Chapter 8 Relational Database Design Database System Concepts
Chapter 8: Relational Database Design Database System Concepts, 6 th Ed. ©Silberschatz, Korth and Sudarshan See www. db-book. com for conditions on re-use
Chapter 8: Relational Database Design n Features of Good Relational Design n Atomic Domains and First Normal Form n Decomposition Using Functional Dependencies n Functional Dependency Theory n Algorithms for Functional Dependencies n Boyce Codd Normal Form (BCNF) n Third Normal Form (3 NF) Database System Concepts - 6 th Edition 8. 2 ©Silberschatz, Korth and Sudarshan
Combine Schemas? n Suppose we combine instructor and department into inst_dept l (No connection to relationship set inst_dept) n Result is possible repetition of information Database System Concepts - 6 th Edition 8. 3 ©Silberschatz, Korth and Sudarshan
A Combined Schema Without Repetition n Consider combining relations l sec_class(sec_id, building, room_number) and l section(course_id, sec_id, semester, year) into one relation l section(course_id, sec_id, semester, year, building, room_number) n No repetition in this case Database System Concepts - 6 th Edition 8. 4 ©Silberschatz, Korth and Sudarshan
What About Smaller Schemas? n Suppose we had started with inst_dept. How would we know to split up (decompose) it into instructor and department? n Write a rule “if there were a schema (dept_name, building, budget), then dept_name would be a candidate key” n Denote as a functional dependency: dept_name building, budget n In inst_dept, because dept_name is not a candidate key, the building and budget of a department may have to be repeated. l This indicates the need to decompose inst_dept n Not all decompositions are good. Suppose we decompose employee(ID, name, street, city, salary) into employee 1 (ID, name) employee 2 (name, street, city, salary) n The next slide shows how we lose information -- we cannot reconstruct the original employee relation -- and so, this is a lossy decomposition. Database System Concepts - 6 th Edition 8. 5 ©Silberschatz, Korth and Sudarshan
A Lossy Decomposition Database System Concepts - 6 th Edition 8. 6 ©Silberschatz, Korth and Sudarshan
Example of Lossless-Join Decomposition n Lossless join decomposition n Decomposition of R = (A, B, C) R 1 = (A, B) A B C A B B C 1 2 1 2 A B A, B(r) r A (r) R 2 = (B, C) B (r) Database System Concepts - 6 th Edition A B C 1 2 B, C(r) A B 8. 7 ©Silberschatz, Korth and Sudarshan
First Normal Form n Domain is atomic if its elements are considered to be indivisible units l Examples of non-atomic domains: 4 Set of names, composite attributes 4 Identification numbers like CS 101 that can be broken up into parts n A relational schema R is in first normal form if the domains of all attributes of R are atomic n Non-atomic values complicate storage and encourage redundant (repeated) storage of data l Example: Set of accounts stored with each customer, and set of owners stored with each account l We assume all relations are in first normal form (and revisit this in Chapter 22: Object Based Databases) Database System Concepts - 6 th Edition 8. 8 ©Silberschatz, Korth and Sudarshan
First Normal Form (Cont’d) n Atomicity is actually a property of how the elements of the domain are used. l Example: Strings would normally be considered indivisible l Suppose that students are given roll numbers which are strings of the form CS 0012 or EE 1127 l If the first two characters are extracted to find the department, the domain of roll numbers is not atomic. l Doing so is a bad idea: leads to encoding of information in application program rather than in the database. Database System Concepts - 6 th Edition 8. 9 ©Silberschatz, Korth and Sudarshan
Goal — Devise a Theory for the Following n Decide whether a particular relation R is in “good” form. n In the case that a relation R is not in “good” form, decompose it into a set of relations {R 1, R 2, . . . , Rn} such that l each relation is in good form l the decomposition is a lossless-join decomposition l Preferably, the decomposition should be dependency preserving n Our theory is based on: l functional dependencies Database System Concepts - 6 th Edition 8. 10 ©Silberschatz, Korth and Sudarshan
Functional Dependencies n Constraints on the set of legal relations. n Require that the value for a certain set of attributes determines uniquely the value for another set of attributes. n A functional dependency is a generalization of the notion of a key. Database System Concepts - 6 th Edition 8. 11 ©Silberschatz, Korth and Sudarshan
Functional Dependencies (Cont. ) n Let R be a relation schema R and R n 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 . That is, t 1[ ] = t 2 [ ] n Example: Consider r(A, B ) with the following instance of r. 1 1 3 4 5 7 n On this instance, A B does NOT hold, but B A does hold. Database System Concepts - 6 th Edition 8. 12 ©Silberschatz, Korth and Sudarshan
Functional Dependencies (Cont. ) n K is a superkey for relation schema R if and only if K R n K is a candidate key for R if and only if l K R, and l for no K, R n Functional dependencies allow us to express constraints that cannot be expressed using superkeys. Consider the schema: inst_dept (ID, name, salary, dept_name, building, budget ). We expect these functional dependencies to hold: dept_name building and ID building but would not expect the following to hold: dept_name salary Database System Concepts - 6 th Edition 8. 13 ©Silberschatz, Korth and Sudarshan
Functional Dependencies cont. n 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. l For example, a specific instance of instructor may, by chance, satisfy name ID. Database System Concepts - 6 th Edition 8. 14 ©Silberschatz, Korth and Sudarshan
Functional Dependencies (Cont. ) n A functional dependency is trivial if it is satisfied by all instances of a relation l l Example: 4 ID, name ID 4 name In general, is trivial if Database System Concepts - 6 th Edition 8. 15 ©Silberschatz, Korth and Sudarshan
Closure of a Set of Functional Dependencies n Given a set F of functional dependencies, there are certain other functional dependencies that are logically implied by F. l For example: If A B and B C, then we can infer that A C n The set of all functional dependencies logically implied by F is the closure of F. n We denote the closure of F by F+. n F+ is a superset of F. Database System Concepts - 6 th Edition 8. 16 ©Silberschatz, Korth and Sudarshan
Functional-Dependency Theory n We now consider the formal theory that tells us which functional dependencies are implied logically by a given set of functional dependencies. n We then develop algorithms to generate lossless decompositions into BCNF and 3 NF n We then develop algorithms to test if a decomposition is dependency- preserving Database System Concepts - 6 th Edition 8. 17 ©Silberschatz, Korth and Sudarshan
Closure of a Set of Functional Dependencies n Given a set F set of functional dependencies, there are certain other functional dependencies that are logically implied by F. l For e. g. : If A B and B C, then we can infer that A C n The set of all functional dependencies logically implied by F is the closure of F. n We denote the closure of F by F+. Database System Concepts - 6 th Edition 8. 18 ©Silberschatz, Korth and Sudarshan
Closure of a Set of Functional Dependencies n We can find F+, the closure of F, by repeatedly applying Armstrong’s Axioms: l if , then (reflexivity) l if , then (augmentation) l if , and , then (transitivity) n These rules are l sound (generate only functional dependencies that actually hold), and l complete (generate all functional dependencies that hold). Database System Concepts - 6 th Edition 8. 19 ©Silberschatz, Korth and Sudarshan
Closure of Functional Dependencies (Cont. ) n Additional rules: l If holds and holds, then holds (union) l If holds, then holds and holds (decomposition) l If holds and holds, then holds (pseudotransitivity) The above rules can be inferred from Armstrong’s axioms. Database System Concepts - 6 th Edition 8. 20 ©Silberschatz, Korth and Sudarshan
Example n R = (A, B, C, G, H, I) F={ A B A C CG H CG I B H} n some members of F+ l A H 4 by l transitivity from A B and B H AG I 4 by augmenting A C with G, to get AG CG and then transitivity with CG I Database System Concepts - 6 th Edition 8. 21 ©Silberschatz, Korth and Sudarshan
Procedure for Computing F+ (*skip) n To compute the closure of a set of functional dependencies F: F+=F repeat for each functional dependency f in F+ apply reflexivity and augmentation rules on f add the resulting functional dependencies to F + for each pair of functional dependencies f 1 and f 2 in F + if f 1 and f 2 can be combined using transitivity then add the resulting functional dependency to F + until F + does not change any further NOTE: We shall see an alternative procedure for this task later Database System Concepts - 6 th Edition 8. 22 ©Silberschatz, Korth and Sudarshan
Closure of Attribute Sets n 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 n 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 Database System Concepts - 6 th Edition 8. 23 ©Silberschatz, Korth and Sudarshan
Example of Attribute Set Closure n R = (A, B, C, G, H, I) n F = {A B A C CG H CG I B H} n (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) n Is AG a candidate key? 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 1. Database System Concepts - 6 th Edition 8. 24 ©Silberschatz, Korth and Sudarshan
Uses of Attribute Closure There are several uses of the attribute closure algorithm: n Testing for superkey: l To test if is a superkey, we compute +, and check if + contains all attributes of R. n Testing functional dependencies n l To check if a functional dependency holds (or, in other words, is in F+), just check if +. l That is, we compute + by using attribute closure, and then check if it contains . l Is a simple and cheap test, and very useful (*skip) n Computing closure of F n For each R, we find the closure +, and for each S +, we output a functional dependency S. Database System Concepts - 6 th Edition 8. 25 ©Silberschatz, Korth and Sudarshan
Canonical Cover n Sets of functional dependencies may have redundant dependencies that can be inferred from the others l For example: A C is redundant in: {A B, B C, A C} l Parts of a functional dependency may be redundant 4 E. g. : on RHS: {A B, B C, A CD} can be simplified to 4 E. g. : to {A B, B C, A D} on LHS: {A B, B C, AC D} can be simplified {A B, B C, A D} n 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 Database System Concepts - 6 th Edition 8. 26 ©Silberschatz, Korth and Sudarshan
Extraneous Attributes n Consider a set F of functional dependencies and the functional dependency in F. l Attribute A is extraneous in if A and F logically implies (F – { }) {( – A) }. l Attribute A is extraneous in if A and the set of functional dependencies (F – { }) { ( – A)} logically implies F. Database System Concepts - 6 th Edition 8. 27 ©Silberschatz, Korth and Sudarshan
Testing if an Attribute is Extraneous n Consider a set F of functional dependencies and the functional dependency in F. n To test if attribute A is extraneous in (Left hand side check) 1. 2. n compute ({ } – A)+ using the dependencies in F check that ({ } – A)+ contains ; if it does, A is extraneous in To test if attribute A is extraneous in (Right hand side check) 1. 2. compute + using only the dependencies in F’ = (F – { }) { ( – A)}, check that + contains A; if it does, A is extraneous in Database System Concepts - 6 th Edition 8. 28 ©Silberschatz, Korth and Sudarshan
Canonical Cover n A canonical cover for F is a set of dependencies Fc such that F logically implies all dependencies in Fc, and l Fc logically implies all dependencies in F, and l n l No functional dependency in Fc contains an extraneous attribute, and l Each left side of functional dependency in Fc is unique. (*skip) n To compute a canonical cover for F: repeat Use 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 /* Note: test for extraneous attributes done using Fc, not F*/ If an extraneous attribute is found, delete it from until F does not change n Note: Union rule may become applicable after some extraneous attributes have been deleted, so it has to be re-applied Database System Concepts - 6 th Edition 8. 29 ©Silberschatz, Korth and Sudarshan
Lossless-join Decomposition n For the case of R = (R 1, R 2), we require that for all possible relations r on schema R r = R 1 (r ) R 2 (r ) n A decomposition of R into R 1 and R 2 is lossless join if at least one of the following dependencies is in F+: l R 1 R 2 R 1 l R 1 R 2 n The above functional dependencies are a sufficient condition for lossless join decomposition; the dependencies are a necessary condition only if all constraints are functional dependencies Database System Concepts - 6 th Edition 8. 30 ©Silberschatz, Korth and Sudarshan
Dependency Preservation n Let Fi be the set of dependencies F 4 + that include only attributes in Ri. A decomposition is dependency preserving, if (F 1 F 2 … Fn )+ = F + 4 If it is not, then checking updates for violation of functional dependencies may require computing joins, which is expensive. Database System Concepts - 6 th Edition 8. 31 ©Silberschatz, Korth and Sudarshan
Example (*skip) n R = (A, B, C) F = {A B, B C) l Can be decomposed in two different ways n R 1 = (A, B), R 2 = (B, C) l Lossless-join decomposition: R 1 R 2 = {B} and B BC l Dependency preserving n R 1 = (A, B), R 2 = (A, C) l Lossless-join decomposition: R 1 R 2 = {A} and A AB l Not dependency preserving (cannot check B C without computing R 1 Database System Concepts - 6 th Edition 8. 32 R 2 ) ©Silberschatz, Korth and Sudarshan
Boyce-Codd Normal Form 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: n is trivial (i. e. , ) n is a superkey for R Example schema not in BCNF: instr_dept (ID, name, salary, dept_name, building, budget ) because dept_name building, budget holds on instr_dept, but dept_name is not a superkey Database System Concepts - 6 th Edition 8. 33 ©Silberschatz, Korth and Sudarshan
Decomposing a Schema into BCNF n Suppose we have a schema R and a non-trivial dependency causes a violation of BCNF. We decompose R into: R 1=( U ) • • R 2=( R - ( - ) ) n In our example, = dept_name l = building, budget and inst_dept is replaced by l R 1=( U ) = ( dept_name, building, budget ) l l R 2=( R - ( - ) ) = ( ID, name, salary, dept_name ) l (*skip) l If R 1 or R 2 are not BCNF, decompose again. Database System Concepts - 6 th Edition 8. 34 ©Silberschatz, Korth and Sudarshan
BCNF and Dependency Preservation n Constraints, including functional dependencies, are costly to check in practice unless they pertain to only one relation n If it is sufficient to test only those dependencies on each individual relation of a decomposition in order to ensure that all functional dependencies hold, then that decomposition is dependency preserving. n Because it is not always possible to achieve both BCNF and dependency preservation, we consider a weaker normal form, known as third normal form. Database System Concepts - 6 th Edition 8. 35 ©Silberschatz, Korth and Sudarshan
Example of BCNF Decomposition(*skip) n R = (A, B, C ) F = {A B B C} Key = {A} n R is not in BCNF (B C but B is not superkey) n Decomposition l R 1 = (B, C) l R 2 = (A, B) Database System Concepts - 6 th Edition 8. 36 ©Silberschatz, Korth and Sudarshan
Example (*skip) n R = (A, B, C ) F = {A B B C} Key = {A} n R is not in BCNF n Decomposition R 1 = (A, B), R 2 = (B, C) l R 1 and R 2 in BCNF l Lossless-join decomposition l Dependency preserving Database System Concepts - 6 th Edition 8. 37 ©Silberschatz, Korth and Sudarshan
BCNF and Dependency Preservation(*skip) It is not always possible to get a BCNF decomposition that is dependency preserving n R = (J, K, L ) F = {JK L L K} Two candidate keys = JK and JL n R is not in BCNF n Any decomposition of R will fail to preserve JK L This implies that testing for JK L requires a join Database System Concepts - 6 th Edition 8. 38 ©Silberschatz, Korth and Sudarshan
Third Normal Form: Motivation n There are some situations where l BCNF is not dependency preserving, and l efficient checking for FD violation on updates is important n Solution: define a weaker normal form, called Third Normal Form (3 NF) l Allows some redundancy (with resultant problems; we will see examples later) l But functional dependencies can be checked on individual relations without computing a join. l There is always a lossless-join, dependency-preserving decomposition into 3 NF. Database System Concepts - 6 th Edition 8. 39 ©Silberschatz, Korth and Sudarshan
Third Normal Form n A relation schema R is in third normal form (3 NF) if for all: in F+ at least one of the following holds: l is trivial (i. e. , ) l is a superkey for R l Each attribute A in – is contained in a candidate key for R. (NOTE: each attribute may be in a different candidate key) n If a relation is in BCNF it is in 3 NF (since in BCNF one of the first two conditions above must hold). n Third condition is a minimal relaxation of BCNF to ensure dependency preservation (will see why later). Database System Concepts - 6 th Edition 8. 40 ©Silberschatz, Korth and Sudarshan
3 NF Decomposition Algorithm Let Fc be a canonical cover for F; i : = 0; for each functional dependency in Fc do if none of the schemas Rj, 1 j i contains then begin i : = i + 1; Ri : = end if none of the schemas Rj, 1 j i contains a candidate key for R then begin i : = i + 1; Ri : = any candidate key for R; end /* Optionally, remove redundant relations */ repeat if any schema Rj is contained in another schema Rk then /* delete Rj */ Rj = R; ; i=i-1; return (R 1, R 2, . . . , Ri) Database System Concepts - 6 th Edition 8. 41 ©Silberschatz, Korth and Sudarshan
3 NF Decomposition Algorithm (Cont. ) n Above algorithm ensures: l each relation schema Ri is in 3 NF l decomposition is dependency preserving and lossless-join Database System Concepts - 6 th Edition 8. 42 ©Silberschatz, Korth and Sudarshan
Comparison of BCNF and 3 NF n It is always possible to decompose a relation into a set of relations that are in 3 NF such that: l the decomposition is lossless l the dependencies are preserved n It is always possible to decompose a relation into a set of relations that are in BCNF such that: l the decomposition is lossless l it may not be possible to preserve dependencies. Database System Concepts - 6 th Edition 8. 43 ©Silberschatz, Korth and Sudarshan
Design Goals n Goal for a relational database design is: l BCNF. l Lossless join. l Dependency preservation. n If we cannot achieve this, we accept one of l Lack of dependency preservation l Redundancy due to use of 3 NF n Interestingly, SQL does not provide a direct way of specifying functional dependencies other than superkeys. Can specify FDs using assertions, but they are expensive to test, (and currently not supported by any of the widely used databases!) n Even if we had a dependency preserving decomposition, using SQL we would not be able to efficiently test a functional dependency whose left hand side is not a key. Database System Concepts - 6 th Edition 8. 44 ©Silberschatz, Korth and Sudarshan
End of Chapter Database System Concepts, 6 th Ed. ©Silberschatz, Korth and Sudarshan See www. db-book. com for conditions on re-use
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