heaths theorem Suppose A, B and C are sets of attributes of a relation R, such that A B and heading(R) = A B C then R = {A, B} JOIN {A, C}

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Normal forms - 1NF, 2NF, 3NF 13 Pragmatic principles when you apply Heaths theorem look for a maximum set of attributes B this will minimise the number of tables in the database try to maintain a one to one correspondence with real life entities

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Normal forms - 1NF, 2NF, 3NF 14 2NF - update anomalies consider the relation Module insert insert that a 3sem module is worth 1.5cu delete exercise modify (update) exercise

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Normal forms - 1NF, 2NF, 3NF 15 FD for 2NF transitive dependency

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Normal forms - 1NF, 2NF, 3NF 16 3NF A relation (with just one CK) is in 3NF if and only if it is in 2NF and every non-key attribute is non-transitively dependent on primary key alternatively … if there are no FDs between non-key attributes

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Normal forms - 1NF, 2NF, 3NF 17 Modules - decomposed (non-loss)

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Normal forms - 1NF, 2NF, 3NF 18 Concluding remarks all definitions given were for relations with just 1CK any relation can be decomposed into a set of 3NF relations 3NF is always achievable are all non-loss decompositions equivalent? what happens in the case of relations with more than 1CK?

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