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Normalization

Normalization. Canonical Cover. Sets of functional dependencies may have redundant dependencies that can be inferred from the others For example: A  C is redundant in: { A  B , B  C } Parts of a functional dependency may be redundant

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Normalization

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  1. Normalization

  2. Canonical Cover • Sets of functional dependencies may have redundant dependencies that can be inferred from the others • For example: A  C is redundant in: {AB, BC} • Parts of a functional dependency may be redundant • E.g.: on RHS: {AB, BC, ACD} can be simplified to {A B, BC, AD} • E.g.: on LHS: {A B, BC, ACD} can be simplified to {A B, BC, AD} • Intuitively, a canonical cover of F is a “minimal” set of functional dependencies equivalent to F, having no redundant dependencies or redundant parts of dependencies

  3. Extraneous Attributes • Consider a set F of functional dependencies and the functional dependency   in F. • Attribute A is extraneous in  if A   and F logically implies (F – {})  {( – A) }. • Attribute A is extraneous in  if A  and the set of functional dependencies (F – {})  {(– A)} logically implies F. • Note: implication in the opposite direction is trivial in each of the cases above, since a “stronger” functional dependency always implies a weaker one • Example: Given F = {AC, ABC } • B is extraneous in AB C because {AC, AB C} logically implies AC (I.e. the result of dropping B from AB C). • Example: Given F = {AC, ABCD} • C is extraneous in ABCD since AB D can be inferred even after deleting C

  4. Testing if an Attribute is Extraneous • Consider a set F of functional dependencies and the functional dependency   in F. • To test if attribute A   is extraneousin • compute ({} – A)+ using the dependencies in F • check that ({} – A)+ contains ; if it does, A is extraneous • To test if attribute A  is extraneous in  • compute + using only the dependencies in F’ = (F – {})  {(– A)}, • check that + contains A; if it does, A is extraneous

  5. Canonical Cover • A canonical coverfor F is a set of dependencies Fc such that • F logically implies all dependencies in Fc, and • Fclogically implies all dependencies in F, and • No functional dependency in Fc contains an extraneous attribute, and • Each left side of functional dependency in Fcis unique. • To compute a canonical cover for F:repeatUse the union rule to replace any dependencies in F11 and 12 with 112 Find a functional dependency  with an extraneous attribute either in  or in  If an extraneous attribute is found, delete it from until F does not change • Note: Union rule may become applicable after some extraneous attributes have been deleted, so it has to be re-applied

  6. Computing Canonical Cover • R = (A, B, C)F = {A BC, B C, A B, ABC} • Combine A BC and A B into A BC • Set is now {A BC, B C, ABC} • A is extraneous in ABC • Check if the result of deleting A from ABC is implied by the other dependencies • Yes: in fact, BC is already present! • Set is now {A BC, B C} • C is extraneous in ABC • Check if A C is logically implied by A B and the other dependencies • Yes: using transitivity on A B and B  C. • Can use attribute closure of A in more complex cases • The canonical cover is: A B, B C

  7. Decomposition 1. Decomposing the schema R = ( bname, bcity, assets, cname, lno, amt) R = R1 U R2 R1 = (cname, lno, amt) R1 = (bname, bcity, assets, cname) 2. Decomposing the instance

  8. Goals of Decomposition 1. Lossless Joins Want to be able to reconstruct big (e.g. universal) relation by joining smaller ones (using natural joins) (i.e. R1 R2 = R) 2. Dependency preservation Want to minimize the cost of global integrity constraints based on FD’s ( i.e. avoid big joins in assertions) 3. Redundancy Avoidance Avoid unnecessary data duplication (the motivation for decomposition) Why important? LJ : information loss DP: efficiency (time) RA: efficiency (space), update anomalies

  9. Lossy Decomposition Spurious Tuples JOIN

  10. Dependency Goal #1: lossless joins A bad decomposition: = Problem: join adds meaningless tuples “lossy join”: by adding noise, have lost meaningful information as a result of the decomposition

  11. Dependency Goal #1: lossless joins Is the following decomposition lossless or lossy? Ans: Lossless: R = R1 R2, it has 4 tuples

  12. Ensuring Lossless Joins A decomposition of R : R = R1 U R2 Is lossless iff R1  R2  R1, or R1  R2  R2 (i.e., intersecting attributes must for a superkey for one of the resulting smaller relations)

  13. Lossless Decomposition Theorem A decomposition of R into R1 and R2 is lossless join wrt FDs F, if and only if at least one of the following dependencies is in F+: • R1  R2 R1 • R1  R2 R2 In other words, R1  R2 forms a superkey of either R1 or R2

  14. Lossy Decomposition

  15. Lossless Decomposition • Observe that S satisfies the FDs: • S#  Status & S#  City • It can not be a coincidence that S is equal to the join of its projections on {S#, Status} & {S#, City} • Heaths’ Theorem: Let R{A,B,C} be a relation, where A, B, & C are sets of attributes. If R satisfies AB & AC, then R is equal to the join of its projections on {A,B} & {A,C} • Observe that in the second decomposition of S the FD, S#  City is lost

  16. Lossless Decomposition • The decomposition of R into R1, R2, …Rn is lossless if for any instance r of R r = R1(r ) R2(r ) …… Rn(r ) • We can replace R by R1 & R2, knowing that the instance of R can be recovered from the instances of R1 & R2 • We can use FDs to show that decompositions are lossless

  17. Decomposition Goal #2: Dependency preservation Goal: efficient integrity checks of FD’s An example w/ no DP: R = ( bname, bcity, assets, cname, lno, amt) bname  bcity assets lno  amt bname Decomposition: R = R1 U R2 R1 = (bname, assets, cname, lno) R2 = (lno, bcity, amt) Lossless but not DP. Why? Ans: bname bcity assets crosses 2 tables

  18. Decomposition Goal #2: Dependency preservation To ensure best possible efficiency of FD checks ensure that only a SINGLE table is needed in order to check each FD i.e. ensure that: A1 A2 ... An  B1 B2 ... Bm Can be checked by examining Ri = ( ..., A1, A2, ..., An, ..., B1, ..., Bm, ...) To test if the decomposition R = R1 U R2 U ... U Rn is DP (1) see which FD’s of R are covered by R1, R2, ..., Rn (2) compare the closure of (1) with the closure of FD’s of R

  19. Decomposition Goal #2: Dependency preservation Example: Given F = { AB, AB D, C D} consider R = R1 U R2 s.t. R1 = (A, B, D) , R2 = (C, D) (1) F+ = { ABD, CD}+ (2) G = {ABD, CD, ...} + (3) F+ = G+ note: G+ cannot introduce new FDs not in F+ Decomposition is DP

  20. Dependency Preservation • Let Fibe the set of dependencies F + that include only attributes in Ri. • A decomposition is dependency preserving, if (F1 F2  …  Fn )+ = F + • If it is not, then checking updates for violation of functional dependencies may require computing joins, which is expensive.

  21. To check if a dependency    is preserved in a decomposition of R into R1, R2, …, Rn we apply the following test (with attribute closure done with respect to F) result = while (changes to result) dofor eachRiin the decompositiont = (result  Ri)+  Riresult = result  t If result contains all attributes in , then the functional dependency    is preserved. We apply the test on all dependencies in F to check if a decomposition is dependency preserving This procedure takes polynomial time, instead of the exponential time required to compute F+and(F1 F2  …  Fn)+ Testing for Dependency Preservation

  22. Example • R = (A, B, C)F = {A B, B C) • Can be decomposed in two different ways • R1 = (A, B), R2 = (B, C) • Lossless-join decomposition: R1  R2 = {B}and B BC • Dependency preserving • R1 = (A, B), R2 = (A, C) • Lossless-join decomposition: R1  R2 = {A}and A  AB • Not dependency preserving (cannot check B C without computing R1 R2)

  23. Decomposition Goal #3: Redudancy Avoidance Example: Redundancy for B=x , y and z (1) An FD that exists in the above relation is: B  C (2) A superkey in the above relation is A, (or any set containing A) When do you have redundancy? Ans: when there is some FD, XY covered by a relation and X is not a superkey

  24. Problems with Decompositions There are three potential problems to consider: • Some queries become more expensive • e.g., What is the price of prop# 1? • Given instances of the decomposed relations, we may not be able to reconstruct the corresponding instance of the original relation! • Fortunately, not in the PLOTS example • Checking some dependencies may require joining the instances of the decomposed relations. • Fortunately, not in the PLOTS example Tradeoff:Must consider these issues vs. redundancy

  25. Example • R = (A, B, C )F = {AB B  C}Key = {A} • R is not in BCNF (B  C but B is not superkey) • Decomposition R1 = (A, B), R2 = (B, C) • R1and R2 in BCNF • Lossless-join decomposition • Dependency preserving

  26. Testing for BCNF • To check if a non-trivial dependency causes a violation of BCNF 1. compute + (the attribute closure of ), and 2. verify that it includes all attributes of R, that is, it is a superkey of R. • Simplified test: To check if a relation schema R is in BCNF, it suffices to check only the dependencies in the given set F for violation of BCNF, rather than checking all dependencies in F+. • If none of the dependencies in F causes a violation of BCNF, then none of the dependencies in F+ will cause a violation of BCNF either. • However, simplified test using only F isincorrectwhen testing a relation in a decomposition of R • Consider R = (A, B, C, D, E), with F = { A  B, BC  D} • Decompose R into R1 =(A,B) and R2 =(A,C,D, E) • Neither of the dependencies in F contain only attributes from (A,C,D,E) so we might be mislead into thinking R2 satisfies BCNF. • In fact, dependency ACD in F+ shows R2 is not in BCNF.

  27. BCNF and Dependency Preservation It is not always possible to get a BCNF decomposition that is dependency preserving • R = (J, K, L )F = {JK L L K }Two candidate keys = JK and JL • R is not in BCNF • Any decomposition of R will fail to preserve JK L This implies that testing for JK L requires a join

  28. Third Normal Form: Motivation • There are some situations where • BCNF is not dependency preserving, and • efficient checking for FD violation on updates is important • Solution: define a weaker normal form, called Third Normal Form (3NF) • Allows some redundancy (with resultant problems; we will see examples later) • But functional dependencies can be checked on individual relations without computing a join. • There is always a lossless-join, dependency-preserving decomposition into 3NF.

  29. Redundancy in 3NF • There is some redundancy in this schema • Example of problems due to redundancy in 3NF • R = (J, K, L)F = {JK L, L K } J L K j1 j2 j3 null l1 l1 l1 l2 k1 k1 k1 k2 • repetition of information (e.g., the relationship l1, k1) • (i_ID, dept_name) • need to use null values (e.g., to represent the relationshipl2, k2 where there is no corresponding value for J). • (i_ID, dept_nameI) if there is no separate relation mapping instructors to departments

  30. Testing for 3NF • Optimization: Need to check only FDs in F, need not check all FDs in F+. • Use attribute closure to check for each dependency   , if  is a superkey. • If  is not a superkey, we have to verify if each attribute in  is contained in a candidate key of R • this test is rather more expensive, since it involve finding candidate keys • testing for 3NF has been shown to be NP-hard • Interestingly, decomposition into third normal form (described shortly) can be done in polynomial time

  31. 3NF Decomposition Algorithm Let Fcbe a canonical cover for F;i := 0;for each functional dependency in Fcdo if none of the schemas Rj, 1  j  i contains then begini := i + 1;Ri := endif none of the schemas Rj, 1  j  i contains a candidate key for Rthen begini := i + 1;Ri := any candidate key for R;end /* Optionally, remove redundant relations */ repeatif any schema Rj is contained in another schema Rkthen /* delete Rj */Rj = R;; i=i-1;return (R1, R2, ..., Ri)

  32. Testing Decomposition for BCNF • To check if a relation Ri in a decomposition of R is in BCNF, • Either test Ri for BCNF with respect to the restriction of F to Ri (that is, all FDs in F+ that contain only attributes from Ri) • or use the original set of dependencies F that hold on R, but with the following test: • for every set of attributes   Ri, check that + (the attribute closure of ) either includes no attribute of Ri- , or includes all attributes of Ri. • If the condition is violated by some   in F, the dependency (+ - )  Rican be shown to hold on Ri, and Ri violates BCNF. • We use above dependency to decompose Ri

  33. BCNF Decomposition Algorithm result := {R };done := false;compute F +;while (not done) do if (there is a schema Riin result that is not in BCNF)then beginlet   be a nontrivial functional dependency that holds on Risuch that  Riis not in F +, and   = ;result := (result – Ri )  (Ri – )  (,  );end else done := true; Note: each Riis in BCNF, and decomposition is lossless-join.

  34. Example of BCNF Decomposition • class (course_id, title, dept_name, credits, sec_id, semester, year, building, room_number, capacity, time_slot_id) • Functional dependencies: • course_id→ title, dept_name, credits • building, room_number→capacity • course_id, sec_id, semester, year→building, room_number, time_slot_id • A candidate key {course_id, sec_id, semester, year}. • BCNF Decomposition: • course_id→ title, dept_name, credits holds • but course_id is not a superkey. • We replace class by: • course(course_id, title, dept_name, credits) • class-1 (course_id, sec_id, semester, year, building, room_number, capacity, time_slot_id)

  35. BCNF Decomposition (Cont.) • course is in BCNF • How do we know this? • building, room_number→capacity holds on class-1 • but {building, room_number} is not a superkey for class-1. • We replace class-1 by: • classroom (building, room_number, capacity) • section (course_id, sec_id, semester, year, building, room_number, time_slot_id) • classroom and section are in BCNF.

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