1. Lossless Decomposition (2) Prof. Sin-Min Lee Department of Computer Science San Jose State University

2. Purpose of Normalization To reduce the chances for anomalies to occur in a database. normalization prevents the possible corruption of databases stemming from what are called "insertion anomalies," "deletion anomalies," and "update anomalies."

3. Normal Forms Each normal form is a set of conditions on a schema that guarantees certain properties (relating to redundancy and update anomalies) First normal form (1NF) is the same as the definition of relational model (relations = sets of tuples; each tuple = sequence of atomic values) Second normal form (2NF) – a research lab accident; has no practical or theoretical value – won’t discuss third normal formBoyce-Codd normal formThe two commonly used normal forms are third normal form (3NF) and Boyce-Codd normal form (BCNF)

5. BCNF Definition: A relation schema R is in BCNF if for every FD X  Y associated with R either –Y  X (i.e., the FD is trivial) or –X is a superkey of R Person1Example: Person1(SSN, Name, Address) –The only FD is SSN  Name, Address Person1 –Since SSN is a key, Person1 is in BCNF

6. (non) BCNF Examples PersonPerson (SSN, Name, Address, Hobby) –The FD SSN  Name, Address does not satisfy requirements of BCNF since the key is (SSN, Hobby) HasAccountHasAccount (AcctNum, ClientId, OfficeId) –The FD AcctNum  OfficeId does not satisfy BCNF requirements since keys are (ClientId, OfficeId) and (AcctNum, ClientId); not AcctNum.

7. Redundancy Suppose R has a FD A  B, and A is not a superkey. If an instance has 2 rows with same value in A, they must also have same value in B (=> redundancy, if the A-value repeats twice) If A is a superkey, there cannot be two rows with same value of A –Hence, BCNF eliminates redundancy SSN  Name, Address SSN Name Address Hobby 1111 Joe 123 Main stamps 1111 Joe 123 Main coins redundancy

8. Third Normal Form A relational schema R is in 3NF if for every FD X  Y associated with R either: –Y  X (i.e., the FD is trivial); or –X is a superkey of R; or –Every A  Y is part of some key of R 3NF is weaker than BCNF (every schema that is in BCNF is also in 3NF) BCNF conditions

9. 3NF Example HasAccountHasAccount (AcctNum, ClientId, OfficeId) –ClientId, OfficeId  AcctNum OK since LHS contains a key –AcctNum  OfficeId OK since RHS is part of a key HasAccountHasAccount is in 3NF but it might still contain redundant information due to AcctNum  OfficeId (which is not allowed by BCNF)

10. 3NF (Non) Example PersonPerson (SSN, Name, Address, Hobby) –(SSN, Hobby) is the only key. –SSN  Name violates 3NF conditions since Name is not part of a key and SSN is not a superkey

11. 3NF but not boyce-codd NF SUPPLIER_PART (supplier#, supplier_name, part#, quantity) –Two candidate keys: (supplier#, part#) and (supplier_name, part#) –(supplier#, part#) -> quantity –(supplier#, part#) -> supplier_name –(supplier#, part#) -> quantity –(supplier#, part#) -> supplier# –supplier_name -> supplier# –supplier# -> supplier_name

12. Another example of boyce-codd NF titleyearlengthfilmTypestudioNamestarName Star Wars1977124colorFoxFisher Star Wars1977124colorFoxHamill Star Wars1977124colorFoxFord Mighty Ducks1991104colorDisneyEsteves Wayne’s World199295colorParamountCarvey Wayne’s World199295colorParamountMeyers

13. Example (cont’d) {title, year, starName} as candidate key title, year  length, filmType, studioName The above FD (Functional Dependency) violates the BCNF condition because title and year do not determine the sixth attribute, starName

14. Example (cont’d) We solve this BCNF violation by decomposing relation Movies into 1. The schema with all the attributes of the FD {title, year, length, filmType, studioName} 2. The schema with all attributes of Movies except the three that appear on the right of the FD {title, year, starName}

15. Decompositions Goal: Eliminate redundancy by decomposing a relation into several relations in a higher normal form losslessDecomposition must be lossless: it must be possible to reconstruct the original relation from the relations in the decomposition We will see why

16. Decomposition Schema R = (R, F) –R is set a of attributes –F is a set of functional dependencies over R Each key is described by a FD decompositionof schemaThe decomposition of schema R is a collection of schemas R i = (R i, F i ) where –R =  i R i for all i (no new attributes) –F i is a set of functional dependences involving only attributes of R i –F entails F i for all i (no new FDs) decomposition of an instanceThe decomposition of an instance, r, of R is a set of relations r i =  R i (r) for all i

17. Example Decomposition Schema (R, F) where R = {SSN, Name, Address, Hobby} F = {SSN  Name, Address} can be decomposed into R 1 = {SSN, Name, Address} F 1 = {SSN  Name, Address} and R 2 = {SSN, Hobby} F 2 = { }

18. Lossless Schema Decomposition A decomposition should not lose information losslessA decomposition (R 1,…,R n ) of a schema, R, is lossless if every valid instance, r, of R can be reconstructed from its components: where each r i =  Ri (r) r = r 1 r2r2 rnrn ……

19. Lossy Decomposition r  r1r  r1 r2r2... rnrn SSN Name Address SSN Name Name Address 1111 Joe 1 Pine 1111 Joe Joe 1 Pine 2222 Alice 2 Oak 2222 Alice Alice 2 Oak 3333 Alice 3 Pine 3333 Alice Alice 3 Pine r  r1r  r1 r2r2 rnrn... r1r1 r2r2 r  The following is always the case (Think why?) : But the following is not always true : Example : The tuples (2222, Alice, 3 Pine) and (3333, Alice, 2 Oak) are in the join, but not in the original

20. Lossy Decompositions: What is Actually Lost? In the previous example, the tuples (2222, Alice, 3 Pine) and (3333, Alice, 2 Oak) were gained, not lost! –Why do we say that the decomposition was lossy? What was lost is information: –That 2222 lives at 2 Oak: In the decomposition, 2222 can live at either 2 Oak or 3 Pine –That 3333 lives at 3 Pine: In the decomposition, 3333 can live at either 2 Oak or 3 Pine

21. Example Schema (R, F) where R = {SSN, Name, Address, Hobby} F = {SSN  Name, Address} can be decomposed into R 1 = {SSN, Name, Address} F 1 = {SSN  Name, Address} and R 2 = {SSN, Hobby} F 2 = { } Since R 1  R 2 = SSN and SSN  R 1 the decomposition is lossless

22. Intuition Behind the Test for Losslessness Suppose R 1  R 2  R 2. Then a row of r 1 can combine with exactly one row of r 2 in the natural join (since in r 2 a particular set of values for the attributes in R 1  R 2 defines a unique row) R 1  R 2 R 1  R 2 …………. a a ………... ………… a b …………. ………… b c …………. ………… c r 1 r 2

23. If R 1  R 2  R 2 then card (r 1 Proof of Lossless Condition r  r 1 r 2 – this is true for any decomposition r 2 ) = card (r 1 ) But card (r)  card (r 1 ) (since r 1 is a projection of r) and therefore card (r)  card (r 1 r 2 ) Hence r = r 1 r2r2 r  r 1 r2r2 (since each row of r 1 joins with exactly one row of r 2 )

24. Dependency Preservation Consider a decomposition of R = (R, F) into R 1 = (R 1, F 1 ) and R 2 = (R 2, F 2 ) –An FD X  Y of F + is in F i iff X  Y  R i –An FD, f  F + may be in neither F 1, nor F 2, nor even (F 1  F 2 ) + Checking that f is true in r 1 or r 2 is (relatively) easy Checking f in r 1 r 2 is harder – requires a join Ideally: want to check FDs locally, in r 1 and r 2, and have a guarantee that every f  F holds in r 1 r 2 dependency preservingThe decomposition is dependency preserving iff the sets F and F 1  F 2 are equivalent: F + = (F 1  F 2 ) + –Then checking all FDs in F, as r 1 and r 2 are updated, can be done by checking F 1 in r 1 and F 2 in r 2

25. Dependency Preservation If f is an FD in F, but f is not in F 1  F 2, there are two possibilities: –f  (F 1  F 2 ) + If the constraints in F 1 and F 2 are maintained, f will be maintained automatically. –f  (F 1  F 2 ) + f can be checked only by first taking the join of r 1 and r 2. This is costly.

26. Example Schema (R, F) where R = {SSN, Name, Address, Hobby} F = {SSN  Name, Address} can be decomposed into R 1 = {SSN, Name, Address} F 1 = {SSN  Name, Address} and R 2 = {SSN, Hobby} F 2 = { } Since F = F 1  F 2 the decomposition is dependency preserving

27. Example Schema: (ABC; F), F = {A  B, B  C, C  B} Decomposition: –(AC, F 1 ), F 1 = {A  C} Note: A  C  F, but in F + –(BC, F 2 ), F 2 = {B  C, C  B} A  B  (F 1  F 2 ), but A  B  (F 1  F 2 ) +. –So F + = (F 1  F 2 ) + and thus the decompositions is still dependency preserving

28. BCNF Decomposition Algorithm Input: R = (R; F) Decomp := R while there is S = (S; F ’ )  Decomp and S not in BCNF do Find X  Y  F ’ that violates BCNF // X isn’t a superkey in S Replace S in Decomp with S 1 = (XY; F 1 ), S 2 = (S - (Y - X); F 2 ) // F 1 = all FDs of F ’ involving only attributes of XY // F 2 = all FDs of F ’ involving only attributes of S - (Y - X) end return Decomp

29. Simple Example HasAccountHasAccount : (ClientId, OfficeId, AcctNum) (ClientId, AcctNum) BCNF (only trivial FDs) Decompose using AcctNum  OfficeId : (OfficeId, AcctNum) BCNF: AcctNum is key FD: AcctNum  OfficeId ClientId,OfficeId  AcctNum AcctNum  OfficeId

30. A Larger Example Given: R = (R; F) where R = ABCDEGHK and F = {ABH  C, A  DE, BGH  K, K  ADH, BH  GE} step 1: Find a FD that violates BCNF Not ABH  C since (ABH) + includes all attributes (BH is a key) A  DE violates BCNF since A is not a superkey (A + =ADE) step 2: Split R into: R 1 = (ADE, F 1 ={A  DE }) R 2 = (ABCGHK; F 1 ={ABH  C, BGH  K, K  AH, BH  G}) Note 1: R 1 is in BCNF Note 2: Decomposition is lossless since A is a key of R 1. Note 3: FDs K  D and BH  E are not in F 1 or F 2. But both can be derived from F 1  F 2 (E.g., K  A and A  D implies K  D) Hence, decomposition is dependency preserving.

31. Example (con’t) Given: R 2 = (ABCGHK; {ABH  C, BGH  K, K  AH, BH  G}) step 1: Find a FD that violates BCNF. Not ABH  C or BGH  K, since BH is a key of R 2 K  AH violates BCNF since K is not a superkey (K + =AH) step 2: Split R 2 into: R 21 = (KAH, F 21 ={K  AH}) R 22 = (BCGK; F 22 ={}) Note 1: Both R 21 and R 22 are in BCNF. Note 2: The decomposition is lossless (since K is a key of R 21 ) Note 3: FDs ABH  C, BGH  K, BH  G are not in F 21 or F 22, and they can’t be derived from F 1  F 21  F 22. Hence the decomposition is not dependency-preserving

32. Properties of BCNF Decomposition Algorithm Let X  Y violate BCNF in R = (R,F) and R 1 = (R 1,F 1 ), R 2 = (R 2,F 2 ) is the resulting decomposition. Then: There are fewer violations of BCNF in R 1 and R 2 than there were in R –X  Y implies X is a key of R 1 –Hence X  Y  F 1 does not violate BCNF in R 1 and, since X  Y  F 2, does not violate BCNF in R 2 either –Suppose f is X ’  Y ’ and f  F doesn’t violate BCNF in R. If f  F 1 or F 2 it does not violate BCNF in R 1 or R 2 either since X ’ is a superkey of R and hence also of R 1 and R 2.

33. Properties of BCNF Decomposition Algorithm A BCNF decomposition is not necessarily dependency preserving But always lossless: since R 1  R 2 = X, X  Y, and R 1 = XY HasAccountBCNF+lossless+dependency preserving is sometimes unachievable (recall HasAccount)

35. Boyce–Codd Normal Form (BCNF) A relation r is in Boyce–Codd normal form if for every (non-trivial) functional dependency X -> Y defined on it, X contains a key K of r. That is, X is a superkey for r. Anomalies and redundancies, as discussed above, do not appear in databases with relations in Boyce–Codd normal form, because the independent pieces of information are separate, one per relation.

38. Decomposition into Boyce–Codd normal form Given a relation that does not satisfy Boyce–Codd normal form, we can often replace it with one or more normalized relations using a process called normalization. We can eliminate redundancies and anomalies for the example relation if we replace it with the three relations, obtained by projections on the sets of attributes corresponding to the three functional dependencies. The keys of the relations we obtain are the left hand side of a functional dependency: the satisfaction of the Boyce–Codd normal form is therefore guaranteed.