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course: Database Systems (A7B36DBS, Y36DBS, X36DBS)

course: Database Systems (A7B36DBS, Y36DBS, X36DBS). Lecturers: RNDr. Irena Mlynkova, Ph.D. & Mgr. Martin Necasky, Ph.D. Acknowledgement: The slides were kindly lent by Doc. RNDr. Tomas Skopal, Ph.D., Department of Software Engineering, Charles University in Prague.

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course: Database Systems (A7B36DBS, Y36DBS, X36DBS)

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  1. course: Database Systems (A7B36DBS, Y36DBS, X36DBS) Lecturers: RNDr. Irena Mlynkova, Ph.D. & Mgr. Martin Necasky, Ph.D. Acknowledgement: The slides were kindly lent by Doc. RNDr. Tomas Skopal, Ph.D., Department of Software Engineering, Charles University in Prague

  2. Today’s lecture outline • motivation • data redundancy and update/insertion/deletion anomalies • functional dependencies • Armstrong’s axioms • attribute and dependency closures • normal forms • 3NF • BCNF Relational design – normal forms (Lect. 8)

  3. Motivation • result of relational design – a set of relational schemas • problems • data redundancy • unnecessary multiple storage of the same data – increased space cost • insert/update/deletion anomalies • insertions and updates must preserve redundant data storage • deletion might cause loss of some data • solution • relational schema normalization Relational design – normal forms (Lect. 8)

  4. Example of “abnormal” schema 1) From functional analysis we know that position determines hourly salary. However, hourly salary data is stored multiple times – redundancy. 2) If we delete employee 6, we lose the information on lecturer salary. 3) If we change the accountant hourly salary, we must do that in three places. Relational design – normal forms (Lect. 8)

  5. How could this even happen? • simply • during “manual” design of relation schemas • badly designed conceptual model • e.g., too many attributes in a class +is owned by 1 +owns 0..* Person- id - address - educationetc. Phone- serial nr.- manufacturer - modeletc. the UML diagram results in 2 tables: Person(id, address, education, ...)Mobil(serial nr., manufacturer, model, ..., id) Relational design – normal forms (Lect. 8)

  6. How could this even happen? Redundancy in attributes Manufacturer, Model, Made in, CertificateWhat happened?ClassPhoneincludes also other classes – Manufacturer, Model,... How to fix it? Two options 1) fix the UMLmodel (design of more classes) 2) alter the already created schemas (see next) Relational design – normal forms (Lect. 8)

  7. Functional dependencies • attribute-based integrity constraints defined by the user (e.g., DB application designer) • a kind of alternative to conceptual modeling (ER and UML invented much later) • functional dependency (FD) X  Y over schema R(A) • mapping fi : Xi Yi, where Xi,Yi A (where i = 1..number of FDs in R(A)) • n-tuple from Xideterminesm-tuple from Yi • m-tuple from Yiis determined by (is dependent on)n-tuple from Xi Relational design – normal forms (Lect. 8)

  8. Functional dependencies • simply, for X  Y, values in X together determine the valuesin Y • if X  Y and Y  X, then X and Y arefunctionally equivalent • could be denoted as X  Y • if X  a, where a  A, then X  a is elementary FD • FDs represent a generalizationof the key concept (identifier) • the key is a special case, see next slides Relational design – normal forms (Lect. 8)

  9. Example – wrong interpretation One might observe from the data, that:Position Hourly salaryand alsoHourly salaryPositionEmpId everythingHours completedeverythingNameeverything (but that is nonsense w.r.t. the natural meaning of the attributes) Relational design – normal forms (Lect. 8)

  10. Example – wrong interpretation newly inserted records Position Hourly salaryEmpId everything Hourly salaryPositionHours completedeverythingNameeverything Relational design – normal forms (Lect. 8)

  11. Example – correct interpretation • at first, after the data analysis the FDs are set “forever”, limiting the content of the tables • e.g. Position Hourly salaryEmpId everything • insertion of the last row is not allowed as it violates both the FDs Relational design – normal forms (Lect. 8)

  12. Armstrong’s axioms Let us haveR(A,F). LetX, Y, Z  A and F is the set of FDs 1) ifY  X, thenX  Y (trivial FD, really axiom) 2) ifX  Y andY  Z, thenX  Z (transitivity, rule) 3) ifX  Y andX  Z, thenX  YZ (composition, rule) 4) ifX  YZ, thenX  Y andX  Z (decomposition, rule) Relational design – normal forms (Lect. 8)

  13. Armstrong’s axioms Armstrong’s axioms: • are correct(sound) • what is derived from F is valid for any instance from R • are complete • all FDs valid in all instances in R (w.r.t. F) can be derived using the axioms • 1,2,3(trivial, transitivity, composition) are independent • removal of any axiom 1,2,3 violates the completeness(decomposition could be derived from trivial FD and transitivity) Proof follows from the FD definition (from the properties of mapping, respectively). Relational design – normal forms (Lect. 8)

  14. Example – deriving FDs R(A,F) A = {a,b,c,d,e}F = {ab  c, ac  d, cd  ed, e  f} We could derive, e.g.,:ab  a (trivial)ab  ac (compositionwithab  c) ab  d (transitivity withac  d)ab  cd (compositionwithab  c)ab  ed (transitivity with cd  ed)ab  e (decomposition)ab  f (transitivity) Relational design – normal forms (Lect. 8)

  15. Example – deriving the decomposition rule R(A,F) A = {a,b,c}F = {a  bc} Deriving:a  bc (assumption) bc b (trivial FD) bc c (trivial FD) a  b (transitivity) a  c (transitivity) i.e.,a  bc a  b  a c Relational design – normal forms (Lect. 8)

  16. Closure of set of FDs • closureF+of FDs set F (FDclosure) is the set of all FDsderivable from F using the Armstrong’s axioms • generallyexponential size w.r.t.|F| • a FDf isredundantin F if(F – {f})+ = F+, i.e., f can be derived from the rest of F Relational design – normal forms (Lect. 8)

  17. Example – closure of set of FDs R(A,F), A = {a,b,c,d}, F = {ab c, cd  b, ad  c} F+ = {a  a, b  b, c  c,ab a, ab b, ab c, cd  b, cd  c, cd  d, ad  a, ad  c, ad  d,abd  a, abd  b, abd  c, abd  d,abd  abcd,...} Relational design – normal forms (Lect. 8)

  18. Cover • cover of a set F is any set of FDs G such that F+=G+ • canonical cover = cover consisting of elementary FDs(decompositions are performed to obtain singleton sets on the right side) • non-redundant cover of F = cover of F after removing all redundant FDs • note the order of removing FDs matters – a redundant FD could become non-redundant FD after removing another redundant FD • implies that there may exist multiple non-redundant covers of F Relational design – normal forms (Lect. 8)

  19. Example – cover R1(A,F), R2(A,G), A = {a,b,c,d}, F ={a  c, b  ac, d  abc},G ={a  c, b  a, d  b}For the check ofG+=F+we do not have to establish the whole covers, it is sufficient to derive F from G, and vice versa, i.e.,F’ = {a  c, b  a, d  b}– decompositionG’ = {a  c, b  ac, d  abc }– transitivity and compositionG+=F+ SchemasR1 andR2 areequivalentbecause G is cover of F, while they share the attribute set A. Moreover, G isminimal cover, whileF is not (for minimal cover see next slides). Relational design – normal forms (Lect. 8)

  20. Example – redundant FDs R1(A,F), R2(A,G), A = {a,b,c,d}, F ={a  c, b  a, b  c, d  a, d  b, d  c}FDsb  c, d  a, d  c are redundantafter their removal F+ is not changed, i.e., they could be derived from the remaining FDsb  c derived using transitivitya  c, b  ad  a derived using transitivityd b, b  ad  c derived using transitivityd b, b  a, a  c Relational design – normal forms (Lect. 8)

  21. Attribute closure, key • attribute closure X+ (w.r.t. F) is a subset of attributes from A determined by X (using F) • consequence: if X+=A, then X is asuper-key • if F contains a FD X  Y andthere exist an attribute a in X such that Y  (X – a)+, then ais anattribute redundant in X  Y • reduced FD is such FD that does not contain any redundant attributes • keyisa set K  Asuch that it is a super-key (i.e., K  A) andK  A ismoreover reduced • there could exist multiple keys (at least one) • if there is no FD in F, it trivially holds A  A, i.e.,the key is the entire set A • key attribute = attribute that is in any key Relational design – normal forms (Lect. 8)

  22. Example – attribute closure R(A,F), A = {a,b,c,d}, F = {a  c, cd  b, ad  c} {a}+ = {a,c}it holdsa  c (+ triviala  a) {b}+ = {b} (trivial b b) {c}+ = {c}(trivial c c) {d}+ = {d}(trivial d d) {a,b}+ = {a,b,c}ab c(+ trivial) {a,d}+ = {a,b,c,d}ad bc(+ trivial) {c,d}+ = {b,c,d}cd b(+ trivial) Relational design – normal forms (Lect. 8)

  23. Example – redundant attribute R(A,F), A = {i,j,k,l,m}, F = {m  k, lm  j, ijk  l, j m, l i, l k} Hypothesis: k is redundant in ijk l, i.e., it holdsij  l Proof: • based on thehypothesis let’s construct FDij  ? • ijk  lremainsin F because weADD new FDij  ?so that we can use ijk  lfor construction of the attribute closure{i,j}+ • we obtain{i,j}+ = {i, j, m, k, l}, i.e.,there existsij  l which we add into F (it is the member ofF+) • now forget how ij  l got into F • becauseijk  l could be trivially derived fromij  l, it is redundant FD and we can remove it from F • so, we removed the redundant attribute kinijk  l In other words, we transformed the problem of removing redundant attribute on the problem of removing redundant FD. Relational design – normal forms (Lect. 8)

  24. FDs: can be redundant “we don’t need it” can have a closure “all derivable FDs” can be elementary “single attribute on the right side” can be reduced “no redundancies on the left side” Attributes: can be redundant “we don’t need it” can have a closure “all derivable attributes” can form (super-)keys FDs vs. attributes

  25. Minimal cover • non-redundant canonical cover that consists of only reduced FDs • i.e. no redundant FDs, no redundant attributes, decomposed FDs • is constructed by removing redundant attributes in FDs followed by removing of redundant FDs (i.e., the order matters) Example: abcd e, e  d, a  b, ac  d Correct order of reduction:Wrong order of reduction: 1. b,d are redundant 1. no redundant FDin abcd  e, i.e., removing them 2. redundant b,d in abcd  e 2. ac  d is redundant (now not a minimal cover, because ac  d is redundant) Relational design – normal forms (Lect. 8)

  26. First normal form (1NF) Every attribute in a relation schema is of simple non-structured type. (1NF is the basiccondition on „flat database“ – a table is really two-dimensional array, not a hidden graphortree) Relational design – normal forms (Lect. 8)

  27. Example – 1NF Person(Id: Integer, Name: String, Birth: Date) is in 1NF Employee(Id: Integer, Subordinate: Person[], Boss: Person) not in 1NF (nestedtable of typePersonin attributeSubordinate, and the Boss attribute is structured) Relational design – normal forms (Lect. 8)

  28. 2nd normal form (2NF) • there do not exist partial dependencies of non-key attributeson (any) key, i.e.,it holdsx  NK KK : KK  x (where NK is the set of non-key attributes and KK is subset of some key) key attribute(s) non-key attribute key Relational design – normal forms (Lect. 8)

  29. Example – 2NF  not in 2NF, becauseHQis determined by a part of key (Company) consequence: redundancyofHQ values Company, DB Server everythingCompany HQ both schemasare in 2NF Company HQ Company, DB Server everything Relational design – normal forms (Lect. 8)

  30. Transitive dependency on key • FD A → B such that A → some key (A is not a super-key), i.e., we get transitivity key → A → B • Transitivities point to probable redundancies, because from the definition of FD as a mapping: • unique values of key are mapped to the same or less unique values of A, and those are mapped to the same or less unique values of B Example in 2NF: ZIPcode → City → Country no redundancymedium redundancyhigh redundancy Relational design – normal forms (Lect. 8)

  31. key Y a a key a key X X a X 3rd normal form (3NF) • non-key attributes are not transitively dependent on key • as the 3NF using the above definition cannot be tested without construction of F+, we use a definition that assumes only R(A,F) • at least one condition holds for each FD X a (where X  A, a A): • FD is trivial • X is super-key • a is part of a key (i.e., a key attribute) Relational design – normal forms (Lect. 8)

  32. Example – 3NF CompanyeverythingZIPcodeHQ is in 2NF, not in 3NF(transitive dependency ofHQonkeythroughZIPcode) consequence: redundancyofHQ values CompanyeverythingZIPcodeeverything both schemas arein 3NF Relational design – normal forms (Lect. 8)

  33. Boyce-Codd normal form (BCNF) • every attribute is (non-transitively) dependent on key • more exactly, in a given schema R(A, F) there holds at least one condition for each FD X a (where X  A, a A): • FD is trivial • X is super-key • the same as 3NF without the last option (a is key attribute) a key a X a key X X Relational design – normal forms (Lect. 8)

  34. Example – BCNF Pilot, Day  everythingPlane, Day  everythingDestination  Pilot is in 3NF, not in BCNF (Pilot is determined by Destination, which is not a super-key) consequence: redundancy of Pilot values Destination  PilotPlane, Day  everything both schemasare in BCNF Relational design – normal forms (Lect. 8)

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