Normal forms - 1NF, 2NF, 3NF

1

Normal forms - 1NF, 2NF and 3NF

(under the assumption of relations with just one candidate key)

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

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

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

Normal forms - 1NF, 2NF, 3NF

5

Relation in 1NF : Students-Info

Students-InfoS_id S_name S_addr M_id M_name Type Value Res

F122 M. Kent UB8 DB1 Databases 2sem 1cu 72%F122 M. Kent UB8 P01 Programming 2sem 1cu 62%F122 M. Kent UB8 SWE SWare Eng 2sem 1cu 65%F122 M. Kent UB8 FYP Fin Year Proj proj 1cu NULLP141 S. Brown SE14 DB1 Databases 2sem 1cu 68%P141 S. Brown SE14 IS01 Info. Sys. 1sem 0.5cu 76%P141 S. Brown SE14 DS Decision Sup 1sem 0.5cu NULLF87 P. Wild MN6 DB2 Databases 1sem 0.5cu NULL… … … … … … … …

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

Normal forms - 1NF, 2NF, 3NF

7

FD diagram for 1NF

S_idRes

S_name

S_addr M_id

TypeM_name Value

not all the determinants are the primary key

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

Normal forms - 1NF, 2NF, 3NF

9

Non-loss decomposition

a decomposition of R into its projections R1, …, Rn is non-loss if and only if the natural join of R1, …, Rn results in the initial R

Normal forms - 1NF, 2NF, 3NF

10

Non-loss decomposition

Students ModulesS_id S_name S_addr M_id M_name Type Value

F122 M. Kent UB8 DB1 Databases 2sem 1cuP141 S. Brown SE14 P01 Programming 2sem 1cuF87 P. Wild MN6 SWE SWare Eng 2sem 1cu… … … FYP Fin Year Proj proj 1cu

IS01 Info. Sys. 1sem 0.5cuDS Decision Sup 1sem 0.5cuDB2 Databases 1sem 0.5cu… … … …

Stud-ModS_id M_id Res

F122 DB1 72%F122 P01 62%F122 SWE 65%F122 FYP NULLP141 DB1 68%P141 IS01 76%P141 DS NULLF87 DB2 NULL… … …

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

Normal forms - 1NF, 2NF, 3NF

12

Heath’s theorem

Suppose A, B and C are sets of attributes of a relation R, such that

• AB and • heading(R) = A B C

then• R = {A, B} JOIN {A, C}

Normal forms - 1NF, 2NF, 3NF

13

Pragmatic principles

when you apply Heath’s 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

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

Normal forms - 1NF, 2NF, 3NF

15

FD for 2NF

M_id

TypeM_name Value

transitive dependency

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

Normal forms - 1NF, 2NF, 3NF

17

Modules - decomposed (non-loss)

Modules-Descr Type-Val-CorrespondenceM_id M_name Type Type ValueDB1 Databases 2sem 1sem 0.5cuP01 Programming 2sem 2sem 1cuSWE SWare Eng 2sem proj 1cuFYP Fin Year Proj proj … …IS01 Info. Sys. 1semDS Decision Sup 1semDB2 Databases 1sem… … …

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?