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Agenda

Agenda. Last time: Normalization Homework 1 due now Project part 2 is up, due on the 19 th (Thurs.) This time: Finish BCNF 3NF 4NF Relational Algebra…. BCNF Review. Q: What’s required for BCNF? Q: What’s the slogan for BCNF? Q: Who are B & C?

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Agenda

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  1. Agenda • Last time: Normalization • Homework 1 due now • Project part 2 is up, due on the 19th (Thurs.) • This time: • Finish BCNF • 3NF • 4NF • Relational Algebra… M.P. Johnson, DBMS, Stern/NYU, Sp2004

  2. BCNF Review • Q: What’s required for BCNF? • Q: What’s the slogan for BCNF? • Q: Who are B & C? • Q: What are the two types of violations? M.P. Johnson, DBMS, Stern/NYU, Sp2004

  3. BCNF Review • Q: How do we fix a non-BCNF relation? • Q: If AsBs violates BCNF, what do we do? • Q: In this case, could the decomposition be lossy? • Q: Under what circumstances could a decomposition be lossy? • Q: How do we combine two relations? M.P. Johnson, DBMS, Stern/NYU, Sp2004

  4. Decomposition algorithm example • R(N,O,R,P) F = {N  O, O  R, R  N} • Key: N,P • Violations of BCNF: N  O, OR, N OR • which kinds of violations are these? • Pick N  OR (on board) • Can we rejoin? (on board) • What happens if we pick N  O instead? • Can we rejoin? (on board) M.P. Johnson, DBMS, Stern/NYU, Sp2004

  5. Lossless BCNF decomposition • Consider simple relation: R(A,B,C) • Only FD: A  B (assume C!A) Key: A,C • Diff vars from text! Also goes through if assumption is false • BCNF violation (which kind?): no key on the left • Thus: Decomposition to BCNF: • Create R1(A,B) and R2(A,C) • Could this be lossy? • We will join R1 and R2 on A to find out Q: Since C ! A, what kind of bad FD do we have? Q: If C  A, then what kind do we have? M.P. Johnson, DBMS, Stern/NYU, Sp2004

  6. Lossless BCNF decomposition • Suppose R contains (b,a,c) and (b’,a,c’) • In projection onto (B,A): • (b,a,c)  (b,a), (b’,a,c’)  (b’,a) • In projection onto (A,C): • (b,a,c)  (a,c), (b’,a,c’)  (a,c’) • In joining, (b’,a), (a,c)  (b’,a,c) • Q: Is/must/can this be correct? • A: Yes! A  B, so b = b’ • So this was lossless • We assumed C!A, but argument also goes through when CA • Moral: BCNF decomp alg is always lossless M.P. Johnson, DBMS, Stern/NYU, Sp2004

  7. BCNF summary • BCNF decomposition is lossless • Can reproduce original by joining • Saw last time: Every 2-attribute relation is in BCNF • Final set of decomposed relations might be different depending on • Order of bad FDs chosen • Saw last time: But all results will be in BCNF M.P. Johnson, DBMS, Stern/NYU, Sp2004

  8. A problem with BCNF • Relation: R(Title, Theater, Neighboorhood) • FDs: • Title,N’hood  Theater • Assume movie can’t play twice in same neighborhood • Theater  N’hood • Keys: • {Title, N’hood} • {Theater, Title} M.P. Johnson, DBMS, Stern/NYU, Sp2004

  9. R1 R2 Theater N’hood Theater Title Angelica Village Angelica City of God Angelica Fog of War A problem with BCNF • BCNF violation: Theater  N’hood • Decompose: • {Theater, N’Hood} • {Theater, Title} • Resulting relations: M.P. Johnson, DBMS, Stern/NYU, Sp2004

  10. (R’) Theater N’hood Title Angelica Village City of God Angelica Village Fog of War Film Forum Village City of God Problem - continued • Suppose we add new rows to R1 and R2: • Their join: R1 R2 A and B could not enforce FD Title,N’hood  Theater M.P. Johnson, DBMS, Stern/NYU, Sp2004

  11. Third normal form: motivation • There are some situations in which • BCNF is not dependency-preserving, and • Efficient checking for FD violation on updates is important • In these cases BCNF is too severe a req. • Solution: define a weaker normal form, called Third Normal Form • in which FDs can be checked on individual relations without performing a join (no inter-relational FDs) • to which relations can be converted, preserving both data and FDs M.P. Johnson, DBMS, Stern/NYU, Sp2004

  12. Third Normal Form • BCNF decomposition is not dependency-preserving! • We now define the (weaker) Third Normal Form • Turns out: this example was already in 3NF A relation R is in 3rd normal form if : For every nontrivial dependency A1, A2, ..., An Bfor R, {A1, A2, ..., An } is a super-key for R, or B is part of a key, i.e., B is prime Tradeoff: BCNF = no FD anomalies, but may lose some FDs 3NF = keeps all FDs, but may have some anomalies M.P. Johnson, DBMS, Stern/NYU, Sp2004

  13. BCNF: vices and virtues • Be clear on the problem just described v. the arg. that BCNF decomp is lossless • BCNF decomp does not lose data • Resulting relations can be rejoined to obtain the original • But: it can can lose dependencies • After decomp, possible to add rows whose corresponding rows would be illegal in (rejoined) original M.P. Johnson, DBMS, Stern/NYU, Sp2004

  14. Recap: goals of normalization • When we decompose a relation R with FDs F into R1..Rn we want: • lossless-join decomposition – no data lost • no/little redundancy: the relations Ri should be in either BCNF or at least 3NF • Dependency preservation: if Fi be the set of dependencies in F+ that include only attributes in Ri: • F is the “sum” of the FDs of the new relations • (F1  F2  F3  …  Fn)+ = F+ • Otherwise checking updates for violation of FDs may require computing joins, which is expensive M.P. Johnson, DBMS, Stern/NYU, Sp2004

  15. Dependency preservation • Saw that last req. didn’t hold in move-theater example • Did it hold in R(N,O,R,P) example? (on board) M.P. Johnson, DBMS, Stern/NYU, Sp2004

  16. Testing for 3NF • For each dependency X  Y, use attribute closure to check if X is a superkey • If X is not a superkey, verify that each attribute in Y is prime • This test is rather more expensive, since it involves finding candidate keys • Testing for 3NF is NP-complete (in what?) • Interestingly, decomposition into 3NF can be done in polynomial time •  Testing for 3NF is harder than decomposing into 3NF! • Optimization: need to check only FDs in F, need not check all FDs in F+ (why?) M.P. Johnson, DBMS, Stern/NYU, Sp2004

  17. 3NF Example • R = (J, K, L) • F = (JK  L, L  K) • Two candidate keys: JK and JL • R is in 3NF • JK  L JK is a superkey • L  K K is prime • BCNF decomposition yields • R1 = (L,K), R2 = (L,J) • testing for JK  L requires a join • There is some redundancy in R M.P. Johnson, DBMS, Stern/NYU, Sp2004

  18. BCNF and 3NF Comparison • Example of problems due to redundancy in 3NF • R = (J, K, L) • F = (JK  L, L  K) • A schema that is in 3NF but not BCNF has the problems of: • redundancy (e.g., the relationship between l1 and k1) • need to use null values (if allowed!), e.g. to represent the relationship between l2 and k2 when there is no corresponding value for attribute J M.P. Johnson, DBMS, Stern/NYU, Sp2004

  19. Comparison of BCNF and 3NF • It is always possible to decompose a relation into relations in 3NF such that: • the decomposition is lossless • the dependencies are preserved • It is always possible to decompose a relation into relations in BCNF such that: • the decomposition is lossless • but it may not be possible to preserve dependencies • But may eliminate more redundancy M.P. Johnson, DBMS, Stern/NYU, Sp2004

  20. The Normal Forms (so far) • 1NF: every attribute has an atomic value • 2NF: 1NF and no partial dependencies • 3NF: for each FD X  Y either • it is trivial, or • X is a superkey, or • Y is a part of some key • BCNF: • 3NF and third 3NF option disallowed • I.e, 2NF and no transitive dependencies M.P. Johnson, DBMS, Stern/NYU, Sp2004

  21. Distinguishing examples • 1NF but not 2NF: R(Name, SSN ,Mailing- address,Phone) • Key: SSN,Phone • Partial: ssn  name, address • 2NF but not 3NF: R(Title,Year,Studio,Pres,Pres-Addr) • Key: Title,Year • Transitive: studio  president • 3NF but not BCNF: R(Title, Theater, N’hood) • Title,N’hood  Theater • Prime-on-right: Theater  N’hood M.P. Johnson, DBMS, Stern/NYU, Sp2004

  22. Design Goals • Goal for a relational database design is: • No redundancy • Lossless Join • Dependency Preservation • If we cannot achieve this, we accept one of • dependency loss • use of more expensive inter-relational methods to preserve dependencies • data redundancy due to use of 3NF • Interesting: SQL does not provide a direct way of specifying FDs other than superkeys • can specify FDs using assertions, but they are expensive to test M.P. Johnson, DBMS, Stern/NYU, Sp2004

  23. 3NF • 3NF means we may have anomalies • Example: TEACH(student, teacher, subject) • student, subject  teacher (students not allowed in the same subject with two teachers) • teacher  subject (each teacher teaches one subject) • Subject is prime, so this is 3NF • But we have anomalies: • Insertion: cannot insert a teacher until we have a student taking his subject • If we convert to BCNF, we lost student, subject  teacher M.P. Johnson, DBMS, Stern/NYU, Sp2004

  24. BCNF and over-normalization • What is the problem? • Schema overload – trying to capture two meanings: • 1) subject X can be taught by teacher Y • 2) student Z takes subject W from teacher V • What to do? • 3NF has anomalies, normalizing to BCNF loses FDs • One soln: keep the 3NF TEACH and another (BCNF) relation SUBJECT-TAUGHT (teacher, subject) • Still (more!) redundancy, but no more insert and delete anomalies M.P. Johnson, DBMS, Stern/NYU, Sp2004

  25. Name SSN Jobs Streets Citys Michael 123 Mayor 111 East 60th Street New York Michael 123 Mayor 222 Brompton Road London Michael 123 CEO 111 East 60th Street New York Michael 123 CEO 222 Brompton Road London Hilary 456 Senator 333 Some Street Chappaqua Hilary 456 Senator 444 Embassy Row Washington Hilary 456 First Lady 333 Some Street Chappaqua Hilary 456 First Lady 444 Embassy Row Washington Hilary 789 Lawyer 333 Some Street Chappaqua Hilary 789 Lawyer 444 Embassy Row Washington New topic: MVDs (3.7) • Consider this relation • People ~ their jobs ~ their residences • Person-address/city: many-many • Person-job: many-many • Address/city-job: independent M.P. Johnson, DBMS, Stern/NYU, Sp2004

  26. Redundancy in BCNF • Lots of redundancy! • Key? All fields • None determined by others! • Non-trivial FDs? None! •  In BCNF? Yes! • Now what? • New concept, leading to another normal form: • Multivalued dependencies M.P. Johnson, DBMS, Stern/NYU, Sp2004

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