1. 1 Normalization Anomalies Boyce-Codd Normal Form 3 rd Normal Form Source: Slides by Jeffrey Ullman

2. 2 Anomalies uGoal of relational schema design is to avoid anomalies and redundancy. wUpdate anomaly : one occurrence of a fact is changed, but not all occurrences. wDeletion anomaly : a valid fact is lost when a tuple is deleted.

3. 3 Example of Bad Design Consumers(name, addr, candiesLiked, manf, favCandy) nameaddrcandiesLikedmanf favCandy JanewayVoyagerTwizzlersHershey Smarties Janeway???SmartiesPete’s ??? SpockEnterpriseTwizzlers??? Twizzlers Data is redundant, because each of the ???’s can be figured out by using the FD’s name -> addr favCandy and candiesLiked -> manf.

4. 4 This Bad Design Also Exhibits Anomalies nameaddrcandiesLikedmanf favCandy JanewayVoyagerTwizzlersHershey Smarties JanewayVoyagerSmartiesNestle Smarties SpockEnterpriseTwizzlersHershey Twizzlers Update anomaly: if Janeway is transferred to Intrepid, will we remember to change each of her tuples? Deletion anomaly: If nobody likes Twizzlers, we lose track of the fact that Hershey manufactures Twizzlers.

5. 5 Problem & Solution uConceptual modeling unnaturally puts attributes of two different entity sets in same relation schema uUse decompositions to break up schema involving such an “unhappy union” into several relation schemas uMust do this decomposition carefully: wVarious normal forms have been proposed for relations wIf a schema is in one of these normal forms, then we now that certain kinds of problems do not occur

6. 6 Properties of a Decomposition uLossless join: any tuple of the original relation can be reconstructed from corresponding tuples of the resulting relations uDependency preservation: we can enforce any constraint on the original relation by enforcing some constraints on the resulting relations

7. 7 When to Decompose? uIf you frequently want the full information available in instances of the original relation, don’t decompose wExpensive to reconstruct original information wInstead, put up with the storage redundancy and put extra checks in apps to avoid anomalies

8. 8 Boyce-Codd Normal Form uWe say a relation R is in BCNF if whenever X ->A is a nontrivial FD that holds in R, X is a superkey. wRemember: nontrivial means A is not a member of set X. wRemember, a superkey is any superset of a key (not necessarily a proper superset).

9. 9 Example Consumers(name, addr, candiesLiked, manf, favCandy) FD’s: name->addr favCandy, candiesLiked->manf uOnly key is {name, candiesLiked}. uIn each FD, the left side is not a superkey. uAny one of these FD’s shows, Consumers is not in BCNF

10. 10 Another Example Candies(name, manf, manfAddr) FD’s: name->manf, manf->manfAddr uOnly key is {name}. uname->manf does not violate BCNF, but manf->manfAddr does.

11. 11 Properties of BCNF uIf a relation is in BCNF, then every attribute of every tuple records a piece of information that cannot be inferred, using FD’s, from any other information in the relation wEliminates redundancy uIf a relation is not in BCNF, there is always a lossless-join decomposition into a collection of BCNF relations wBut may not be a dependency-preserving decomposition

12. 12 Decomposition into BCNF uGiven: relation R with set of FD’s F. uCompute keys for R (sets of attributes that functionally determine all other attributes) uLook among the given FD’s for a BCNF violation X ->B. wIf any FD following from F violates BCNF, then there will surely be an FD in F itself that violates BCNF. uCompute X +. wWon't be all attributes, or else X would be a superkey and X ->B would not be a violation

13. 13 Decompose R Using X -> B uReplace R by relations R 1 and R 2 with schemas: wR 1 : X + (all attributes "reachable" from X) wR 2 : R – (X + – X ) (X plus all attributes not "reachable" from X) uProject given FD’s F onto the two new relations (ignore FD's containing attributes no longer in the new relation). uCheck if procedure must be repeated with R 1 and/or R 2.

14. 14 Decomposition Picture R-X +R-X + XX +-XX +-X R2R2 R1R1 R

15. 15 Example Consumers(name, addr, candiesLiked, manf, favCandy) F = name->addr, name -> favCandy, candiesLiked->manf uPick BCNF violation name->addr. uClose the left side: {name} + = {name, addr, favCandy}. uDecomposed relations: wConsumers1(name, addr, favCandy) wConsumers2(name, candiesLiked, manf)

16. 16 Example, Continued uWe are not done; we need to check Consumers1 and Consumers2 for BCNF. uProjecting FD’s is easy here. uFor Consumers1(name, addr, favCandy), relevant FD’s are name->addr and name->favCandy. wThus, {name} is the only key and Consumers1 is in BCNF.

17. 17 Example, Continued uFor Consumers2(name, candiesLiked, manf), the only FD is candiesLiked -> manf, and the only key is {name, candiesLiked}. wViolation of BCNF. ucandiesLiked + = {candiesLiked, manf}, so we decompose Consumers2 into: wConsumers3(candiesLiked, manf) wConsumers4(name, candiesLiked)

18. 18 Example, Concluded uThe resulting decomposition of Consumers : Consumers1(name, addr, favCandy) Consumers3(candiesLiked, manf) Consumers4(name, candiesLiked) uNotice: Consumers1 tells us about consumers, Consumers3 tells us about candies, and Consumers4 tells us the relationship between consumers and the candies they like.

19. 19 BCNF Decomposition Not Unique uSuppose several dependencies violate BCNF. uDepending on which is used to guide the next decomposition, might get different collections of decomposed relations uNormalization theory does not help us decide among alternatives: designer must consider alternatives and choose based on semantics of the application

20. 20 Third Normal Form - Motivation uThere is one structure of FD’s that causes trouble when we decompose. uAB ->C and C ->B. wExample: A = street address, B = city, C = zip code. uThere are two keys, {A,B } and {A,C }. wThink about why. uC ->B is a BCNF violation (why?), so we must decompose into AC, BC.

21. 21 We Cannot Enforce FD’s uThe problem is that if we use AC and BC as our database schema, we cannot enforce the FD AB ->C by checking FD’s in these decomposed relations. uExample with A = street, B = city, and C = zip on the next slide.

22. 22 An Unenforceable FD uSuppose we have relation Addr with FD's: wstreet city -> zip wzip -> city uKeys are {street, city} and {street, zip} uAfter decomposing, we have relations wAddr1(street,zip) with no FD (!) wAddr2(city,zip) with FD zip -> city uConsider instances on next slide…

23. 23 An Unenforceable FD (cont'd) street zip 545 Tech Sq.02138 545 Tech Sq.02139 city zip Cambridge02138 Cambridge02139 Join tuples with equal zip codes. street city zip 545 Tech Sq.Cambridge02138 545 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. Addr1:Addr2:

24. 24 3NF Lets 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.

25. 25 Example 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. uSo no need to decompose.

26. 26 What 3NF and BCNF Give You uThere are two important properties of a decomposition: 1.Recovery (aka 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 (original) given FD’s are satisfied.

27. 27 3NF and BCNF, Continued uWe can get (1) with a BCNF decomposition. wExplanation needs to wait for relational algebra. 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.