7: Chapter Relational Database Design
Database System Concepts
Silberschatz, Korth and Sudarshan See www.db-book.com for conditions on re-use
Chapter 7: Relational Database Design
First Normal Form Pitfalls in Relational Database Design Functional Dependencies Decomposition Boyce-Codd Normal Form Third Normal Form Overall Database Design Process
WARNING "I am studing IS and one of the courses is Database Concepts. Normailization in the text was like reading Greek."
SOLUTION: Capture the intuition + more exercises
Database System Concepts, 5th Ed. 7.2 Silberschatz, Korth and Sudarshan
First Normal Form
Domain is atomic if its elements are considered to be indivisible units Examples of non-atomic domains: multivalued attributes: e.g., Set of names composite attributes identification numbers like CS101 that can be broken up into parts
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 E.g. Set of accounts stored with each customer, and set of owners stored with each account We assume all relations are in first normal form
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First Normal Form (Contd.)
Atomicity is actually a property of how the elements of the domain are used. E.g. Strings would normally be considered indivisible Suppose that students are given roll numbers which are strings of the form CS0012 or EE1127 If the first two characters are extracted to find the department, the domain of roll numbers is not atomic. Doing so is a bad idea: leads to encoding of information in application program rather than in the database.
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Pitfalls in Relational Database Design
Relational database design requires that we find a "good" collection of relation schemas. A bad design may lead to Repetition of Information. Inability to represent certain information. Avoid redundant data Ensure that relationships among attributes are represented Facilitate the checking of updates for violation of database integrity constraints.
Design Goals:
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Example
Consider the relation schema: Lending-schema = (branch-name, branch-city, assets, customer-name, loan-number, amount)
Redundancy: Data for branch-name, branch-city, assets are repeated for each loan that a branch makes Wastes space Complicates updating, introducing possibility of inconsistency of assets value Cannot store information about a branch if no loans exist Can use null values, but they are difficult to handle.
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Null values
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Decomposition
Decompose the relation schema Lending-schema into: Branch-schema = (branch-name, branch-city, assets) Loan-info-schema = (customer-name, loan-number, branch-name, amount) All attributes of an original schema (R) must appear in the decomposition (R1, R2): R = R1 R2 Lossless-join decomposition. For all possible relations r on schema R r = R1 (r) R2 (r)
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Example of Non Lossless-Join Decomposition
Decomposition of R = (A, B) R1 = (A) R2 = (B) A B 1 2 B(r)
Another example
A
B 1 2 1 r
A(r)
A 1 4 7
B 2 5 2
C 3 6 8
A 1 4 7
B 2 5 2
A (r)
B (r)
A B 1 2 1 2
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A 1 4 7 1 7
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B 2 5 2 2 2
C 3 6 8 8 3
B 2 5 2
C 3 6 8
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Goal -- Devise a Theory for the Following
Decide whether a particular relation R is in "good" form. 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 functional dependencies multivalued dependencies (out of the scope of this course)
Our theory is based on:
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Functional Dependencies
Constraints on the set of legal relations. Require 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.
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Functional Dependencies (Cont.)
Let R be a relation schema R and R The functional dependency
(functionally) determines ; is (functionally) determined by
holds on R if and only if for any legal relations r(R), whenever any two tuples t1 and t2 of r agree on the attributes , they also agree on the attributes . That is, t1[] = t2 [] t1[ ] = t2 [ ] Example: Consider r(A,B) with the following instance of r. A 1 1 3
B 4 5 7
On this instance, A B does NOT hold, but B A does hold.
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Functional Dependencies (Cont.)
K is a superkey for relation schema R if and only if K R K is a candidate key for R if and only if K R, and for no K, R
Functional dependencies allow us to express constraints that cannot be expressed using superkeys. Consider the schema: Loan-info-schema = (customer-name, loan-number, branch-name, amount). We expect this set of functional dependencies to hold: loan-number amount loan-number branch-name loan-number customer-name loan-number will be the key for R(loan-number, amount)
but would not expect the following to hold (in general):
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Use of Functional Dependencies
We use functional dependencies to: test relations to see if they are legal under a given set of functional dependencies. If a relation r is legal under a set F of functional dependencies, we say that r satisfies F. We say that F holds on R if all legal relations on R satisfy the set of functional dependencies F.
specify constraints on the set of legal relations
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 Loan-schema may, by chance, satisfy loan-number customer-name. think of FDs as constraints, such as CHECK(qty > min_qty)
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Functional Dependencies (Cont.)
A functional dependency is trivial if it is satisfied by all instances of a relation E.g. customer-name, loan-number customer-name customer-name customer-name
In general, is trivial if
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Closure of a Set of Functional Dependencies
Given a set F set 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 The set of all functional dependencies logically implied by F is the closure of F. We denote the closure of F by F+. We can find all of F+ by applying Armstrong's Axioms: if , then if , then (reflexivity) (augmentation)
if , and , then (transitivity) sound (generate only functional dependencies that actually hold) and complete (generate all functional dependencies that hold). F+ is usually what we want to enforce a consistent state of the DB
These rules are
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Example
R = (A, B, C, G, H, I) F={ AB AC CG H CG I B H} some members of F+ AH by transitivity from A B and B H by augmenting A C with G, to get AG CG and then transitivity with CG I by augmenting CG I to infer CG CGI, and augmenting of CG H to infer CGI HI, and then transitivity
Database System Concepts, 5th Ed. 7.16 Silberschatz, Korth and Sudarshan
AG I
CG HI
Procedure for Computing F+
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 f1and f2 in F+ if f1 and f2 can be combined using transitivity then add the resulting functional dependency to F+ until F+ does not change any further NOTE: We will see an alternative procedure for this task later
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Closure of Functional Dependencies (Cont.)
We can further simplify manual computation of F+ by using the following additional rules. If holds and holds, then holds (union) If holds, then holds and holds (decomposition) If holds and holds, then holds (pseudotransitivity)
The above rules can be inferred from Armstrong's axioms.
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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 e.g., F = {A B, A C}, then A+ = ABC Semantics: A functionally determines any subset of A+ (i.e., ABC)
and no more other attribute combinations (i.e., closure)
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
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Example of Attribute Set Closure
R = (A, B, C, G, H, I) F = {A B AC CG H CG I B H} (AG)+ 1. result = AG 2. result = AGBC 3. result = AGBCH
(A C and A B) (CG H and CG AGBC) (CG I and CG AGBCH)
4. result = AGBCHI = ABCGHI Is AG a candidate key?
1.
Is AG a super key?
1.
Does AG R? Is (AG)+ R Does A R? Is (A)+ R Does G R? Is (G)+ R
2.
Is any subset of AG a superkey?
1. 2.
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Uses of Attribute Closure
There are several uses of the attribute closure algorithm: Testing for superkey: To test if is a superkey, we compute +, and check if + contains all attributes of R. To check if a functional dependency holds (or, in other words, is in F+), just check if +. That is, we compute + by using attribute closure, and then check if it contains . Is a simple and cheap test, and very useful For each R, we find the closure +, and for each S +, we output a functional dependency S.
Testing functional dependencies
Computing closure of F
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Canonical Cover
Sets of functional dependencies may have redundant dependencies that can be inferred from the others
1. 2.
For example: A C is redundant in: {A B, B C, A C} Parts of a functional dependency may be redundant E.g.: on RHS: {A B, B C, A CD} can be simplified to G {A B, B C, A D} A CD AD A C and A D E.g.: on LHS: to
F
or use attrib closure
A CD
F+ = G+
{A B, B C, AC D} can be simplified {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
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Database System Concepts, 5th Ed.
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.
beware of the direction of the implications
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). C is extraneous in AB CD since AB C can be inferred even after deleting C
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Example: Given F = {A C, AB CD}
Database System Concepts, 5th Ed.
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 extraneous in
1. 2.
compute ({} A)+ using the dependencies in F check that ({} A)+ contains ; if it does, A is extraneous compute + using only the dependencies in F' = (F { }) { ( A)}, check that + contains A; if it does, A is extraneous
To test if attribute A is extraneous in
1.
2.
Example: Given F = {A C, AB C } Example: Given F = {A C, AB CD}
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Canonical Cover
A canonical cover for F is a set of dependencies Fc such that F logically implies all dependencies in Fc, and Fc logically implies all dependencies in F, and No functional dependency in Fc contains an extraneous attribute, and Each left side of functional dependency in Fc is unique.
F+ = Fc+ minimal
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 ; if 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
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Computing a Canonical Cover
R = (A, B, C) F = {A BC BC AB AB C} Combine A BC and A B into A BC Set is now {A BC, B C, 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} 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}
7.26 Silberschatz, Korth and Sudarshan Database System Concepts, 5th Ed.
A is extraneous in AB C
C is extraneous in A BC
Goals of Normalization
Let R be a relation scheme with a set F of functional dependencies. Decide whether a relation scheme R is in "good" form. In the case that a relation scheme R is not in "good" form, decompose it into a set of relation scheme {R1, R2, ..., Rn} such that each relation scheme is in good form the decomposition is a lossless-join decomposition Preferably, the decomposition should be dependency preserving.
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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 and only if at least one of the following dependencies is in F+: R1 R2 R1 R1 R2 R2
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Example
R = (A, B, C) F = {A B, B C) Can be decomposed in two different ways Lossless-join decomposition: R1 R2 = {B} and B BC Dependency preserving Lossless-join decomposition: R1 R2 = {A} and A AB Not dependency preserving cannot check B C without computing R1 R2
R1 = (A, B), R2 = (B, C)
R1 = (A, B), R2 = (A, C)
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Dependency Preservation
Let Fi be 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.
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Testing for Dependency Preservation
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 w.r.t. to F) result = while (changes to result) do for each Ri in the decomposition t = (result Ri)+ Ri result = result t
w.r.t F
Essentially, calculate + in the decomposed schemas without explicitly calculating Fi
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)+
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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 all have the same B value!
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First Normal Form
Domain is atomic if its elements are considered to be indivisible units
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Second Normal Form
R: Sales ( dept, item, price) FD: dept, item price; item price
Anomalies?
KEY dept item Nonfull functional dependency price
A 2NF relation is a 1NF relation whose nonprime attributes are fully and functionally dependent on any of the keys
Sales (dept, item) Iteminfo (item, price)
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Third 5th Normal Form
Iteminfo (item, price,discount) item price; price discount
Anomalies? Relation R with FDs F is in 3NF if, for all X A in F + A X (called a trivial FD), or X contains a key for R, or A is a prime attribute of R (i.e., A is part of some key for R).
KEY item price
discount
Transitive dependency Iteminfo (item, price) Discnt(price,discount)
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What Does 3NF Achieve?
If 3NF violated by X A, one of the following holds: X is a subset of some key K We store (X, A) pairs redundantly. There is a chain of FDs K X A, which means that we cannot associate an X value with a K value unless we also associate an A value with an X value. X is not a proper subset of any key.
But: even if a relation is in 3NF, these problems could arise. e.g., Reserves(S, B, D, C), S C, C S is in 3NF, but for each reservation of sailor S, same (S, C) pair is stored.
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Boyce-Codd Normal Form
SalaryInfo(salary-bracket, status, bonus) Salary-bracket, status bonus; bonus status
KEY Salarybracket status bonus
. (( QQ JJ LL QQ HH HH UU . 0 D Q D J H U . .. 00 DD QQ DD JJ HH UU
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.. . . .
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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: is trivial (i.e., ) is a superkey for R
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Example
R = (A, B, C ) F = {A B B C} A+ = ABC The only key is A R is not in BCNF (because of B C) B+ = BC R is not in BCNF (because B is not a superkey) Decomposition R1 = (A, B), R2 = (B, C) R1 and R2 in BCNF Always! (Why?) Lossless-join decomposition Dependency preserving
or
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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, using only F is incorrect when testing a relation in a decomposition of R Consider R (A, B, C, D, E), with F = { A B, BC D} Not in BCNF (because of either FD) 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.
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Database System Concepts, 5th Ed.
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 (also called "projection") 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 subset 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 (+ - ) Ri can be shown to hold on Ri, and Ri violates BCNF. We use above dependency to decompose Ri
"bad" FD
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BCNF Decomposition Algorithm
result := {R}; done := false; compute F+; while (not done) do if (there is a schema Ri in result that is not in BCNF) then begin let be a nontrivial functional dependency that holds on Ri "bad" FD such that Ri is not in F+, and = ; ensure lossless-join result := (result Ri ) (Ri ) (, ); end else done := true; Note: each Ri is in BCNF, and decomposition is lossless-join.
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Example of BCNF Decomposition
R = (A, B, C ) F = {A B B C} Key = {A} R is not in BCNF (B C but B is not a superkey) Decomposition R1 = (B, C) R2 = (A, B)
i.e., won't choose A B for decomposition
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Example of BCNF Decomposition
Original relation R and functional dependency F R = (branch-name, branch-city, assets, customer-name, loan-number, amount) F = {branch-name assets branch-city loan-number amount branch-name} Key = {loan-number, customer-name} Decomposition
bad FD: branch-name assets branch-city
R1 = (branch-name, branch-city, assets) R2 = (branch-name, customer-name, loan-number, amount) R21 = (branch-name, loan-number, amount) R22 = (customer-name, loan-number)
bad FD: loan-number amount branch-name
Final decomposition R1, R21, R22
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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
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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.
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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. (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 condition is a minimal relaxation of BCNF to ensure dependency preservation (will see why later).
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3NF Example
Relation R: R = (J, K, L) F = {JK L, L K} Two candidate keys: JK and JL R is in 3NF JK L LK
JK is a superkey K is contained in a candidate key
Equivalent to example in book: K L J Banker-schema = (branch-name, customer-name, banker-name) banker-name branch-name branch-name customer-name banker-name
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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 j1 j2 j3 null L l1 l1 l1 l2 K k1 k1 k1 k2
repetition of information (e.g., the relationship l1, k1) need to use null values (e.g., to represent the relationship l2, k2 where there is no corresponding value for J).
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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
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3NF Decomposition/Synthesis 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 return (R1, R2, ..., Ri)
FD preserving
help to ensure lossless-join
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3NF Decomposition Algorithm (Cont.)
Above algorithm ensures: each relation schema Ri is in 3NF decomposition is dependency preserving and lossless-join
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Example
Relation schema: Banker-info-schema = (branch-name, customer-name, banker-name, office-number) The functional dependencies for this relation schema are: banker-name branch-name office-number customer-name branch-name banker-name The key is: {customer-name, branch-name}
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Example (cont'd)
banker-name branch-name office-number
Fc = F
customer-name branch-name banker-name
The for loop in the algorithm causes us to include the following schemas in our decomposition: R1 = (banker-name, branch-name, office-number) R2 = (customer-name, branch-name, banker-name) Since Banker-schema contains a candidate key for Banker-info-schema, we are done with the decomposition process.
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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
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Design Goals
Goal for a relational database design is: BCNF. Lossless join. Dependency preservation. Lack of dependency preservation Redundancy due to use of 3NF
If we cannot achieve this, we accept one of
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 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 ...