normalization 如何將表格切割而不造成資料遺失. the evils of redundancy redundancy is...

Normalization

如何將表格切割而不造成資料遺失

Introduction to Database Systems

The Evils of Redundancy

Redundancy is at the root of several problems associated with relational schemas: redundant storage, insert/delete/update

anomalies( 資料庫管理困難 ) Integrity constraints, in particular

functional dependencies, can be used to identify schemas with such problems and to suggest refinements.


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Main refinement technique: decomposition (replacing ABCD with, say, AB and BCD, or ACD and ABD).

Decomposition should be used judiciously: Is there reason to decompose a relation? What problems (if any) does the

decomposition cause?


Functional Dependencies (FDs)

A functional dependency X Y holds over relation R if, for every allowable instance r of R: t1 r, t2 r, X (t1) = X (t2) implies Y (t1) = Y (t2) i.e., given two tuples in r, if the X values agree, then the

Y values must also agree. (X and Y are sets of attributes.)

An FD is a statement about all allowable relations. Must be identified based on semantics of application. Given some allowable instance r1 of R, we can check if it

violates some FD f, but we cannot tell if f holds over R! K is a candidate key for R means that K R

However, K R does not require K to be minimal!


Example: Constraints on Entity Set

Consider relation obtained from Hourly_Emps: Hourly_Emps (ssn, name, lot, rating, hrly_wages,

hrs_worked) Notation: We will denote this relation schema

by listing the attributes: SNLRWH This is really the set of attributes {S,N,L,R,W,H}. Sometimes, we will refer to all attributes of a

relation by using the relation name. (e.g., Hourly_Emps for SNLRWH)

Some FDs on Hourly_Emps: ssn is the key: S SNLRWH rating determines hrly_wages: R W


Example Problems due to R W :

Update anomaly: Can we change W in just the 1st tuple of SNLRWH?

Insertion anomaly: What if we want to insert an employee and don’t know the hourly wage for his rating?

Deletion anomaly: If we delete all employees with rating 5, we lose the information about the wage for rating 5!

S N L R W H

123-22-3666 Attishoo 48 8 10 40

231-31-5368 Smiley 22 8 10 30

131-24-3650 Smethurst 35 5 7 30

434-26-3751 Guldu 35 5 7 32

612-67-4134 Madayan 35 8 10 40

S N L R H

123-22-3666 Attishoo 48 8 40

231-31-5368 Smiley 22 8 30

131-24-3650 Smethurst 35 5 30

434-26-3751 Guldu 35 5 32

612-67-4134 Madayan 35 8 40

R W

8 10

5 7

Hourly_Emps2

Wages


Reasoning About FDs

Given some FDs, we can usually infer additional FDs: ssn did, did lot implies ssn lot

An FD f is implied by a set of FDs F if f holds whenever all FDs in F hold.F+ = closure of F is the set of all FDs that are implied by F.

Armstrong’s Axioms (X, Y, Z are sets of attributes): Reflexivity: If X Y, then X Y Augmentation: If X Y, then XZ YZ for any Z Transitivity: If X Y and Y Z, then X Z


Reasoning About FDs (Contd.)

Couple of additional rules (that follow from AA): Union: If X Y and X Z, then X YZ Decomposition: If X YZ, then X Y and

X Z


Reasoning About FDs (Contd.)

Computing the closure of a set of FDs can be expensive. (Size of closure is exponential in # attrs!)

Typically, we just want to check if a given FD X Y is in the closure of a set of FDs F. An efficient check: Compute attribute closure of X (denoted X+)

wrt F: Set of all attributes A such that X A is in F+

There is a linear time algorithm to compute this. Check if Y is in X+


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計算 Closure 給一個 FD set F ，找出所有被 Attribute A

所 determine 的所有 attribute ，表示為A+ ；即 B A+ if AB( 含推導而得者）

演算法 result := A while (result changes) do for each FD B C in F if B is in result then result := result ∪ C


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範例 R （ W,X,Y,Z) FD: W Z YZ X WZ Y 計算 WZ +

(1) result = {WZ} (F1) WZ Y result = {WYZ} (2) YZ X result = {WXYZ} 所以 WZ 為一

superkey


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再計算 w +

(1) result = {W}W Z result = {WZ}WZ Y result = {WYZ}YZ X result = {WXYZ}

所以 W 為一 candidate key 而 WZ 不是 Candidate key


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計算 Redundant FD 定義：一條 FD X Y 是一條 redundant FD 如果他可以

被不包括它的 FD 聯合推得演算法

1. Choose a candidate FD, say X Y , remove it from F2. Result := {x}; while (result changes and Y is not contained in result)

do for each FD, A B, remaining in the reduced set of FDs if A is a subset of result then result := result ∪ B end3. If Y is a subset of result then, FD X Y is redundant


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範例 FDs (1)W Z (2) W Y (3)YZ X (4)WZ YCheck rule W Y {W} {WZ} {WYZ} {WXYZ} W 在 {WXYZ} 中 , 所以 (2) W Y 是一條

redundant rule (remove it)


Normal Forms Back to schema refinement… Q1: is any refinement is needed??! If a relation is in a normal form (BCNF,

3NF etc.): we know that certain problems are

avoided/minimized. helps decide whether decomposing a relation

is useful.


Boyce-Codd Normal Form (BCNF)

Reln R with FDs F is in BCNF if, for all X A in F+

A X (called a trivial FD), or X contains a key for R.


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I.e.: R is in BCNF if the only non-trivial FDs over R are key constraints. No dependency in R that can be predicted using

FDs alone. If we are shown two tuples that agree upon the X

value, we cannot infer the A value in one tuple from the A value in the other, unless X is a key. If example relation is in BCNF, and X A, the 2 tuples must be identical (since X is a key).


Third Normal Form (3NF)

Reln 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 part of some key (not superkey!) for R.

(A is prime)


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Minimality of a key is crucial in third condition above!

If R is in BCNF, obviously in 3NF. If R is in 3NF, some redundancy is possible.

It is a compromise, used when BCNF not achievable (e.g., no ``good’’ decomp, or performance considerations). Lossless-join, dependency-preserving

decomposition of R into a collection of 3NF relations always possible.


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Key review Key: minimal subset of the fields

(attributes) of a relation that is a unique identifier for a tuple.

Super key: a set of fields (attributes) that includes the key


Decomposition of a Relation Scheme

Suppose that relation R contains attributes A1 ... An. A decomposition of R consists of replacing R by two or more relations such that: Each new relation scheme contains a subset of the

attributes of R (and no attributes that do not appear in R), and

Every attribute of R appears as an attribute of one of the new relations.


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Intuitively, decomposing R means we will store instances of the relation schemes produced by the decomposition, instead of instances of R.

E.g., Can decompose SNLRWH into SNLRH and RW.

{ssn, name, lot, rating, hourly_wages, hours_worked}


Example Decomposition Decompositions should be used only

when needed. SNLRWH has FDs S SNLRWH and R W Second FD causes violation of 3NF; W values

repeatedly associated with R values. Easiest way to fix this is to create a relation RW to store these associations, and to remove W from the main schema:

i.e., we decompose SNLRWH into SNLRH and RW Q: potential problems of decomposition?


Problems with Decompositions

There are three potential problems to consider: Some queries become more expensive.

e.g., How much did sailor Joe earn? (salary = W*H)

May be impossible to reconstruct the original relation!

Fortunately, not in the SNLRWH example. Dependency checking may require joins.

Fortunately, not in the SNLRWH example.


Lossless Join Decompositions

Decomposition of R into X and Y is lossless-join w.r.t. a set of FDs F if, for every instance r that satisfies F: (r) (r) = r

It is always true that r (r) (r) In general, the other direction does not hold! If it does,

the decomposition is lossless-join. Definition extended to decomposition into 3 or

more relations in a straightforward way. It is essential that all decompositions used to deal

with redundancy be lossless! (Avoids Problem (2).)

X Y X

Y


More on Lossless Join The decomposition of R

into X and Y is lossless-join wrt F if and only if the closure of F contains: X Y X, or X Y Y

In particular, the decomposition of R into UV and R - V is lossless-join if U V holds over R.

A B C1 2 34 5 67 2 81 2 87 2 3

A B C 1 2 3 4 5 6 7 2 8

A B1 24 57 2

B C2 35 62 8


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驗證 Lossless Join preservation Simple rule

R decompose into R1, R1 is lossless if1. R1∩R2 R1 (R2 含有 R1 的 Foreign

Key), 或2. R1∩R2 R2 (R1 含有 R2 的 Foreign

Key) 切割成多個可以以兩個的 case 推論


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正式演算法給定 Functional dependency set F;

R(A1,A2,…,An) decompose 為 R1,R2,…,Rm, 決定是否是 Lossless decomposition 的方法1. Construct an m by n tables s; where

columns for attributes and rows for the decomposed relations

2. For each cell s(i, j) of s, if the attribute for column Aj is in row Ri, put a(j) in the cell else put b(I,j) in the cell. (Initialization)


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3. Repeat the following until no changes on sFor each X Y in F找出所有 X attribute 的 column 內為 a 的 row （切

割 ) 者 , 如果有任一個 row 其 column Y 內為 a, 則這堆 row 的 Y column 都放 a

4. 如果做完 step ３之後有任一個 row 的所有column 都是 a, 則為 lossless, 否則為lossy.


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範例 R(A,B,C,D,E) FD: A C --- (1) ;AB

D ---(2); D E ---(3); 切割成 R1(A,C) R2(A,B,D) R3(D,E)

Step 1&2 建表Step3 考慮 rule 1; R1, R2, 在 column

A 上都是 a而 R1 在 C 的位置是 a(3); 所以 R2 在

column C 的位置放 a(3); 同理 rule 3 使我們在 R2 的 column E 放 a(5)


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A B C D E

R1(A,C) a(1) b(2,1)

a(3) b(1,4)

b(1,5)

R2(A,B,D)

a(1) a(2) b(2,3)

a(4) b(2,5)

R3(D,E) b(3,1)

b(3,2)

b(3,3)

a(4) a(5)


Dependency Preserving Decomposition

Consider CSJDPQV, C is key, JP C and SD P.

{contractid, supplierid, projectid,deptid,partid, qty, value} BCNF decomposition: CSJDQV and SDP Problem: Checking JP C requires a join!


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Dependency preserving decomposition (Intuitive): If R is decomposed into X, Y and Z, and we

enforce the FDs that hold on X, on Y and on Z, then all FDs that were given to hold on R must also hold. (Avoids Problem (3).)

Projection of set of FDs F : If R is decomposed into X, the projection of F on X (denoted FX ) is the set of FDs U V in F+ (closure of F ) such that all of the attributes U, V are in X.


Dependency Preserving Decompositions (Contd.)

Decomposition of R into X and Y is dependency preserving if (FX FY ) + = F

+

i.e., if we consider only dependencies in the closure F + that can be checked in X without considering Y, and in Y without considering X, these imply all dependencies in F +.


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Important to consider F + in this definition: ABC, A B, B C, C A, decomposed into

AB and BC. Is this dependency preserving? Is C A

preserved????? Dependency preserving does not imply

lossless join: ABC, A B, decomposed into AB and BC.


Decomposition into BCNF

Consider relation R with FDs F. If X Y violates BCNF, decompose R into R - Y and XY. Repeated application of this idea will give us a

collection of relations that are in BCNF; lossless join decomposition, and guaranteed to terminate.

e.g., CSJDPQV, key C, JP C, SD P, J S {contractid, supplierid, projectid,deptid,partid, qty, value} To deal with SD P, decompose into SDP, CSJDQV. To deal with J S, decompose CSJDQV into JS and

CJDQV


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In general, several dependencies may cause violation of BCNF. The order in which we ``deal with’’ them could lead to very different sets of relations!


BCNF and Dependency Preservation

In general, there may not be a dependency preserving decomposition into BCNF. e.g., CSZ, CS Z, Z C Can’t decompose while preserving 1st FD;

not in BCNF.


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Similarly, decomposition of CSJDPQV into SDP, JS and CJDQV is not dependency preserving (w.r.t. the FDs JP C, SD P and J S).

{contractid, supplierid, projectid,deptid,partid, qty, value} However, it is a lossless join decomposition. In this case, adding JPC to the collection of

relations gives us a dependency preserving decomposition.

JPC tuples stored only for checking FD! (Redundancy!)


Decomposition into 3NF

Obviously, the algorithm for lossless join decomp into BCNF can be used to obtain a lossless join decomp into 3NF (typically, can stop earlier).

To ensure dependency preservation, one idea: If X Y is not preserved, add relation XY.

Refinement: Instead of the given set of FDs F, use a minimal cover for F.


Minimal Cover for a Set of FDs

Minimal cover G for a set of FDs F: Closure of F = closure of G. Right hand side of each FD in G is a single

attribute. If we modify G by deleting an FD or by

deleting attributes from an FD in G, the closure changes.


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Minimal cover Put the FDs in a standard form: single

attribute on the right side ( using decomposition axiom)

A->BC change to A->B, A->C Minimize the left side of each FD ABC->D and B->C ; change ABC->D to AB->D Delete redundant FDs


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Intuitively, every FD in G is needed, and ``as small as possible’’ in order to get the same closure as F.

e.g., A B, ABCD E, EF GH, ACDF EG has the following minimal cover: A B, ACD E, EF G and EF H


Summary of Schema Refinement

BCNF: free of redundancies detectable by FDs. ensuring BCNF is a good heuristic.

Not in BCNF? Try decomposing into BCNF relations. Must consider whether all FDs are preserved!


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Lossless-join, dependency preserving decomposition into BCNF impossible? Consider 3NF. Same if BCNF decomp is unsuitable

for typical queries Decompositions should be carried out

and/or re-examined while keeping performance requirements in mind.


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習題14.26 Consider the universal relation R = {A,

B, C, D, E, F, G, H, I} and the set of

functional dependencies F = { {A, B} -> {C}, {A} -> {D, E}, {B} -> {F}, {F} ->

{G, H}, {D} -> {I, J} }. What is the key for R? Decompose R into 2NF, then 3NF

relations.


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14.27 Repeat exercise 14.26 for the following different set of functional dependencies

G = { {A, B} -> {C}, {B, D} -> {E, F}, {A, D} -> {G, H}, {A} -> {I}, {H} -> {J} }.

Answer:


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14.30 Consider the relation R, which has attributes that hold schedules of courses and

sections at a university; R = {CourseNo, SecNo, OfferingDept, CreditHours,CourseLevel, InstructorSSN, Semester, Year, Days_Hours, RoomNo,NoOfStudents}. Suppose that the following functional dependencies hold on

R:{CourseNo} -> {OfferingDept, CreditHours, CourseLevel}{CourseNo, SecNo, Semester, Year} ->{Days_Hours, RoomNo, NoOfStudents, InstructorSSN}{RoomNo, Days_Hours, Semester, Year} -> {InstructorSSN, CourseNo,

SecNo}Try to determine which sets of attributes form keys of R. How would younormalize this relation?


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14.19 Consider the following two sets of functional dependencies F= {A ->C, AC ->D,

E ->AD, E ->H} and G = {A ->CD, E ->AH}. Check whether or not they are equivalent.


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15.29 consider the following decomposition for the relation schema R of Exercise 14.26. Determine whether each decomposition has (i) the dependency preservation property, and (ii) the lossless join property, with respect to F. Also determine which normal form each relation in the decomposition is in.

a. D1 = {R1, R2, R3, R4, R5}; R1 = {A,B,C}, R2={A,D,E}, R3={B,F}, R4={F,G,H}, R5={D,I,J}

b. D2={R1,R2,R3}; R1={A,B,C,D,E}, R2={B,F,G,H}, R3={D,I,J}

c. D3= {R1,R2,R3,R4,R5}; R1={A,B,C,D}, R2={D,E}, R3={B,F}, R4={F,G,H}, R5={D,I,J}


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15.30 Consider the following relation REFRIG(Model#, Year, Price, Manu_Plant, Color), which is abbreviated as REFRIG(M,Y,P,MP,C), and with the following set F of functional dependencies:F = {M MP, {M,Y} P, MP C}

a. Evaluate each of the following as a candidate key for REFRIG , given reasons why it can or cannot be a key:{M}, {M,Y},{M,C}.

b. Based on the above key determination, state whether the relation REFRIG is n 3NF and in BCNF, giving proper reasons.

c. Consider the decomposition of REFRIG into D={R1(M,Y,P), R2(M,MP,C)}. Is this decomposition lossless? Show why.

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