Chapter 7: Relational Database Design : Chapter 7: Relational Database Design
Chapter 7: Relational Database Design : Chapter 7: Relational Database Design Features of Good Relational Design
Atomic Domains and First Normal Form
Decomposition Using Functional Dependencies
Functional Dependency Theory
Algorithms for Functional Dependencies
Decomposition Using Multivalued Dependencies
More Normal Form
Database-Design Process
Modeling Temporal Data
The Banking Schema : The Banking Schema branch = (branch_name, branch_city, assets)
customer = (customer_id, customer_name, customer_street, customer_city)
loan = (loan_number, amount)
account = (account_number, balance)
employee = (employee_id. employee_name, telephone_number, start_date)
dependent_name = (employee_id, dname)
account_branch = (account_number, branch_name)
loan_branch = (loan_number, branch_name)
borrower = (customer_id, loan_number)
depositor = (customer_id, account_number)
cust_banker = (customer_id, employee_id, type)
works_for = (worker_employee_id, manager_employee_id)
payment = (loan_number, payment_number, payment_date, payment_amount)
savings_account = (account_number, interest_rate)
checking_account = (account_number, overdraft_amount)
Combine Schemas? : Combine Schemas? Suppose we combine borrow and loan to get
bor_loan = (customer_id, loan_number, amount )
Result is possible repetition of information (L-100 in example below)
A Combined Schema Without Repetition : A Combined Schema Without Repetition Consider combining loan_branch and loan
loan_amt_br = (loan_number, amount, branch_name)
No repetition (as suggested by example below)
What About Smaller Schemas? : What About Smaller Schemas? Suppose we had started with bor_loan. How would we know to split up (decompose) it into borrower and loan?
Write a rule “if there were a schema (loan_number, amount), then loan_number would be a candidate key”
Denote as a functional dependency:
loan_number amount
In bor_loan, because loan_number is not a candidate key, the amount of a loan may have to be repeated. This indicates the need to decompose bor_loan.
Not all decompositions are good. Suppose we decompose employee into
employee1 = (employee_id, employee_name)
employee2 = (employee_name, telephone_number, start_date)
The next slide shows how we lose information -- we cannot reconstruct the original employee relation -- and so, this is a lossy decomposition.
A Lossy Decomposition : A Lossy Decomposition
First Normal Form : 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
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
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 (and revisit this in Chapter 9)
First Normal Form (Cont’d) : First Normal Form (Cont’d) Atomicity is actually a property of how the elements of the domain are used.
Example: 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.
Goal — Devise a Theory for the Following : 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
Our theory is based on:
functional dependencies
multivalued dependencies
Functional Dependencies : 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.
Functional Dependencies (Cont.) : Functional Dependencies (Cont.) 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 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.
On this instance, A B does NOT hold, but B A does hold. 4
1 5
3 7
Functional Dependencies (Cont.) : 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:
bor_loan = (customer_id, loan_number, amount ).
We expect this functional dependency to hold:
loan_number amount
but would not expect the following to hold:
amount customer_name
Use of Functional Dependencies : 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.
specify constraints on the set of legal relations
We say that F holds on R if all legal relations on R satisfy the set of functional dependencies F.
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 may, by chance, satisfy amount customer_name.
Functional Dependencies (Cont.) : Functional Dependencies (Cont.) A functional dependency is trivial if it is satisfied by all instances of a relation
Example:
customer_name, loan_number customer_name
customer_name customer_name
In general, is trivial if
Closure of a Set of Functional Dependencies : 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+.
F+ is a superset of F.
Boyce-Codd Normal Form : 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: Example schema not in BCNF:
bor_loan = ( customer_id, loan_number, amount )
because loan_number amount holds on bor_loan but loan_number is not a superkey
Decomposing a Schema into BCNF : Decomposing a Schema into BCNF Suppose we have a schema R and a non-trivial dependency causes a violation of BCNF.
We decompose R into:
(U )
( R - ( - ) )
In our example,
= loan_number
= amount
and bor_loan is replaced by
(U ) = ( loan_number, amount )
( R - ( - ) ) = ( customer_id, loan_number )
BCNF and Dependency Preservation : BCNF and Dependency Preservation Constraints, including functional dependencies, are costly to check in practice unless they pertain to only one relation
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.
Because it is not always possible to achieve both BCNF and dependency preservation, we consider a weaker normal form, known as third normal form.
Third Normal Form : 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).
Goals of Normalization : 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.
How good is BCNF? : How good is BCNF? There are database schemas in BCNF that do not seem to be sufficiently normalized
Consider a database
classes (course, teacher, book )
such that (c, t, b) classes means that t is qualified to teach c, and b is a required textbook for c
The database is supposed to list for each course the set of teachers any one of which can be the course’s instructor, and the set of books, all of which are required for the course (no matter who teaches it).
How good is BCNF? (Cont.) : There are no non-trivial functional dependencies and therefore the relation is in BCNF
Insertion anomalies – i.e., if Marilyn is a new teacher that can teach database, two tuples need to be inserted
(database, Marilyn, DB Concepts) (database, Marilyn, Ullman) course teacher book database
database
database
database
database
database
operating systems
operating systems
operating systems
operating systems Avi
Avi
Hank
Hank
Sudarshan
Sudarshan
Avi
Avi
Pete
Pete DB Concepts
Ullman
DB Concepts
Ullman
DB Concepts
Ullman
OS Concepts
Stallings
OS Concepts
Stallings classes How good is BCNF? (Cont.)
How good is BCNF? (Cont.) : Therefore, it is better to decompose classes into: course teacher database
database
database
operating systems
operating systems Avi
Hank
Sudarshan
Avi
Jim teaches course book database
database
operating systems
operating systems DB Concepts
Ullman
OS Concepts
Shaw text This suggests the need for higher normal forms, such as Fourth Normal Form (4NF), which we shall see later. How good is BCNF? (Cont.)
Functional-Dependency Theory : Functional-Dependency Theory We now consider the formal theory that tells us which functional dependencies are implied logically by a given set of functional dependencies.
We then develop algorithms to generate lossless decompositions into BCNF and 3NF
We then develop algorithms to test if a decomposition is dependency-preserving
Closure of a Set of Functional Dependencies : 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 (reflexivity)
if , then (augmentation)
if , and , then (transitivity)
These rules are
sound (generate only functional dependencies that actually hold) and
complete (generate all functional dependencies that hold).
Example : Example R = (A, B, C, G, H, I)F = { A B A C CG H CG I B H}
some members of F+
A H
by transitivity from A B and B H
AG I
by augmenting A C with G, to get AG CG and then transitivity with CG I
CG HI
by augmenting CG I to infer CG CGI,
and augmenting of CG H to infer CGI HI,
and then transitivity
Procedure for Computing F+ : Procedure for Computing F+ To compute the closure of a set of functional dependencies F:
F + = Frepeat 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 shall see an alternative procedure for this task later
Closure of Functional Dependencies (Cont.) : 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.
Closure of Attribute Sets : Closure of Attribute Sets Given a set of attributes a, define the closure of a under F (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
Example of Attribute Set Closure : 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?
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
Uses of Attribute Closure : 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.
Testing functional dependencies
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
Computing closure of F
For each R, we find the closure +, and for each S +, we output a functional dependency S.
Canonical Cover : Canonical Cover Sets of functional dependencies may have redundant dependencies that can be inferred from the others
For example: A C is redundant in: {A B, B C}
Parts of a functional dependency may be redundant
E.g.: on RHS: {A B, B C, A CD} can be simplified to {A B, B C, A D}
E.g.: on LHS: {A B, B C, AC D} can be simplified to {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
Extraneous Attributes : 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.
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).
Example: Given F = {A C, AB CD}
C is extraneous in AB CD since AB C can be inferred even after deleting C
Testing if an Attribute is Extraneous : 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
compute ({} – A)+ using the dependencies in F
check that ({} – A)+ contains A; if it does, A is extraneous
To test if attribute A is extraneous in
compute + using only the dependencies in F’ = (F – { }) { ( – A)},
check that + contains A; if it does, A is extraneous
Canonical Cover : 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.
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 an extraneous attribute is 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
Computing a Canonical Cover : Computing a Canonical Cover R = (A, B, C)F = {A BC B C A B AB C}
Combine A BC and A B into A BC
Set is now {A BC, B C, AB C}
A is extraneous in 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}
C is extraneous in A BC
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
Lossless-join Decomposition : 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
Example : Example R = (A, B, C)F = {A B, B C)
Can be decomposed in two different ways
R1 = (A, B), R2 = (B, C)
Lossless-join decomposition:
R1 R2 = {B} and B BC
Dependency preserving
R1 = (A, B), R2 = (A, C)
Lossless-join decomposition:
R1 R2 = {A} and A AB
Not dependency preserving (cannot check B C without computing R1 R2)
Dependency Preservation : 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.
Testing for Dependency Preservation : 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 with respect to F)
result = while (changes to result) do for each Ri in the decomposition t = (result Ri)+ Ri result = result t
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)+
Example : Example R = (A, B, C )F = {A B B C}Key = {A}
R is not in BCNF
Decomposition R1 = (A, B), R2 = (B, C)
R1 and R2 in BCNF
Lossless-join decomposition
Dependency preserving
Testing for BCNF : 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}
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.
Testing Decomposition for BCNF : 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 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 set 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 (+ - ) Rican be shown to hold on Ri, and Ri violates BCNF.
We use above dependency to decompose Ri
BCNF Decomposition Algorithm : 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 such that Ri is not in F +, and = ; result := (result – Ri ) (Ri – ) (, ); end else done := true;
Note: each Ri is in BCNF, and decomposition is lossless-join.
Example of BCNF Decomposition : 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 superkey)
Decomposition
R1 = (B, C)
R2 = (A,B)
Example of BCNF Decomposition : 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
R1 = (branch_name, branch_city, assets )
R2 = (branch_name, customer_name, loan_number, amount )
R3 = (branch_name, loan_number, amount )
R4 = (customer_name, loan_number )
Final decomposition R1, R3, R4
BCNF and Dependency Preservation : BCNF and Dependency Preservation 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 It is not always possible to get a BCNF decomposition that is
dependency preserving
Third Normal Form: Motivation : 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.
3NF Example : 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 JK is a superkey L K K is contained in a candidate key
Redundancy in 3NF : Redundancy in 3NF 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). There is some redundancy in this schema
Example of problems due to redundancy in 3NF
R = (J, K, L)F = {JK L, L K }
Testing for 3NF : 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
3NF Decomposition Algorithm : 3NF 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 := endif 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)
3NF Decomposition Algorithm (Cont.) : 3NF Decomposition Algorithm (Cont.) Above algorithm ensures:
each relation schema Ri is in 3NF
decomposition is dependency preserving and lossless-join
Proof of correctness is at end of this file (click here)
Example : Example Relation schema:
cust_banker_branch = (customer_id, employee_id, branch_name, type )
The functional dependencies for this relation schema are: customer_id, employee_id branch_name, type employee_id branch_name
The for loop generates:
(customer_id, employee_id, branch_name, type )
It then generates
(employee_id, branch_name)
but does not include it in the decomposition because it is a subset of the first schema.
Comparison of BCNF and 3NF : 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.
Design Goals : Design Goals 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
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 a key.
Multivalued Dependencies (MVDs) : Multivalued Dependencies (MVDs) Let R be a relation schema and let R and R. The multivalued dependency
holds on R if in any legal relation r(R), for all pairs for tuples t1 and t2 in r such that t1[] = t2 [], there exist tuples t3 and t4 in r such that:
t1[] = t2 [] = t3 [] = t4 [] t3[] = t1 [] t3[R – ] = t2[R – ] t4 [] = t2[] t4[R – ] = t1[R – ]
MVD (Cont.) : MVD (Cont.) Tabular representation of
Example : Example Let R be a relation schema with a set of attributes that are partitioned into 3 nonempty subsets.
Y, Z, W
We say that Y Z (Y multidetermines Z )if and only if for all possible relations r (R )
< y1, z1, w1 > r and < y2, z2, w2 > r
then
< y1, z1, w2 > r and < y2, z2, w1 > r
Note that since the behavior of Z and W are identical it follows that
Y Z if Y W
Example (Cont.) : Example (Cont.) In our example:
course teacher course book
The above formal definition is supposed to formalize the notion that given a particular value of Y (course) it has associated with it a set of values of Z (teacher) and a set of values of W (book), and these two sets are in some sense independent of each other.
Note:
If Y Z then Y Z
Indeed we have (in above notation) Z1 = Z2The claim follows.
Use of Multivalued Dependencies : Use of Multivalued Dependencies We use multivalued dependencies in two ways:
1. To test relations to determine whether they are legal under a given set of functional and multivalued dependencies
2. To specify constraints on the set of legal relations. We shall thus concern ourselves only with relations that satisfy a given set of functional and multivalued dependencies.
If a relation r fails to satisfy a given multivalued dependency, we can construct a relations r that does satisfy the multivalued dependency by adding tuples to r.
Theory of MVDs : Theory of MVDs From the definition of multivalued dependency, we can derive the following rule:
If , then
That is, every functional dependency is also a multivalued dependency
The closure D+ of D is the set of all functional and multivalued dependencies logically implied by D.
We can compute D+ from D, using the formal definitions of functional dependencies and multivalued dependencies.
We can manage with such reasoning for very simple multivalued dependencies, which seem to be most common in practice
For complex dependencies, it is better to reason about sets of dependencies using a system of inference rules (see Appendix C).
Fourth Normal Form : Fourth Normal Form A relation schema R is in 4NF with respect to a set D of functional and multivalued dependencies if for all multivalued dependencies in D+ of the form , where R and R, at least one of the following hold:
is trivial (i.e., or = R)
is a superkey for schema R
If a relation is in 4NF it is in BCNF
Restriction of Multivalued Dependencies : Restriction of Multivalued Dependencies The restriction of D to Ri is the set Di consisting of
All functional dependencies in D+ that include only attributes of Ri
All multivalued dependencies of the form
( Ri)
where Ri and is in D+
4NF Decomposition Algorithm : 4NF Decomposition Algorithm result: = {R};done := false;compute D+;Let Di denote the restriction of D+ to Ri
while (not done) if (there is a schema Ri in result that is not in 4NF) then begin
let be a nontrivial multivalued dependency that holds on Ri such that Ri is not in Di, and ; result := (result - Ri) (Ri - ) (, ); end else done:= true;
Note: each Ri is in 4NF, and decomposition is lossless-join
Example : Example R =(A, B, C, G, H, I)
F ={ A B
B HI
CG H }
R is not in 4NF since A B and A is not a superkey for R
Decomposition
a) R1 = (A, B) (R1 is in 4NF)
b) R2 = (A, C, G, H, I) (R2 is not in 4NF)
c) R3 = (C, G, H) (R3 is in 4NF)
d) R4 = (A, C, G, I) (R4 is not in 4NF)
Since A B and B HI, A HI, A I
e) R5 = (A, I) (R5 is in 4NF)
f)R6 = (A, C, G) (R6 is in 4NF)
Further Normal Forms : Further Normal Forms Join dependencies generalize multivalued dependencies
lead to project-join normal form (PJNF) (also called fifth normal form)
A class of even more general constraints, leads to a normal form called domain-key normal form.
Problem with these generalized constraints: are hard to reason with, and no set of sound and complete set of inference rules exists.
Hence rarely used
Overall Database Design Process : 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).
Normalization breaks R into smaller relations.
R could have been the result of some ad hoc design of relations, which we then test/convert to normal form.
ER Model and Normalization : 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 department_number and department_address, and a functional dependency department_number department_address
Good design would have made department an entity
Functional dependencies from non-key attributes of a relationship set possible, but rare --- most relationships are binary
Denormalization for Performance : Denormalization for Performance May want to use non-normalized schema for performance
For example, displaying customer_name along with account_number and balance requires join of account with depositor
Alternative 1: Use denormalized relation containing attributes of account as well as depositor 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 account depositor
Benefits and drawbacks same as above, except no extra coding work for programmer and avoids possible errors
Other Design Issues : Other Design Issues Some aspects of database design are not caught by normalization
Examples of bad database design, to be avoided:
Instead of earnings (company_id, year, amount ), use
earnings_2000, earnings_2001, earnings_2002, etc., all on the schema (company_id, earnings).
Above are in BCNF, but make querying across years difficult and needs new table each year
company_year(company_id, earnings_2000, earnings_2001, earnings_2002)
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
Modeling Temporal Data : Modeling Temporal Data Temporal data have an association time interval during which the data are valid.
A snapshot is the value of the data at a particular point in time.
Adding a temporal component results in functional dependencies like
customer_id customer_street, customer_city
not to hold, because the address varies over time
A temporal functional dependency holds on schema R if the corresponding functional dependency holds on all snapshots for all legal instances r (R )
End of Chapter : End of Chapter
Proof of Correctness of 3NF Decomposition Algorithm : Proof of Correctness of 3NF Decomposition Algorithm
Correctness of 3NF Decomposition Algorithm : Correctness of 3NF Decomposition Algorithm 3NF decomposition algorithm is dependency preserving (since there is a relation for every FD in Fc)
Decomposition is lossless
A candidate key (C ) is in one of the relations Ri in decomposition
Closure of candidate key under Fc must contain all attributes in R.
Follow the steps of attribute closure algorithm to show there is only one tuple in the join result for each tuple in Ri
Correctness of 3NF Decomposition Algorithm (Cont’d.) : Correctness of 3NF Decomposition Algorithm (Cont’d.) Claim: if a relation Ri is in the decomposition generated by the
above algorithm, then Ri satisfies 3NF.
Let Ri be generated from the dependency
Let B be any non-trivial functional dependency on Ri. (We need only consider FDs whose right-hand side is a single attribute.)
Now, B can be in either or but not in both. Consider each case separately.
Correctness of 3NF Decomposition (Cont’d.) : Correctness of 3NF Decomposition (Cont’d.) Case 1: If B in :
If is a superkey, the 2nd condition of 3NF is satisfied
Otherwise must contain some attribute not in
Since B is in F+ it must be derivable from Fc, by using attribute closure on .
Attribute closure not have used . If it had been used, must be contained in the attribute closure of , which is not possible, since we assumed is not a superkey.
Now, using (- {B}) and B, we can derive B
(since , and B since B is non-trivial)
Then, B is extraneous in the right-hand side of ; which is not possible since is in Fc.
Thus, if B is in then must be a superkey, and the second condition of 3NF must be satisfied.
Correctness of 3NF Decomposition (Cont’d.) : Correctness of 3NF Decomposition (Cont’d.) Case 2: B is in .
Since is a candidate key, the third alternative in the definition of 3NF is trivially satisfied.
In fact, we cannot show that is a superkey.
This shows exactly why the third alternative is present in the definition of 3NF.
Q.E.D.
Figure 7.5: Sample Relation r : Figure 7.5: Sample Relation r
Figure 7.6 : Figure 7.6
Figure 7.7 : Figure 7.7
Figure 7.15: An Example of Redundancy in a BCNF Relation : Figure 7.15: An Example of Redundancy in a BCNF Relation
Figure 7.16: An Illegal R2 Relation : Figure 7.16: An Illegal R2 Relation
Figure 7.18: Relation of Practice Exercise 7.2 : Figure 7.18: Relation of Practice Exercise 7.2