1. Closure The closure of {B 1 …B k } under the set of FDs S, denoted by {B 1 …B k } +, is defined as follows: {B 1 …B k } + = {B | any relation satisfies S will also satisfies B 1 …B k  B}

2. Computing the closure Given: the set S and {A 1,…,A n } Compute: {A 1,…,A n } + - denote this set by X Step 1: X = {A 1,…,A n } Step 2: find a FD B 1 …B k  B in S such that {B 1,…,B k }  X and B  X, then X=X  {B} Step 3: repeat step 2 until nothing more can be added to X, then go to step 4 Step 4: return X

3. Example S = {AB  C, BC  AD, D  E, CF  B} Compute {A,B} + Step 1: X = {A,B} Step 2: X = X  {C}={A,B,C} because AB  C Step 3 back to step 2: X = X  {D} because BC  AD Step 3 back to step 2: X = X  {E} because D  E Step 3 back to step 2: nothing more Step 3 go to step 4: return {A,B,C,D,E}

4. Correctness of closure algorithm It computes true functional dependencies –proof: show that if B belongs to {A 1,…,A n } + then A 1 …A n  B holds. By induction over the number of steps (n) used in adding an attribute B into the set X n=0 then B belongs to {A 1,…,A n } and so A 1 …A n  B is a trivial functional dependency n  n+1: if B is added to X in the step n+1, then A 1 …A n  B j for all j by inductive hypothesis; this, together with B 1 …B k  B, implies that A 1 …A n  B It computes all functional dependencies –proof: show that if B does not belong to {A 1,…,A n } + then A 1 …A n  B does not hold. By constructing an instance I of the relation R such that the FD does not hold. A in the closure Others 111 … 111 000 … 000 111 … 111

5. Simple questions What is {A 1,…,A n } + if {A 1,…,A n } is a key of the relation? Can {A 1,…,A n } + ={A 1,…,A n }? Does {B 1,…,B m }  {A 1,…,A n } imply {B 1,…,B m } +  {A 1,…,A n } + ?

6. Transitive Rules GivenA 1 …A n  B 1 …B m B 1 …B m  C 1 …C k then A 1 …A n  C 1 …C k

7. Closing sets of FDs Given a set of FDs we can derive some other FDs using the rules about FDs (e.g. combining, splitting, and transitive) For a relation R, a set of FD is called a basis for R if all other FDs of R can be derived form it. A basis is minimal if none of its proper subsets is a basis.

8. Projecting FDs B’s R S Given: –R with a set of FDs F –S (a new relation) is obtained by removing the attributes {B 1,…,B m } from R Questions: What are the FDs of S? Answer: if A 1 …A n  C 1 …C k is a FD of R and none of the Bs appears on the left or right side ({B 1,…,B m }  {A 1,…,A n,C 1,…,C k }=  ) is a FD of S

9. Projecting - Example Given R(A,B,C,D) with the FDs A  B, B  C, and C  D. Remove the attribute B from R, we obtain a new relation S(A,C,D). What are the FDs of S? –A  C? –A  D? –C  D? We can compute this by: Compute all the closure of every subset of {A,C,D} by using the FDs of R that do not contain B.

10. Homework 3.5.1 Consider a relation with schema R(A,B,C,D) and FD’s AB  C, C  D, and D  A. –What are all the nontrivial FD’s that follow from the given FD’s? List only the FDs with one attribute on the right? (5pt) –What are the keys of R? (5pt) –What are the superkeys but not keys? (5pt) 3.5.3 Show that the following rule holds: (5pt) if A 1 …A n  B 1 …B m and C 1 …C k  D 1 …D t hold then A 1 …A n C 1 …C k  B 1 …B m D 1 …D t also holds.

11. For those whole like fun: 3.5.4 Does the following hold: –if A  B then B  A –if AB  C and A  C then B  C 3.5.8 A set of attributes is closed if X + =X. What are the FDs of a relation R(A,B,C,D) if –all sets of four attributes are closed –the only closed sets are {} and {A,B,C,D} –the closed sets are {}, {A,B}, {A,B,C,D} (note: the cases are considered separate) Stars: try the exercises with stars.

12. Design of Relational Database Schema titleyearlengthstudioNamestarNamefilmType Star Wars1977124FoxMark Hamillcolor Star Wars1977124FoxHarrison Fordcolor Star Wars1977124FoxCarrie Fishercolor Mighty Ducks1991104DisneyEmilio Estevezcolor Wayne’s World199295ParamountDana Carveycolor Wayne’s World199295ParamountMike Meyerscolor Some observations: value of studioName is the same in several tuples value of filmType is also repeated What wrong with it? redundancy  store the same value unnecessary several time update anormalies  an update might require several changes deletion anormalies  losing information if delete a value CAN WE AVOID THESE ANORMALIES?

13. Possible ways to avoid anormalies (Intuition) The bad way: start again (Oh, no!) The natural way: try to decompose the given relation into two or more relations that –contain the same information –avoid the anormalies

14. Example titleyearlengthstudioNamestarNamefilmType Star Wars1977124FoxMark Hamillcolor Star Wars1977124FoxHarrison Fordcolor Star Wars1977124FoxCarrie Fishercolor Mighty Ducks1991104DisneyEmilio Estevezcolor Wayne’s World199295ParamountDana Carveycolor Wayne’s World199295ParamountMike Meyerscolor titleyearlengthstudioNamefilmType Star Wars1977124Foxcolor Mighty Ducks 1991104Disneycolor Wayne’s World 199295Paramountcolor titleyearstarName Star Wars1977Mark Hamill Star Wars1977Harrison Ford Star Wars1977Carrie Fisher Mighty Ducks1991Emilio Estevez Wayne’s World1992Dana Carvey Wayne’s World1992Mike Meyers MovieStudioStar(title, year, length, studioName, starName, filmType) is decomposed into 2 relations MovieStudio(title, year, length, studioName, filmType) and StarsIn(title, year, starName)

15. Decomposition Given a relation R with schema {A 1,…,A n }. A decomposition of R into two relations S and T with schemas {B 1,…,B m } and {C 1,…,C k }, respectively, such that 1.{A 1,…,A n } = {B 1,…,B m }  {C 1,…,C k } 2.The tuples in S are the projections onto {B 1,…,B m } of all the tuples in R. 3.The tuples in T are the projections onto {C 1,…,C k } of all the tuples in R.

16. Example – Projections titleyearlengthstudioNamestarNamefilmType Star Wars1977124FoxMark Hamillcolor Star Wars1977124FoxHarrison Fordcolor Star Wars1977124FoxCarrie Fishercolor Mighty Ducks1991104DisneyEmilio Estevezcolor Wayne’s World199295ParamountDana Carveycolor Wayne’s World199295ParamountMike Meyerscolor titleyearlengthstudioNamefilmType Star Wars1977124Foxcolor Mighty Ducks 1991104Disneycolor Wayne’s World 199295Paramountcolor titleyearstarName Star Wars1977Mark Hamill Star Wars1977Harrison Ford Star Wars1977Carrie Fisher Mighty Ducks1991Emilio Estevez Wayne’s World1992Dana Carvey Wayne’s World1992Mike Meyers MovieStudioStar(title, year, length, studioName, starName, filmType) is decomposed into 2 relations MovieStudio(title, year, length, studioName, filmType) and StarsIn(title, year, starName) How do we come up with this decomposition?

17. Boyce-Codd Normal Form (BCNF) BCNF: a relation R is in BCNF iff: whenever there is a nontrivial FD A 1 …A n  B for R, it is the case that {A 1,…,A n } is a superkey for R. Why this definition? Answer: if a relation is in BCNF then there is no anormaly. Example: MovieStudioStar(title, year, length, studioName, starName, filmType): not in BCNF MovieStudio(title, year, length, studioName, filmType): in BCNF StarsIn(title, year, starName): in BCNF

18. Decomposition into BCNF Suppose that we decompose a relation R into two relations S and T which are in BCNF. The requirements for S and T: –S and T is a decomposition of R –it is possible to reconstruct R from S and T Will every decomposition of R satisfy these two conditions? What are the FDs of the new relations?

19. Algorithm Given a relation R with the attributes {A 1,…,A n }. Step 1: For every nontrivial FD B 1 …B m  B if {B 1,…,B m } is a superkey then returns R (no decomposition is needed) Step 2: Takes a nontrivial FD B 1 …B m  B such that {B 1,…,B m } is not a superkey, then decomposes R into two relations S and T with the following schema: –S’s schema: {B 1,…,B m } + –T’s schema: {B 1,…,B m }  ({A 1,…,A n }\{B 1,…,B m } + ) Repeat Step 1&2 for S and T until no decomposition is needed for every new relation; return the set of new relations as the result

20. Example The ‘new’ movie relation with the following attributes: {title,year,studioName,president,presAddress} (we call this set ALL) with the FDs: {title year  studioName, studioName  president, president  presAddress} Only one key: {title,year} studioName  president violated BCNF Step 2: takes studioName  president, decomposes into –S with the schema {studioName} + ={studioName,president,presAddress} –T with the schema {studioName,title,year}={studioName}  (ALL\ {studioName} + ) Check: {studioName,title,year} is in BCNF (the first two FDs) {studioName,president,presAddress} is not in BCNF Continue with the decomposition of S using president  presAddress and we get the following two relation schemas: {president,presAddress} and {president,studioName} both are in BCNF. The final result: {studioName,title,year}, {president,presAddress},{president,studioName}

21. Recovering information from a decomposition Suppose that R with the schema {A 1,…,A n } is decomposed into two relations S and T according to the algorithm whose attributes are {B 1,…,B m } + and {B 1,…,B m }  ({A 1,…,A n }\{B 1,…,B m } + ) The tuples of R can be obtained by joining all possible pairs of S and T where {B 1,…,B m } have the same values.

22. Recovering … t (R) t’ (S) t’’ (T) the B’s the rest of the closure others Projection Join {B1,…,Bm} {A1,…,An}\{B1,…,Bm} + {B1,…,Bm} + \ {B1,…,Bm}

23. Example – Decomposition and Recovering titleyearlengthstudioNamestarNamefilmType Star Wars1977124FoxMark Hamillcolor Star Wars1977124FoxHarrison Fordcolor Star Wars1977124FoxCarrie Fishercolor Mighty Ducks1991104DisneyEmilio Estevezcolor Wayne’s World199295ParamountDana Carveycolor Wayne’s World199295ParamountMike Meyerscolor titleyearlengthstudioNamefilmType Star Wars1977124Foxcolor Mighty Ducks 1991104Disneycolor Wayne’s World 199295Paramountcolor titleyearstarName Star Wars1977Mark Hamill Star Wars1977Harrison Ford Star Wars1977Carrie Fisher Mighty Ducks1991Emilio Estevez Wayne’s World1992Dana Carvey Wayne’s World1992Mike Meyers MovieStudioStar(title, year, length, studioName, starName, filmType) is not in BCNF is decomposed into 2 relations that are in BCNF: MovieStudio(title, year, length, studioName, filmType) and StarsIn(title, year, starName)

24. Some remarks The algorithm will stop and output a set of BCNF relations. Not every decomposition according to the algorithm is good The FD’s for the new relations are determined by ‘projecting’. If a decomposition is based on FDs (according to the algorithm) then the recovering process will give us exactly the original relation. If a decomposition is not based on FDs then we might not be able to recover the original relation from the new ones: –Example: R(A,B,C) with A  B and we decompose it into S(A,B) and T(B,C): ABC 123 425 AB 12 42 BC 23 25 ABC 123 125 423 425

25. Third Normal Form (3NF) So far: if a relation is not in BCNF then anormalies arise. Given a relation Bookings with the attributes: –title: name of the movie –theater: name of the theater where the movie is being shown –city: the city where the theater is located (a tuple (m,t,c): represents the fact that movie m is shown at theater t in city c)

26. Bookings(title,theater,city) The FDs of the relations: –theater  city –title city  theater theater  city violates the BCNF condition, why? decomposition yields: {theater,city} and {theater,title} Consider the relations: theatercity GuildMenlo ParkMenlo theatertitle GuildNet ParkNet theatertitlecity GuildNetMenlo ParkNetMenlo recovering Violate the FD title city  theater Possible relations according to the FDs of each schema

27. 3NF A relaxation of the BCNF condition: a relation R is in 3NF if: whenever there is a nontrivial FD A 1 …A n  B, either {A 1,…,A n } is a superkey or B is a member of some key. Bookings(title,theater,city) is in 3NF

28. Checking BCNF and 3NF Given R(A,B,C,D) with FDs AB  C, C  D, D  A. Question: Indicate the BCNF violations and 3NF violations. Steps in answering the question: –Step 1: compute all nontrivial FDs (right side: one att) –Step 2: find all keys –Step 3: find all the violations Step 1: AB  C, C  D, D  A, AB  D, C  A, DB  C, AC  D Step 2: Keys – {A,B}, {C,B}, and {D,B} Step 3: –BCNF violation: C  D, D  A, C  A, AC  D and their trivial extensions (e.g. CD  D, DA  A,…) –3NF violation: none