chapter 8: relational database design
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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?. - PowerPoint PPT PresentationTRANSCRIPT
8.1Database System Concepts - 6th Edition
Chapter 8: Relational Database DesignChapter 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
8.2Database System Concepts - 6th Edition
Larger Schemas?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
8.3Database System Concepts - 6th Edition
Disadvantage of Large SchemasDisadvantage 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
8.4Database System Concepts - 6th Edition
What About Smaller Schemas?What About Smaller Schemas?
Suppose we had started with inst_dept. How would we know to split up (decompose) it into instructor and department?
Intuition: put “strongly-related” attributes in the same relation
Write a rule: “every department (identified by its department name) must have only one building and one budget value”.
Denote as a functional dependency:
dept_name building, budget
In inst_dept, because dept_name is not a candidate key, the building and budget of a department may have to be repeated.
This indicates the need to decompose inst_dept
Not all decompositions are good. Suppose we decompose employee(ID, name, street, city, salary) into
employee1 (ID, name)
employee2 (name, street, city, salary)
The next slide shows how we lose information -- we cannot reconstruct the original employee relation -- and so, this is a lossy decomposition.
8.5Database System Concepts - 6th Edition
A Lossy DecompositionA Lossy Decomposition
8.6Database System Concepts - 6th Edition
Example of Lossless-Join DecompositionExample of Lossless-Join Decomposition
Lossless join decomposition
Decomposition of R = (A, B, C)R1 = (A, B) R2 = (B, C)
A B
12
A
B
12
r B,C(r)
A (r) B (r)A B
12
C
AB
B
12
C
AB
C
AB
A,B(r)
8.7Database System Concepts - 6th Edition
Goal of NormalizationGoal 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.
8.8Database System Concepts - 6th Edition
First Normal FormFirst 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.
ID Name Phone_number
1 John {6601, 6602}
ID Name
1 John
ID Phone_number
1 6601
1 6602
8.9Database System Concepts - 6th Edition
ID Phone_number
2222222222
456-7890123-4567
ID Phone_number
22222 {456-7890,123-4567}
8.10Database System Concepts - 6th Edition
Functional Dependencies Functional Dependencies
Let R be a relation schema
R and R The functional dependency
holds on R if and only if for any legal relations r(R), whenever any two tuples t1 and t2 of r agree on the attributes , they also agree on the attributes .
Example: Consider R(A,B, C, D ) with the following instance of r.
On this instance, A C is satisfied, but C A is not satisfied.
8.11Database System Concepts - 6th Edition
Functional Dependencies (Cont.)Functional Dependencies (Cont.)
Note: A specific instance of a relation schema may satisfy a functional dependency even if the functional dependency does not hold on all legal instances.
For example, a specific instance of classroom may, by chance, satisfy room_number capacity.
We usually use functional dependencies to specify constraints on all legal relations
We say that F holds on R if all legal relations on R satisfy the set of functional dependencies F.
8.12Database System Concepts - 6th Edition
PracticePractice
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:
8.13Database System Concepts - 6th Edition
Functional Dependencies (Cont.)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.
8.14Database System Concepts - 6th Edition
Functional Dependencies (Cont.)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.
8.15Database System Concepts - 6th Edition
Closure of Attribute SetsClosure of Attribute Sets
Given a set of attributes define the closure of under F (denoted by +) as the set of attributes that are functionally determined by under F
Algorithm to compute +, the closure of under F
result := ;while (changes to result) do
for each in F dobegin
if result then result := result end
PS: result, result , =>
8.16Database System Concepts - 6th Edition
Example of Attribute Set ClosureExample of Attribute Set Closure R = (A, B, C, G, H, I) F = {A B
A 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)
1. Is AG a super key?
1. Does AG R? == Is (AG)+ R
2. Is any subset of AG a superkey?
1. Does A R? == Is (A)+ R
2. Does G R? == Is (G)+ R
8.17Database System Concepts - 6th Edition
Lossless-join DecompositionLossless-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 R1 or R2, the decomposition of R is a lossless decomposition.
8.18Database System Concepts - 6th Edition
Boyce-Codd Normal FormBoyce-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, budgetholds on instr_dept, but dept_name is not a superkey
8.19Database System Concepts - 6th Edition
Testing for BCNFTesting for BCNF
To check if a non-trivial dependency causes a violation of BCNF
1. compute + (the attribute closure of ), and
2. verify that it includes all attributes of R, that is, it is a superkey of R.
For a non-BCNF Schema R and a non-trivial dependency causeing a violation of BCNF, we decompose R into:
• (U )
• ( R - )
• 注意 : 課本用 R - ( - ) ,是當 和有交集時 , 需要減掉。 In our example, “dept_name building, budget” causes the violation
= dept_name = building, budget
and inst_dept (ID, name, salary, dept_name, building, budget )
is replaced by (U ) = (dept_name, building, budget ) <- department relation ( R - ) = (ID, name, salary, dept_name ) <- instructor relation
8.20Database System Concepts - 6th Edition
BCNF Decomposition AlgorithmBCNF Decomposition Algorithm
Given F and R;
result := {R };done := false;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 and cause Ri not in BCNF; result := (result – Ri ) (Ri – ) (, );
endelse done := true;
Note: each Ri is in BCNF, and decomposition is lossless-join.
8.21Database System Concepts - 6th Edition
Example of BCNF DecompositionExample 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)
8.22Database System Concepts - 6th Edition
Example of BCNF Decomposition (Cont.)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.
8.23Database System Concepts - 6th Edition
PracticePractice 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}
8.24Database System Concepts - 6th Edition
BCNF and Dependency PreservationBCNF 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.
8.25Database System Concepts - 6th Edition
Third Normal FormThird 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.
8.26Database System Concepts - 6th Edition
3NF Example3NF 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
8.27Database System Concepts - 6th Edition
Redundancy in 3NFRedundancy 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_IDs1
s2
s3
s1null
i_IDl1
l1
I1
l3l2
dept_nameCS
CS
CSMAEE
s_ID
s1
s2
s3
s1
i_ID
l1l1I1
l3
i_ID
l1l3l2
dept_name
CSMAEE
3 N F
B C N F
8.28Database System Concepts - 6th Edition
Testing for 3NFTesting 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.
8.29Database System Concepts - 6th Edition
3NF Decomposition Algorithm3NF Decomposition Algorithm
Given F and R;
Result := {R };done := false;
while (not done) doif (there is a schema Ri in result that is not in 3NF) then begin let be a nontrivial functional dependency that
holds on Ri such that is not a superkey of Ri , and some attribute A in is not contained in any candidate key for R;
result := (result – Ri ) (Ri – ) (, ); endelse done := true;
Note: This algorithm does not always achieve dependency preserving. There exists an algorithm which ensures:
each relation schema Ri is in 3NF
decomposition is dependency preserving and lossless-join
8.30Database System Concepts - 6th Edition
Example of 3NF DecompositionExample 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
8.31Database System Concepts - 6th Edition
※ ※ Dependency-preserving lossless-join Dependency-preserving lossless-join decomposition into 3NFdecomposition into 3NF
Let Fc be a canonical cover for F;i := 0;for each functional dependency in Fc do
i := 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 */
repeatif any schema Rj is contained in another schema Rk
then /* delete Rj */ Rj := Ri; I := i-1;
until no more Rj can be deleted
return (R1, R2, ..., Ri)
Note: This algorithm is also called Note: This algorithm is also called 3NF Synthesis Algorithm3NF Synthesis Algorithm
8.32Database System Concepts - 6th Edition
※ ※ 3NF Decomposition: An Example3NF 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!
8.33Database System Concepts - 6th Edition
Comparison of BCNF and 3NFComparison 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
8.34Database System Concepts - 6th Edition
Overall Database Design ProcessOverall 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.
8.35Database System Concepts - 6th Edition
ER Model and NormalizationER 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
8.36Database System Concepts - 6th Edition
Denormalization for PerformanceDenormalization for Performance
May want to use non-normalized schema for performance
For example, displaying prereqs along with course_id and title requires join of course with prereq
Alternative 1: Use denormalized relation containing attributes of course as well as prereq with all above attributes
faster lookup
extra space and extra execution time for updates
extra coding work for programmer and possibility of error in extra code
Alternative 2: use a materialized view defined as course prereq
Benefits and drawbacks same as above, except no extra coding work for programmer and avoids possible errors
8.37Database System Concepts - 6th Edition
Other Design IssuesOther 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
8.38Database System Concepts - 6th Edition
※ ※ 2NF (2NF ( 完全函數相依完全函數相依 )) 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.
8.39Database System Concepts - 6th Edition
※ ※ 3NF3NF ((不可遞移函數相依)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.