Normalizations
- 1. Schema Refinement and Normalization R&G – Chapter 19 Lecture 14 Nobody realizes that some people expend tremendous energy merely to be normal. Albert Camus
- 2. Functional Dependencies (Review) • A functional dependency X Y holds over relation schema R if, for every allowable instance r of R: t1 r, t2 r, pX (t1) = pX (t2) implies pY (t1) = pY (t2) (where t1 and t2 are tuples;X and Y are sets of attributes) • In other words: X Y means Given any two tuples in r, if the X values are the same, then the Y values must also be the same. (but not vice versa) • Can read “” as “determines”
- 3. Normal Forms • Back to schema refinement… • Q1: is any refinement is needed??! • If a relation is in a normal form (BCNF, 3NF etc.): – we know that certain problems are avoided/minimized. – helps decide whether decomposing a relation is useful. • Role of FDs in detecting redundancy: – Consider a relation R with 3 attributes, ABC. • No (non-trivial) FDs hold: There is no redundancy here. • Given A B: If A is not a key, then several tuples could have the same A value, and if so, they’ll all have the same B value! • 1st Normal Form – all attributes are atomic • 1st 2nd (of historical interest) 3rd Boyce-Codd …
- 4. Boyce-Codd Normal Form (BCNF) • Reln R with FDs F is in BCNF if, for all X A in F+ – A X (called a trivial FD), or – X is a superkey for R. • In other words: “R is in BCNF if the only non-trivial FDs over R are key constraints.” • If R in BCNF, then every field of every tuple records information that cannot be inferred using FDs alone. – Say we know FD X A holds this example relation: X Y A x y1 a x y2 ? • Can you guess the value of the missing attribute? •Yes, so relation is not in BCNF
- 5. Decomposition of a Relation Schema • If a relation is not in a desired normal form, it can be decomposed into multiple relations that each are in that normal form. • Suppose that relation R contains attributes A1 ... An. A decomposition of R consists of replacing R by two or more relations such that: – Each new relation scheme contains a subset of the attributes of R, and – Every attribute of R appears as an attribute of at least one of the new relations.
- 6. Example (same as before) • SNLRWH has FDs S SNLRWH and R W • Q: Is this relation in BCNF? S N L R W H 123-22-3666 Attishoo 48 8 10 40 231-31-5368 Smiley 22 8 10 30 131-24-3650 Smethurst 35 5 7 30 434-26-3751 Guldu 35 5 7 32 612-67-4134 Madayan 35 8 10 40 Hourly_Emps No, The second FD causes a violation; W values repeatedly associated with R values.
- 7. Decomposing a Relation • Easiest fix is to create a relation RW to store these associations, and to remove W from the main schema: S N L R H 123-22-3666 Attishoo 48 8 40 231-31-5368 Smiley 22 8 30 131-24-3650 Smethurst 35 5 30 434-26-3751 Guldu 35 5 32 612-67-4134 Madayan 35 8 40 R W 8 10 5 7 Hourly_Emps2 Wages •Decompositions should be used only when needed. –Q: potential problems of decomposition? •Q: Are both of these relations are now in BCNF?
- 8. Problems with Decompositions • There are three potential problems to consider: 1) May be impossible to reconstruct the original relation! (Lossiness) • Fortunately, not in the SNLRWH example. 2) Dependency checking may require joins. • Fortunately, not in the SNLRWH example. 3) Some queries become more expensive. • e.g., How much does Guldu earn? Tradeoff: Must consider these issues vs. redundancy.
- 9. Lossless Decomposition (example) S N L R H 123-22-3666 Attishoo 48 8 40 231-31-5368 Smiley 22 8 30 131-24-3650 Smethurst 35 5 30 434-26-3751 Guldu 35 5 32 612-67-4134 Madayan 35 8 40 R W 8 10 5 7 S N L R W H 123-22-3666 Attishoo 48 8 10 40 231-31-5368 Smiley 22 8 10 30 131-24-3650 Smethurst 35 5 7 30 434-26-3751 Guldu 35 5 7 32 612-67-4134 Madayan 35 8 10 40 =
- 10. Lossy Decomposition (example) A B C 1 2 3 4 5 6 7 2 8 1 2 8 7 2 3 A B 1 2 4 5 7 2 B C 2 3 5 6 2 8 A B C 1 2 3 4 5 6 7 2 8 A B; C B A B 1 2 4 5 7 2 B C 2 3 5 6 2 8 =
- 11. Lossless Join Decompositions • Decomposition of R into X and Y is lossless-join w.r.t. a set of FDs F if, for every instance r that satisfies F: (r) (r) = r • It is always true that r (r) (r) – In general, the other direction does not hold! If it does, the decomposition is lossless-join. • Definition extended to decomposition into 3 or more relations in a straightforward way. • It is essential that all decompositions used to deal with redundancy be lossless! (Avoids Problem #1) p X p Y p X p Y
- 12. More on Lossless Decomposition • The decomposition of R into X and Y is lossless with respect to F if and only if the closure of F contains: X Y X, or X Y Y I.E.: decomposing ABC into AB and BC is lossy, because intersection (i.e., “B”) is not a key of either resulting relation. • Useful result: If W Z holds over R and W Z is empty, then decomposition of R into R-Z and WZ is loss-less.
- 13. Lossless Decomposition (example) A C 1 3 4 6 7 8 B C 2 3 5 6 2 8 A B C 1 2 3 4 5 6 7 2 8 A B; C B A B C 1 2 3 4 5 6 7 2 8 But, now we can’t check A B without doing a join! B C 2 3 5 6 2 8 = A C 1 3 4 6 7 8
- 14. Dependency Preserving Decomposition • Dependency preserving decomposition (Intuitive): – If R is decomposed into X, Y and Z, and we enforce the FDs that hold individually on X, on Y and on Z, then all FDs that were given to hold on R must also hold. (Avoids Problem #2 on our list.) • Projection of set of FDs F : If R is decomposed into X and Y the projection of F on X (denoted FX ) is the set of FDs U V in F+ (closure of F , not just F ) such that all of the attributes U, V are in X. (same holds for Y of course)
- 15. Dependency Preserving Decompositions (Contd.) • Decomposition of R into X and Y is dependency preserving if (FX FY ) + = F + – i.e., if we consider only dependencies in the closure F + that can be checked in X without considering Y, and in Y without considering X, these imply all dependencies in F +. • Important to consider F + in this definition: – ABC, A B, B C, C A, decomposed into AB and BC. – Is this dependency preserving? Is C A preserved????? • note: F + contains F {A C, B A, C B}, so… • FAB contains A B and B A; FBC contains B C and C B • So, (FAB FBC)+ contains C A
- 16. Decomposition into BCNF • Consider relation R with FDs F. If X Y violates BCNF, decompose R into R - Y and XY (guaranteed to be loss-less). – Repeated application of this idea will give us a collection of relations that are in BCNF; lossless join decomposition, and guaranteed to terminate. – e.g., CSJDPQV, key C, JP C, SD P, J S – {contractid, supplierid, projectid,deptid,partid, qty, value} – To deal with SD P, decompose into SDP, CSJDQV. – To deal with J S, decompose CSJDQV into JS and CJDQV – So we end up with: SDP, JS, and CJDQV • Note: several dependencies may cause violation of BCNF. The order in which we ``deal with’’ them could lead to very different sets of relations!
- 17. BCNF and Dependency Preservation • In general, there may not be a dependency preserving decomposition into BCNF. – e.g., CSZ, CS Z, Z C – Can’t decompose while preserving 1st FD; not in BCNF. • Similarly, decomposition of CSJDPQV into SDP, JS and CJDQV is not dependency preserving (w.r.t. the FDs JP C, SD P and J S). • {contractid, supplierid, projectid,deptid,partid, qty, value} – However, it is a lossless join decomposition. – In this case, adding JPC to the collection of relations gives us a dependency preserving decomposition. • but JPC tuples are stored only for checking the f.d. (Redundancy!)
- 18. Third Normal Form (3NF) • Reln R with FDs F is in 3NF if, for all X A in F+ A X (called a trivial FD), or X is a superkey of R, or A is part of some candidate key (not superkey!) for R. (sometimes stated as “A is prime”) • Minimality of a key is crucial in third condition above! • If R is in BCNF, obviously in 3NF. • If R is in 3NF, some redundancy is possible. It is a compromise, used when BCNF not achievable (e.g., no ``good’’ decomp, or performance considerations). – Lossless-join, dependency-preserving decomposition of R into a collection of 3NF relations always possible.
- 19. What Does 3NF Achieve? • If 3NF violated by X A, one of the following holds: – X is a subset of some key K (“partial dependency”) • We store (X, A) pairs redundantly. • e.g. Reserves SBDC (C is for credit card) with key SBD and S C – X is not a proper subset of any key. (“transitive dep.”) • There is a chain of FDs K X A • So we can’t associate an X value with a K value unless we also associate an A value with an X value (different K’s, same X implies same A!) – problem with initial SNLRWH example. • But: even if R is in 3NF, these problems could arise. – e.g., Reserves SBDC (note: “C” is for credit card here), S C, C S is in 3NF (why?), but for each reservation of sailor S, same (S, C) pair is stored. • Thus, 3NF is indeed a compromise relative to BCNF. – You have to deal with the partial and transitive dependency issues in your application code!
- 20. Decomposition into 3NF • Obviously, the algorithm for lossless join decomp into BCNF can be used to obtain a lossless join decomp into 3NF (typically, can stop earlier) but does not ensure dependency preservation. • To ensure dependency preservation, one idea: – If X Y is not preserved, add relation XY. Problem is that XY may violate 3NF! e.g., consider the addition of CJP to `preserve’ JP C. What if we also have J C ? • Refinement: Instead of the given set of FDs F, use a minimal cover for F.
- 21. Minimal Cover for a Set of FDs • Minimal cover G for a set of FDs F: – Closure of F = closure of G. – Right hand side of each FD in G is a single attribute. – If we modify G by deleting an FD or by deleting attributes from an FD in G, the closure changes. • Intuitively, every FD in G is needed, and ``as small as possible’’ in order to get the same closure as F. • e.g., A B, ABCD E, EF GH, ACDF EG has the following minimal cover: – A B, ACD E, EF G and EF H • M.C. implies Lossless-Join, Dep. Pres. Decomp!!! – (in book)
- 22. Summary of Schema Refinement • BCNF: each field contains information that cannot be inferred using only FDs. – ensuring BCNF is a good heuristic. • Not in BCNF? Try decomposing into BCNF relations. – Must consider whether all FDs are preserved! • Lossless-join, dependency preserving decomposition into BCNF impossible? Consider 3NF. – Same if BCNF decomp is unsuitable for typical queries – Decompositions should be carried out and/or re-examined while keeping performance requirements in mind. • Note: even more restrictive Normal Forms exist (we don’t cover them in this course, but some are in the book.)