Schema Refinement and Normal Forms Chapter 19 Database

Schema Refinement and Normal Forms Chapter 19 Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 1

Chapter Goal The goal of this chapter is to use functional dependency constraints to refine a conceptual schema (one that was derived from an ER model). v The idea is to eliminate or reduce redundancy. v Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 2

Outline Overview of schema refinement approach v Functional Dependencies v Reasoning about FDs to infer more FDs v Introduce Normal Forms v Decompositions and properties v Decompositions of relations with redundancies into smaller relations but without redundancy v Schema Refinement discussion v Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 3

Overview of the approach v To give an overview of the approach in this chapter, we start with an example to illustrate the issues, notations, terminology, and potential solution. Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 4

Example: Constraints on Entity Set v Consider relation obtained from Hourly_Emps: § v Notation: We will denote this relation schema by listing the attributes: SNLRWH § § v Hourly_Emps (ssn, name, lot, rating, hrly_wages, hrs_worked) This is really the set of attributes {S, N, L, R, W, H}. Sometimes, we will refer to all attributes of a relation by using the relation name. (e. g. , Hourly_Emps for SNLRWH) Examples of FD constraints on Hourly_Emps: § § ssn is the key: S SNLRWH rating determines hrly_wages: R Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke W 5

Example (cont-d): Redundancy Issues v Problems due to R W: § Update anomaly: Can we change W in just the 1 st tuple of SNLRWH? § Insertion anomaly: What if we want to insert an employee and don’t know the hourly wage for his rating? § Deletion anomaly: If we delete all employees with rating 5, we lose the information about the wage for rating 5! Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 6

Example (Cont-d. ): Will two tables help? Hourly_Emps 2 Wages v Now let’s check again: § Update anomaly: Can we change W in just the 1 st tuple of Wages? § Insertion anomaly: What if we want to insert an employee in Hourly_Emps 2 and don’t know the hourly wage for his rating? § Deletion anomaly: If we delete all employees with rating 5 in Hour; _Emps 2, do we lose the information about the wage for rating 5? Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 7

Problems with decomposition 1. 2. 3. v Query Performance may degrade Decomposed tables may be lossy, ie may not be able to reconstruct original table. Performance issue with constraint Checking: May require a join across split tables. These issues will be discussed further in this chapter, along with special decompositions (BCNF, 3 NF) that may address some of these issues. Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 8

In summary … v Redundancy is at the root of several problems associated with relational schemas: § redundant storage, insert/delete/update anomalies Integrity constraints, in particular functional dependencies, can be used to identify schemas with such problems and to suggest refinements. v Main refinement technique: decomposition (replacing ABCD with, say, AB and BCD, or ACD and ABD). v Decomposition should be used judiciously: v § § Is there reason to decompose a relation? What problems (if any) does the decomposition cause? Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 9

Outline Overview of schema refinement approach v Functional Dependencies v Reasoning about FDs to infer more FDs v Introduce Normal Forms v Decompositions and properties v Decompositions of relations with redundancies into smaller relations but without redundancy v Schema Refinement discussion v Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 10

Functional Dependencies (FDs) v A functional dependency X Y holds over relation R if, for every allowable instance r of R: § § v An FD is a statement about allowable relations. § § v t 1 r, t 2 r, (t 1) = (t 2) implies (t 1) = (t 2) i. e. , given two tuples in r, if the X values agree, then the Y values must also agree. (X and Y are sets of attributes. ) Must be identified based on semantics of application. Given some allowable instance r 1 of R, we can check if it violates some FD f, but we cannot tell if f holds over R! K is a candidate key for R means that K § However, K R R does not require K to be minimal! Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 11

FD Example Include example here v Illustrate : v § We can check if a table instance violate a certain FD. § But we can not infer from an instance the FDs or that the FD holds all the time. § Note that FD constraints are given from the semantics of the problem. Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 12

Outline Overview of schema refinement approach v Functional Dependencies v Reasoning about FDs to infer more FDs v Introduce Normal Forms v Decompositions and properties v Decompositions of relations with redundancies into smaller relations but without redundancy v Schema Refinement discussion v Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 13

Reasoning About FDs v Given some FDs, we can usually infer additional FDs: § v lot implies ssn lot = closure of F is the set of all FDs that are implied by F. Armstrong’s Axioms (X, Y, Z are sets of attributes): § § § v did, did An FD f is implied by a set of FDs F if f holds whenever all FDs in F hold. § v ssn Reflexivity: If X Y, then Y X Augmentation: If X Y, then XZ YZ for any Z Transitivity: If X Y and Y Z, then X Z These are sound (ie AA generate FDs only in F+) and complete (repeated application of AA will generate all FDs in F+) inference rules for FDs! Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 14

Reasoning About FDs (Contd. ) v Couple of additional rules (that follow from AA): § § v Union: If X Y and X Z, then X Decomposition: If X YZ, then X Example: and: § § § YZ Y and X Z Contracts(cid, sid, jid, did, pid, qty, value), C is the key: C CSJDPQV (jid==project ID) Project purchases each part usingle contract: JP Dept purchases at most one part from a supplier: SD C P JP C, C CSJDPQV imply JP CSJDPQV v SD P implies SDJ JP v SDJ JP, JP CSJDPQV imply SDJ 15 v Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke

Reasoning About FDs (Contd. ) Computing the closure of a set of FDs can be expensive. (Size of closure is exponential in # attrs!) v Typically, we just want to check if a given FD X Y is in the closure of a set of FDs F. An efficient check: v § § v Compute attribute closure of X (denoted ) wrt F: • Set of all attributes A such that X A is in Check if Y is in Does F = {A E? § i. e, is A ? B, B C, C D E in the closure Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke E } imply A ? Equivalently, is E in 16

Attribute Closure Ex. v F={SSN ENAME, PNUMBER {PNAME, PLOCATION}, {SSN, PNUMBER} HOURS} Find the following closure sets: {SSN}+ {SSN, ENAME} ? {PNUMBER}+ {PNUMBER, PNAME, ? PLOCATION} {SSN, PNUMBER}+ {SSN, PNUMBER, ENAME, PLOCATION, HOURS} ? Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 17

Example problems (1) 1. Consider the relation schema R(A, B, C), which has the FD B → C. If A is a candidate key for R, is it possible for R to be in BCNF? If so, under what conditions? If not, explain why not. 2. Suppose we have a relation schema R(A, B, C) representing a relationship between two entity sets with keys A and B, respectively, and suppose that R has (among others) the FDs A → B and B → A. Explain what such a pair of dependencies means (i. e. , what they imply about the relationship that the relation models). ANS: ONE-TO-ONE RELATION Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 18

Example problems (2) Consider a relation R with five attributes ABCDE. You are given the following dependencies: A → B, BC → E, and ED → A. 1. List all keys for R. ANS: BCD, CDE, ACD Approach: Start with the listed dependencies, and see how they can be augmented to get dependencies implied on all attributes. Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 19

Outline Overview of schema refinement approach v Functional Dependencies v Reasoning about FDs to infer more FDs v Introduce Normal Forms v Decompositions and properties v Decompositions of relations with redundancies into smaller relations but without redundancy v Schema Refinement discussion v Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 20

Normal Forms Returning to the issue of schema refinement, the first question to ask is whether any refinement is needed! v If a relation is in a certain normal form (BCNF, 3 NF etc. ), it is known that certain kinds of problems are avoided/minimized. This can be used to help us decide whether decomposing the relation will help. v Role of FDs in detecting redundancy: v § Consider a relation R with 3 attributes, ABC. • No FDs hold: There is no redundancy here. • Given A B: Several tuples could have the same A value, and if so, they’ll all have the same B value! • But if A is superkey (BCNF), no repetition. Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 21

Characterizing Normal Forms BCNF: Left side of FD is a superkey v 3 NF: Right side of FD part of key v 2 NF: Left side and right side are not parts of a key v 1 NF: (None of the above. ) Every field contains only atomic values, that is no lists or sets. v Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 22

Boyce-Codd Normal Form (BCNF) v Reln R with FDs F is in BCNF if, for all X § § v A in A X (called a trivial FD), or X contains a key for R. In other words, R is in BCNF if the only non-trivial FDs that hold over R are key constraints. No redundancy in R. Example: § If example relation is in BCNF, the 2 tuples must be identical (since X is a key), and this can not happen, so no redundancy. § Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 23

Third Normal Form (3 NF) v Reln R with FDs F is in 3 NF if, for all X § § § A in A X (called a trivial FD), or X contains a key for R, or A is part of some key for R. Minimality of a key is crucial in third condition above! v If R is in BCNF, obviously in 3 NF. v If R is in 3 NF, some redundancy is possible. It is a compromise, used when BCNF not achievable (e. g. , no ``good’’ decomp, or performance considerations). v § Lossless-join, dependency-preserving decomposition of R into a collection of 3 NF relations always possible. Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 24

Redundancy in 3 NF - Example v v v v Reserves(Sailors, Boat, Day, credit card) represented as SBDC FD: S C ; states that a sailor uses a unique credit card to pay for a reservation. The key is SBD. Is R in 3 NF? Assume that the credit card also uniquely defines the Sailor i. e. C S Therefore, CBD is also a key. Is R now in 3 NF? But in the table Reserves every tuple with same S value should have the same C value some redundancy Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 25

2 NF and 1 NF 2 NF: Left side and right side are not parts of a key v 1 NF: (None of the above. ) Every field contains only atomic values, that is no lists or sets. v Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 26

Examples – Normal Forms Given R(students, course, instructor) where FD 1: {student, course} instructor and FD 2: instructor course Is this relation in BCNF? Is it in 3 NF? Ans: IS and SC are keys. From SC, we can conclude it is in 3 NF Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 27

Example problems (2) Consider a relation R with five attributes ABCDE. You are given the following dependencies: A → B, BC → E, and ED → A. 2. Is R in 3 NF? 3. Is R in BCNF? ANS: List the keys (see earlier example): BCD, CDE, ACD We conclude it is in 3 NF Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 28

Outline Overview of schema refinement approach v Functional Dependencies v Reasoning about FDs to infer more FDs v Introduce Normal Forms v Decompositions and properties v Decompositions of relations with redundancies into smaller relations but without redundancy v Schema Refinement discussion v Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 29

Decomposition of a Relation Scheme v Suppose that relation R contains attributes A 1. . . 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 no attributes that do not appear in R), and Every attribute of R appears as an attribute of one of the new relations. Decomposing R means we will store instances of the relation schemes produced by the decomposition, instead of instances of R. v E. g. , Can decompose SNLRWH into SNLRH and 30 v Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke

Example Decomposition v Decompositions should be used only when needed. § § v SKIP SNLRWH has FDs S SNLRWH and R W Second FD causes violation of 3 NF; W values repeatedly associated with R values. Easiest way to fix this is to create a relation RW to store these associations, and to remove W from the main schema: • i. e. , we decompose SNLRWH into SNLRH and RW The information to be stored consists of SNLRWH tuples. If we just store the projections of these tuples onto SNLRH and RW, are there any potential problems that we should be aware of? Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 31

Problems with Decompositions v There are three potential problems to consider: 1. Some queries become more expensive. • e. g. , How much did sailor Joe earn? (salary = W*H) 2. Given instances of the decomposed relations, we may not be able to reconstruct the corresponding instance of the original relation! (lossy decomposition) • Fortunately, not in the SNLRWH example. 3. Checking some dependencies may require joining the instances of the decomposed relations. • Fortunately, not in the SNLRWH example. v Tradeoff: Must consider these issues vs. redundancy. Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 32

Example with Lossy Decomposition Decomposing ABC into AB and BC: Trying to recover ABC from joining AB and BC: We do not get ABC! See the additional rows below… Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 33

Lossless Join Decompositions v 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: § v (r) = r It is always true that r § v (r) (r) In general, the other direction does not hold! If it does, the decomposition is lossless-join. It is essential that all decompositions used to deal with redundancy be lossless! (Avoids Problem (2). ) Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 34

More on Lossless Join v The decomposition of R into X and Y is lossless-join wrt F if and only if the closure of F contains: § § v X X Y Y X, or Y In particular, the decomposition of R into UV and R - V is lossless-join if U V holds over R. Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 35

Example 1 Lossless Join Lossless Example : • Decompose SNLRWH into SNLRH and RW. • We get full recovery from the join Note that R W causes violation of 3 NF (it is 2 NF). Since R is common, the lossless join condition is satisfied, and that explains why we get full recovery. Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 36

Example 2 – Lossless Join Consider R(M, Y, P, MP, C) AND v Functional dependencies: F={M MP, {M, Y} P, MP C} 1. Which of the following is a candidate key {M}, {M, Y}, {M, C} 2. Is R in BCNF? Is it in 3 NF? Neither 3. Assume that R is decomposed into R 1(M, Y, P) and R 2(M, MP, C). Is this decomposition lossless? v ANS: Intersection = M MP C; therefore M R 2 So lossless! Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 37

Example – Illustrating issue with Dependency Preserving Decomposition v Consider CSJDPQV, C is key, JP P. § § v C and SD BCNF decomposition: CSJDQV and SDP. (Check: JP is a super key) Problem: Checking JP C requires a join! Want Dependency preserving decomposition (Intuitive): If R is decomposed into X, Y and we enforce the FDs that hold on X, and on Y , then all FDs that were given to hold on R must also hold. (Avoids Problem (3). ) Projection of set of FDs F: If R is decomposed into X, . . . projection of F onto X (denoted FX ) is the set of FDs U V in F+ (closure of F ) such that U, V are in X. § v Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 38

Dependency Preserving Decompositions SKIP v Decomposition of R into X and Y is dependency preserving if (FX union FY ) + = F + § § v 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 +. i. e. All the FDs are either in X+ or Y+ but not across. Important to consider F +, not F, in this definition: Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 39

Example - Dependency Preserving Decompositions does not imply Lossless § § v Dependency preserving does not imply lossless join: § v ABC, A B, B C, C A, decomposed into AB and BC. Is this dependency preserving? Is C A preserved? ? ? ABC, A No! B, decomposed into AB and BC. Lossless? And vice-versa! Recall Earlier Example: § Consider CSJDPQV, C is key, JP C and SD P. • BCNF decomposition: CSJDQV and SDP. (Check: JP is a super key) • Dependency preserving? NO: Checking JP • 3 ed, R. Ramakrishnan and J. Gehrke Database Management Systems, C requires a join! 40

Outline Overview of schema refinement approach v Functional Dependencies v Reasoning about FDs to infer more FDs v Introduce Normal Forms v Decompositions and properties v Decompositions of relations with redundancies into smaller relations but without redundancy v Schema Refinement discussion v Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 41

Decomposition into BCNF v Consider relation R with FDs F. If X Y violates BCNF, decompose R into R - Y and XY. § § v 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 To deal with SD P, decompose into SDP, CSJDQV. To deal with J S, decompose CSJDQV into JS and CJDQV In general, several dependencies may cause violation of BCNF. The order in which we ``deal with’’ them could lead to very different sets of relations! 42 Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke

BCNF and Dependency Preservation – Fix creates redundancy v In general, there may not be a dependency preserving decomposition into BCNF. § § v e. g. , CSZ, CS Z, Z C (note that Z is not a superkey) Can’t decompose while preserving 1 st FD; not in BCNF. Similarly, decomposition of CSJDPQV into SDP, JS and CJDQV is not dependency preserving (w. r. t. the FDs: SD P, J S, and JP C). § § However, it is a lossless join decomposition. In this case, adding JPC to the collection of relations gives us a dependency preserving decomposition. • JPC tuples stored only for checking FD! (Redundancy!)43 Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke

Decomposition into 3 NF Obviously, the algorithm for lossless join decomp into BCNF can be used to obtain a lossless join decomp into 3 NF (typically, can stop earlier). v To ensure dependency preservation, one idea: v § § v If X Y is not preserved, add relation XY. Problem is that XY may violate 3 NF! 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. Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 44

Minimal Cover for a Set of FDs v 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. v e. g. , A B, ABCD E, EF GH, ACDF EG has the following minimal cover: v § v A B, ACD E, EF G and EF H M. C. ® Lossless-Join, Dep. Pres. Decomp. Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 45

Outline Overview of schema refinement approach v Functional Dependencies v Reasoning about FDs to infer more FDs v Introduce Normal Forms v Decompositions and properties v Decompositions of relations with redundancies into smaller relations but without redundancy v Schema Refinement discussion v Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 46

How are FDs used in practice? – Refining ER Diagrams and Conceptual Designs v Let’s consider the following examples to help in: § Describe constraints in entity sets (other than key constraints) using FD and then use Normalization to adjust design. § Describe constraints on a relationship set using FD then use Normalization to adjust design. § Using FD and Normalization to then conclude best placement of attributes. § Adjust the design in the “internet shop”case study Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 47

Constraint on an Entity set v Consider the relation SNLRWH: Ssn ssn, name, lot, rating, hourly_wages, hours_worked v v v Note that the FD: R W can not be expressed in the ER diagram. Since ER diagram can only capture key constraints for an entity. Following the capture of the FD, we can conclude that we need to decompose the relation to eliminate redundancy: SNLRH and RW. This leads to the adjusted design of having two relations. Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 48

Constraints on a relationship Set v v v Recall the ternary relationship Contract that describes a relationship among: Parts, Suppliers, and Departments all together. Refer to this schema as CQPSD: A contract with contract id C specifies that a supplier S will supply some quantity Q of a part P to department D. Assume that we have a policy that a department purchases at most one part from a given supplier: DS P This dependency can not be captured in the ER diagram. Following Normalization, we can decompose the relation into: SQSD and SDP. Which eliminates redundancy. Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 49

Identifying Attributes of Entities v 1 st diagram translated: Workers(S, N, L, D, S) Departments(D, M, B) § v v v Before: since name ssn dname lot did budget Lots associated with workers. Employees Suppose all workers in a dept are assigned the same lot: D L After: Redundancy; fixed by: Workers 2(S, N, D, S) name Dept_Lots(D, L) ssn Can fine-tune this: Employees Workers 2(S, N, D, S) Departments(D, M, B, L) Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke Works_In budget since dname did Works_In Departments lot Departments 50

Summary of Schema Refinement If a relation is in BCNF, it is free of redundancies that can be detected using FDs. Thus, trying to ensure that all relations are in BCNF is a good heuristic. v If a relation is not in BCNF, we can try to decompose it into a collection of BCNF relations. v § § Must consider whether all FDs are preserved. If a losslessjoin, dependency preserving decomposition into BCNF is not possible (or unsuitable, given typical queries), should consider decomposition into 3 NF. Decompositions should be carried out and/or re-examined while keeping performance requirements in mind. Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 51

Backup: Additional Examples Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 52

What Does 3 NF Achieve? v If 3 NF violated by X § § v X is a subset of some key K • We store (X, A) pairs redundantly. X is not a proper subset of any key. • There is a chain of FDs K X A, which means that we cannot associate an X value with a K value unless we also associate an A value with an X value. But: even if reln is in 3 NF, these problems could arise. § v A, Redundancy may happen: e. g. , Reserves SBDC, S C, C S is in 3 NF, but for each reservation of sailor S, same (S, C) pair is stored. Recall: (SBDC === Sailor, Boat, Day, Credit Card) 3 NF is a compromise relative to BCNF. It eliminates 53 the first two problems, but not the third. Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke

Review - Characterizing Normal Forms BCNF: Left side of FD is a superkey v 3 NF: Right side of FD part of key v 2 NF: Left side and right side are not parts of a key (except for the key FD) v 1 NF: None of the above v Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 54

Forms and Decomposition Example (1) v Given the following relation: R(A, B, C, D, E), A → B, C → D. § State the strongest normal form that the relation is in § If it is not in BCNF decompose it into a collection of BCNF v Solution: § To determine form, need to determine key. (Starting with fewest attributes, we try to determine which makes up the key): ACE § From above FDs, we conclude that R is in 1 NF. § To decompose into BCNF, apply (R – right side) & left-right relation. Build the decomposition tree. Result: AB, CD, ACE. Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 55

Forms and Decomposition Example (2) v Consider the following relation R(A, B, C, D), and the functional dependencies: C → D, C → A, B → C. § Identify the candidate key(s) § Identify the best normal form that R satisfies § If R is not in BCNF, decompose it into set of BCNF relations that preserve dependencies. v Solution: § To determine form, need to determine key. (Starting with fewest attributes, we try to determine which makes up the key): B § From above FDs, we conclude that R is in 2 NF. § To decompose into BCNF, apply (R – right side) & left-right relation. Build the decomposition tree. Result: CD, AC, BC. § Check if Functional dependencies preserved in decomposition: This one checks ok. Database Management Systems, 3 ed, R. Ramakrishnan and J. Gehrke 56