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3 Decompositions uDo we need to decompose a relation? wSeveral normal forms for relations. If schema in these normal forms certain problems don’t arise wThe two commonly used normal forms are third normal form (3NF) and Boyce-Codd normal form (BCNF) uWhat problems does decomposition cause? wLossless-join property: get original relation by joining the resulting relations wDependency-preservation property: enforce constraints on original relation by enforcing some constraints on resulting relations wQueries may require a join of decomposed relations!

4 Finding All Implied FD’s in a Projected Schema uMotivation: “normalization,” the process where we break a relation schema into two or more schemas. uExample: ABCD with FD’s AB ->C, C ->D, and D ->A. wDecompose into ABC, AD wWhat FD’s hold in ABC ? Not only AB ->C, but also C ->A ! uSame as F + method, but restrict to those FD’s that involve only attributes of the projected schema

5 Example: Projecting FD’s uABC with FD’s A ->B and B ->C. Project onto AC. wA + =ABC ; yields A ->B, A ->C. We do not need to compute AB + or AC +. wB + =BC ; yields B ->C. wC + =C ; yields nothing. wBC + =BC ; yields nothing. uResulting FD’s: A ->B, A ->C, andB ->C. uProjection onto AC : A ->C. wOnly FD that involves a subset of {A,C }.

6 Boyce-Codd Normal Form uWe say a relation R is in BCNF if whenever X ->Y is a nontrivial FD that holds in R, X is a superkey. wRemember: nontrivial means Y is not contained in X. wRemember, a superkey is any superset of a key (not necessarily a proper superset).

7 Example Drinkers(name, addr, beersLiked, manf, favBeer) FD’s: name->addr favBeer, beersLiked->manf uOnly key is {name, beersLiked}. uIn each FD, the left side is not a superkey. uAny one of these FD’s shows Drinkers is not in BCNF

8 Another Example Beers(name, manf, manfAddr) FD’s: name->manf, manf->manfAddr uOnly key is {name}. uname->manf does not violate BCNF, but manf->manfAddr does.

9 Decomposition into BCNF uGiven: relation R with FD’s F. uLook among the given FD’s for a BCNF violation X ->Y. wIf any FD following from F violates BCNF, then there will surely be an FD in F itself that violates BCNF. uCompute X +. wNot all attributes, or else X is a superkey.

10 Decompose R Using X -> Y uReplace R by relations with schemas: 1. R 1 = X R 2 = X (R – X + ) uProject given FD’s F onto the two new relations.

11 Example: BCNF Decomposition Drinkers(name, addr, beersLiked, manf, favBeer) F = name->addr, name -> favBeer, beersLiked->manf uPick BCNF violation name->addr. uClose the left side: {name} + = {name, addr, favBeer}. uDecomposed relations: 1.Drinkers1(name, addr, favBeer) 2.Drinkers2(name, beersLiked, manf)

12 Example -- Continued uWe are not done; we need to check Drinkers1 and Drinkers2 for BCNF. uProjecting FD’s is easy here. uFor Drinkers1(name, addr, favBeer), relevant FD’s are name->addr and name->favBeer. wThus, {name} is the only key and Drinkers1 is in BCNF.

13 Example -- Continued uFor Drinkers2(name, beersLiked, manf), the only FD is beersLiked->manf, and the only key is {name, beersLiked}. wViolation of BCNF. ubeersLiked + = {beersLiked, manf}, so we decompose Drinkers2 into: 1.Drinkers3(beersLiked, manf) 2.Drinkers4(name, beersLiked)

14 Example -- Concluded uThe resulting decomposition of Drinkers : 1.Drinkers1(name, addr, favBeer) 2.Drinkers3(beersLiked, manf) 3.Drinkers4(name, beersLiked) uNotice: Drinkers1 tells us about drinkers, Drinkers3 tells us about beers, and Drinkers4 tells us the relationship between drinkers and the beers they like.

15 Third Normal Form -- Motivation uThere is one structure of FD’s that causes trouble when we decompose. uAB ->C and C ->B. wE.g: A = street address, B = city, C = zip code. uThere are two keys, {A,B } and {A,C } (why?) uC ->B is a BCNF violation, so we must decompose into AC, BC.

16 street zip 545 Tech Sq Tech Sq city zip Cambridge02138 Cambridge02139 Join tuples with equal zip codes. street city zip 545 Tech Sq.Cambridge Tech Sq.Cambridge02139 Although no FD’s were violated in the decomposed relations, FD street city -> zip is violated by the database as a whole. ABC C We Cannot Enforce FD’s we cannot enforce the FD AB ->C by checking FD’s in these decomposed relations C ->B

17 3NF Let’s Us Avoid This Problem u3 rd Normal Form (3NF) modifies the BCNF condition so we do not have to decompose in this problem situation. uAn attribute is prime if it is a member of any key. uX ->A violates 3NF if and only if X is not a superkey, and also A is not prime or X ->A satisfies 3NF if and only if X is a superkey or A is prime

18 Example: 3NF uIn our problem situation with FD’s AB ->C and C ->B, we have keys AB and AC. uThus A, B, and C are each prime. uAlthough C ->B violates BCNF, it does not violate 3NF.

19 What 3NF and BCNF Give You uThere are two important properties of a decomposition: 1.Lossless Join : it should be possible to project the original relations onto the decomposed schema, and then reconstruct the original. 2.Dependency Preservation : it should be possible to check in the projected relations whether all the given FD’s are satisfied.

20 3NF and BCNF -- Continued uWe can get (1) with a BCNF decomposition. uWe can get both (1) and (2) with a 3NF decomposition. uBut we can’t always get (1) and (2) with a BCNF decomposition. wstreet-city-zip is an example.

21 3NF Synthesis Algorithm uWe can always construct a decomposition into 3NF relations with a lossless join and dependency preservation. uNeed minimal basis for the FD’s: 1.Right sides are single attributes. 2.No attribute can be removed from a left side. 3.No FD can be removed.

22 Computing Minimal Basis ustep 1: RHS of each FD is a single attribute. ustep 2: Eliminate unnecessary attributes from LHS. wAlgorithm: If FD XB  A  F (where B and A are single attributes) and X  A is entailed by F, then B was unnecessary ustep 3: Delete unnecessary FDs from F wAlgorithm: If F - {f} entails f, then f is unnecessary. If f is X  A then check if A  X + F-{f}

23 3NF Synthesis – (2) uOne relation for each FD in the minimal basis. wSchema is the union of the left and right sides. uIf no key is contained in an FD, then add one relation whose schema is some key.

24 Example u{A → B, ABCD → E, EF → GH, ACDF → EG} u Make RHS a single attribute: {A → B, ABCD → E, EF → G, EF → H, ACDF → E, ACDF → G} u Minimize LHS: ACD → E instead of ABCD → E u Eliminate redundant FDs w Can ACDF → G be removed? w Can ACDF → E be removed? u Final answer: {A → B, ACD → E, EF → G, EF → H}

25 Example u{A → B, ABCD → E, EF → GH, ACDF → EG}

26 Synthesizing a 3NF Schema ustep 1: Compute a minimal cover U, of F. The decomposition is based on U, but since U + = F + the same functional dependencies will apply ustep 2: Group all FDs in U with a common LHS, and then Partition U into sets U 1, U 2, … U n ustep 3: For each U i form a schema R i = set of attributes named in U i wEach FD of U will apply to some R i. Hence the decomposition is dependency preserving ustep 4: If no R i is a key of R, add schema R 0 = a key of R. wR 0 might be needed since not all attributes are necessarily contained in R 1  R 2 …  R n wThis guarantees lossless decomposition

27 Example uRelation: R=ABCDEFGH uFDs: {A → B, ABCD → E, EF → GH, ACDF → EG} uFind minimal cover: {A → B, ACD → E, EF → G, EF → H} u Combine LHS: {A → B, ACD → E, EF → GH} u New relations: AB, ACDE, EFGH u Is any of these a key? u No, so add one (e.g., ACDF) u Final tables: AB, ACDE, EFGH, ACDF

28 Why It Works uPreserves dependencies: each FD from a minimal basis is contained in a relation, thus preserved. uLossless Join: use the chase to show that the row for the relation that contains a key can be made all- unsubscripted variables. u3NF: hard part – a property of minimal bases.

29 BCNF via 3NF Synthesis uTry 3NF synthesis to decompose the original relation R uThis is loss-less and dependency preserving uIf all sub-relations in BCNF, done! uIf not, use the BCNF decomposition on the violating sub-relation uIf some FDs are not preserved, at least we tried hard!