1 / 39

Chapter 8: Relational Database Design

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. Larger Schemas?.

brooke
Télécharger la présentation

Chapter 8: Relational Database Design

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. 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. 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. 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. 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 decomposeemployee(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. A Lossy Decomposition

  6. Example of Lossless-Join Decomposition • Lossless join decomposition • Decomposition of R = (A, B, C) R1 = (A, B) R2 = (B, C) A B C A B B C   1 2 A B   1 2 1 2 A B r B,C(r) A,B(r) A B C A (r) B (r)   1 2 A B

  7. 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 {R1, R2, ..., Rn} 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. 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.

  9. Phone_number ID Phone_number ID 22222 22222 456-7890 123-4567 22222 {456-7890,123-4567}

  10. Functional Dependencies • Let R be a relation schema   R and   R • The functional dependency  holds onR if and only if for any legal relations r(R), whenever any two tuples t1and t2 of ragree 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. 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 Fholds onR if all legal relations on R satisfy the set of functional dependencies F.

  12. 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. 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. Functional Dependencies (Cont.) • A functional dependency is trivial if it is satisfied by all instances of a relation • Example: (see 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) • 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. Closure of Attribute Sets • Given a set of attributes a, define the closureof aunderF (denoted by a+) as the set of attributes that are functionally determined by a under F • Algorithm to compute a+, the closure of a under F result := a;while (changes to result) do for each in F do begin if  result then result := result end • PS:a result, result, => a

  16. Example of Attribute Set Closure • R = (A, B, C, G, H, I) • F = {A BA C CG HCG IB 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) • Is AG a super key? • Does AG R? == Is (AG)+  R • Is any subset of AG a superkey? • Does AR? == Is (A)+  R • Does GR? == Is (G)+  R

  17. Lossless-join Decomposition • For the case of R = (R1, R2), we require that for all possible relations r on schema R r = R1(r ) R2(r ) • A decomposition of R into R1 and R2 is lossless join if at least one of the following dependencies is in F + : (see examples) • R1 R2R1 • R1 R2R2 • In other words, if R1 R2 forms a superkey of either R1or R2, the decomposition of R is a lossless decomposition.

  18. 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. To check if a non-trivial dependency causes a violation of BCNF 1. compute + (the attribute closure of ), and 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 Testing for BCNF

  20. BCNF Decomposition Algorithm Given F and R; result := {R };done := false;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 Riand cause Rinot in BCNF;result := (result – Ri )  (Ri – )  (,  );end else done := true; Note: each Riis in BCNF, and decomposition is lossless-join.

  21. 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. 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. 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. 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. 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. 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. 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 3NF • repetition of information (e.g., the relationship “l1, CS”) • need to use null values (e.g., to represent the relationship “l2, EE” where there is no corresponding value for s_ID). s_ID i_ID dept_name s_ID i_ID i_ID dept_name s1 s2 s3 s1 null l1 l1 I1 l3 l2 CS CS CS MA EE s1 s2 s3 s1 l1 l1 I1 l3 l1 l3 l2 CS MA EE BCNF 3NF

  28. 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. • The 3NF decomposition algorithm is more complex than the BCNF decomposition algorithm.

  29. 3NF Decomposition Algorithm Given F and R; Result := {R };done := false;while (not done) do if (there is a schema Riin result that is not in 3NF)then beginlet   be a nontrivial functional dependency that holds on Risuch that is not a superkey ofRi, and some attribute A in  is not contained in any candidate key for R;result := (result – Ri )  (Ri – )  (,  );end else done := true; Note: • This algorithm does not always achieve dependency preserving. • There exists an algorithm which ensures: • each relation schema Riis in 3NF • decomposition is dependency preserving and lossless-join

  30. 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. ※ Dependency-preserving lossless-join decomposition into 3NF Let Fcbe a canonical cover for F;i := 0;for each functional dependency in Fcdoi := i + 1;Ri := if none of the schemas Rj, 1  j  i contains a candidate key for Rthen i := i + 1;Ri := any candidate key for R;/* Optionally, remove redundant relations */ repeat if any schema Rjis contained in another schema Rkthen /* delete Rj*/Rj := Ri; I := i-1;until no more Rj can be deleted return (R1, R2, ..., Ri) Note: This algorithm is also called 3NF Synthesis Algorithm

  32. ※ 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. 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. 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. 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. 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 ascourseprereq • Benefits and drawbacks same as above, except no extra coding work for programmer and avoids possible errors

  37. 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. ※ 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. ※ 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.

More Related