LECTURE 5 Schema Refinement and Normal Forms SOME

LECTURE 5: Schema Refinement and Normal Forms SOME OF THESE SLIDES ARE BASED ON YOUR TEXT BOOK n 12/2/2020 1

n Anomalies n Insertion anomalies n n Deletion anomalies n n Cannot record film. Type without star. Name If we delete the last star, we also lose the movie info. Modification (update) anomalies

DECOMPOSITION

Is there a possibility of an update, deletion, and insertiona anomalies in the table below? Explain with examples Student. Num Course. Num Student Name Address Course S 21 9201 Jones Edinburgh Accounts 1 S 21 9267 Jones Edinburgh physics S 24 9267 Smith Glasgow physics S 30 9201 Richards Manchester Accounts 1 S 30 9322 Richards Manchester Maths 12/2/2020 4

Set valued attributes: example n Consider a database of customer transactions n n n You have customers represented by cid, and transactions represented by tid, where each transaction has multiple items, represented by itemid, together with name, quantity and the cost of the item Design a database to keep track of the transactions of customers (set valued, vs flat design) Example queries: n n 12/2/2020 what is the total sales in dollars per item What is the total sales per customer 5

Schema Refinement n n functional dependencies, can be used to identify schemas with problems and to suggest refinements. Decomposition is used for schema refinement. 12/2/2020 6

Example FD n Database of beer drinkers

Example FD n n title year length film. Type studio. Name

Functional Dependencies (FDs) n A functional dependency X Y holds over relation R if, for every allowable instance r of R: n n t 1 t 2 t 1 r, t 2 (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. ) X 1 2 Y a b r, Z p q r p

Functional Dependencies (FDs) Does the following relation instance satisfy X->Y ? X 1 2 Y a b a c Z p q r p

Functional Dependencies (FDs) n A functional dependency X Y holds over relation R if, for every allowable instance r of R: n n n 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. ) An FD is a statement about allowable relations. n n n t 1 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 n However, K R R does not require K to be minimal!

Functional Dependencies (FDs) Does the following relation instance satisfy X->Y ? X 1 2 1 3 12/2/2020 Y a b Z p q r p 12

Functional Dependencies (FDs) If X is a candidate key, then X -> Y Z ! X 1 2 1 3 12/2/2020 Y a b Z p q r p 13

Functional Dependencies (FDs) If Y Z -> X can we say YZ is a candidate key? X 1 2 1 3 12/2/2020 Y a b Z p q r p 14

Example: Constraints on Entity Set n n Consider relation obtained from Hourly_Emps: n Hourly_Emps (ssn, name, lot, rating, hrly_wages, hrs_worked) Notation: We will denote this relation schema by listing the attributes: SNLRWH n This is really the set of attributes {S, N, L, R, W, H}. n Sometimes, we will refer to all attributes of a relation by using the relation name. (e. g. , Hourly_Emps for SNLRWH) Hourly_Emps S N L R W H

Example: Constraints on Entity Set n Some FDs on Hourly_Emps: n n ssn is the key: S SNLRWH rating determines hrly_wages: R W Did you notice anything wrong with the following instance ? S N L R W 1 100 2 200 3 250 2 300 H

Example: Constraints on Entity Set n Some FDs on Hourly_Emps: n n ssn is the key: S SNLRWH rating determines hrly_wages: R W Salary should be the same for a given rating! S N L R W 1 100 2 200 3 250 2 200 H

Example n Problems due to R n n n 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!

Hourly_Emps 2 Wages

Refining an ER Diagram n 1 st diagram translated: Workers(S, N, L, D, S) Departments(D, M, B) n n Lots associated with workers. Suppose all workers in a dept are assigned the same lot: D since name ssn Employees lot dname did Works_In budget Departments L

Refining an ER Diagram n Suppose all workers in a dept are assigned the same lot: D Redundancy; fixed by: Workers 2(S, N, D, S) Dept_Lots(D, L) n Can fine-tune this: Workers 2(S, N, D, S) Departments(D, M, B, L) n name ssn Employees budget since dname did Works_In lot Departments L

Reasoning About FDs n n Given some FDs, we can usually infer additional FDs: n ssn did, did lot implies ssn lot An FD f is implied by a set of FDs F if f holds whenever all FDs in F hold. n = closure of F is the set of all FDs that are implied by F. Armstrong’s Axioms (X, Y, Z are sets of attributes): n Reflexivity: If X Y, then Y X (a trivial FD) n Augmentation: If X Y, then XZ YZ for any Z n Transitivity: If X Y and Y Z, then X Z These are sound and complete inference rules for FDs!

Reasoning About FDs S N L R W H For example, in the above schema S N -> S is a trivial FD since {S, N} is a superset of {S}

Reasoning About FDs S N L R W H For example, in the above schema If S N -> R W, then S N L -> R W L (by augmentation)

Reasoning About FDs (Contd. ) n Couple of additional rules (that follow from AA): n Union: If X -> Y and X -> Z, then X -> YZ Proof: n n From X -> Y, we have XX -> XY (by augmentation) Note that XX is X, therefore X -> XY From X -> Z, we have XY -> YZ (by augmentation) From X -> XY and XY -> YZ, we have X -> YZ (by transitiviy)

Reasoning About FDs (Contd. ) n Couple of additional rules (that follow from AA): n Decomposition: If X -> YZ, then X -> Y and X -> Z n Try to prove it at home/dorm/IC/Vitamin/DD/Bus!

Reasoning About FDs (Contd. ) n Example: n n n Contracts(cid, sid, jid, did, pid, qty, value), and: C is the key: C CSJDPQV Project purchases each part usingle contract: JP Dept purchases at most one part from a supplier: SD JP SD SDJ C P C, C CSJDPQV imply JP CSJDPQV P implies SDJ JP JP, JP CSJDPQV imply SDJ CSJDPQV

Reasoning About FDs (Contd. ) n n Computing the closure of a set of FDs ( ) can be expensive. (Size of closure is exponential in # attrs!) Example: n A database with 3 attributes (A, B, C) n F = {A->B, B->C} n Find the closure of F denoted by F+

Example n Suppose we are given a relation scheme R = (A, B, C, G, H, I), and the set of functional dependencies, F provided below: A→B A→C CG → H CG → I B→H n Is A → H logically implied by F? n Is AG → I logically implied by F? 12/2/2020 29

Reasoning About FDs (Contd. ) n n Computing the closure of a set of FDs ( ) can be expensive. (Size of closure is exponential in # attrs!) 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: n Compute attribute closure of X (denoted n n n ) wrt F: Set of all attributes A such that X A is in There is a linear time algorithm to compute this. For each FD Y -> Z in F, if is a superset of Y then add Y to

Reasoning About FDs (Contd. ) n Does F = {A n i. e, is A n. Lets B, B C, C D E in the closure E } imply A ? Equivalently, is E in E? ? compute A+ n. Initialize A+ to {A} : A+ = {A} n. From A -> B, we can add B to A+ : A+ = {A, B} n. From B -> C, we can add C to A+ : A+ = {A, B, C} n. We can not add any more attributes, and A + does not contain E therefore A -> E does not hold.

DB Design Guidelines n n 12/2/2020 Design a relation schema with a clearly defined semantics Design the relation schemas so that there is not insertion, deletion, or modification anomalies. If there may be anomalies, state them clearly Avoid attributes which may frequently have null values as much as possible. Make sure that relations can be combined by keyforeign key links 32

Normal Forms n n n Normal forms are standards for a good DB schema (introduced by Codd in 1972) If a relation is in a certain normal form (such as BCNF, 3 NF etc. ), it is known that certain kinds of problems are avoided/minimized. Normal forms help us decide if decomposing a relation helps.

Normal Forms n First Normal Form: No set valued attributes (only atomic values) sid name phones 1 ali {5332344568, 2165533561} 2 veli … 3 ayse … 4 fatma …

Normal Forms (Contd. ) n Role of FDs in detecting redundancy: n Consider a relation R with 3 attributes, ABC. n n 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!

Normal Forms (Contd. ) n n Second Normal Form : Every non-prime attribute should be fully functionally dependent on every key (i. e. , candidate keys). In other words: “No non-prime attribute in the table is functionally dependent on a proper subset of any candidate key” n n n Prime attribute: any attribute that is part of a key Non-prime attributes: rest of the attributes Ex: If AB is a key, and C is a non-prime attribute, then if A->C holds then A partially determines C (there is a partial functional dependency to a key)

Third Normal Form (3 NF) n Reln R with FDs F is in 3 NF if, for all X n n A in A X (called a trivial FD), or X contains a key for R, or A is part of some key for R. If R is in 3 NF, some redundancy is possible.

What Does 3 NF Achieve? n If 3 NF violated by X -> A, one of the following holds: n X is a subset of some key K n n X is not a proper subset of any key. n n 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. n n We store (X, A) pairs redundantly. 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. There is a stricter normal form (BCNF).

Boyce-Codd Normal Form (BCNF) n Reln R with FDs F is in BCNF if, for all X n n n A in A X (called a trivial FD), or X contains a key for R. (i. e. , X is a superkey) In other words, R is in BCNF if the only non-trivial FDs that hold over R are key constraints. n n n No dependency in R that can be predicted using FDs alone. If we are shown two tuples that agree upon the X value, we cannot infer the A value in one tuple from the A value in the other. If example relation is in BCNF, the 2 tuples must be identical (since X is a key).

Normal Forms Contd. n n Person(SSN, Name, Address, Hobby) F = {SSN Hobby -> Name Address, SSN ->Name Address} SSN Name Address Hobby 111111 Celalettin Sabanci D. Stamps 111111 Celalettin Sabanci D. Coins 555555 Elif Mutlukent Skating 555555 Elif Mutlukent Surfing 666666 Sercan Esentepe Math Is the above relation in 2 nd normal form ? 12/2/2020 40

Normal Forms Contd. n Ex: R = ABCD, F={AB->CD, AC->BD} n n n What are the (candidate) keys for R? Is R in 3 NF? Is R in BCNF? A B C D 1 1 3 4 2 1 3 4 Is there a redundancy in the above instance With respect to F ? 12/2/2020 41

Example n n An employee can be assigned to at most one project, but many employees participate in a project EMP_PROJ(ENAME, SSN, ADDRESS, PNUMBER, PNAME, PMGRSSN) PMGRSSN is the SSN of the manager of the project Is this a good design? 12/2/2020 42

Decomposition of a Relation Scheme n Suppose that relation R contains attributes A 1. . . An. A decomposition of R consists of replacing R by two or more relations such that: n n 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. Intuitively, decomposing R means we will store instances of the relation schemes produced by the decomposition, instead of instances of R. E. g. , Can decompose SNLRWH into SNLRH and RW.

Decomposition of a Relation Scheme n We can decompose SNLRWH into SNL and RWH. S S N N L L R W R H W H

Example Decomposition n n SNLRWH has FDs S -> SNLRWH and R -> W n Is this in 3 NF? n R->W violates 3 NF (W values repeatedly associated with R values) In order to fix the problem, we need to create a relation RW to store the R>W associations, and to remove W from the main schema: n S i. e. , we decompose SNLRWH into SNLRH and RW N L R H R W

Problems with Decompositions n There are potential problems to consider: n Some queries become more expensive. n e. g. , How much did Ali earn ? (salary = W*H) S N L R H R W

Problems with Decompositions n Given instances of the decomposed relations, we may not be able to reconstruct the corresponding instance of the original relation!

Lossless Join Decompositions n 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: n n n ( r) = r It is always true that r n n ( 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 (2). )

More on Lossless Join n The decomposition of R into X and Y is lossless-join wrt F if and only if the closure of F contains: n n n 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.

n n Person(SSN, Name, Address, Hobby) F = {SSN Hobby -> Name Address, SSN ->Name Address} SSN Name Address Hobby 111111 Celalettin Sabanci D. Stamps 111111 Celalettin Sabanci D. Coins 555555 Elif Mutlukent Skating 555555 Elif Mutlukent Surfing 666666 Sercan Esentepe Math Person 1 Hobby SSN Name Address SSN Hobby 111111 Celalettin Sabanci D. 111111 Stamps 555555 Elif Mutlukent 111111 Coins 666666 Sercan Esentepe 555555 Skating 555555 Surfing 666666 Math

n n Person(SSN, Name, Address, Hobby) F = {SSN Hobby -> Name Address, SSN ->Name Address} SSN Name Address SSN Hobby 111111 Celalettin Sabanci D. 111111 Stamps 555555 Elif Mutlukent 111111 Coins 666666 Sercan Esentepe 555555 Skating 555555 Surfing 666666 Math SSN Name Address Hobby 111111 Celalettin Sabanci D. Stamps 111111 Celalettin Sabanci D. Coins 555555 Elif Mutlukent Skating 555555 Elif Mutlukent Surfing 666666 Sercan Esentepe Math

exercise n n F 1 = {AB->C, BC->AD, D->E, CF->B} Does AB ->D hold wrt F 1? Does D->A hold wrt F 1? Is AB a key wrt F 1? 12/2/2020 52

Projecting FDs n n 12/2/2020 Suppose that we have R(A, B, C, D) with F = {A->B, B->C, C->D} We would like to project the FDs to R 1(A, C, D) 53

Projecting FDs Need to compute the closure of n F = {A->B, B->C, C->D} A simple exponential algorithm is: n 1. 2. 3. 12/2/2020 For each set of attributes X, compute X +. Add X ->A for all A in X + - X. Finally, use only FD’s involving projected attributes. 54

Projecting FDs n n n Suppose that we have R(A, B, C, D) with = {A->B, B->C, C->D} We would like to project the FDs to R 1(A, C, D) Using the algorithm in the previous slide: n n n 12/2/2020 F {A}+ = {A, B, C, D}, therefore A->C, and A->D hold in F 1 {C}+ = {C, D}, therefore only C->D is in F 1 {D}+ = {D} No need to check the supersets of A since {A}+ includes all the attributes For {C, D}+ = {C, D} we have only {A->C, A->D, C->D} in the projected FDs 55

Problems with Decompositions (Contd. ) n Checking some dependencies may require joining the instances of the decomposed relations. 12/2/2020 56

Dependency Preserving Decomposition n Consider CSJDPQV, C is key, JP -> C and SD -> P. n n n BCNF decomposition: CSJDQV and SDP Problem: Checking JP -> C requires a join! Dependency preserving decomposition: n n n A dependency X->Y that appear in F should either appear in one of the sub relations or should be inferred from the dependencies in one of the subrelations. 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. Ex: R=ABC, F={ A -> B, B -> C, C -> A} n F+ includes FDs, {A->B, B->C, C->A, B->A, A->C, C->B } n FAB= {A->B, B->A}, FAC={C->A, A->C}

Dependency Preserving Decompositions (Contd. ) n Decomposition of R into X and Y is dependency preserving if (FX union FY ) + = F + n n 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 +, not F, in this definition: n n ABC, A -> B, B -> C, C -> A, decomposed into AB and BC. Is this dependency preserving? Is C ->A preserved? ? ? n F+ includes FDs, {A->B, B->C, C->A, B->A, A->C, C->B } n FAB= {A->B, B->A}, FBC={B->C, C->B}, n FAB U FBC = {A->B, B->A, B->C, C->B} n Does the closure of FAB U FBC imply C->A ?

Dependency Preserving Decompositions (Contd. ) n Dependency preserving does not imply lossless join: n n Ex: ABC, A -> B, decomposed into AB and BC, is lossy. And vice-versa! (Example? )

Decomposition into BCNF n Consider relation R with FDs F. If X -> Y violates BCNF, decompose R into R - Y and XY. n n Repeated application of this idea will give us a collection of relations that are in n BCNF; n lossless join decomposition, n and guaranteed to terminate. 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!

Example: Decomposition into BCNF n n n R= ABCDEFGH with FDs n ABH->C n A->DE n BGH->F n F->ADH n BH->GE Is R in BCNF? Which FD violates the BCNF ? n ABH -> C ? n No, since ABH is a superkey n A->DE violates BCNF n Since attribute closure of A is ADE and therefore A is not a superkey n Decompose R=ABCDEFGH into R 1=ADE and R 2=ABCFGH

Example: Decomposition into BCNF n n R= ABCDEFG with FDs n ABH->C n A->DE n BGH->F n F->ADH n BH->GE R 1=ADE F 1= {A->DE} R 2=ABCFGH F 2= {ABH->C, BGH->F, F->AH, BH->G} n Note that the new FDs are obtained by projecting the original FDs on the attributes in the new relations. n For example BH->GE is decomposed into {BH->G, BH->E} and BH->E is not included in F 1 or F 2, BH->G is included into R 2. n Is the decomposition of R into R 1 and R 2 dependency preserving? R 1 is in BCNF, but we need to apply the algorithm on R 2 since it is not in BCNF

BCNF and Dependency Preservation n In general, there may not be a dependency preserving decomposition into BCNF. n n n e. g. , CSZ, CS -> Z, Z -> C Can’t decompose while preserving 1 st FD; not in BCNF. Similarly, decomposition of CSJDQV into SDP, JS and CJDQV is not dependency preserving (w. r. t. the FDs JP -> C, SD -> P and J -> S). n n However, it is a lossless join decomposition. In this case, adding JPC to the collection of relations gives us a dependency preserving decomposition. n JPC tuples stored only for checking FD! (Redundancy!)

Decomposition into 3 NF n n The algorithm for lossless join decomp into BCNF can be used to obtain a lossless join decomp into 3 NF (typically, can stop earlier). To ensure dependency preservation, one idea: n n n 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.

Minimal Cover for a Set of FDs n Minimal cover G for a set of FDs F: n n n 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: n A -> B, ACD -> E, EF -> G and EF -> H

Obtaining the Minimal Cover n Algorithm Steps: 1. Put the FDs in standard form (single attribute on the right hand side) 2. Minimize the left hand side of each FD 3. Delete the redundant FDs n The steps should be performed in the above order (think why) n Is there a unique minimal cover for a given set of FDs?

Obtaining the Minimal Cover n Example: F = {ABCD->E, E->D, A->B, AC->D} n n Notice that the right hand sides have a single attr (if not we had to decompose the right hand sides first) Can we remove B from the left hand side of ABCD->E? n Check if ACD->E is implied by F n In order to do this, find the attribute closure ACD wrt F n If B is in the attribute closure, then ACD->E is implied by F, and therefore we can replace ABCD->E with ACD->E (note that given ACD->E, we have ABCD->E) Can we remove D from ACD->E n Check if AC->E is implied by F’ (obtained by replacing ABCD->E in F with ACD->E) F’’ = {AC->E, E->D, A->B, AC->D} n Can we drop any FDs in F’’ ? n Could we drop any FDs in F before minimizing the left hand sides?

Dependency Preserving Decomposition into 3 rd. NF n n n Let R be the relation to be decomposed into 3 rd NF and F be the FDs that is a minimal cover Algorithm Steps n Perform lossless-join decomposition of R into R 1, R 2, . . Rn n Project the FDs in F into F 1, F 2, …, Fn (that correspond to R 1, R 2, …, Rn) n Identify the set of FDs that are not preserved (I. e. , that are not in the closure of the union of F 1, F 2, …, Fn) n For each FD X->A that is not preserved, create a relation schema XA and it to the decomposition.

Example n Consider the relation R = ABCDE with FDs: n n n A->B BC->E ED->A Lets first find all the keys Lets check if R is in 3 NF Lets check if R is in BCNF 12/2/2020 69