FUNCTIONAL DEPENDENCIES
Chapter Outline 1 Informal Design Guidelines for Relational Databases 1.1Semantics of the Relation Attributes 1.2 Redundant Information in Tuples and Update Anomalies 1.3 Null Values in Tuples 1.4 Spurious Tuples 2 Functional Dependencies (FDs) 2.1 Definition of FD 2.2 Inference Rules for FDs 2.3 Equivalence of Sets of FDs 2.4 Minimal Sets of FDs
Chapter Outline(contd.) 3 Normal Forms Based on Primary Keys 3.1Normalization of Relations 3.2 Practical Use of Normal Forms 3.3 Definitions of Keys and Attributes Participating in Keys 3.4First Normal Form 3.5Second Normal Form 3.6Third Normal Form 4 General Normal Form Definitions (For Multiple Keys) 5 BCNF (Boyce-Codd Normal Form)
1 Informal Design Guidelines for Relational Databases (1) What is relational database design? The grouping of attributes to form "good" relation schemas Two levels of relation schemas The logical "user view" level The storage "base relation" level Design is concerned mainly with base relations What are the criteria for "good" base relations?
1.1 Semantics of the Relation Attributes GUIDELINE 1: Informally, each tuple in a relation should represent one entity or relationship instance. Attributes of different entities (EMPLOYEEs, DEPARTMENTs, PROJECTs) should not be mixed in the same relation Only foreign keys should be used to refer to other entities Entity and relationship attributes should be kept apart as much as possible. Bottom Line: Design a schema that can be explained easily relation by relation. The semantics of attributes should be easy to interpret.
Figure 10.1 A simplified COMPANY relational database schema Note: The above figure is now called Figure 10.1 in Edition 4
1.2 Redundant Information in Tuples and Update Anomalies Mixing attributes of multiple entities may cause problems Information is stored redundantly wasting storage Problems with update anomalies Insertion anomalies Deletion anomalies Modification anomalies
EXAMPLE OF AN UPDATE ANOMALY (1) Consider the relation: EMP_PROJ ( Emp#, Proj#, Ename, Pname, No_hours) Update Anomaly: Changing the name of project number P1 from “Billing” to “Customer-Accounting” may cause this update to be made for all 100 employees working on project P1.
EXAMPLE OF AN UPDATE ANOMALY (2) Insert Anomaly: Cannot insert a project unless an employee is assigned to. Inversely - Cannot insert an employee unless an he/she is assigned to a project. Delete Anomaly: When a project is deleted, it will result in deleting all the employees who work on that project. Alternately, if an employee is the sole employee on a project, deleting that employee would result in deleting the corresponding project.
Figure 10.3 Two relation schemas suffering from update anomalies
Figure 10.4 Example States for EMP_DEPT and EMP_PROJ Note: The above figure is now called Figure 10.4 in Edition 4
Guideline to Redundant Information in Tuples and Update Anomalies GUIDELINE 2: Design a schema that does not suffer from the insertion, deletion and update anomalies. If there are any present, then note them so that applications can be made to take them into account
1.3 Null Values in Tuples GUIDELINE 3: Relations should be designed such that their tuples will have as few NULL values as possible Attributes that are NULL frequently could be placed in separate relations (with the primary key) Reasons for nulls: attribute not applicable or invalid attribute value unknown (may exist) value known to exist, but unavailable
1.4 Spurious Tuples Bad designs for a relational database may result in erroneous results for certain JOIN operations The "lossless join" property is used to guarantee meaningful results for join operations
GUIDELINE 4: The relations should be designed to satisfy the lossless join condition. No spurious tuples should be generated by doing a natural-join of any relations.
Consider EMP_PROJ relation of Case II, and its relation state
Let us split EMP_PROJ into 2 relation EMP_PROJ1 and EMP_LOCS as shown below. A tuple in EMP_LOCS means that the employee whose name is ENAME works on some project whose location is PLOCATION A tuple in EMP_PROJ1 means that the employee whose social security number is SSN works HOURS per week on the project whose name, number, and location are PNAME, PNUMBER, and PLOCATION
Now produce tuples for EMP_LOCS by applying project operation to EMP_PROJ relation state, So we get tuples as shown below Tuples for EMP_LOCS obtained by applying Project operation to EMP_PROJ
Now produce tuples for EMP_PROJ1 by applying project operation to EMP_PROJ relation state, So we get tuples as shown below Tuples for EMP_PROJ1 obtained by applying Project operation to EMP_PROJ
Now apply natural join operation (*) on EMP_PROJ1 and EMP_LOCS to get the original relation EMP_PROJ. EMP_PROJ1 * EMP_LOCS
Additional tuples that were not in EMP_PROJ are called spurious tuples because they represent spurious or wrong information that is not valid. The spurious tuples are marked by asterisks (*). Decomposing EMP_PROJ into EMP_LOCS and EMP_PROJ ! is undesirable because, when we JOIN them back using NATURAL JOIN, we do not get the correct original information. This is because PLOCATION is neither a primary key nor a foreign key in either EMP_LOCS or EMP_PROJ1
Spurious Tuples (2) There are two important properties of decompositions: (a) non-additive or losslessness of the corresponding join (b) preservation of the functional dependencies. Note :property (a) is extremely important and cannot be sacrificed. Property (b) is less stringent and may be sacrificed.
Functional Dependencies
2.1 Functional Dependencies (1) Functional dependencies (FDs) are used to specify formal measures of the "goodness" of relational designs FDs and keys are used to define normal forms for relations FDs are constraints that are derived from the meaning and interrelationships of the data attributes X->Y : A set of attributes X functionally determines a set of attributes Y if the value of X determines a unique value for Y.
Functional Dependencies (2) X -> Y holds if whenever two tuples have the same value for X, they must have the same value for Y For any two tuples t1 and t2 in any relation instance r(R): If t1[X]=t2[X], then t1[Y]=t2[Y] X -> Y in R specifies a constraint on all relation instances r(R) Written as X -> Y; can be displayed graphically on a relation schema as in Figures. ( denoted by the arrow->). FDs are derived from the real-world constraints on the attributes
Examples of FD constraints (1) social security number determines employee name SSN -> ENAME project number determines project name and location PNUMBER -> {PNAME, PLOCATION} employee ssn and project number determines the hours per week that the employee works on the project {SSN, PNUMBER} -> HOURS
From the semantics of the attributes, we know that the following functional dependencies should hold: FD1 : {SSN, PNUMBER} HOURS FD2 : SSN ENAME FD3 : PNUMBER {PNAME, PLOCATION} DIAGRAMMATIC NOTATION FOR DISPLAYING FDs
A FD cannot be inferred automatically from a given relation state r; but must be defined explicitly by someone who knows the semantics of the attributes of R. Example: TEXT(Author) COURSE(Subject). Is it true for all legal states of TEACH? YES But we can definitely say TEACHER COURSE is not true. FDs – NOT A PROPERTY OF A RELATION STATE
Important…. An FD is a property of the attributes in the schema R The constraint must hold on every relation instance r(R) If K is a key of R, then K functionally determines all attributes in R (since we never have two distinct tuples with t1[K]=t2[K])
2.2 Inference Rules for FDs (1) Given a set of FDs F, we can infer additional FDs that hold whenever the FDs in F hold Armstrong's inference rules: IR1. (Reflexive) If Y subset-of X, then X -> Y IR2. (Augmentation) If X -> Y, then XZ -> YZ (Notation: XZ stands for X U Z or {X, Z}) IR3. (Transitive) If X -> Y and Y -> Z, then X -> Z IR1, IR2, IR3 form a sound and complete set of inference rules
Inference Rules for FDs (2) Some additional inference rules that are useful: (Decomposition) If X -> YZ, then X -> Y and X -> Z (Union) If X -> Y and X -> Z, then X -> YZ (Psuedotransitivity) If X -> Y and WY -> Z, then WX -> Z The last three inference rules, as well as any other inference rules, can be deduced from IR1, IR2, and IR3 (completeness property)
GivenR = (A, B, C, G, H, I)F = { A B A C CG H CG I B H } Based on Inference Rules, some members of F + would be A H (transitivity from A B and B H) AG I (augmenting A C with G, to get AG CG and then transitivity with CG I ) CG HI (addingCG I and CG H to infer CG HI) INFERENCE RULES - EXAMPLE
Inference Rules for FDs (3) Closure of a set F of FDs is the set F + of all FDs that can be inferred from F Closure of a set of attributes X with respect to F is the set X + of all attributes that are functionally determined by X X + can be calculated by repeatedly applying IR1, IR2, IR3 using the FDs in F
2.3 Equivalence of Sets of FDs Two sets of FDs F and G are equivalent if: - every FD in F can be inferred from G, and - every FD in G can be inferred from F Hence, F and G are equivalent if F + =G + Definition: F covers G if every FD in G can be inferred from F (i.e., if G + subset-of F + ) F and G are equivalent if F covers G and G covers F
1. Consider the following two sets of functional dependencies: F = {A C, AC D, E AD, E H} G = {A CD, E AH}. Check whether they are equivalent. 2. Consider two sets of FDs, F and G, F = {A B, B C, AC D} and G = {A B, B C, A D} Are F and G equivalent? 3. F = {A B, A C} G = {A B, B C} Are F and G equivalent? PRACTICE - EQUIVALENCE OF SETS OF FUNCTIONAL DEPENDENCIES
Closure of a Set of Functional Dependencies For a set F of functional dependencies, we call the closure of F, noted F+, the set of all the functional dependencies that can be derived from F (by the application of Armstrong’s axioms). Intuitively, F+ is equivalent to F, but it contains some additional FDs that are only implicit in F. Consider the relation scheme R(A,B,C,D) with F = {{A} {B},{B,C} {D}} F+ = { {A} {A}, {B} {B}, {C} {C}, {D} {D}, {A,B} {A,B}, […], {A} {B}, {A,B} {B}, {A,D} {B,D}, {A,C} {B,C}, {A,C,D} {B,C,D}, {A} {A,B}, {A,D} {A,B,D}, {A,C} {A,B,C}, {A,C,D} {A,B,C,D}, {B,C} {D}, […], {A,C} {D}, […]}
Algorithm for Computing the Closure of a Set of Attributes Input: R a relation scheme F a set of functional dependencies X R (the set of attributes for which we want to compute the closure) Output: X+ the closure of X w.r.t. F X (0) := X Repeat X (i+1) := X (i) Z, where Z is the set of attributes such that there exists Y Z in F, and Y X (i) Until X (i+1) := X (i) Return X (i+1)
Closure of a Set of Attributes: Example R = {A,B,C,D,E,G} F = { {A,B} {C}, {C} {A}, {B,C} {D}, {A,C,D} {B}, {D} {E,G}, {B,E} {C}, {C,G} {B,D}, {C,E} {A,G}} X = {B,D} Find X + X (0) = {B,D} {D} {E,G}, X (1) = {B,D,E,G}, {B,E} {C} X (2) = {B,C,D,E,G}, {C,E} {A,G} X (3) = {A,B,C,D,E,G} X (4) = X (3)
Uses of Attribute Closure There are several uses of the attribute closure algorithm: Testing for superkey To test if X is a superkey, we compute X +, and check if X + contains all attributes of R. X is a candidate key if none of its subsets is a key. Testing functional dependencies To check if a functional dependency X Y holds (or, in other words, is in F + ), just check if Y X +. Computing the closure of F For each subset X R, we find the closure X +, and for each Y X +, we output a functional dependency X Y. Computing if two sets of functional dependencies F and G are equivalent, i.e., F+ = G+ For each functional dependency Y Z in F Compute Y+ with respect to G If Z Y+ then Y Z is in G+ And vice versa
Redundancy of FDs Sets of functional dependencies may have redundant dependencies that can be inferred from the others {A} {C} is redundant in: {{A} {B}, {B} {C},{A} {C}} Parts of a functional dependency may be redundant Example of extraneous/redundant attribute on RHS: {{A} {B}, {B} {C}, {A} {C,D}} can be simplified to {{A} {B}, {B} {C}, {A} {D}} (because {A} {C} is inferred from {A} {B}, {B} {C}) Example of extraneous/redundant attribute on LHS: {{A} {B}, {B} {C}, {A,C} {D}} can be simplified to {{A} {B}, {B} {C}, {A} {D}} (because of {A} {C})
2.4 Minimal Sets of FDs (1) A set of FDs is minimal if it satisfies the following conditions: (1) Every dependency in F has a single attribute for its RHS. (2) We cannot remove any dependency from F and have a set of dependencies that is equivalent to F. (3) We cannot replace any dependency X -> A in F with a dependency Y -> A, where Y proper-subset-of X ( Y subset-of X) and still have a set of dependencies that is equivalent to F.
Minimal Sets of FDs (2) Every set of FDs has an equivalent minimal set There can be several equivalent minimal sets There is no simple algorithm for computing a minimal set of FDs that is equivalent to a set F of FDs
Canonical Cover A canonical cover for F is a set of dependencies F c such that F and F c, are equivalent F c contains no redundancy Each left side of functional dependency in F c is unique. For instance, if we have two FD X Y, X Z, we convert them to X Y Z. Algorithm for canonical cover of F: repeat Use the union rule to replace any dependencies in F X 1 Y 1 and X 1 Y 2 with X 1 Y 1 Y 2 Find a functional dependency X Y with an extraneous attribute either in X or in Y If an extraneous attribute is found, delete it from X Y 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
Example of 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 because of B C. Set is now {A BC, B C} C is extraneous in A BC because of A B and B C. The canonical cover is: A B B C
Keys and FDs
A FD is a generalization of the notion of a key. For Student (sid, name, supervisor_id, specialization), we write: {sid} {name, supervisor_id, specialization} The sid determines all attributes (i.e., the entire record) If two tuples in the relation student have the same sid, then they must have the same values on all attributes. In other words they must be the same tuple (since the relational model does not allow duplicate records)
Superkeys and Candidate Keys A set of attributes that determine the entire tuple is a superkey {sid, name} is a superkey for the student table. Also {sid, name, supervisor_id} etc. A minimal set of attributes that determines the entire tuple is a candidate key {sid, name} is not a candidate key because I can remove the name. sid is a candidate key If there are multiple candidate keys, the DB designer chooses designates one as the primary key.
Finding Keys Example: Consider the relation scheme R(A,B,C,D) with functional dependencies {A} {C} and {B} {D}. Is {A,B} a candidate key? For {A,B} to be a candidate key, it must determine all attributes (i.e., be a superkey) be minimal {A,B} is a superkey because: {A} {C} {A,B} {A,B,C} (augmentation by AB) {B} {D} {A,B,C} {A,B,C,D} (augmentation by A,B,C) We obtain {A,B} {A,B,C,D} (transitivity) {A,B} is minimal because neither {A} nor {B} alone are candidate keys
Pitfalls in Relational Database Design Functional dependencies can be used to refine ER diagrams or independently (i.e., by performing repetitive decompositions on a "universal" relation that contains all attributes). Relational database design requires that we find a “good” collection of relation schemas. A bad design may lead to Repetition of Information. Inability to represent certain information. Design Goals: Avoid redundant data Ensure that relationships among attributes are represented Facilitate the checking of updates for violation of database integrity constraints.
Example of Bad Design Consider the relation schema: Lending-schema = (branch-name, branch-city, assets, customer- name, loan-number, amount) where: {branch-name} {branch-city, assets} Bad Design Wastes space. Data for branch-name, branch-city, assets are repeated for each loan that a branch makes Complicates updating, introducing possibility of inconsistency of assets value Difficult to store information about a branch if no loans exist. Can use null values, but they are difficult to handle.
Usefulness of FDs Use functional dependencies to 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 {R 1, R 2,..., R n } such that each relation is in good form the decomposition is a lossless-join decomposition if possible, preserve dependencies In our example the problem occurs because there FDs ({branch-name} {branch-city, assets}) where the LHS is not a key Solution: decompose the relation schema Lending-schema into: Branch-schema = (branch-name, branch-city,assets) Loan-info-schema = (customer-name, loan-number, branch-name, amount)