Download presentation
Presentation is loading. Please wait.
1
CS4433 Database Systems Relational Design
2
Why Do We Learn This? We have designed ER diagram, and translated it into a relational database schema R (or a set of R1, R2, ...) Now what? We can do the following specify all relevant constraints over R implement R in SQL start using it, making sure the constraints always remain valid However, R may not be well-designed, thus causing us a lot of problems Well designed: Eliminate Data Redundancy: the same piece of data shall not be stored in more than one place. This is because duplicate data not only waste storage spaces but also easily lead to inconsistencies. Ensure Data Integrity and Accuracy:
3
Is This a Good Design? Potential problems Redundancy Update anomalies
SSN Address Phone 10 Green (201) (206) 431 Purple (908) (212) Potential problems Redundancy Update anomalies Deletion anomalies
4
How do We Obtain a Good Design?
Start with the original database schema R Transform it until we get a good design R* Desirable properties for R* must preserve the information of R must have minimal amount of redundancy must be dependency-preserving if R is associated with a set of constraints C, then it should be easy to also check C over R* (must also give good query performance)
5
Normal Forms OK, But …… How do we recognize a good design R*?
How do we transform R into R*? What we need is the “theory” of … Normal Forms
6
Normal Forms DB gurus have developed many normal forms
Most important ones Boyce-Codd (BCNF), 3rd (3NF), and 4th(4NF) normal forms If R* is in one of these forms, then R* is guaranteed to achieve certain good properties e.g., if R* is in Boyce-Codd NF, it is guaranteed to not have certain types of redundancy DB gurus have also developed algorithms to transform R into R* that is in some of these normal forms
7
Normal Forms DB gurus have also discussed trade-offs among normal forms Thus, all we have to do is learn these forms transform R into R* in one of these forms carefully evaluate the trade-offs Many of these normal forms are defined based on various constraints functional dependencies and keys
8
Behind the Scene: Know Whom We Should Blame?
9
Our Plan Motivation Functional dependencies & keys
Reasoning with FDs and keys Desirable properties of schema refinement Various normal forms and the trade-offs BCNF, 3rd normal form, 4th normal form, etc. Putting all together: how to design DB schema
10
Better Designs Exist SSN Address Phone 123-321-99 10 Green
(201) (206) 431 Purple (908) (212) SSN Phone (201) (206) (908) (212) SSN Address 10 Green 431 Purple
11
Functional Dependencies
A form of constraint (hence, part of the schema) a relationship between two attributes, typically between the PK and other non-key attributes within a table. For any relation R, attribute Y is functionally dependent on attribute X (usually the PK), if for every valid instance of X, that value of X uniquely determines the value of Y. This relationship is indicated by the representation below : X ———–> Y The left side of the above FD diagram is called the determinant, and the right side is the dependent. Finding them is part of the database design Used heavily in schema refinement
12
Example EmpID Name Phone Position E0045 Smith 1234 Clerk E1847 John
9876 Sales rep. E1111 E9999 Mary Lawyer EmpID Name, Phone, Position Position Phone but Phone Position
13
Composite Determinants
Composite determinant = a determinant of a functional dependency that consists of more than one attribute (StudentName, ClassName) (Grade)
14
Functional Dependency Rules
If A (B, C), then A B and A C. If (A,B) C, then neither A nor B determines C by itself.
15
Functional Dependencies
An FD is a statement about all allowable relations Must be identified based on semantics of application Given some instance r1of R, we can check if r1 violates some FD f, but we cannot determine if f holds over R Question: How is FD related to keys? if “K all attributes of R” then K is a superkey for R does not require K to be minimal FDs are a generalization of keys
16
Functional Dependencies
Q: From this, can you conclude phone SSN? SSN Phone (201) (206) (908) (212)
17
Relation Keys After defining FDs, we can now define keys
Key of a relation R is a set of attributes that functionally determines all attributes of R NONE of its subsets determines all attributes of R A composite key a key that consists of two or more columns. Superkey a set of attributes that contains a key We will need to know the keys of the relations in a DB schema, so that we can refine the schema
18
Candidate and Primary Keys
A candidate key is a key that determines all of the other columns in a relation. A primary key is a candidate key selected as the primary means of identifying rows in a relation. There is only one primary key per relation. The primary key may be a composite key. The ideal primary key is short, numeric, and never changes.
19
Foreign Keys A foreign key is the primary key of one relation that is placed in another relation to form a link between the relations. A foreign key can be a single column or a composite key. The term refers to the fact that key values are foreign to the relation in which they appear as foreign key values.
20
The Referential Integrity Constraint
A referential integrity constraint is a statement that limits the values of the foreign key to those already existing as primary key values in the corresponding relation. SKU_DATA (SKU, SKU_Description, Department, Buyer) ORDER_ITEM (OrderNumber, SKU, Quantity, Price, ExtendedPrice) Where ORDER_ITEM.SKU must exist in SKU_DATA.SKU
21
Foreign Key with a Referential Integrity Constraint
NOTE: The primary key of the relation is underlined and any foreign keys are in italics in the relations below: SKU_DATA (SKU, SKU_Description, Department, Buyer) ORDER_ITEM (OrderNumber, SKU, Quantity, Price, ExtendedPrice) Where ORDER_ITEM.SKU must exist in SKU_DATA.SKU
22
Finding the Keys of a Relation
Given a relation constructed from an E/R diagram, what is its key? If the relation comes from an entity set, the key of the relation is the set of attributes which is the key of the entity set If the relation comes from a many-many relationship, the key of the relation include the set of all attribute keys in the relations corresponding to the entity sets (and additional attributes if necessary) Many-one relationship, weak-entity set, multi-way relationship ……
23
Functional Dependencies in the SKU_DATA Table
SKU (SKU_Description, Department, Buyer) SKU_Description (SKU, Department, Buyer) Buyer Department
24
Functional Dependencies in the ORDER_ITEM Table
(OrderNumber, SKU) (Quantity, Price, ExtendedPrice) (Quantity, Price) (ExtendedPrice)
25
Example 3.1.1 Consider a relation about people in the United States, including their name, Social Security number, street address, city, state, ZIP code, area code, and phone number (7 digits).What FD's would you expect to hold? What are the keys for the relation? To answer this question, you need to know something about the way these numbers are assigned. For instance, can an area code straddle two states? Can a ZIP code straddle two area codes? Can two people have the same Social Security number? Can they have the same addressor phone number Consider a relation about people in the United States, including their name, Social Secu- rity number, street address, city, state, ZIP code, area code, and phone number (7 digits). What FD's would you expect to hold? What are the keys for the relation? To answer this Consider a relation about people in the United States, including their name, Social Secu- question, you need to know something about the way these numbers are assigned. For rity number, street address, city, state, ZIP code, area code, and phone number (7 digits). instance, can an area code straddle two states? Can a ZIP code straddle two area codes? What FD's would you expect to hold? What are the keys for the relation? To answer this Can two people have the same Social Security number? Can they have the same address question, you need to know something about the way these numbers are assigned. For or phone number? instance, can an area code straddle two states? Can a ZIP code straddle two area codes? Can two people have the same Social Security number? Can they have the same address or phone number
26
Solution Social Security number name Area code state
Street address, city, state zipcode personal number → name #a person could have several homes, and several phones phone number → zipcode #a phone is generally only signed to one person street address, city → county, zipcode city → county Possible keys are any attributes on the left hand side of the arrows {Social Security number, street address, city, state, area code, phone number} Need street address, city, state to uniquely determine location. A person could have multiple addresses. The same is true for phones. These days, a person could have a landline and a cellular phone personal number → name #a person could have several homes, and several phones phone number → personal number #a phone is generally only signed to one person street address, city → county, postal number city → county Possible keys are any attributes on the left hand side of the arrows
27
Closure of FD sets Given a relation schema R & a set S of FDs
is a new FD f logically implied by S? Example R = {A,B,C,G,H,I} S = A B, A C, CG H, CG I, B H would A H be logically implied? yes (you can prove this, using the definition of FD) Closure of S: S+ = all FDs logically implied by S How to compute S+? we can use Armstrong's axioms
28
Armstrong's Axioms Reflexivity rule A1A2...An a subset of A1A2...An
ABC AB ABC AC ABC A … Augmentation rule A1A2...An B1B2...Bm, then A1A2...An C1C2..Ck B1B2...Bm C1C2...Ck A B AC BC Transitivity rule A1A2...An B1B2...Bm and B1B2...Bm C1C2...Ck, then A1A2...An C1C2...Ck A B and B C then A C
29
Inferring S+ using Armstrong's Axioms
S+ := S Loop foreach f in S, apply reflexivity and augmentation rules add the new FDs to S+ foreach pair of FDs in S, apply the transitivity rule add the new FD to S+ Until S+ does not change any further
30
Additional Rules Union rule X Y and X Z, then X YZ
(X, Y, Z are sets of attributes) Decomposition rule X YZ, then X Y and X Z Pseudo-transitivity rule X Y and YZ U, then XZ U These rules can be inferred from Armstrong's axioms XZ YZ(augmentation) XZ YZ U XZ U (Transitivity) Union: XZ then XXXZ, i.e., XXZ XY then XZYZ, So X YZ Decomposition: YZ Y, YZZ Because XYZ So XY and XZ Pseudo-Transitivity: XY So XZYZ Because YZU So XZU
31
Closure of a Set of Attributes
Given a set of attributes {A1, …, An} and a set of dependencies S find all attributes B such that any relation which satisfies S also satisfies: A1, …, An B The closure of {A1, …, An}, denoted {A1, …, An}+ , is the set of all such attributes B Usage Test if X= {A1, …, An} is a superkey compute X+, and check if X+ contains all attributes of R Check if X Y holds by checking if Y is contained in X+
32
Algorithm Start with X={A1, …, An} Repeat until X doesn’t change do:
if is in S, and are all in X, and C is not in X then add C to X B , B , … B B , B , … B n C 1 2 n 1 2
33
For Relation R(A,B,C,D,E,F) with FD’s
A D E B D A F B What is the closure of {A,B} and {A,F}: Closure of {A,B}: X = {A, B, C, D, E} Closure of {A, F}: X = {A, F, B, D, C, E}
34
Example R = {A, B, C, D, E} F = { B CD, D E, B A, E C, AD B }
Is B E in F+ ? B+ = B B+ = BCD B+ = BCDA B+ = BCDAE … Yes! and B is a key for R too! Is D a key for R? D+ = D D+ = DE D+ = DEC … Nope! Is AD a key for R? AD+ = AD AD+ = ABD and B is a key, so Yes! And… A+ = A, D+ = DEC … A,D not keys, so Yes! Is ADE a candidate key for R? … No! AD is a key, so ADE is a superkey, but not a candidate key What is F+
35
Why do we study F.D.s and Keys Together?
Keys are (special) F.D.s FSUID Name, Age, Department “Good” F.D.s (like keys) and “bad” F.D.s Key (good) FSUID Name, Department, Chair F.D. but not key (bad) – make redundancy Department Chair FSUID Name Age Department FSUID Name Department Chair A01 Alex CS Xin B03 Bob C05 Chris EE Edwin
36
Normal Forms Desirable Properties of Schema Refinement
minimize redundancy avoid info loss preserve dependency ensure good query performance Convert to better design Explore problem Decomposition Convert to normal forms Normal Forms first normal form = all attributes are atomic second normal form (2NF) = old and obsolete Boyce Codd normal form (BCNF) third normal form (3NF) others……
37
Boyce-Codd Normal Form
A simple condition for removing anomalies from relations: A relation R is in BCNF if and only if: Whenever there is a nontrivial FD A1, A2, …, An B for R , it is the case that {A1, A2, …, An} is a super-key for R In English (though a bit vague): Whenever a set of attributes of R is determining another attribute, it should determine all attributes of R
38
Example What are the dependencies? SSN Name (non-key f.d.)
Phone Fred (201) (206) Joe (908) (212) What are the dependencies? SSN Name (non-key f.d.) What are the keys? *Phone Is it in BCNF? *No, SSN does not determine Phone.
39
Decompose It Into BCNF SSN Name: SSN is key No F.D. at all !
Fred Joe No F.D. at all ! SSN Phone (201) (206) (908) (212) Question: How can I find the phone information of Joe?
40
What about this? Name Price, Category Name Price Category Gizmo $20
Gadgets OneClick $25 Camera Name Price, Category
41
BCNF Decomposition Decompose: Continue until there are no
Find a functional dependency that violates the BCNF condition A1, A2, …, An B1, B2, …, Bn Heuristics: choose B1 , B2 , … Bn “as large as possible” Decompose: Continue until there are no BCNF violations left. B’s A’s Others R1 R2
42
Example Decomposition
SSN Name Age EyeColor PhoneNumber SSN Name, Age, EyeColor BCNF decomposition: Person (SSN, Name, Age, EyeColor) Phone-Info (SSN, PhoneNumber) What if we also had an attribute Draft-worthy, and the FD: Age Draft-worthy
43
BCNF Decomposition: The Algorithm
Input: relation R, set S of FDs over R 1) Compute S+ 2) Compute keys for R (from ER or from S+) 3) Use S+ and keys to check if R is in BCNF, if not: a) pick a violating FD f: A B b) expand B as much as possible, by computing A+ c) create R1 = A union B, R2 = A union (others in R) d) compute all FDs over R1, using R and S+, then compute keys for R1. Repeat similarly for R2 e) Repeat Step 3 for R1 and R2 4) Stop when all relations are BCNF, or are two-attributes
44
Properties of BCNF BCNF removes certain types of redundancy
those caused by adding many-many or one-many relations For examples of redundancy that it cannot remove, see "multivalued redundancy" BCNF avoids information loss the decomposition is dependency preserving Questions Is BCNF unique? Does BCNF always exist? No Example: R(A,B,C) F = {AB C, C B}
45
BCNF Decomposition Not Unique
Suppose several dependencies violate BCNF. Depending on which is used to guide the next decomposition, might get different collections of decomposed relations Normalization theory does not help us decide among alternatives: designer must consider alternatives and choose based on semantics of the application
46
Lossless Decompositions
A decomposition is lossless if we can recover: R(A,B,C) { R1(A,B) , R2(A,C) } R’(A,B,C) = R(A,B,C) Decompose Recover R’ is in general larger than R. Must ensure R’ = R
47
Lossy Decomposition: An Example
B C 1 2 3 4 5 6 7 8 FD’s: A B; C B Decompose A B 1 2 4 5 7 B C 2 3 5 6 8 Recover A B C 1 2 3 4 5 6 7 8
48
Lossless Decomposition: An Example
B C 1 2 3 4 5 6 7 8 FD’s: A B; C B Decompose A C 1 3 4 6 7 8 B C 2 3 5 6 8 Recover A B C 1 2 3 4 5 6 7 8 But, now we can’t check A B without doing a join!
49
Preserving Dependencies
What if, when a relation is decomposed, X of an XY ends up only in one of the new relations and Y ends up only in another? Such a decomposition is not “dependency-preserving” Goal: Always have FD-preserving decompositions BCNF is not always dependency preserving In fact, some times we cannot find a BCNF decomposition that is dependency preserving Can handle this situation using 3NF
50
Example FD’s: Unit Company; Company, Product Unit
So, there is a BCNF violation, and we decompose Unit Company Unit Company No FD’s Unit Product
51
So, What’s the Problem? Unit Company SQL Server Microsoft ACCESS Unit Product SQL Server Database ACCESS No problem so far. All local FD’s are satisfied. Let’s put all the data back into a single table again: Unit Company Product SQL Server Microsoft Database ACCESS Violates the dependency: company, product unit!
52
Solution: 3rd Normal Form (3NF)
A simple condition for removing anomalies from relations A relation R is in 3rd normal form if : Whenever there is a nontrivial dependency A1, A2, ..., An B for R , then {A1, A2, ..., An } is a super-key for R, or B is part of a key.
53
3NF (General Definition)
A relation is in Third Normal Form (3NF) if whenever XA holds, either X is a superkey, or A is a prime attribute (part of a candidate key) Informally: everything depends on the key or is in a key Despite the thorny technical definitions that lead up to it, 3NF is intuitive and not hard to achieve Aim for it in all designs unless you have strong reasons otherwise
54
3NF vs. BCNF R is in BCNF if whenever XA holds, then X is a superkey
Slightly stronger than 3NF Example: R(A,B,C) with {A,B}C, CA 3NF but not BCNF Guideline: Aim for BCNF and settle for 3NF
55
Decomposing R into 3NF Get a “minimal cover” G 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 2. Find a lossless-join decomposition of R (which might miss dependencies) BCNF decomposition 3. Add additional relations to the decomposition to cover any missing FDs of the cover Result will be lossless and dependency-preserving
56
Example Answer: R is not in 3NF: 3NF: R1 = (banker, bname, office)
R = (bname, cname, banker, office) FD’s = {banker bname office, cname bname banker} Q1: candidate keys of R {cname, bname} or {cname, banker} Q2: Is R in 3NF? If not, decompose R into 3NF Answer: R is not in 3NF: banker bname office, {bname, office} not a subset of a candidate key 3NF: R1 = (banker, bname, office) R2 = (cname, bname, banker)
57
Confused by Normal Forms ?
BCNF 4NF In practice: (1) 3NF is enough, (2) don’t overdo it !
58
Normalization Summary
1NF: usually part of the woodwork 2NF: usually skipped 3NF: a biggie always aim for this BCNF and 4NF: tradeoffs start here in re: dependency preserving and losslessness 5NF: You can say you've heard of it...
59
Caveat Normalization is not the be-all and end-all of DB design
Example: suppose attributes A and B are always used together, but normalization theory says they should be in different tables. decomposition might produce unacceptable performance loss (extra disk reads) Plus -- there are constraints other than FDs and MVDs
60
Current Trends Object DBs and Object-Relational DB’s
may permit complex attributes 1st normal form unnecessary Data Warehouses huge historical databases, seldom or never updated after creation joins expensive or impractical argues against normalization Everyday relational DBs aim for BCNF, settle for 3NF
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.