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  1. Schema Refinement and Normal Forms Chapter 19

  2. The Evils of Redundancy • Redundancyis at the root of several problems associated with relational schemas: • redundant storage, insert/delete/update anomalies • Integrity constraints, in particularfunctional dependencies, can be used to identify schemas with such problems and to suggest refinements. • Main refinement technique: decomposition (replacing ABCD with, say, AB and BCD, or ACD and ABD). • Decomposition should be used judiciously: • Is there a reason to decompose a relation? • What problems (if any) does the decomposition cause?

  3. Example in page 607: Hourly_Emps(ssn, name, lot, rating, hourly_wages, hours_worked) suppose hourly_wages attribute is determined by the rating attribute. For a given rating value, there is only one permissible hourly_wages value. This integrity constraint (IC) is an example of functional dependency. See problems in next page

  4. Figure 19.1 • Problems due to R W : • Redundant storage • Update anomaly: Can we change W in just the 1st tuple of SNLRWH? • Insertion anomaly: What if we want to insert an employee and don’t know the hourly wage for his rating? • Deletion anomaly: If we delete all employees with rating 5, we lose the information about the wage for rating 5!

  5. Can null values help? • Null values cannot help eliminate redundant storage or update anomalies • The insertion anomaly in the previous page may be solved by using null values, null values cannot address all insertion anomalies: we cannot record the hourly wages for a rating unless there is an employee with that rating (primary key field cannot store null) • Null values can not solve deletion anomalies either

  6. Use of Decompositions Decomposing Hourly_Emps into two relations: page 609 Are all problems solved?

  7. Problems related to decomposition • Queries over the original relation may require us to join the decomposed relations. If such queries are common, the performance penalty of decomposing the relation may not be acceptable. • In this case we may choose to live with some of the problems of redundancy and not decompose the relation. • It is important to be aware of the potential problems caused by such residual redundancy in the design and to take steps to avoid them (e.g., by adding some checks to application code). • A good DB designer should have a firm grasp of normal forms and what problems they do (or do not) alleviate, the technique of decomposition, and potential problems with decompositions.

  8. Functional Dependencies (FDs) • A functional dependencyX Y holds over relation R if, for every allowable instance r of R: • t1 r, t2 r, (t1) = (t2) implies (t1) = (t2) • i.e., given two tuples in r, if the X values agree, then the Y values must also agree. (X and Y are sets of attributes.) • An FD is a statement about all allowable relations. • Must be identified based on semantics of application. • Given some allowable instance r1 of R, we can check if it violates some FD f, but we cannot tell if f holds over R! • K is a candidate key for R means that K R • However, K R does not require K to be minimal!

  9. Examples motivating schema refinement • ER design CAN generate some schemas with redundancy problems, because it is a complex, subjective process, and certain constraints are NOT expressible in terms of ER diagrams. • Constraints on an entity set • Constraints on a relationship set • Identifying attributes of entities • Identifying entity sets

  10. Constraints on Entity Set • Consider relation obtained from Hourly_Emps: • Hourly_Emps (ssn, name, lot, rating, hrly_wages, hrs_worked) • Notation: We will denote this relation schema by listing the attributes: SNLRWH • This is really the set of attributes {S,N,L,R,W,H}. • Sometimes, we will refer to all attributes of a relation by using the relation name. (e.g., Hourly_Emps for SNLRWH) • Some FDs on Hourly_Emps: • ssn is the key: S SNLRWH • rating determines hrly_wages: R W

  11. Example (Contd.) • Problems due to R W : • Redundant storage • Update anomaly: Can we change W in just the 1st tuple of SNLRWH? • Insertion anomaly: What if we want to insert an employee and don’t know the hourly wage for his rating? • Deletion anomaly: If we delete all employees with rating 5, we lose the information about the wage for rating 5! This FD (RW) can NOT be expressed in terms of the ER model. Only FDs that determine all attributes of a relation (i.e., key constraints) can be expressed in the ER model. Therefore we can NOT detect it when we considered Hourly_Emps as an entity set during ER modeling.

  12. Example (Contd.) Wages Hourly_Emps2

  13. since name dname ssn lot did budget Works_In Employees Departments budget since name dname ssn did lot Works_In Employees Departments Refining an ER Diagram Before: • 1st diagram translated: Workers(S,N,L,D,S) Departments(D,M,B) • Lots associated with workers. • Suppose all workers in a dept are assigned the same lot: D L • Redundancy; fixed by: Workers2(S,N,D,S) Dept_Lots(D,L) • Can fine-tune this: Workers2(S,N,D,S) Departments(D,M,B,L) After:

  14. Reasoning About FDs • Given some FDs, we can usually infer additional FDs: • ssn did, did lot implies ssn lot • An FD f is implied bya set of FDs F if f holds whenever all FDs in F hold. • = closure of F is the set of all FDs that are implied by F. • Armstrong’s Axioms (X, Y, Z are sets of attributes): • Reflexivity: If X  Y, then X Y • Augmentation: If X Y, then XZ YZ for any Z • Transitivity: If X Y and Y Z, then X Z • These are sound and completeinference rules for FDs!

  15. Reasoning About FDs (Contd.) • Couple of additional rules (that follow from AA): • Union: If X  Y and X  Z, then X  YZ (XX XY and XYYZ; XXYZ) • Decomposition: If X  YZ, then X  Y and X  Z (YZY and YZ Z) • Example: Contracts(cid,sid,jid,did,pid,qty,value), and: • C is the key: C  CSJDPQV • A project purchases a given part using a single contract: JP  C • A dept purchases at most one part from a supplier: SD  P • JP  C, C  CSJDPQV imply JP  CSJDPQV • SD  P implies SDJ  JP • SDJ  JP, JP  CSJDPQV imply SDJ  CSJDPQV

  16. Reasoning About FDs (Contd.) • Computing the closure of a set of FDs can be expensive. (Size of closure is exponential in # attrs!) • Typically, we just want to check if a given FD X Y is in the closure of a set of FDs F. An efficient check: • Compute attribute closureof X (denoted ) wrt F: • Set of all attributes A such that X A is in • There is a linear time algorithm to compute this. • Check if Y is in • Does F = {A B, B C, C D E } imply A E? • i.e, is A E in the closure ? Equivalently, is E in ?

  17. Apply algorithm in Figure 19.4 to compute attribute closure for A+ closure = A; repeat until there is no change: { if there is an FD U  V in F s.t. U  closure, then set closure = closure  V } U V closure = A FD: A B closure = AB FD: B C closure = ABC A+ = {A,B,C} and E  A+ , hence A E  F+

  18. Another Example ― 習題 19.2 第一小題 Given a relation R with 5 attributes ABCDE and the following FDs: A  B, BC  E, and ED  A. We want to find all keys for R. Any key K for R must satisfy K  ABCDE. There are 5 + 10 + 10 + 5 + 1 = 31 different possibilities. Let’s first try K = A; we need to show A  ABCDE closure = A; U V closure = A FD: A B closure = AB A+ = {A,B} and ABCDE  A+ , hence A ABCDE  F+. In other words, A is not a key for R.

  19. Another Example ― 習題 19.2 第一小題 Now you try K = BCD; you need to show BCD  ABCDE

  20. Another Example ― 習題 19.2 第一小題 Show BCD  ABCDE closure = BCD; U V closure = BCD FD: BC  E closure = BCDE FD: ED  A Closure=ABCDE BCD+ = {A,B,C,D,E} and ABCDE  BCD+ , hence BCD ABCDE  F+ . In other words, BCD is a key for R.

  21. Another Example ― 習題 19.2 第一小題 You try K = ACD; you need to show ACD  ABCDE You try K = CDE; you need to show CDE  ABCDE

  22. Another Example ― 習題 19.2 第一小題 Hence, there are three keys for R: CDE, ACD, BCD

  23. Normal Forms • Returning to the issue of schema refinement, the first question to ask is whether any refinement is needed! • If a relation is in a certain normal form(BCNF, 3NF etc.), it is known that certain kinds of problems are avoided/minimized. This can be used to help us decide whether decomposing the relation will help. • Role of FDs in detecting redundancy: • Consider a relation R with 3 attributes, ABC. • No FDs hold: There is no redundancy here. • Given A B: Several tuples could have the same A value, and if so, they’ll all have the same B value!

  24. Normal Forms • Normal forms based on FDs: • First normal form (1NF) • second normal form (2NF) • third normal form (3NF) • Boyce-Codd normal form (BCNF) • 1NF: every field contains only atomic values, that is, no lists or sets.

  25. Boyce-Codd Normal Form (BCNF) • Reln R with FDs F is in BCNF if, for every X A (single attribute) in F • A X (called a trivial FD), or • X contains a key for R (i.e., X is a superkey). • In other words, R is in BCNF if the only non-trivial FDs that hold over R are key constraints (see Figure 19.5). • Given a Reln R with a set F of FDs, to show whether R is in BCNF, it is sufficient to check whether the left side of each FD in F is a superkey

  26. Another Example ― 習題 19.2 第三小題 Given a relation R with 5 attributes ABCDE and the following FDs: A  B, BC  E, and ED  A. Is R in BCNF? We have shown that there are 3 keys for R: CDE, ACD, BCD. R is not in BCNF because none of A, BC, and ED contain a key.

  27. Boyce-Codd Normal Form (BCNF) • BCNF ensures that no redundancy can be detected using FD information alone (because key is unique). • It is thus the most desirable normal form (from the point of view of redundancy) if we take into account only FD information.

  28. Third Normal Form (3NF) • Reln R with FDs F is in 3NF if, for every X A in F • A X (called a trivial FD), or • X contains a key for R (i.e., X is a superkey), or • A is part of some key for R. • Minimality of a key is crucial in third condition above! • If R is in BCNF, obviously in 3NF. • If R is in 3NF, some redundancy is possible. It is a compromise, used when BCNF not achievable (e.g., no ``good’’ decomp, or performance considerations). • Lossless-join, dependency-preserving decomposition of R into a collection of 3NF relations always possible.

  29. Another Example ― 習題 19.2 第二小題 Given a relation R with 5 attributes ABCDE and the following FDs: A  B, BC  E, and ED  A. Is R in 3NF? We have shown that there are 3 keys for R: CDE, ACD, BCD. R is in 3NF because the attribute on the right hand side of each FD, i.e., B, E, and A, are all parts of keys.

  30. Suppose a dependency X  A causes a violation of 3NF (2 cases to consider) • X is a proper subset of some key K (see Fig. 19.7): this is called partial dependency. • In this case, we store (X,A) pairs redundantly. • As an example, consider the Reserves relation with attribute SBDC where S is sid, B is bid, D is day, and C denotes the credit card to which the reservation is charged. The only key is SBD and we have S  C (sid determines credit card). In this case, we store the credit card number for a sailor as many times as there are reservations for that sailor.

  31. Suppose a dependency X  A causes a violation of 3NF (the second case) • X is not a proper subset of any key (see Fig. 19.8): such a dependency is called a transitive dependency. • As an example, consider the Hourly_Emps relation with attributes SNLRWH. The only key is S, but there is a FD R  W, which gives rise to the chain S  R  W. • The consequence is that we cannot record the fact that employee S has rating R without knowing the hourly wage for that rating. • This condition leads to insertion, deletion, and update anomalies.

  32. Redundancy is possible with 3NF • Let us revisit the Reserves relation with attributes SBDC and the FD S  C, which states a sailor uses a unique credit card to pay for reservations. • S is not a key, and C is not part of a key (the only key is SBD). Hence, this relation is not in 3NF; (S,C) pairs stored redundantly.

  33. Redundancy is possible with 3NF • If we also know that credit cards uniquely identify the owner, we have the FD C S, which means that CBD is also a key for Reserves. • Therefore, the dependency S  C does not violate 3NF, and Reserves is in 3NF. • Nonetheless, in all tuples containing the same S value, the same (S,C) pair is redundantly recorded.

  34. Second Normal Form (2NF) • 2NF are relations in which partial dependencies are not allowed. • Thus, if a relation is in 3NF (which precludes both partial and transitive dependencies), it is also in 2NF.

  35. 習題 19.1 第3小題 Given a set of FDs for the relation schema R(A,B,C,D) with Primary key AB under which R is in 1NF but not in 2NF.

  36. 習題 19.1 第3小題 Given a set of FDs for the relation schema R(A,B,C,D) with Primary key AB under which R is in 1NF but not in 2NF. Answer: A  C or A  D or B  C or B  D.

  37. 習題 19.1 第4小題 Given a set of FDs for the relation schema R(A,B,C,D) with Primary key AB under which R is in 2NF but not in 3NF.

  38. 習題 19.1 第4小題 Given a set of FDs for the relation schema R(A,B,C,D) with Primary key AB under which R is in 2NF but not in 3NF. Answer: D  C or C  D or AC  D or AD  C or BC  D or BD  C

  39. 3NF but not BCNF Given a set of FDs for the relation schema R(A,B,C,D) with Primary key AB under which R is in 3NF but not in BCNF.

  40. 3NF but not BCNF Given a set of FDs for the relation schema R(A,B,C,D) with primary key AB under which R is in 3NF but not in BCNF. Answer: C  A or C  B or D  A or D  B or CD  A or CD  B or CDA  B or CDB  A

  41. BCNF Given a set of FDs for the relation schema R(A,B,C,D) with Primary key AB under which R is in BCNF.

  42. BCNF Given a set of FDs for the relation schema R(A,B,C,D) with Primary key AB under which R is in BCNF. Answer: AB  ABCD

  43. 習題 19.1 第5小題 Consider the relation schema R(A,B,C), which has the FD B  C. If A is a candidate key for R, is it possible for R to be in BCNF? If so, under what conditions? If not, explain why not.

  44. 習題 19.1 第5小題 Consider the relation schema R(A,B,C), which has the FD B  C. If A is a candidate key for R, is it possible for R to be in BCNF? If so, under what conditions? If not, explain why not. Answer: If B is also a key, then R is in BCNF.

  45. Decomposition of a Relation Scheme • Suppose that relation R contains attributes A1 ... An. A decompositionof R consists of replacing R by two or more relations such that: • Each new relation scheme contains a subset of the attributes of R (and no attributes that do not appear in R), and • Every attribute of R appears as an attribute of one of the new relations. • Intuitively, decomposing R means we will store instances of the relation schemes produced by the decomposition, instead of instances of R.

  46. Example Decomposition • Decompositions should be used only when needed. • SNLRWH has FDs S SNLRWH and R W • Second FD causes violation of 3NF; W values repeatedly associated with R values. Easiest way to fix this is to create a relation RW to store these associations, and to remove W from the main schema: • i.e., we decompose SNLRWH into SNLRH and RW

  47. Example (Contd.) Wages Hourly_Emps2

  48. Example Decomposition • The information to be stored consists of SNLRWH tuples. If we just store the projections of these tuples onto SNLRH and RW, are there any potential problems that we should be aware of?

  49. Problems with Decompositions • There are three potential problems to consider: • Some queries become more expensive. • e.g., How much did employee Guldu earn? (salary = W*H) • Given instances of the decomposed relations, we may not be able to reconstruct the corresponding instance of the original relation! • Fortunately, not in the SNLRWH example. • Checking some dependencies may require joining the instances of the decomposed relations. • Fortunately, not in the SNLRWH example. • Tradeoff: Must consider these issues vs. redundancy.

  50. Lossless Join Decompositions • Decomposition of R into X and Y is lossless-join w.r.t. a set of FDs F if, for every instance r that satisfies F: • (r) (r) = r • It is always true that r (r) (r) • In general, the other direction does not hold! If it does, the decomposition is lossless-join. • Definition extended to decomposition into 3 or more relations in a straightforward way. • It is essential that all decompositions used to deal with redundancy be lossless! (Avoids Problem (2).)