RELATIONAL DESIGN THEORY / LECTURE 4 TIM KRASKA

Transcription

1 RELATIONAL DESIGN THEORY / LECTURE 4 TIM KRASKA

2 RECAP Physical Independence Logical Independence

3 Simplified Zoo Relations animals name age species cageno keptby feedtime mike 13 giraffe :00am sam 3 salam :00am sally 1 student 1 2 1:00pm keepers keeper name 1 jenny 2 joe Primary Key Foreign Key cages cageno bldg

4 KEYS AND RELATIONS Kinds of keys Superkeys: set of attributes of table for which every row has distinct set of values Candidate keys: minimal superkeys Primary keys: DBA-chosen candidate key (marked in schema by underlining) ISBN Title Author Edition Publisher Price Harry Potter J.K. Rowling 1 Scholastic $ Mockingjay Suzanne Collins 1 Scholastic $7.39

5 Relational Design Theory =Assess the quality of a schema =redundancy =integrity constraints =Quality seal: normal forms (1-3, BCNF/3.5) =Improve the quality of a schema =synthesis algorithm =decomposition algorithm =Construct a (high-quality) schema =start with universal relation =apply synthesis or decomposition algorithms 5

6 Bad Schemas ProfLecture PersNr Name Level Room NB Title CP 2125 Tim AP G DB Systems Tim AP G Algorithms Tim AP G Logik Raul AP German Mike FP ML 4 = Update-Anomaly =What happens when Tim moves to a different room? = Insert-Anomaly =What happens if Raul is elected as a new professor? = Delete-Anomaly =What happens if Tim does not teach this semester? 6

7 Functional Dependencies = Schema: R = {A:D A, B:D B, C:D C, D:D D } = Instance: R = Let a Í R, b Í R = a b iff "r, s Î R: r.a = s.a Þ r.b = s.b = (There is a function f: D a D b ) R A B C D a4 b2 c4 d3 a1 b1 c1 d1 a1 b1 c1 d2 a2 b2 c3 d2 a3 b2 c4 d3 {A} à {B} {C, D } à {B} Not: {B} à {C} Convention: CD à B 8

8 Example Family Tree Child Father Mother Grandma Grandpa Sofie Alfons Susan Zoe Kevin Sofie Alfons Susan Isabella Mike Mark Alfons Susan Zoe Kevin Mark Alfons Susan Isabella Mike Zoe Martha 9

9 Example Family Tree Child Father Mother Grandma Grandpa Sofie Alfons Susan Zoe Kevin Sofie Alfons Susan Isabella Mike Mark Alfons Susan Zoe Kevin Mark Alfons Susan Isabella Mike Zoe Martha = Childà Father,Mother = Child, Grandpa à Grandma = Child, Grandma à Grandpa 10

10 Analogy to functions = f1 : Child à Father =E.g., f1(mark) = Alfons = f2: Child à Mother =E.g., f2(mark) = Susan = f3: Child x Grandpa à Grandma = FD: Child à Father, Mother =represents two functions (f1, f2) =Comma on right side indicates multiple functions = FD: Child, Grandpa à Grandma =Comma on the left side indicates Cartesian product 11

11 Decomposition of Relations = Bad relations combine several concepts = decompose them so that each concept in one relation = R R 1,.., R n 1. Lossless Decomposition R = R 1. R R n 2. Preservation of Dependencies FD(R) = (FD(R 1 ) È... È FD(R n )) 12

12 Example Drinker Pub Guest Beer Kowalski Kemper Pils Kowalski Eickler Hefeweizen Innsteg Kemper Hefeweizen 14

13 Lossy Decomposition Drinker Pub Guest Beer Kowalski Kemper Pils Kowalski Eickler Hefeweizen Innsteg Kemper Hefeweizen P Pub, Guest P Guest, Beer Visitor Pub Guest Kowalski Kemper Kowalski Eickler Innsteg Kemper Drinks Guest Beer Kemper Pils Eickler Hefeweizen Kemper Hefeweizen 15

14 Drinker Kneipe Gast Bier Kowalski Kemper Pils Kowalski Eickler Hefeweizen Innsteg Kemper Hefeweizen Visitor Drinks Pub Guest P... Guest Beer Kowalski Kowalski Innsteg Kemper Eickler Kemper Kemper Eickler Kemper Pils Hefeweizen Hefeweizen VisitorA Drinks Pub Guest Beer Kowalski Kemper Pils Kowalski Kemper Hefeweizen Kowalski Eickler Hefeweizen Innsteg Kemper Pils Innsteg Kemper Hefeweizen 16

15 Preservation of Dependencies = Let R be decomposed into R1,..., Rn = F R = (F R1 È... È F Rn ) = ZipCodes: {[Street, City, State, Zip]} = Functional dependencies in ZipCodes ={Zip} à {City, State} ={Street, City, State} à {Zip} = What about this decomposition? =Streets: {[Zip, Street]} = Cities: {[Zip, City, State]} = Clicker: Is it lossless? Does it preserve functional dependencies? Answer A: Yes, Yes Answer C: No, Yes Answer B: Yes, No Answer D: No, No 17

16 Decomposition of ZipCodes ZipCodes City State Street Zip Cambridge MA Vassar St Cambridge MA Main St Cambridge TX Vassar St P Zip,Street P City,State,Zip Streets Zip Street Vassar St Vassar St Main St Cities City State Zip Cambridge MA Cambridge MA Cambridge TX {Street, City, State} à {Zip} not checkable in decomp. schema It is possible to insert inconsistent tuples 18

17 Violation of City,State,StreetàZip ZipCodes City State Street Zip Cambridge MA Vassar St Cambridge MA Main St Cambridge TX Vassar St P Zip,Street Streets Zip Street Vassar St Vassar St Main St Vassar St P City,State,Zip Cities City State Zip Cambridge MA Cambridge MA Cambridge TX Cambridge TX

18 Violation of City,State,StreetàZip ZipCodes City State Street Zip Cambridge MA Vassar St Cambridge MA Main St Cambridge TX Vassar St Cambridge TX Vassar St Streets Zip Street Vassar St Vassar St Main St Vassar St Cities City State Zip Cambridge MA Cambridge MA Cambridge TX Cambridge TX

19 First Normal Form = Only atomic domains (as in SQL 92) Parents Father Mother Children Johann Martha {Else, Lucie} Johann Maria {Theo, Josef} Heinz Martha {Cleo} vs. Parents Father Mother Child Johann Martha Else Johann Martha Lucie Johann Maria Theo Johann Maria Josef Heinz Martha Cleo 21

20 Second Normal Form =No non-prime attribute is dependent on any subset of any candidate key of the relation StundentAttends Student-ID Course-Nb Name Semester Fichte Schopenhauer Schopenhauer Carnap Carnap Carnap Carnap

21 Second Normal Form =No non-prime attribute is dependent on any subset of any candidate key of the relation StundentAttends Student-ID Course-Nb Name Semester Fichte Schopenhauer Schopenhauer Carnap Carnap Carnap Carnap =StudentAttends is not in 2NF!!! ={Student-ID} à {Name, Semester} 23

22 Third Normal Form = A table is in 3NF if and only if, for each of its functional dependencies X A, at least one of the following conditions holds: =X A is trivial: X contains A =X is a superkey =Each attribute in A -X (i.e, the set difference of between A and X) is contained in some candidate key 24

23 Third Normal Form = A table is in 3NF if and only if, for each of its functional dependencies X A, at least one of the following conditions holds: =X A is trivial: X contains A =X is a superkey =Each attribute in A -X (i.e, the set difference of between A and X) is contained in some candidate key Alternative: =The relation R (table) is in second normal form (2NF) =Every non-prime attribute of R is non-transitively dependent on every key of R. 25

24 Is The Following Table in 3NF? Drinker Pub Guest Beer Kowalski Kemper Hefeweizen Kowalski Eickler Pils Innsteg Kemper Hefeweizen {Pub}à{Guest} {Guest}à{Beer} 26

25 IS The Following Table in 3NF? Drinker Pub Guest Beer Kowalski Kemper Hefeweizen Kowalski Eickler Pils Innsteg Kemper Hefeweizen {Pub,Guest}à{Beer} 27

26 Boyce-Codd-Normal Form (BCNF ) = R is in BCNF iff for all a B in R at least one condition holds: =B Î a (i.e., a B is trivial) =a is a superkey of R = R in BCNF implies R in 3NF =Proof trivial from definition 28

27 Decomposition Algorithm (BCNF) While some relation R is not in BCNF: Find an FD F=XàY that violates BCNF on R Split R into: R1 = (X U Y) R2 = R Y

28 BCNFify Example for Hobbies Iter 1 Schema (S,H,N,A,C) key S = SSN, H = Hobby, N = Name, A = Addr, C = Cost FDs S,H à N,A,C S à N, A H à C violates bcnf Iter 2 Schema (S, N,A) Schema (S,H, C) Iter 3 FDs S à N, A FDs S,H à C H à C Schema (H, C) violates bcnf FDs H à C Schema (S,H) FDs

29 ZipCodes(Street, State, City, Zip) =ZipCodes is not in BCNF ={Zip} à {State, City} ={Street, State, City} à {Zip} // evil // okay =Redundancy in ZipCodes =(Vassar St, MA, Cambridge, 02139) =(Main St, MA, Cambridge, 02139) =(Mass. Ave, MA, Cambridge, 02139) =stores several times that belongs to MA 31

30 Decomposition of ZipCodes =ZipCodes: {[Street, City, State, Zip]} ={Zip} à {City, State} // evil ={Street, City, State} à {Zip} // okay =Applying the decomposition algorithm... =Street: {[Zip, Street]} =Cities: {[Zip, City, State]} =Assessment =decomposition is lossless =decomposition does not preserve dependencies 32

31 Boyce-Codd-Normal Form (BCNF ) = R is in BCNF iff for all a B in R at least one condition holds: =B Î a (i.e., a B is trivial) =a is a superkey of R = R in BCNF implies R in 3NF =Proof trivial from definition =Result =any schema can be decomposed losslessly into BCNF =but, preservation of dependencies cannot be guaranteed =need to trade correctness for efficiency =that is why 3NF is so important in practice 33