1. © D. Wong 2003 1 Functional Dependencies (FD)  Given: relation schema R(A1, …, An), and X and Y be subsets of (A1, … An). FD : X  Y means X functionally determines Y e.g. A 1 A 2 …A n  B 1 B 2 …B m  Significance: –Key: 1.Entity set R (A1, …, An), and X is a subset of A1, …, An that form a key for R, then may assert: X  Y where Y is any subset of {A1, …, An} 2.R(A1, …, An) is a many-one relationship from entity set E1 to entity set E2, and among the Ai’s are attributes that form a key X for E1 and a key Y for E2, then may assert: X  Y

2. © D. Wong 2003 2 Keys of Relations K is a key for relation R if: 1.K  all attributes of R 2.For no proper subset of K is (1) true (i.e. a key must be minimal) 3.If K at least satisfies (1), then K is a superkey Example: {title, year, starName) forms a key for the Movie relation Superkeys: A set of attributes that contains a key is called a superkey

3. © D. Wong 2003 3 FD Rules  Splitting / Combining rule  Trivial Dependencies  Armstrong’s Axioms: 1.Reflexivity 2.Augmentatioin 3.Transitivity  Computing closure of attributes  Finding all implied FD’s ( Ref. section: 3.5.7)

4. © D. Wong 2003 4 Given vs. Derived FD’s  Given FD’s: stated initially for a relation. FD’s These are known to hold for a relation R.  Derived FD’s are FD’s that follow logically from the given FD’s (using the inference rules) or by the closure algorithm of attributes.  Basis : any set of given FD’s from which all FD’s for a relation can be inferred.  Minimal basis: if no proper subset of the FD’s in a basis can also derive the complete set of FD’s.  Example: 3.22 – pp. 98

5. © D. Wong 2003 5 Closure of attributes  Let {A 1, A 2, …,A n } is a set of attributes, S is a set of FD’s  Closure of {A 1, A 2, …,A n } under the dependencies in S = set of attributes B such that every relation that satisfies all the dependencies in set S also satisfies A 1 A 2 …A n  B. I.e. B is the set of attributes functionally determined by {A 1, A 2, …,A n }  Notation: {A 1, A 2, …,A n } +  Allow trivial dependencies, so A 1 A 2 …A n are always in {A 1, A 2, …,A n } +

6. © D. Wong 2003 6 Closure Algorithm Let X be {A 1, A 2, …,A n } + 1.X = {A 1, A 2, …,A n }// initialize X 2.Search for some FD B 1 B 2 …B m  C where B 1 B 2 …B m  X, but C is not in X. If found, add C to X 3.Repeat step 2 until no more attributes can be added to X. 4.The resulting X is the correct value of {A 1, A 2, …,A n } +

7. © D. Wong 2003 7 Attributes Closure Example. 3.19 pp. 93 Given: R(A, B, C, D, E, F) and FD’s AB  C BC  AD D  E CF  B To find: {A, B} + Procedure: 1.Let X = {A, B} 2.Add C to X // AB  C,  X = {A, B, C} 3.Add D to X// BC  AD,  X = {A, B, C, D} 4.Add E to X// D  E,  X = {A, B, C, D, E} 5.End, {A, B} + = {A, B, C, D, E}

8. © D. Wong 2003 8 To infer if a FD follows a set of FD in a relation  AB  D follows from dependencies in example 3.19? –Compute {A, B} +, if D ends up in the closure, then AB  D follows from the dependencies –Since {A, B} + = {A, B, C, D, E},  AB  D follows.  D  A follows from dependencies in example 3.19? –Compute {D} +, if A ends up in the closure, then D  A follows from the dependencies –But {D} + = {D, E},  D  A does NOT follows.

9. © D. Wong 2003 9 Attributes Closures and Keys To test if A 1,A 2,…,A n is a key for a relation R:  Compute {A 1, A 2, …,A n } + 1.Check if {A 1, A 2, …,A n } + is the set of all attributes of R 2.If yes, then check no subset of A 1,A 2,…,A n, say S, such that S + is the set of all attributes of R  If A 1,A 2,…,A n only satisfies 1 but not 2, then A 1,A 2,…,A n is a superkey

10. © D. Wong 2003 10 Finding all implied FD’s  Motivation: Given relation R with FDs F. When projecting R to form new relation S, want to know what FD’s hold in S  Method: compute all FD’s that 1.Follow from F 2.Involve only attributes of S  Problem: no. of FDs may be large (could be exponential in the number of attributes of S)  So, make simplifications in computing attribute closure: –Ignore empty set and set of all attributes (trivial FD’s) –Drop XY  A if X  A holds. –If closure of some set X is all attributes, then ignore supersets of X.

11. © D. Wong 2003 11 Algorithm to find all implied FD’s 1. For each set of attributes X compute X +. Start with closures of singleton set, and then move onto doubleton if necessary. 2. Add X  A for each A in X + - X 3. Ignore “obvious” dependencies that follow from others as described in the simplification guidelines.

12. © D. Wong 2003 12 Finding all implied FD’s Example  Given: R(ABCD), and FD’s F: AB  C, C  D, D  A  What FD’s follow from F? A + = A; B + = B // nothingA + = A; B + = B // nothing C + = ACD // add C  AC + = ACD // add C  A D + = DA// nothing newD + = DA// nothing new (AB) + = ABCD// add AB  D; skip all supersets of AB(AB) + = ABCD// add AB  D; skip all supersets of AB (BC) + = ABCD// nothing new; skip all supersets of BC(BC) + = ABCD// nothing new; skip all supersets of BC (BD) + = ABCD// add BD  C; skip all supersets of BD(BD) + = ABCD// add BD  C; skip all supersets of BD (AC) + = ACD ; (AD) + = (AD); (CD) + = ACD // nothing new(AC) + = ACD ; (AD) + = (AD); (CD) + = ACD // nothing new (ACD)+ = ACD// nothing new;(ACD)+ = ACD// nothing new; All other sets contain AB, BC, or BD, so skipAll other sets contain AB, BC, or BD, so skip  the only interesting FD’s that follow from F are :  the only interesting FD’s that follow from F are : C  A, AB  D, BD  C

13. © D. Wong 2003 13 Normalization  Purpose: process to eliminate redundancy in relations due to functional or multi-valued dependencies.  Normal forms: –Boyce-Codd Normal Form (BCNF) –Third Normal Form (3NF) –Fourth Normal Form (4NF)

14. © D. Wong 2003 14 Boyce-Codd Normal Form (BCNF)  Definition: A relation R is in BCNF iff whenever there is a nontrivial dependency A 1 A 2 …A n  B holds for R, {A 1, A 2, …,A n } must be a superkey for R  Alternative definition after applying the combining rule to all FD’s with a common L.S.: A relation R is in BCNF iff whenever nontrivial dependency A 1 A 2 …A n  B 1 B 2 …B m holds for R, {A 1, A 2, …,A n } must be a superkey for R  Guarantees no redundancy, prevents update, deletion, and insertion anomalies  E.g. title year  length filmType studioName // violates BCNF because key of Movie is {title, year, starName}, so Movie is not in BCNF

15. © D. Wong 2003 15 Decomposition to BCNF Given: relation R, FD’s F. Decomposition strategy: 1.look for a non-trivial FD A 1 A 2 …A n  B 1 B 2 …B m that violates BCNF. (Heuristic: add to the RS as many attributes as are functionally determined by A 1 A 2 …A n ) 2.Compute {A 1, A 2, …,A n } + // all the attributes involved in the violating dependency 3.Decompose R into 1. 1. {A 1, A 2, …,A n } +, and 2. 2. (R - {A 1, A 2, …,A n } + )  {A 1, A 2, …,A n } // LS + all attributes not involved in the violating dependency Others A’s B’s

16. © D. Wong 2003 16 BCNF Decomposition Example 3.24 pp 104  Relation: Movie(title, year, length, filmType, studioName, starName)  Key: {title, year, starName}  FD’s: title year  length filmType studioName is a BCNF violation, so Movie not in BCNF  Decomposition: Schema 1: {title, year, length, filmType, studioName} Schema 2: {title, year, starName}  To obtain the new relations, project the schemas onto Movie  To recover information (I.e. Movie) from the new relations: natural join the new relations. Does not lose information.

17. © D. Wong 2003 17 Additional points about BCNF  A relation may have more than 1 keys  Need some key in the LHS of any nontrivial FD  Only nontrivial FD’s are potential BCNF violation candidates  Any 2-attributes relations is in BCNF  Decomposition must be based on FD that holds in the relation, otherwise can’t recover the original relation from the new relations.  Eliminates all redundancies  Some decomposition may not preserve the FD’s (e.g. 3.32 pp 114)

18. © D. Wong 2003 18 Third Normal Form (3NF)  Definition: A relation R is in 3NF if: whenever there is a nontrivial dependency A 1 A 2 …A n  B holds for R, either {A 1, A 2, …,A n } is a superkey for R, or B is a member of some key (i.e. B is prime).  BCNF with a relaxed condition  Some redundancy might be left in 3NF if resulting relations are not in BCNF  Preserves all the FD’s  Process similar to BCNF with the added condition  To recover information (I.e. Movie) from the new relations: natural join the new relations. Does not lose information.