1. 8.1Database System Concepts - 6 th Edition Chapter 8: Relational Database Design Features of Good Relational Design Lossless join decomposition Functional Dependency Theory Atomic Domains and First Normal Form Normal Forms: BCNF and 3NF Decomposition Algorithms Using Functional Dependencies Database-Design Issues

2. 8.2Database System Concepts - 6 th Edition Larger Schemas? Two schemas: instructor (ID, name, dept_name, salary) department (dept_name, building, budget) Suppose we combine instructor and department into inst_dept as follows: Result is possible repetition of information (the amount of budget) 刪掉就無 EE 資訊 重複

3. 8.3Database System Concepts - 6 th Edition Disadvantage of Large Schemas Update anomalies: Modifying the budget in one tuple but not all tuples leads to inconsistency. Cannot insert a new department until the first instructor is hired  ID is part of PK. Deleting the last instructor will lose the department information  刪掉 Kim 就無 EE 資訊

4. 8.4Database System Concepts - 6 th Edition What About Smaller Schemas? Suppose we had started with inst_dept. How would we know to split up (decompose) it into instructor and department? Intuition: put “strongly-related” attributes in the same relation Write a rule: “every department (identified by its department name) must have only one building and one budget value”. Denote as a functional dependency: dept_name  building, budget In inst_dept, because dept_name is not a candidate key, the building and budget of a department may have to be repeated. This indicates the need to decompose inst_dept Not all decompositions are good. Suppose we decompose employee(ID, name, street, city, salary) into employee1 (ID, name) employee2 (name, street, city, salary) The next slide shows how we lose information -- we cannot reconstruct the original employee relation -- and so, this is a lossy decomposition.

5. 8.5Database System Concepts - 6 th Edition A Lossy Decomposition

6. 8.6Database System Concepts - 6 th Edition Example of Lossless-Join Decomposition Lossless join decomposition Decomposition of R = (A, B, C) R 1 = (A, B)R 2 = (B, C) AB  1212 A  B 1212 r  B,C (r)  A (r)  B (r) AB  1212 C ABAB B 1212 C ABAB C ABAB  A,B (r)

7. 8.7Database System Concepts - 6 th Edition Goal of Normalization Decide whether a particular relation R is in “good” form. 1NF, 2NF, 3NF, BCNF, etc In the case that a relation R is not in “good” form, decompose it into a set of relations {R 1, R 2,..., R n } such that each relation is in good form the decomposition is a lossless-join decomposition The normalization theory is based on functional dependencies Constraints requiring that the value for a certain set of attributes determines uniquely the value for another set of attributes. A functional dependency is a generalization of the notion of a key.

8. 8.8Database System Concepts - 6 th Edition First Normal Form Domain is atomic if its elements are considered to be indivisible units Examples of non-atomic domains:  Set of names, composite attributes A relational schema R is in first normal form if the domains of all attributes of R are atomic Non-atomic values complicate storage and encourage redundant (repeated) storage of data Example: Set of accounts stored with each customer, and set of owners stored with each account We assume all relations are in first normal form. (It is not required in Object Based Databases (see page 1.18)) Non-1NF should be normalized as discussed in Chapter 7. IDNamePhone_number 1John{6601, 6602} IDName 1John IDPhone_number 16601 16602

9. 8.9Database System Concepts - 6 th Edition ID Phone_number 22222 456-7890 123-4567 ID Phone_number 22222{456-7890,123-4567}

10. 8.10Database System Concepts - 6 th Edition Functional Dependencies Let R be a relation schema   R and   R The functional dependency    holds on R if and only if for any legal relations r(R), whenever any two tuples t 1 and t 2 of r agree on the attributes , they also agree on the attributes . Example: Consider R(A,B, C, D ) with the following instance of r. On this instance, A  C is satisfied, but C  A is not satisfied.  

11. 8.11Database System Concepts - 6 th Edition Functional Dependencies (Cont.) Note: A specific instance of a relation schema may satisfy a functional dependency even if the functional dependency does not hold on all legal instances. For example, a specific instance of classroom may, by chance, satisfy room_number  capacity. We usually use functional dependencies to specify constraints on all legal relations We say that F holds on R if all legal relations on R satisfy the set of functional dependencies F.

12. 8.12Database System Concepts - 6 th Edition Practice inst_dept (ID, name, salary, dept_name, building, budget ). For each constraint, write the corresponding functional dependency Every instructor (ID) has only one name and one salary. Ans: Every department (dept_name) is located in only one building and has only one budget. Ans: Every instructor belongs to only one department. Ans: Explain why each of the following functional dependency doesn’t make sense. dept_name  ID Ans: building  dept_name Ans:

13. 8.13Database System Concepts - 6 th Edition Functional Dependencies (Cont.) Functional dependencies allow us to express constraints that cannot be expressed using superkeys. K is a superkey for relation schema R if and only if K  R Example: employee = (ID, name, street, city, salary), {ID} is a superkey iff ID-> ID, name, street, city, salary {ID, name} is a superkey iff ID, name -> ID, name, street, city, salary K is a candidate key for R if and only if K  R, and for no   K,   R In other words, K is a minimal set of attributes Example: {ID} is a candidate key, but {ID, name} is NOT.

14. 8.14Database System Concepts - 6 th Edition Functional Dependencies (Cont.) A functional dependency is trivial if it is satisfied by all instances of a relation Example: (see examples)examples  dept_name, building  dept_name In general,    is trivial if    Given a set F of functional dependencies, there are certain other functional dependencies that are logically implied by F. For example: If A  B and B  C, then we can infer that A  C (see examples)examples The set of all functional dependencies logically implied by F is the closure of F. We denote the closure of F by F +. F + is a superset of F.

15. 8.15Database System Concepts - 6 th Edition Closure of Attribute Sets Given a set of attributes  define the closure of  under F (denoted by  + ) as the set of attributes that are functionally determined by  under F Algorithm to compute  +, the closure of  under F result :=  ; while (changes to result) do for each    in F do begin if   result then result := result   end PS:   result, result  ,    =>   

16. 8.16Database System Concepts - 6 th Edition Example of Attribute Set Closure R = (A, B, C, G, H, I) F = {A  B A  C CG  H CG  I B  H} (AG) + 1.result = AG 2.result = ABCG(A  C and A  B) 3.result = ABCGH(CG  H and CG  AGBC) 4.result = ABCGHI(CG  I and CG  AGBCH) Is AG a candidate key? (see definition)definition 1. Is AG a super key? 1. Does AG  R? == Is (AG) +  R 2. Is any subset of AG a superkey? 1. Does A  R? == Is (A) +  R 2. Does G  R? == Is (G) +  R

17. 8.17Database System Concepts - 6 th Edition Lossless-join Decomposition For the case of R = (R 1, R 2 ), we require that for all possible relations r on schema R r =  R1 (r )  R2 (r ) A decomposition of R into R 1 and R 2 is lossless join if at least one of the following dependencies is in F + : (see examples)examples R 1  R 2  R 1 R 1  R 2  R 2 In other words, if R 1  R 2 forms a superkey of either R 1 or R 2, the decomposition of R is a lossless decomposition.

18. 8.18Database System Concepts - 6 th Edition Boyce-Codd Normal Form     is trivial (i.e.,    )  is a superkey for R A relation schema R is in BCNF with respect to a set F of functional dependencies if for all functional dependencies in F + of the form     where   R and   R, at least one of the following holds: 注意 : 在後續的討論中, 為了簡化起見, 我們直接使用 F 而不是 F +. Example schema : instr_dept (ID, name, salary, dept_name, building, budget ) It is not in BCNF because dept_name  building, budget holds on instr_dept, but dept_name is not a superkey

19. 8.19Database System Concepts - 6 th Edition 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. For a non-BCNF Schema R and a non-trivial dependency    causeing a violation of BCNF, we decompose R into: (   U  ) ( R -  ) 注意 : 課本用 R - (  -  ) ，是當  和  有交集時,  需要減掉  。 In our example, “dept_name  building, budget” causes the violation  = dept_name  = building, budget and inst_dept (ID, name, salary, dept_name, building, budget ) is replaced by (   U  ) = (dept_name, building, budget ) <- department relation ( R -  ) = (ID, name, salary, dept_name ) <- instructor relation

20. 8.20Database System Concepts - 6 th Edition BCNF Decomposition Algorithm Given F and R; result := {R }; done := false; while (not done) do if (there is a schema R i in result that is not in BCNF) then begin let     be a nontrivial functional dependency that holds on R i and cause R i not in BCNF; result := (result – R i )  (R i –  )  ( ,  ); end else done := true; Note: each R i is in BCNF, and decomposition is lossless-join.lossless-join

21. 8.21Database System Concepts - 6 th Edition 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 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)

22. 8.22Database System Concepts - 6 th Edition Example of 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. Final decomposition: course, classroom, section.

23. 8.23Database System Concepts - 6 th Edition Practice Decompose inst_dept(ID, name, dept_name, salary, building, budget, SID) based on the following set of functional dependency: {dept_name  building budget, ID  name salary, SID dept_name  ID}

24. 8.24Database System Concepts - 6 th Edition BCNF and Dependency Preservation Sometimes, to achieve BCNF, we will force attributes in one functional dependency represented in different schemas. (This is called the violation of dependency preservation). Example: Given schema: dept_advisor(s_ID, i_ID, dept_name), and functional dependencies: {i_ID->dept_name ； s_ID, dept_name -> i_ID} PS: {s_ID, dept_name } is a primary key. 一個學生在一個系只有一個指導老師 ( 學生可以多系 ), 老師只能一系. After the BCNF decomposition, we get (s_ID, i_ID) and (i_ID, dept_name) Note that s_ID, dept_name, i_ID are represented in two schemas. Because it is not always possible to achieve both BCNF and dependency preservation, we consider a weaker normal form, known as third normal form.

25. 8.25Database System Concepts - 6 th Edition Third Normal Form A relation schema R is in third normal form (3NF) if for all:    in F + at least one of the following holds:    is trivial (i.e.,    )  is a superkey for R Each attribute A in  is contained in a candidate key for R.  此處的意思是, A 被  所決定, 也被某個 CK 所決定.  PS: 課本是  - . (NOTE: each attribute may be in a different candidate key) If a relation is in BCNF, it is in 3NF (since in BCNF one of the first two conditions above must hold). Third Normal Form (3NF) Allows some redundancy (see the next two pages) There is always a lossless-join, dependency-preserving decomposition into 3NF.

26. 8.26Database System Concepts - 6 th Edition 3NF Example Relation dept_advisor: dept_advisor: (s_ID, i_ID, dept_name) F = {s_ID, dept_name  i_ID ； i_ID  dept_name} Two candidate keys: {s_ID, dept_name} and {i_ID, s_ID} R is in 3NF  s_ID, dept_name  i_ID – s_ID, dept_name is a superkey  i_ID  dept_name – dept_name is contained in a candidate key

27. 8.27Database System Concepts - 6 th Edition Redundancy in 3NF Example dept_advisor (s_ID, i_ID, dept_name) F = {s_ID, dept_name  i_ID ; i_ID  dept_name} Example of problems due to redundancy in 3NFproblems repetition of information (e.g., the relationship “l 1, CS”) need to use null values (e.g., to represent the relationship “l 2, EE” where there is no corresponding value for s_ID). s_ID s 1 s 2 s 3 s 1 null i_ID l1l1I1l3l2l1l1I1l3l2 dept_name CS MA EE s_ID s1s2s3s1s1s2s3s1 i_ID l1l1I1l3l1l1I1l3 l1l3l2l1l3l2 dept_name CS MA EE 3 N F3 N F B C N FB C N F

28. 8.28Database System Concepts - 6 th Edition Testing for 3NF 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  Note: We can get all the candidate keys based on the previous definition and attribute closure. definition The 3NF decomposition algorithm is more complex than the BCNF decomposition algorithm.

29. 8.29Database System Concepts - 6 th Edition 3NF Decomposition Algorithm Given F and R; Result := {R }; done := false; while (not done) do if (there is a schema R i in result that is not in 3NF) then begin let     be a nontrivial functional dependency that holds on R i such that   is not a superkey of R i, and some attribute A in  is not contained in any candidate key for R; result := (result – R i )  (R i –  )  ( ,  ); end else done := true; Note: This algorithm does not always achieve dependency preserving. There exists an algorithm which ensures:algorithm  each relation schema R i is in 3NF  decomposition is dependency preserving and lossless-join

30. 8.30Database System Concepts - 6 th Edition Example of 3NF 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}. 3NF decomposition course_id→ title, dept_name, credits holds  but course_id is not a superkey.  title (or dept_name, credits) is not contained in the candidate key 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) The following step is similar, and the result is the same as that of BCNF decomposition

31. 8.31Database System Concepts - 6 th Edition ※ Dependency-preserving lossless-join decomposition into 3NF Let F c be a canonical cover for F; i := 0; for each functional dependency    in F c do i := i + 1; R i :=   if none of the schemas R j, 1  j  i contains a candidate key for R then i := i + 1; R i := any candidate key for R; /* Optionally, remove redundant relations */ repeat if any schema R j is contained in another schema R k then /* delete R j */ R j : = R i ; I := i-1; until no more R j can be deleted return (R 1, R 2,..., R i ) Note: This algorithm is also called 3NF Synthesis Algorithm

32. 8.32Database System Concepts - 6 th Edition ※ 3NF Decomposition: An Example 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 canonical cover ! A candidate key {course_id, sec_id, semester, year}. 3NF Decomposition: course (course_id, title, dept_name, credits) classroom (building, room_number, capacity) section (course_id, sec_id, semester, year, building, room_number, time_slot_id) → contains a candidate key of the original schema, done!

33. 8.33Database System Concepts - 6 th Edition Comparison of BCNF and 3NF It is always possible to decompose a relation into a set of relations that are in 3NF such that: the decomposition is lossless the dependencies are preserved It is always possible to decompose a relation into a set of relations that are in BCNF such that: the decomposition is lossless it may not be possible to preserve dependencies. Goal for a relational database design is: BCNF. Lossless join. Dependency preservation. If we cannot achieve this, we accept one of Lack of dependency preservation Redundancy due to use of 3NF

34. 8.34Database System Concepts - 6 th Edition Overall Database Design Process We have assumed schema R is given R could have been generated when converting E-R diagram to a set of tables. R could have been a single relation containing all attributes that are of interest (called universal relation). R could have been the result of some ad hoc design of relations. Normalization breaks R into smaller relations.

35. 8.35Database System Concepts - 6 th Edition ER Model and Normalization When an E-R diagram is carefully designed, identifying all entities correctly, the tables generated from the E-R diagram should not need further normalization. However, in a real (imperfect) design, there can be functional dependencies from non-key attributes of an entity to other attributes of the entity Example: an employee entity with attributes ID, department_name and building, and a functional dependency department_name  building The schema is employee( ID, department_name, building), which is not in 3NF or BCNF. Good design would have made department an entity

36. 8.36Database System Concepts - 6 th Edition Denormalization for Performance May want to use non-normalized schema for performance For example, displaying prereqs along with course_id and title requires join of course with prereq Alternative 1: Use denormalized relation containing attributes of course as well as prereq with all above attributes faster lookup extra space and extra execution time for updates extra coding work for programmer and possibility of error in extra code Alternative 2: use a materialized view defined as course prereq Benefits and drawbacks same as above, except no extra coding work for programmer and avoids possible errors

37. 8.37Database System Concepts - 6 th Edition Other Design Issues Some aspects of database design are not caught by normalization Examples of bad database design, to be avoided: Instead of total_inst (dept_name, year, size ), use total_inst _2007(dept_name, size ), total_inst _2008(dept_name, size ), total_inst _2009(dept_name, size ), etc.,  Above are in BCNF, but make querying across years difficult and needs new table each year dept_year (dept_name, total_inst _2007, total_inst _2008, total_inst _2009)  Also in BCNF, but also makes querying across years difficult and requires new attribute each year.  Is an example of a crosstab, where values for one attribute become column names  Used in spreadsheets, and in data analysis tools

38. 8.38Database System Concepts - 6 th Edition ※ 2NF ( 完全函數相依 ) A functional dependency α → β is called a partial dependency if there is a proper subset γ of α such that γ → β. We say that β is partially dependent on α. A relation schema R is in second normal form (2NF) if each attribute A in R meets one of the following criteria: It appears in a candidate key. It is not partially dependent on a candidate key.

39. 8.39Database System Concepts - 6 th Edition ※ 3NF （不可遞移函數相依） Let a prime attribute be one that appears in at least one candidate key. Let α and β be sets of attributes such that α → β holds, but β → α does not hold. Let A be an attribute that is not in α, is not in β, and for which β → A holds. We say that A is transitively dependent on α. A relation schema R is in 3NF with respect to a set F of functional dependencies if there are no nonprime attributes A in R for which A is transitively dependent on a key for R.