normalisation Goals: Suppose we have a db scheme: is it good? define precise notions of the qualities of a relational database scheme


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1 Goals: Suppose we have a db scheme: is it good? Suppose we have a db scheme derived from an ER diagram: is it good? define precise notions of the qualities of a relational database scheme define algorithms to generate good relational database schemes When is a db scheme good anyway? theory 1 2 The Holy Grail of : input: the set of all relevant attributes a set of constraints based on the attribute semantics Requirement 1: Your decomposition should be lossless Requirement 2: Your decomposition should avoid redundancy output: an optimal relational scheme for these attributes 3 4
2 Formalisation of avoiding redundancy: Suppose we have a scheme R and a set FDs F. R is in BCNF (w.r.t. F) iff each left side of a nontrivial FD is a superkey BCNF: BoyceCodd Normal Form Algorithm: INPUT: a universe R, a set FDs F OUTPUT: a lossless BCNFdecomposition van R METHOD: while there is a scheme S not in BCNF do suppose the villain has left side X let Y = X + without X let Z be the remaining attributes split S(XY Z) into S 1 (XY ), S 2 (XZ) od 5 6 properties of BCNF: BCNF represents the ultimate level of redundancy prevention caused by FDs implementation: in a BCNF schema, the left sides of FDs are keys, so you are able to enforce FDs by defining indexes observation 1: the BCNF decomposition algorithm is not deterministic observation 2: the number of possible BCNF decompositions may be very large observation 3: some BCNF decompositions turn out to be preferable to other BCNF decompositions 7 8
3 The closure of a set FDs F, denoted F +, is the set of all FDs that can be derived from F. Suppose we have a set FD s F and an FD X Y on a scheme R. Question: X Y F +? Method 1: Calculate F + and check whether X Y F +. Expensive! Method 2: Calculate X + and check whether Y X +. This can be done efficiently! 9 Observation: dependency sets may contain redundancy redundant FD: {A B, A C, B C} A C is redundant reducable left side: {A B, AB C} B is superfluous in AB C reducable right side: {A BC, B C} C is superfluous in A BC 10 Two FDsets F, G are equivalent if F + = G +. We say that these sets cover each other. G is a minimal cover of F if F and G are equivalent and 1. G does not contain redundant FDs 2. all right sides in G are minimal 3. all left sides in G are minimal 11 12
4 Algorithm: INPUT: a set FDs F OUTPUT: a minimal cover for F METHOD: 1. split each FD into single right side FDs 2. reduce left sides 3. eliminate redundant FDs 4. combine FDs with identical left sides Details: 2. reduce left sides: for each FD α β in F do for each attribute A in α do let α = α {A}; if α β can be derived from F then replace α β with α β fi od od Details: Problems with BCNF (continued) 3. eliminate redundant FDs: for each FD α β in F do let F = F {α β}; if α β can be derived from F then delete α β from F od; The projection of FD α β on scheme R is: 1. α β, if αβ attr(r) 2. void, if one of the attributes (left or right) is not in attr(r) 15 16
5 Suppose we have a scheme R and a set FDs F. A decomposition of R into R 1, R 2 is called dependency preserving (DP) if: (F 1 F 2 ) + = F + Examples: R = (ABCDE) F = {A BC, D E} R 1 = (ABC), R 2 = (ADE) Is this decomposition lossless, BCNF, DP? where F i is the projection of F + on R i. R = (ABCDE) F = {A BC, D E} R 1 = (AB), R 2 = (AC), R 3 = (ADE) Is this decomposition lossless, BCNF, DP? 17 R = (ABCDE) F = {A BCDE, C A, D E} R 1 = (ABC), R 2 = (CD), R 3 = (DE) Is this decomposition lossless, BCNF, DP? 18 observation 4: a DP decomposition is preferable to a non DP decomposition, because it enables efficient FD checking observation 5: there does not always exist a DP/BCNF decomposition observation 6: the BCNF decomposition algorithm does not produce always DP decompositions (even if they do exist) Making a choice: Minimize the level of redundancy: BCNF. Accept that the DP property may be lost. or: Allow a bit more redundancy (3NF in stead of BCNF). Enforce the DP property
6 An attribute is prime if it is contained in a candidate key. A relation scheme is in 3NF if for each non trivial FD X A the following condition holds: X is a superkey or A is prime BCNF versus 3NF ADDRESS Street Number Zipcode Ooievaarspad AM Ooievaarspad AM Meerkoetweide AK Meerkoetweide AK Meerkoetweide AL Meerkoetweide AL The following non trivial FDs hold: Street, Number Zipcode Zipcode Street ADDRESS is in 3NF ADDRESS is not in BCNF (redundancy) StrZip Street Zipcode O pad 3403 AM M weide 3403 AK M weide 3403 AL NoZip Number Zipcode AM AM AK AK AL AL StrZip and NoZip are in BCNF. This decomposition is not DP. An operation insert(6, 3403 AK ) on NoZip is incorrect! Detecting this violation requires a join with StrZip. Insert operations should be real time! 3NFAlgorithm: INPUT: a universe R, a set FDs F OUTPUT: a lossless DP 3NFdecomposition of R METHOD: create a minimal cover G from F; generate for each FD X A 1,..., A n a scheme (XA 1 A 2..A n ); if there is a scheme containing a global key K for R, then you are finished else add a global key as an extra relation scheme 23 24
7 overview normal forms normal form feasible complexity 3NF + DP always polynomial BCNF always polynomial BCNF + DP not always NPhard Requirement 2: you should strive for a high normal form Requirement 3: you should strive for a DP decomposition 25 Multivalued dependencies (MVD) Entity (target) with two unrelated multivalued attributes (company, car) target Charles Charles Andrew Andrew Private investigations target company car Charles Camilla Lamborghini Charles Claudia Maserati Charles Camilla Maserati Charles Claudia Lamborghini Andrew Koo Lamborghini Andrew Claudia Bugatti Andrew Koo Bugatti Andrew Claudia Lamborghini company Camilla Claudia Koo Claudia target car Charles Lamborghini Charles Maserati Andrew Lamborghini Andrew Bugatti 26 Compare: target is spotted with company in car Caught in the act target company car Charles Claudia Lamborghini Charles Camilla Maserati Andrew Koo Bugatti Andrew Claudia Maserati Intuition: A relation r with scheme R(XY Z) obeys the multivalued dependency (MVD) X Y if it is the case that Y and Z each have some connection with X, but do not have anything to do with each other. A relation r with scheme R(XY Z) obeys the multivalued dependency (MVD) X Y if the presence of t 1 and t 2 guarantees the presence of t 3 MVD X Y Z t 1 x y 1 z t 2 x y 2 z t 3 x y 1 z
8 Theorem: Suppose we have a scheme R(XY Z). X Y the decomposition R 1 (XY ), R 2 (XZ) is lossless observation: Suppose we have a scheme R(XY Z). X Y X Y Suppose we have a scheme R(XY Z). We call an MVD X Y trivial if Y X or if Z =. Suppose we have a relation r over a scheme R(XY Z); r is in 4NF if each left side of a non trivial MVD in R is a superkey. observation: Suppose we have a scheme R(XY Z). X Y X Z 29 30
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